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A in y with <1>(0) E u. But since , p) E C in statements (1) - (4) of Proposition 3.4 and does not cluster to p can be replaced by (, p) does not belong to C. Then we have: THEOREM 3.4 Let C be a cluster class on a set X and for each A eX let CI(A) be the set of all x E X with ( cA. Then' 'Ct" is a closure operator on X and (, x) E C if and only if :X ~ veX) by 4>(x) = {4>)..(x)} for each x E X. It is easily shown [see Exercise 4] that ).. (y) for some A E A. To see that 4> is a uniform homeomorphism, we show that )..(Y) E 4>)..(Un+!)' Then there exists x',y' E U n+! such that (x), a(x)} = x* and a(y)} = y* so x*, y* E 4>(Un)' Therefore, 1t~! (4))..(U~+!)) < (X)) = 0. Then S(1t)..(y), (X») = 0. But 1t)..(CI(4>(X») = X/!!>).. and S(1t)..(y), 4>)..(U~» C X/!!>).. so we have a A(X)- ",» = Cl I3X [U?'=l 0(<1>,,)]. If Y E V" (x) then ~ I<1>,,(x) - 0 for some j E {I ... J A(F) =1= 0 which implies (<1>~rl(V)nF =1= 0. But then Sex, Un)nF =1= 0 for each positive mteger n. Therefore, the family {S(x, Un)nF} has the finite intersection property. For each positive integer n put Kn = n;=l CI(S(x,Un)nF). Then {Kn} is a decreasing sequence of non-empty closed sets in X such that Kn c Sex, Un) for each positive integer fl, so by (M), nKn =f. 0. Let 2 E nK n. Then Z E F and z E Sex, Un) for each fl. Consequently. y* = <1>A(x) = <1>1.(2) E ~h~(x/ )1) = ~(X/ct>A) - X/ct>A' Hence X/ct>)1 = «<1>~h-l (X/ct>A) for each "A. E A'. Then (<1>ar l (X/ct>)1) = (<1>ar l «( ,,) = X/c.l>" we have the following diagram: ,,) = X/c.l>w Then f = h © n I such that Clx(O(n» c Wn for each n. Then CIBx(O(n»nC = 0 for each
To show CI(CI(A» = CI(A) suppose a E CI(CI(A» which implies the existence of a transfinite sequence \jI:Y -t CI(A) with (\jI, a) E C. For each e E Y, \jI(e) E CI(A) which implies there is a transfinite sequence \jIe:Ye ---7 A such that (\jIe,\jI(e» E C. But then the sum L of the transfinite sequences {\jIe} and the point a form a pair (L, a) E C by statement (4) of Proposition 3.4. Therefore a E CI(A) since LeA. Consequently CI(CI(A» c CI(A) and hence CI(CI(A» = CI(A). To show CI(AuB) = CI(A)uCI(B) first note that a E CI(A) implies there is a pair (\jI,a) E C with \jI c A and hence \jI c AuB so a E CI(AuB). Similarly a E CI(B) implies a E CI(AuB) so CI(A)uCI(B) c CI(AuB). Finally, p E CI(AuB) implies the existence of a pair (\jI, p) E C with \jI c AuB. Let M = YIl\jl-l(A) and N = YIl\jl-l(B) where \jI:Y-t X. Then Y= MuN and by statement (2) of Proposition 3.4, either (\jIM,P) or (\jIN,P) E C. Since'VM c A and \jIN c B either P E Cl(A) or P E CI(B). Hence P E CI(A)uCI(B). We conclude that CI(AuB) = CI(A)uCI(B).
3.3 Transfinite Sequences and Topologies
81
It remains to show that ('I', p) E C if and only if 'I' clusters to a relative to the topology associated with Cl which will now be referred to as 'to First suppose ('I', a) E C but 'I' does not cluster to a relative to 'to Then there is an open neighborhood U of a such that 'I' is not frequently in U which implies a residual R c y with 'I'(R) c X - U where 'I':y --t X. By statement (2) of Proposition 3.4, ('I'R, a) E C which implies a E CI(X - U). But this is a contradiction since U is open. Therefore 'I' clusters to a relative to 'to
Conversely, suppose 'I':y --t X clusters to a. Then a E CI(Me) for each e E y where Me = {'I'(~) Ie :<::; ~). Therefore there is a transfinite sequence 'l'e:')'e ---7 Me with ('I'e, a) E C for each e E y. By statement (3) of Proposition 3.4 we must then have ('I', a) E c.Proposition 3.4 and Theorem 3.5 set up a one-to-one correspondence between the various topologies a set can have and the cluster classes on the set. It is clear from the definition of clustering that if C J and C 2 are two cluster classes and 'tJ and't2 are the associated topologies, that C J C C 2 if and only if 't2 C 'tJ. Moreover, it is interesting to note that if (C J nC 2) denotes the smallest cluster class that is larger than each of C J and C 2 then (C J nC 2) is the cluster class associated with 'tJ n'!2' We conclude this chapter with a characterization of continuous functions in terms of transfinite sequences. Various other interesting mappings (e.g., open mappings, closed mappings, quotient mappings, etc.) can be characterized in a similar manner but this will be left for the exercises. PROPOSITION 3.5 Let f:X --t Y be a function from a space X into a space Y. Thenfis continuous if and only iffor each transfinite sequence {xa) in X that clusters to some p E X. {f(x a ) I clusters to f(p). Proof: Assume f is continuous and suppose {x a) is a transfinite sequence in X that clusters to some point p E X. Let U be a neighborhood of f(P) in Y and pick a neighborhood V of p such that fey) c U. Then {x a) frequently in V implies {f(x a )) is frequently inf(V) c U, so {f(x a )) clusters tof(P).
Conversely, assume that for each transfinite sequence {x a) in X that clusters to some point p in X, {f(xa)) clusters to f(P). Suppose f is not continuous. Then there is apE X and a neighborhood U of f(P) such thatf(V) is not contained in U for each neighborhood V of p. By Theorem 3.4 there exists a well ordered neighborhood base {Val a < y) of p such that < is compatible with the partial ordering of set inclusion. For each a E A pick x a E Va such that f(x a) is not contained in U. Then {x a) clusters to p but {f(x a)) does not cluster to f(P) which is a contradiction. We conclude that f must be continuous. -
82
3. Transfinite Sequences
EXERCISES
1. Show that X is T I if and only if for each pair of distinct points in X there are two transfinite sequences clustering to the two points respectively but neither clustering to the other point. 2. A function is said to be open if the image of each open set is again an open set. Show that an onto function f is open if and only if for each transfinite sequence {y ~} in Y that clusters to some p E Y and for each q E rl (P) there is a transfinite sequence {xa} in u{f-l(Yj3)} that clusters to q. 3. A function is said to be closed if the image of each closed set is a closed set. Show that a function f is closed if and only if whenever a transfinite sequence {y ~} clusters to some p E Y there is a transfinite sequence Ix a} that is a subset ofrl({y~}) that clusters tor1(p). 4. A function is said to be a quotient if whenever the inverse image of a set is open, then the set itself must be open. Show that a function f is a quotient if and only if for each transfinite sequence {y~} clustering to some p E Y there is a transfinite sequence {x a} contained in the inverse image of {y /3} under f that is frequently in each open inverse image of an open set in Y that contains r 1(P). 5. Show that Corollary 3.2 still holds when the assumption of complete regularity is removed from the hypothesis; i.e., show that any space is compact if and only if each transfinite sequence clusters.
Chapter 4 COMPLETENESS, COFINAL COMPLETENESS AND UNIFORM PARACOMPACTNESS
4.1 Introduction In 1915, A paper by E. H. Moore appeared in the Proceedings of the National Academy of Science U.S.A. titled Definition of limit in general integral analysis. This study of unordered summability of sequences led to a theory of convergence by Moore and H. L. Smith titled A general theory of limits which appeared in the American Journal of Mathematics in 1922. In 1937, G. Birkhoff applied the Moore-Smith theory to general topology in an article titled Moore-Smith convergence in general topology, which appeared in the Annals of Mathematics, No. 38, pp. 39-56. In 1940, J. W. Tukey made extensive use of the theory in his monograph titled Convergence and uniformity in topology published in the Annals of Mathematics Studies series. Tukey worked with objects that were generalizations of sequences that he referred to as phalanxes. They were a special case of the objects that are usually called nets today. An equivalent theory of convergence using objects called filters emerged in the thirties from the Bourbaki group in France. The theory of filters is the convergence theory of choice for many topologers. There are many things to recommend it but it is also very awkward to use in certain situations. For instance, in the treatment of hyperspaces (Chapter 5), the objects of the hyperspace are subsets of the original space. Filters, which are themselves collections of subsets of a space, become awkward to construct in the hyperspace because they are collections of collections of subsets. For some proofs, filters carry along too much baggage. In proofs about the completions of uniform spaces it is sometimes desirable to pick a convergence object from the original space that is arbitrarily close to a convergence object in the completion. Then conclusions are made about the convergence object in the original space and these conclusions are then shown to hold for the convergence object in the completion due to its proximity to the object in the original space. A natural way to attempt this with filters is to restrict each subset in the completion filter to the original space, make conclusions about the restricted filter, then take the closures of the members of the restricted filter to get back to the completion. The problem here is that taking the closure picks up too many
84
4. Completeness, Cofinal Completeness and Uniform Compactness
points. Since the original filter in the completion is not always obtained, it may not be possible to draw conclusions about the original filter in the completion. Surely with time, one should be able to find filter type proofs for the theorems we will be presenting in the sequel, but this will not be our viewpoint. Instead, in some of the areas where the filter type proofs are especially nice, the recasting of the results into filter terminology will be suggested as exercises.
4.2 Nets A non-void set D is said to be directed by the binary relation the following conditions hold:
~
provided that
(1) if m,n and d E D with m ~ nand n ~ d then m ~ d, (2) d ~ d for each d in D and (3) if m,n E D then there exists a d in D with m ~ d and n ~ d.
Clearly a directed set is a particular type of partially ordered set. Condition (3) in the definition of a directed set is what sets the directed set apart from an ordinary partially ordered set. Also, it should be noticed that well ordered sets are necessarily directed sets. Some useful examples of directed sets that are not necessarily well ordered are given below. Let X be a topological space and let B be a local neighborhood base for a point p E X. Define the relation ~ on B as follows: U ~ V if U,V E B and V c U. Clearly (B, ~) satisfies conditions (1) and (2) above so (B, ~) is a partially ordered set. To show (B,~) also satisfies (3) assume U,V E B. Then W = UnV belongs to Band W c U and W c V. Thus U ~ Wand V ~ W, so (B,~) satisfies (3).
Another directed set that we will employ frequently is the uniformity itself. Let (X. f.!) be a uniform space. The relations < (refinement) and <* (star refinement) have already been defined for f.! (see Section 2.1). Clearly (f.!, <) and (11, <*) satisfy conditions (1) and (2) defined above so that both are partially ordered sets. That < also satisfies condition (3) above follows from condition (1) of Section 2.1. At this point we are careful to remind the reader that we consider these orderings as "proceeding toward the left" in the sense that in proving condition (3) for <* we will show that if U,V E 11 then there exists aWE 11 such that W* < U and W* < V. By (1) of Section 2.1 there exists a Z E 11 such that Z < U and Z < V. By (3) there is a Win 11 with W <* Z. Therefore W <* U and W <* V. Consequently both (11, <) and (11, <*) are directed sets. A subset R 'F- 0 of a directed set (D, ~) is said to be residual if whenever m,n E D with mER and m ~ n then n E R. IfC c D such that whenever m E D there is an n E C with m ~ n we say C is cofinal in D. A net is a function \jI:D
4.2 Nets
85
-t X from a directed set D into a space X. In the event D is well ordered then '" is simply a transfinite sequence with which we are already familiar. For each ex E D let x a = ",(ex). We will often identify a net", with its range ",(D) = {x a I ex E D). If the set D is understood we simply write {xa)' We say a net {xa) is
frequently in U c X if there is a cofinal C cD with {x~ I ~ E C) cU. We say {x a) is eventually in U if there is a residual ReD with (x y I 'Y E R) cU. {x a) is said to converge to a point p in X if it is eventually in each neighborhood of p and to cluster to p if it is frequently in each neighborhood of p. PROPOSITION 4.1 A subset U of a space X is open net in X - U converges to a point of U.
if and only if no
Proof: Assume U is open in X and let {x a) be a net that converges to some p E U. From the definition of convergence {x a) must eventually be in U. But then {x a) cannot lie entirely in X - U. Conversely, assume no net in X - U converges to a point of U and suppose U is not open. Then there is apE U every neighborhood of which meets X - U. Let D be a local basis for p and let :s; be the partial ordering of set inclusion on D. We have already seen that D is directed with respect to this ordering. For each V E D pick Xv E Vn(X - U). Then {xv) is a net in X - U. It is easily shown that {xv) converges to p. Indeed, let WED and put R(W) = {V E D IV c W). Then R(W) is residual in D and Xv E W for each V E R(W). But this is a contradiction since {xv} lies in X - U. We conclude that U must be open. PROPOSITION 4.2 A point p of a space X belongs to the closure of E c X
if and only if there is a net in E that converges to p.
Proof: Assume p E CI(E) and let D be a local basis for p. Then D is directed by the partial ordering of set inclusion. For each V E D pick Xv E V nE. Then {vv) is a net in E. An argument similar to the one in Proposition 4.1 shows that {xv} converges to p. Conversely, assume (x a) is a net in E converging to a point p E X. Let U be an open set containing p. Then {x a ) is eventually in U which implies U nE *" 0. Thus p E CI(E). PROPOSITION 4.3 A point p of a space X is a limit point of E c X and only if there is a net in E - {p) that converges to p.
if
Proof: Let p be a limit point of E c X and let D be a local basis for p. For each V E D pick Vv E VneE - {p). Then {xv) is a net in E - {p) converging to p. Conversely, if {x a} is a net in E - {p) converging to p E X then for each open set U containing p, {x a) is eventually in U. But then Un(E - {p) *" 0 so p is a limit point of E. -
86
4. Completeness, Cofinal Completeness and Uniform Compactness
PROPOSITION 4.4 A space X is Hausdorff if and only converges to at most one point.
if each net in X
Proof: Let X be a Hausdorff space and assume p and q are distinct points of X. Since X is Hausdorff, there are disjoint open sets U and V such that p E U and q E Y. If {x a} is a net in X that converges to p then {x a} is eventually in U so it cannot eventually be in Y. Thus {x a } cannot converge to q. We conclude that a net can converge to at most one point.
Conversely, assume each net in X can converge to at most one point and suppose X is not Hausdorff. Then there exists two distinct points p and q such that for each pair of neighborhoods U of p and V of q, Un V oF 0. Let B(P) and B(q) be local bases for p and q respectively and put D = B(P) x B(q). Define $ on D as follows: for each a,c E D where a = (A, B) and c = (U,V), put a $ c if U c A and V c B. Then (D, $) is a directed set. For each c = (U,V) E D pick Xc E Un Y. Then {xc} is a net in X. To show that {xc} converges to both p and q let U E B(P) and V E B(q). Put d = (U,V) and let R(d) = {a E Did $ a}. Then R(d) is residual in D. If a E R(d) then a = (A, B) for some A E B(P) and B E B(q) such that A c U and BeY. Moreover, Xa E AnB cUnY. Therefore, for each a E R(d), Xa E U and Xa E V, so {xa} is eventually in both U and V. We conclude {xa} converges to both p and q which is a contradiction. Hence X must be Hausdorff. A net <jl:E ~ X is called a subnet of the net 'II:D function A:E ~ D such that:
~
X if there exists a
(l) <jl = 'V © A and (2) for each d E D there is an e E E such that if e $ c then d $ A(e).
Such a function A is called a cofinal function. It is easily shown that if C is a cofinal subset of D that the identity function i:C ~ D is a cofinal function and hence a net 'II:D ~ X restricted to a cofinal subset C is a subnet of'll. However, in the theory of convergence of nets, we will see that such subnets are usually not very useful. Normally we will be interested in subnets whose domains are not subdomains of the domain of the original net. It should be noted that a subnet is a more complex object than a subsequence of a transfinite sequence and that a subsequence of a transfinite sequence is a sub net when the transfinite sequence is considered as a net. The principle advantage of nets is that one works with convergence rather than clustering (compare Proposition 4.4 with Proposition 3.3).
PROPOSITION 4.5 If B is a family of subsets of the space X with the property that the intersection of two members of B contains a member of Band if the net {x a) is frequently in each member of B then {x a} has a subnet that is eventually in each member of B.
87
4.2 Nets
Proof: Let {xa I a E A} be the net in the hypothesis above. Since the intersection of any two members of B contains a member of B, B is directed hy set inclusion. Then E = {(a, F)
E
A x B Ix a
E
F}
is directed by the ordering :c; defined as follows: if (aI, F I) and (a2' F 2) E E then (a2, F 2) :c; (aI, F I) if a2 :c; al and FIe F 2. Now define a function A:E ~ A by A(a, F) = a for each element (a, F) of E. Clearly A is cofinal so the net lYe leE E} defined by Ye = xA(e) for each e E E is a subnet of {x a }. Next let H E B. Since {xa} is frequently in H, there exists apE A such that x~ E H. Hence e' = (P, H) E E. Let e = (a, F) be an arbitrary member of E such that e' :c; e. Then P:c; a and F c H so Ye
= XA(a,F) = Xa E
F c H.
Thus {x a} is eventually in H.PROPOSITION 4.6 A net clusters to a point subnet that converges to the point.
if and
only
if it
has a
Proof: Let p be a cluster point of the net {x a Ia E A} and let B be a local basis for p. Then the intersection of any two members of B contains a member of B and {xa} is frequently in each member of B. Hence by Proposition 4.5, there exists a sub net {y~} of {xa} that is eventually in each member of B. Hence {y~} converges to p.
Conversely, assume that the net {x a } has a subnet {y~} that converges to p. Let D be the domain of {y ~} and A the cofinal function from D to A that defines {Y~}. Let U he a neighborhood of p and let () E A. Since A is cofinal there is an element E E 0 such that A(P) ~ () for each P ~ E. Since {y~} converges to p there is an element E' ~ E with YE' E U. Now let ()' = A(E'). Then we have ()' ~ () and Xli' =XA(E') = Yf' E U. Hence {xa} clusters to p.PROPOSITION 4.7 A space is compact convergent subnet.
if and only if each
net has a
Proof: By Proposition 4.6 it suffices to show that a space is compact if and only if each net clusters. For this let x = {x a I a E A} be a net in the compact space X. For each a E A let M a be set set of all x~ such that p ~ a. Since A is directed by ~, {Mal a E Al has the finite intersection property so the family {CI(M a) Ia E A I also has the finite intersection property. Since X is compact, by Proposition 0.35, there is apE X which belongs to CI(M a) for each a E A. We will show that x clusters to p. For this let U be a neighborhood of p and let a E A. Since p E CI(M a), it follows that ManU "# 0. Hence there is a f3 E A
88
4. Completeness, Cofinal Completeness and Uniform Compactness
with ~ ~ a and x~ E U. This implies x is frequently in U. We conclude x clusters to p. Conversely, assume every net in X clusters. Let F be a family of closed sets in X with the finite intersection property. Let G denote the family of finite intersections of members of F. Then G also has the finite intersection property. Since F c G, it suffices to show that the intersection of all members of G is nonempty to show that X is compact. Now G is directed by set inclusion c. For each FE G pick Xp E F. Then the assignment F -7 Xp defines a net x = {xp} in X. By hypothesis, x clusters to some p E X. Let H, KEG such that K c H. Then XK EKe H, so x is eventually in the closed set H. By Propositions 4.l and 4.6, p E H. Hence p E H for each HE G. We conclude that X is compact.
-
PROPOSITION 4.8 A function f:X only j(p).
-7
Y is continuous at p
E
X if and
if for each net {x a} in X converging to p. the net (f(x a)} in Y converges to
Proof' Suppose f:X -7 Y is continuous at p and let V be a neighborhood of f(P) in Y. By definition of continuity at p, there is a neighborhood U of p in X such thatfCU) c V. Since {x a } converges to p, there is a residual R in the domain of {x a } such that x~ E U for each ~ E R. Thus f(x~) E feU) c V for each ~ E R, so (f(x a)} converges to f(P). Conversely, assume f is not continuous at p. Then there is an open neighborhood V of f(P) in Y such that every neighborhood of p meets 1 (Y -V). Let B be a local basis for p. Then B is directed by set inclusion. For each U E B pick Xu in Unj-l (Y - V). The assignment U -7 Xu defines a net {xu} in X that converges to p. Now the composition (f(xu)} is a net in Y - V. Since V is open and f(P) E V, (f(xu)} cannot converge to f(P). -
r
In the last chapter, cluster classes of transfinite sequences were discussed and it was shown that the various topologies a space can have correspond to the various cluster classes of transfinite sequences in the space . A similar situation holds for nets, only now it is possible to use convergence instead of clustering. In order to define the concept of a convergence class, it is first necessary to have a result on iterated limits. Iterated limits are usually first encountered in calculus where the limits are the limits of ordinary sequences. Iterated limits of nets can be defined in an analogous manner as in what follows. If (D,
4.2 Nets
89
e if d a
°
Next, consider the case where D is a directed set and for each E D, E 8 is another directed set. Let ~ = {(o,£) 1o E D and £ E E 8}. Consider a function 'I':~ ~ X where X is a topological space. Then for each EDit is possible that the net '1'8 = {'I'(O,£) 1£ E E 8} converges to some point p 8 E X. We also denote p 8 by lima'l'(O,a) and say that the limit of '1'8 exists and equals p 8. Furthermore, it is possible that the net <1> = {p 81o ED} also converges to some point p E X. In this case we say that p is the iterated limit lim8Iima'l'(0,a) of 'I' with respect to and a. Let P = D x fl{E810 ED}.
°
°
°
THEOREM 4.1 Let D be a directed set and for each E Diet E8 be another directed set. Let ~ and P be defined as above. Then there exists a net A:P ~ ~ such that for any function f:~ ~ X for which the iterated limit lim8Iima!(0.a) exists,f© A converges to the iterated limit. Proof: Define A:P ~ ~ as follows: for each (0, d) E P put A(O, d) = (0, d(O». Suppose lim8lima! (o,a) = q and U is an open neighborhood of q. We must find a member (0, d) of P such that if (0, d) < (~, g) then f © A(~, g) E Pick y E D
u.
such that liml3f(~,v) E U for each ~ following y and then, for each such ~ pick d(~) E EI3 such thatf(~.v) E U for all v following d(~) in EI3' If ~ is a member of D which does not follow y let d(~) be an arbitrary member of E 13' If (~, g) > (y, d), then ~ > y, hence liml3f(~,v) E U, and since g(~) > d(~) we have f © A(~,g) = f(~, g(~» E U. We conclude thatf© A converges to q.Let C be a collection of pairs (<1>, p) of nets <1> in the space X and points p E X. We say C is a convergence class for X if it satisfies the following conditions: (1) if <1> is a constant net such that <1>( a) = p for each a then
(<1>, p) E C, (2) if (<1>, p) E C then so does ('I', p) for each subnet 'I' of <1>, (3) if (<1>. p) does not belong to C then there is a subnet 'I' of <1> such that (~, p) does not belong to C for each subnet ~ of 'I' and (4) let D be a directed set and for each E Diet E 8 be another directed set. Let ~ and P and A be defined as in Theorem 4.1. Then for each functionf:~ ~ X for which lim8lima! (o,a) exists, (j© A,p) E C.
°
THEOREM 4.2 Let C be a convergence class for a set X and for each A c X let CI(A) be the set of all p E X with ('I'. p) E C and 'I' c A. Then "Cl" is a closure operator on X and ('I'. p) E C if and only if 'I' converges to p relative to the topology associated with Ct.
The proof of Theorem 4.2 is left as an exercise (see Exercise 2). A net in a space X is said to be a universal net if for each A c X the net is eventually in A
90
4. Completeness, Cofinal Completeness and Uniform Compactness
or eventually in X-A. Universal nets have the property that if they are frequently in a set that they must eventually be in the set. Consequently, universal nets converge to each of their cluster points. PROPOSITION 4.9 Every net has a universal subnet. Proof: Let x = Ix 0.1 a. ED} be a net in X and let
(1) x is frequently in each member of
if each
universal nef
Proof: Let X be compact and let let x be a universal net in X. By Propositions 4.6 and 4.7, x must cluster. But a universal net that clusters must also converge. Hence x converges. Therefore compactness implies that each universal net converges. Conversely, if every universal net in X converges, then by Proposition 4.9, every net has a universal subnet so every net in X has a convergent subnet. Then by Proposition 4.7, X must be compact. -
Let (X, f..l.) be a unifonn space. A net {x a.} in X is said to be Cauchy if it is eventually in some sphere of radius U for each U E f..l.. {x a.} is said to be
91
4.2 Nets
cofinally Cauchy if it is frequently in some sphere of radius U for each U E !l. The proofs of the following two propositions are left as exercises (Exercise 3). PROPOSITION 4.10 A net {xa} in a uniform space (X,!l) is Cauchy if and only iffor each U E !l there is a U E U such that {x a} is eventually in U PROPOSITION 4.11 A net {x a J in (X, !l) is cofinally Cauchy if and only iffor each U E !l there is aU E U such that {xa} isfrequently ill U
EXERCISES
1. Let {xa I a. E OJ be a net in the pseudo-metric space (X, d). Show that {xa J converges to p E X if and only if the net {Ya I a. E 0 J, defined by Ya = d(p, x a) for each a. E 0, converges to zero. 2. Prove Theorem 4.2 (see Theorem 3.4).
3. Prove Propositions 4.10 and 4.11. 4. Let {xa Ia. E O} be a net in R (the reals). If a. < ~ in 0 impliesxa ~ XI3 (xa ~ xl3), then {x a } is said to be monotone increasing (decreasing). Show that a
monotone increasing net that is bounded above, or a monotone decreasing net that is bounded below converges. FILTERS 5. A filter on a set X is a collection F of subsets of X satisfying the following properties: (1) F is closed under finite intersections, (2) the empty set does not belong to F, and (3) every subset of X containing a member of F belongs to F. Show that the following three examples of filters satisfy properties (1) through (3) above. The Neighborhood Filter: Let X be a space and let p
E
X. Let F denote the
neighborhood base at p. The Filter Associated with a Net: Let {xa I a. EO} be a net. Put F E F for each ~ E R for some residual ReO}. The Intersection Filter: Let {F a I a. = nF a.
E
= {F c
X IXI3
A J be a non-empty family of filters on X.
Then put F
6. A filter is said to converge to a point p E X if each neighborhood of p belongs to F. F is said to cluster to p if every neighborhood of p meets every member of F. A filter base is a collection B of subsets of X satisfying
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properties (1) and (2) in the definition of a filter above. Clearly, the collection F of all subsets of X containing a member of B is a filter called the filter generated by B. The filter base B is said to converge to p E X if the filter generated by B converges to p and to cluster to p if the filter generated by B clusters to p. Show the following: (a) U c X is open if and only if no filter base in X - U converges to a point of U. (b) p E Cl(E) if and only if there is a filter base in E that converges to p. (c) p is a limit point of E c X if and only if there is a filter base in E - {p} that converges to p. 7. Show that a space is Hausdorff if and only if each filter converges to at most one point. 8. Filters can be compared in the following way: Let F and G be two filters on X. F is said to be finer than G and G is said to be coarser than F if G c F. In addition if F :f:. G then F is said to be strictly finer than G and G is strictly coarser than F. Two filters are said to be comparable if one is finer than the other. Show the following: (a) If a filter F clusters to a point p E X then there is a filter G finer than F that converges to p. (b) A space is compact if and only if each filter clusters. (c) A space is compact if and only if each filter is contained in a convergent filter.
4.3 Completeness, Cofinal Completeness and Uniform Paracompactness We have already encountered the concept of completeness in metric spaces (Chapter 1, Section 6). Just as with metric spaces, completeness plays a fundamental role in the theory of uniform spaces. The concept of completeness in metric spaces generalizes in a natural way to uniform spaces. The next three propositions show that Cauchy nets and cofinally Cauchy nets behave in a manner analogous to Cauchy sequences and cofinally Cauchy sequences in metric spaces. Their proofs are also left as exercises (Exercises 1 and 2). PROPOSITION 4.12 A convergent net is Cauchy. PROPOSITION 4.13 A net that clusters is cofinally Cauchy. PROPOSITION 4.14 If a Cauchy net clusters to P then it also converges to p.
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The following theorem is the net version of Theorems 3.1, 3.2 and 3.3. It appeared in the 1971 paper titled On Completeness (Pacific Journal of Mathematics, Volume 38, Number 2, pp. 431-440). The proof is similar to the proofs of Theorems 3.l, 3.2 and 3.3, so we leave it as an exercise (Exercise 6). THEOREM 4.4 (N. Howes, 1971) A space is paracompact (resp. LindelOf or compact) if and only if each net that is cofinally Cauchy with respect to the u (resp. e or~) uniformity clusters.
A uniform space is said to be complete if each Cauchy net converges. It is said to be cofinally complete if each cofinally Cauchy net clusters. COROLLARY 4.1 A cofinally complete uniform space is complete.
The Lebesgue property for metric spaces also generalizes in a natural way to unifonn spaces. We say (X, 11) has the Lebesgue property if for each open covering V of X, there is a U E 11 such that V can be refined by the covering consisting of spheres of radius U. An equivalent characterization of the Lebesgue property is the following: PROPOS1T10N 4.15 (X, 11) has the Lebesgue property if and only if each open covering of X has a refinement in 11 (i.e., 11 is fine and X is paracompact). Proof: Assume each open covering V of X has a refinement say U in 11. Let W <* U. Then the spheres of radius W also refine V so (X, 11) has the Lebesgue property. Conversely, if X has the Lebesgue property and V is an open covering of X, pick U E 11 such that the spheres of radius U refine V. Clearly then U refines V.· PROPOS1T10N 4.16 A compact Hausdorff uniform space has the Lebesgue property. Proof: Let (X, 11) be a compact Hausdorff unifonn space. Then by Theorem 2.7 and Lemma 3.7 we have 11 = ~ is fine. Since X is paracompact, it has the Lebesgue property by Proposition 4.l5.· PROPOSITION 4.17 A uniform space with the Lebesgue property is cofinally complete. Proof: Let (X, 11) be a unifonn space with the Lebesgue property. Suppose (X,ll) is not cofinally complete. Then there exists a cofinally Cauchy net (x u I that does not cluster. For each p E X let U(P) be an open set containing p such that {xu} is eventually in X-U(P) and put U= {U(P)lpE Xl. Since (X, 11) has the Lebesgue property, by Proposition 4.15, U has a refinement V t:: 11. Since
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{x a } is cofinally complete it is frequently in V for some V E V. But V c U(P) for some P E X so {x a} cannot be eventuall y in X - U (P) which is a contradiction. Thus (X, Il) is cofinally complete. •
A uniform space eX, Il) is said to be precompact or totally bounded if each uniform covering has a finite subcovering. Clearly each precompact metric space is a precompact uniform space. THEOREM 4.5 A uniform space is precompact if and only if each net has a Cauchy subnet. Proof: Assume each net in the uniform space (X, Il) has a Cauchy subnet and suppose X is not precompact. Then there is a U E Il such that {S(x, U) I x EX} has no finite subcovering. Pick PI, P 2 E X such that p 2 does not belong to S(p 1, U). Next assume {p 1 . . . Pn} has been defined such that for each pair i,j with i < j ::::; n we have Pj does not belong to S(Pi, U). Since u{S(P 1, U) ... S(pn, U)} cf. X we can choose Pn+l E X with Pn+l not belonging to S(p" U) for each i = 1 ... n. Consequently it is possible (by induction) to construct a sequence {Pn} c X such that for each pair of positive integers i,j with i , j we have Pj does not belong to S(Pi, U).
Let x = {x a I ex ED} be such a Cauchy subnet of the sequence {Pn} for some cofinal function A:D ~ N such that x a = PAra) for each ex E D. Let VEil with V <* U. Since x is Cauchy there is a residual ReD and a POE X such that PAra) E S(p o,v) for each ex E R. Since A is a cofinal function, A(R) is cofinal in N which implies {Pn} is frequently in S(po,v). Pick positive integers k,m with k < m such thatpkoPm E S(Po,V). Then S(PkoV)US(PO,V) c S(Pko U). But then Pm E S(Pko U) which is a contradiction. We conclude (X, Il) is precompact. Conversely, assume (X, Il) is precompact and let x = {x a}, ex E D, be a net in X. If U E Il, then U has a finite subcovering, say lUI ... UN}' Since Ix a} cannot be eventually in X - Uj for j = 1 ... N, it must be frequently in one of them. Therefore, every net in (X, Il) is cofinally Cauchy. By Proposition 4.9, {x a } has a universal subnet IXA(~)} where A:B ~ D is a cofinal function from a directed set B into D. Let VEil. Since IXA(~)} is cofinally Cauchy, there exists a V E V such that {XA(~)} is frequently in V. But then {XA(~)} must eventually be in V, so IXA(~)} is Cauchy.· PROPOSITION 4.18 A closed subspace of a complete uniform space is complete. PROPOSITION 4.19 A uniform space is compact if and only if it is complete and precompact.
4.3 Completeness, Cofinal Completeness and Uniform Paracompactness
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The proofs of the two propositions above are essentially the same as the proofs of Propositions 1.23 through 1.25 with sequences being replaced by nets. They are left as an exercise (Exercise 5). In a 1977 paper titled A note on uniform paracompactness that appeared in the Proceedings of the American Mathematical Society (Volume 62, Number 2, pp. 359-362), M. Rice introduced the concept of uniform paracompactness which is defined as the property that every open covering has a uniformly locally finite open refinement. A refinement V of a covering U is said to be uniformly locally finite if there exists a uniform covering W such that each member of W meets only finitely many members of V. THEOREM 4.6 A uniform space is cofinally complete if and only if it is uniformly paracompact. Proof: We will prove that a uniform space (X, Il) is cofinally complete if and only if each directed open covering of X is uniform and then use the equivalent fonn of unifonn paracompactness given in Exercise 7(b) to finish the argument. To prove the sufficiency. assume each directed open covering is uniform and suppose (X, Il) is not cofinally complete. Then there exists a cofinally Cauchy net 'V:D ~ X that does not cluster. For each a E D put H a = {'V(o) I 0 ~ a} and let Fa = et(H a). Then nF a = 0 since'll does not cluster. For each a E D put U a = X - Fa. Then U = {U a} is a directed open covering of X, so U E Il. But for each a E D, 'V is eventually in X - U a which is a contradiction. Therefore, (X, Il) is cofinally complete. Conversely, assume (X, Il) is cofinally complete. Let U = {U a Ia ED} be a directed open covering of X where (D, <) is a directed set. If X = U a for some a E D then U E Il so assume Fa = X - U a =f. 0 for each a. Put E = {(F a,x) Ia E D and x E Fa} and for each a let O. But Fe = {\jI(Fe, x)lx E Fo},soVcU e . Therefore, V< U,SO UE Il.-
EXERCISES 1. Prove Propositions 4.12 and 4.13. 2. Prove Proposition 4.14. 3. Let (X, Il) be a uniform space and let U E Il. A c X is said to be II-small if A c U for some U E U. A is U-Iarge if its complement is U-small. A
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collection H of proper subsets of X is heavy if for each U E ~ there is an HE H that is U-large. Show that X is complete if and only if each heavy covering has a finite subcovering. 4. A collection U of proper subsets of X is said to be bound to another collection V if for each finite subcollection W of U that does not cover X. neither does Wu {V} for each V ::j; X in V. A collection that is bound to a heavy collection is called heavily bound. Show that a uniform space is cofinally complete if and only if each heavily bound open covering has a finite subcovering. 5. Prove Propositions 4.18 and 4.19. 6. Prove Theorem 4.4. UNIFORMLY PARACOMPACT SPACES 7. [M. Rice, 1977] The following statements are equivalent: (a) X is uniformly paracompact. (b) Each directed open covering is uniform. (c) If U is an open covering of X, there exists a uniform covering V such that U I V has a finite subcovering for each V E V. 8. [M. Rice, 1977] A locally compact uniform space is uniformly paracompact if and only if it is uniformly locally compact. A uniform space (X, ~) is said to be uniformly locally compact if there exists a U E ~ consisting of compact sets. UNIFORML Y PARACOMPACT METRIC SPACES 9. [M. Rice, 1977] The collection of points of a uniformly paracompact metric space that admit no compact neighborhood is compact. 10. [A. Hohti, 1981] A necessary and sufficient condition for a metric space to be uniformly paracompact is that there exists a compact K c X with X Star(K,c:) is uniformly locally compact for each c: > O. 11. [A. Hohti, 1981] If (X, d) and (Y, 8) are uniformly paracompact metric spaces, then (X x Y, d x 8) is uniformly paracompact if and only if one of the following conditions hold: (a) either (X, d) or (Y, 8) is compact or (b) both (X, d) and (Y, 8) are locally compact. 12. [A. Hohti, 1981] Let a be an infinite ordinal. Put H(a)
= ([0,1] x a)/E Where
4.4 The Completion ofa Uniform Space
=
= = =
97
xEy if and only if x y or P I (x) 0 PI (y). Define a metric d on H(a) by d(x,y) Ip I (x) - PI (y) 1if P2(X) P2(Y) or d(x, y) P I (x) + PI (y) otherwise. Then (H(a), d) is the hedgehog metric space with a spines. H(a) is a uniformly
=
=
paracompact metric space that is not locally compact. It is therefore a counter example to a statement by P. Fletcher and W. Lindgren in 1978 that the product of a uniformly locally compact space with a C-complete (uniformly paracompact) space is C-complete.
4.4 The Completion of a Uniform Space As seen in Chapter 2 (Section 2.1, Exercise 1), every metric space is a uniform space. Our first proposition below states that a complete metric space is also complete when considered as a uniform space. This may not seem surprising at first glance, but it might if we rephrase it in the following manner: In a metric space, the convergence of all Cauchy sequences forces the convergence of all Cauchy nets (of all cardinalities). The proof of this proposition is left as an exercise (Exercise 1). PROPOSITION 4.20 A complete metric space is also complete when considered as a uniform space.
Our next result will show that Proposition 4.20 can be generalized to uniform spaces; i.e., that the convergence of all Cauchy nets on a certain ordered set forces the convergence of all Cauchy nets. We will then use these specialized nets to construct a completion of the uniform space from equiValence classes of these specialized nets in a manner similar to the construction of the metric completion in Chapter 1. Our first step will be to define these specialized nets. For this let (X, J.l) be a uniform space and let v be a cofinal subset of J.l of least cardinality. Then v is a basis for J.l. Next, well order v by some ordering < such that the cardinality of each initial interval is less than the cardinality of v. From the proof of the Ordering Lemma, it can be seen that there exists a cofinal subset A of v (with respect to the ordering <*) such that the well ordering < is compatible with <* on A (i.e., U <* V implies that V < U). Then A is a well ordered basis for J.l of least cardinality such that the well ordering < is compatible with the directed ordering <* and the cardinality of each initial interval (with respect to <) is less than the cardinality of A. We call A a fundamental basis for J.l and a net {xa 1 a E A} C X a fundamental net with respect to A. Let {x a Ia ED} be a net in X and let (E, <) be a directed set. A function ~:E ~ D is called a compatible function if whenever a, ~ E E with a < ~ then ~(~) is not strictly less than ~(a) in D. A net {y 13 ~ E E} c X where y 13 = x ~(13) for each ~ E E is said to be contained in the net {x a}. Note that a cofinal function is a compatible function so a subnet of a net is contained in the 1
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net. Also note that a compatible function need not be cofinal so that a net {y j3 } contained in a net Ix a} need not span the net {x a} as a subnet would.
THEOREM 4.7 Each Cauchy net 4> contains afundamental Cauchy net \jf such that 4> converges to a point p if and only if\jf converges to p. Proof: (Construction of \jf) Let 4> = {x a 1a ED} where D is ordered by =:;. Well order D by < such that the cardinality of each initial interval is less than the cardinality of D. Then there is a cofinal E c D with respect to =:; such that < is compatible with =:; on E. Now 4>£ = {x a 1a E E} is a Cauchy subnet of 4>. For each a E A there is a Va E a and a residual Ra c E with respect to =:; such that 4>£(R a) C Va' Put Va = Star{Ua. a) and let U = {Va} and V = {Va}. Then U has the finite intersection property and for each Va E U. if V is another member of a with 4>£ eventually in V then V c Va' Also, V is directed by set inclusion; i.e.,ifa,bE AthereisacE AwithVccVanVb • ForeachaE AdefineEa={D E E 14>£(8) E Va} and for each b E A let ~(b) be the first element of Eb (with respect to <). Then ~:A ~ E is a compatible function and \jf = 4>£ © ~ is a fundamental net contained in 4>. Let a E A and pick bE A with b <* a. Then Vb c W for some WE a so \jf(b) = Xt;(b) E W. Also, if c E A such that c <* b then Vc c Vb' Thus \jf(c) E W. Put R = {c E AI c <* a}. Then R is residual in A and \jf(R) c W so \jf is Cauchy. (Proof 4> converges if \jf converges) Assume \jf converges to p E X and N is a neighborhood of p. Then there are a,b E A with a <* band Star(p,b) eN. Also, there is aCE A with c <* a and \jf(c) E Star(p,a). Then p and \jf(c) both belong to some Wa E a. Also \jf(c) E Vc and Vc C Va' Hence Star(p,a) c Star(Wa , a) c Wb for some Wb E b and Va C Star(Wa, a) c Wb. Now 4>(Ec) = 4>£(Ec) c Vc C Va so 4>(RJ c Va' Since Rc is residual in E with respect to =:;, it is cofinal in D with respect to =:;, so 4> is frequently in Va C Wb c Star(p,b) eN. Since 4> is Cauchy, 4> converges to p. (Proof \jf converges if 4> converges) Assume 4> converges to p and N is a neighborhood of p. Then there are a,b E A with a <* band Star(p,b) c N. Also, there is a residual ReD with respect to =:; with 4>(R) c Wa for some Wa E a such that p E Wa' Then RnE is residual in E and 4>£(RnE) c Wa' Since Ra is residual in E so is RnR a. Thus there is a 8 E RnRa with Xc = 4>£(8) E Wa' But Xc E Va, so WanVa i= 0. Also, \jf(a) = Xt;(a) implies \jf(a) E 4>£(Ea) eVa. Moreover, if C E A with c <* a then Vc C Va so \jf(C) E Va' Thus \jf is eventually in Va. Now a <* b implies there is a Wb E b with WauVa C Wb so P E Wb which implies Va C Star(p,b) eN. Hence \jf converges to p.Until now, we have avoided the definition of a subspace of a uniform space. Subspaces are one of the fundamental constructions to be dealt with in the next chapter. The approach so far has been to avoid these constructions (i.e., building new uniform spaces from old ones) as long as possible in order to examine the behavior of the generalizations of sequences and convergence
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(from metric spaces) and some of their topological consequences in uniform spaces as soon as possible. Strictly speaking, the completion of a uniform space can be introduced without introducing the concept of a subspace, but the most useful way of thinking of the completion is, that it is a (perhaps) larger uniform space that contains the original uniform space as a subspace. Consequently, we now give a simple definition of a uniform subspace, but will revisit the concept in the next chapter in a more formal setting. A subset A of a uniform space (X, 11) can be given a uniform structure that is derived from the uniformity 11 in the following way: for each U E 11 put UA = {UnAIU E U}. Then let IlA = {U A IU E Ill. It is easily shown that IlA is a uniformity on A. IlA is called the uniformity induced on A by the uniformity 11. IlA is also said to be the uniformity of 11 relativized to A. It should be noted that the uniformity IlA causes the identity mapping iA:A ~ X to be uniformly continuous. Our next task is to construct a completion for (X, 11) from the fundamental Cauchy nets. For this let L be the set of all fundamental Cauchy nets in X and define an equivalence relation - on L by - \jf if for each U E A there is a U E U such that and \jf are both eventually in U. Let X' be the set of all equivalence classes with respect to - and for each U E 11 put U~ = {U~ IU E U} where U~
= {' E x'i is eventually in U for each E '}.
Then Il~ = {U~ I U Ell} is a basis for a uniformity for X'. Let 11' be the uniformity generated by Il~. Then (X', 11) is a uniform space. Note that if x E X there is a unique <1>' E X' that consists of all fundamental Cauchy nets that converge to x. For each x E X let i(x) denote the unique equivalence class <1>' whose members converge to x. Then i:X ~ X' is well defined and since X is Hausdorff, i is one-to-one. Let U Eiland U E U. It is easily shown that i(U) = U~ni(X). Define i(U) = Ii(U) I U E U} for each U E 11. Then for each U' E 11' there is a U~ E Il~ that refines U' so U~ni(X) = {U~ni(X) I U E U} = i(U) refines U'ni(X). But then U refines j-l(Uni(X)) so i-1(U'ni(X)) E 11. Consequently i(X) is a uniform subspace. It is easily shown that i(X) is dense in X'. To show X' is Hausdorff let <1>' and \jf' be distinct elements of X'. Then there exists E <1>' and \jf E \jf' such that is not equivalent to \jf. Hence there is a U E 11 such that either or \jf is not eventually in U for each U E U. Note this implies that if VEil with V < U then either or \jf is not eventually in V for each V E V. Pick WEll with W** < U. Since both and \jf are Cauchy there are WI, W 2 E W with eventually in WI and \jf eventually in W 2. Now W* < U implies W InW 2 = 0 so W;nW; = 0.
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Let 8 E q>'. There is aWE Wand residual P,Q C A with 8(P), q>(Q) c W. Also, there is a residual RCA with q>(R) C WI' Then S = PnQnR is residual in A, 8(S) and q>(S) c Wand q>(S) c WI, so there is a V I E W* with W c V I. Thus q>' E V;. Similarly, there is a V 2 E W* with 'Jf' E V;. W** < U implies V I n V 2 = 0 so V; n V; = 0. Since V; and V; are neighborhoods of q>' and 'Jf' respectively in X', X' is Hausdorff. It remains to show (X', fJ.') is complete. Let {q>~ IU E A} be a fundamental net in i(X). Then for each U E A, there is an Xu E X with q>~ = i(xu)' Put 'Jf(U) =Xu' Then 'Jf:A ~ X is a fundamental net. It is easily seen that 'Jf is Cauchy. Let U' be an open set in X' containing 'Jf', Then there is a UAE fJ.A with 'Jf' E Star('Jf', U c U' so there is a VA E U' with 'Jf' E VA C U'. Then 'Jf = {Xu} is eventually in U so q>~ E U'. Thus {q>~ I converges to 'Jf'. A
)
Next let {q>~ IU E AI be a fundamental Cauchy net in X'. For each a E A there is a residual Ra c A with {q>~ I U E Ra) c u~ for some u~ E aA. Let V~ = Star(U~, a A) and put U' = {U~} and V = {V~}. Just as in the proof of the preceding theorem, for each a E A, if UAE a and {q>~} is eventually in U A, then UAnU~ "F 0 so U AC V~, and if a,b E A with a <* b then V~ c V~. A
Since i(X) is dense in X', for each a E A there is an Xa E X with i(xa ) EVa' Then {i(x a )) is a fundamental net in i(X). It is easily shown that {i(xa)} is Cauchy and consequently converges to some p' E X'. Let N be a neighborhood of p' and b E A with Star(p', b A) c N. Pick a E A with a <* b. Since {i(xa)} converges to p' there is aCE A with c <* a and i(xJ E Star(p', a A). Thus p' and i(xJ both belong to some W A E aA. Now i(xc) E V; and {q>~} is eventually in U; c V;. Hence there is a W~ E a A with V; c W~ so W~nU~ "F 0. Since i(xc ) E WAnW~ we have W A c Star(W~, a A) c W~ for some W~ E bA. Since U~ c W~, {q>~} is eventually in W~ c Star(p', bA) c N and we conclude {<1>~} converges to p'. Then by the above theorem, (X', fJ.') is complete. (X', fJ.) is called the completion of (X, fJ.). Since the function i is one-toone and uniformly continuous we often identify X with the dense uniform subspace i(X) of X'. In this case, i(x) = X for each X E X; i.e., the function i is the identity mapping on X. Using this identification, we notice that for each U E fJ. there is a U' E fJ.' such that U = i-I (U'), so U = U' nX = {U'nX I U' E U'}. Also, if V' E fJ.' then V' nX E fJ.. We record these results as: THEOREM 4.8 Every uniform space has a completion; i.e., ij(X, fJ.) is a uniform space, there is a complete uniform space (X', fJ.') and a one-to-one uniformly continuous function i:X ~ X' such that i(X) is a dense uniform subspace of X',
Theorem 4.8 corresponds to Theorem 1.15 for metric spaces. Theorems 1.16 and l.17 can also be generalized to uniform spaces as follows:
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THEOREM 4.9 iff is a uniformly continuous function on a subset A of a uniform space X into a complete uniform space Y then f has a unique uniformly continuous extensionj' to CI(A). ProoF Let!l be the uniformity on X and v the uniformity on Y. For each x A let {x a} be the constant fundamental Cauchy net (x a = x for each ex) that converges to x and for each x E CI(A) - A pick a fundamental Cauchy net {x a } c A that converges to x. Let V E v. Since f is uniformly continuous, there exists a U E 11 such that whenever a,b E A with a,b E U for some U E U, then f(a),f(b) E V for some V E V. But then the net {f(x a )} is Cauchy in Y for each x E CI(A). Since Y is complete, {f(x a )} converges to some x' E Y. Define j':CI(A) -7 Y by j'(x) = x' for each x E CI(A). Clearly,j'(a) = f(a) for each a E A so j' is and extension of f. E
To show j' is well defined, let x E CI(A) and let {Ya} be another fundamental Cauchy net converging to x. Suppose x' -j:. y'. Then there exists a V E v with S(x',v)nS(y',v) = 0. By the uniform continuity of f, there exists a U E 11 such that a,b E U E U implies f(a), feb) E V for some V E V. Since both {x a} and {Ya} converge to x, there exists a ~ such that ex > ~ implies x a,Ya E U for some U E U which in turn implies f(x a), f(Ya) E V for some V E V. But this is impossible since {f(x a)} is eventually in S(x',V) and {f(Ya)} is eventually in S(y',v). Hence x' = y' so j' is well defined. To show j' is uniformly continuous on CI(A), let W E v and pick V E v with V* < W. Since f is uniformly continuous on A, there exists a U E 11 such that whenever a,b E A with a,b E U for some U E U, then f(a),f(b) E V for some V E V. Let Z E 11 with Z* < U. Let X,Y E CI(A) such that X,Y E Z for some Z E Z. Since {xa} converges to x, {Yu} converges to Y, {f(x a )} converges to x', and {fey a)} converges to y', it is possible to choose a ~ such that ex > ~ implies x a,x E Z I, and YaS E Z2 for some Z j, Z2 E Z, and x',f(x a ) E VI and y',f(ya) E V2 for some Vj, V2 E V. Then Xa,Ya E Star (Z,Z) c U for some U E U. Hence f(x a), f(Ya) E V for some V E V. But then x',y' E Star(V,V) c W for some WE W. Therefore,x,y E Z which impliesj'(x),j'(y) E W, so j' is uniformly continuous. To show j' is unique, suppose F is another uniformly continuous extension of ffrom A to CI(A). Let x E CI(A) - A. Then {x a } converges to x. Since both j' and F are continuous, it follows that {j'(x a )} converges to j'(x) and {F(xa)} converges to F(x). But {xa} c A implies {j'(x a )) = {f(x u )} = {F(xa)}' Since a net in a Hausdorff space has (at most) a unique limit, j'(x) = F(x), so j' is unique. THEOREM 4./0 For each uniform space (X, 11), its completion (X', 11') is unique; i.e., if (X', 11') is another completion of (X, 11) then there is a uniform homeomorphism h:X' -7 X' that keeps each point of X fixed.
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4. Completeness, Cofinal Completeness and Uniform Compactness
Proof: Let j:X ~ X' denote the uniform homeomorphism that maps X into the completion X'. By Theorem 4.9 there exists a unique uniformly continuous extension }':X' ~ X'. Since}' is an extension of j we have J = © i where i:X
r
~
X' denotes the uniform homeomorphism that maps X into its completion X'. Similarly, there is a unique uniformly continuous extension t:X' ~ X" of i. Then i = (© j. For each x' E X' put g'(x') = i'(j'(x')). Then g' = }' © ( so g':X' ~ X' is uniformly continuous. If x E X then g'(x) = {(j'(x» = {(j (x» = i (x) = x so g' is an extension of the identity map i:X ~ X'. But by Theorem 4.9, this extension is unique so g' = i' (the identity map on X'). Hence}' © { = i' so { = (j'r i . But then X' and X' are uniformly homeomorphic. -
EXERCISES 1. Prove Proposition 4.20. 2. Show that the completion of a unifonn space is compact if and only if the uniformity is precompacr. 3. Show that a complete subspace of a Hausdorff uniform space is closed. 4. [K. Morita, 1951] Let 11* be the uniformity of!J.X. For each U = {U a I in 11, put U' = {U~ I where U~ = !J.X - Clf1X(X - U a) for each a. Then {U'I U E III is a basis for 11*. CAUCHY FILTERS AND WEAKLY CAUCHY FILTERS A filter F in a uniform space (X, 11) is said to be Cauchy if for each U E 11, there exists an F E F such that FeU for some U E U. It is said to be weakly Cauchy if for each U E 11, there exists aU E U with UnF of- 0 for each F E F. 5. Show that (X, 11) is complete if and only if each Cauchy filter in X converges. 6. [N. Howes, 1971] (X, 11) is cofinally complete if and only if each weakly Cauchy filter in (X, u) clusters. 7. [H. Corson, 1958] X is paracompact if and only if each weakly Cauchy filter in (X, u) clusters.
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4.5 The Cofinal Completion or Uniform Paracompactification Since each uniform space has a unique completion. it is natural to define a unifonn space (X'. f..l') to be a cofinal completion of the uniform space (X. f..l) if (X'. f..l') is cofinally complete and (X. f..l) is uniformly homeomorphic to a dense uniform subspace of (X'. f..l') and to ask: "When does (X. f..l) have a cofinal completion. and if a cofinal completion exists. is it unique?" Providing answer~ to these questions is the objective of this section.
It turns out these same techniques can be used to provide answers to a variety of other questions asked by K. Morita and H. Tamano. In the late fifties. Tamano considered the following question: "What is a necessary and sufficient condition for a uniform space to have a paracompact completion?" Although he did not arrive at a solution. he obtained elegant characterizations of completeness. paracompactness and the structure of the completion by means of a concept called the radical of a uniform space. These results were published in the Journal of the Mathematical Society ofJapan in 1960 (Volume 12, No.1, pp. 104-117) under the title Some Properties of the Stone-Cech Compactijication. We will analyze these results in Chapter 6. In 1970, K. Morita presented a paper titled Topological completions and M-spaces at an international Topology Conference held at the University of Pittsburgh. In Section 7 of that paper, five unsolved problems were listed including a special case of Tamano's question mentioned above and the question: "What is a necessary and sufficient condition for a Tychonoff space to have a LindelOf topological completion?" By a topological completion we mean the completion with respect to the finest uniformity. In this section and in later chapters, we provide answers to all these questions. All of the solutions are in terms of the cofinally Cauchy nets and their behavior with respect to various uniformities. Since some of these questions ask if the completion of a unifonn space has a certain topological property, one might be interested in knowing if there are solutions in terms of topological properties, or if topological properties exist that are only necessary or only sufficiem. Since the literature is rich with characterizations of paracompactness and the LindeWf property, one might expect a variety of solutions to these problems. However, at the present time, this area is largely unexplored. We define a uniform space to be preparacompact if each cofinally Cauchy net has a Cauchy subnet. Recall that a uniform space is precompact if every net has a Cauchy subnet. Consequently, preparacompactness is a generalization of precompactness. To continue the parallel. recall that complete precompact spaces are compact and the completion of a precompact space is compact. We will see that complete preparacompact spaces are paracompact and that the completion of a preparacompact space is paracompact.
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4. Completeness, Cofinal Completeness and Uniform Compactness
A unifonn space will be called countably bounded if each unifonn covering has a countable subcovering. If (X, Jl) is a unifonn space and (X*, Jl*) its completion, let u* denote the universal unifonnity for X* and let v be the uniformity induced on X by u*. The v unifonnity is called the uniformity derived from u* or simply the derived uniformity. A directed set D is said to be co directed or countably directed if for each countable {d;} c D there is a d E D such that d; ~ d for each i. A net {xa I a E D} is co directed if D is an co directed set. LEMMA 4.1 A completely regular space is LindelOf if and only if each co directed net clusters. Proof: Assume that each co directed net clusters in the completely regular space X. Suppose X is not Linde16f. Then there is a covering U of X having no countable sUbcovering. For each countable V c U put G(V) = u {u IU E V} and F(V) = X - G(V). Let S be the set of all countable subsets of U and for each V E S pick Xv E F(V). The assignment V -7 Xv defines an co directed net {xv IV E S} in X that clusters to some p E X where S is directed by set inclusion. Let U E U such that p E U. Then {xv} is eventually in F( {U}) and hence cannot be frequently in U which is a contradiction. Consequently X is Linde16f.
Conversely assume X is LindeWf and let {x a Ia ED} be an co directed net in X. Let U be a member of the e unifonnity of X. Then there is a countable subcovering V = {Vd such that VEe and V refines U. Suppose {x a} is not frequently in some member of V and put D; = {O E D Ix Ii E V;}. Then there exists a 0; E D such that 0 ~ 0; for each 0 E D; or else {Xli 10 E D;} would be frequently in V;. Since {x a} is CO directed there is a 0 E D such that 0; ~ 0 for each i. Now Xli E Vj for some j since V covers X. Hence 0 E D j which implies 0< OJ. But OJ ~ 0 which is a contradiction. Therefore Xa must frequently be in some member of V. Since V was chosen arbitrarily, {xa} is cofinally Cauchy with respect to e and therefore clusters by Theorem 4.4. • The following theorem appeared in an article titled On completeness in the Pacific Journal of Mathematics in 1971 (Volume 38, pp. 431-440). THEOREM 4.11 (N. Howes, 1970) Let (X, Jl) be a uniform space and v the derived uniformity. Then: (1) (X, Jl) has a paracompact completion if and only if (X,v) is preparacompact and (2) (X, Jl) has a LindelOf completion if and only if (X,v) is countably bounded and preparacompact, and (3) (X, Jl) has a compact completion if and only if (X,v) is precompact. Proof of (1): Let (X', Jl') be the completion of (X, Jl) and let u' be the universal unifonnity for X'. Then (X,v) is a dense unifonn subspace of (X', U). Assume (X,v) is preparacompact and that ",:D -7 X' is a cofinally Cauchy net with
4.5 The Cofinal Completion or Uniform Paracompactification
105
respect to u'. Since (X', Il') is complete, so is (X', u'). Let E = D x u' and define on E by (d, U') $ (e,v') if d $ e and V' <* U'. For each (d, U') E E put S(d,U,) = a for some a E X such that a and 'V(d) both belong to some U' E U'. Then the correspondence (d, U') ---+ Sed, U') defines a net S:E ---+ X. $
Let U' E u' and pick V' E u' with V' <* U'. Since 'V is cofinally Cauchy there is a cofinal C c D with 'V(C) c V' for some V' E V'. Put A = {(d,W')ldE C and W'<* V'}. Then A is cofinal in E. Let (d,W') E A. Then S(d,W') = Y E X such that y and 'V(d) both belong to some W' E W'. Since (d,W') E A, dEC which puts 'V(d) in V'. Consequently we have:
YE Star(y',W') c Star(Y',V') c U' for some U' E U'. Therefore SeA) c U' which implies S is cofinally Cauchy in (X', u'). But SeE) c X implies S is cofinally Cauchy in (X,v). Consequently S has a Cauchy sub net ~. But then ~ converges to some x' E X'. Therefore S clusters to x'. It remains to show that 'V clusters to x'. For this let 0 be an open set containing x'. Then there is a U' E u' such that x' E Star (x',U') cO where the members of U' are open sets. Pick V' E u' such that V' <* U'. Let S be cofinal in E such that S(S) c V' for some V' E V' containing x'. Put D(S)
=
{d E DI (d,W') E S for some W' <* V'}.
Then D(S) is cofinal in D. For each d E D(S), 'V(d) and S(d,W') are contained in some W' E W' for some (d,W,) E S where W' <* V' which implies 'V(d) and S(d,W,) are contained in some Y'j E V' for each (d,W') E S. But (d,W') E S implies S(d,W') E V'. Hence 'V(d)
E
V'uY~ c Star(Y',V') c U'
for some U' E U' . But x' E V' implies x' E U/. Then U' c 0 since Star (x',U') c O. Consequently 'V(d) E 0 for each dE D(S) which implies 'V clusters to x'. Therefore each cofinally Cauchy net in X' clusters which implies (X', u') is cofinaIly complete. But then X' is paracompact by Theorem 4.4. Conversely suppose X' is paracompact. By Theorem 4.4 (X', u') is cofinally complete. Let 'V be a cofinally Cauchy net with respect to v. Since (X,v) is a uniform subspace of (X', u,), we know that 'V is cofinaIly Cauchy in (X', u'). Also since (X', u') is cofinally complete, \jI clusters to some p EX'. But then 'V has a subnet e that converges to p. Then e is Cauchy in (X', u'). But SeX and therefore e is Cauchy in (X,v). Consequently, each cofinaIly Cauchy net in (X,v) has a Cauchy subnet so that (X,v) is preparacompact.
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4. Completeness, Cofinal Completeness and Uniform Compactness
Proof of (2): Assume first that X' is LindelOf. Then X' is paracompact and hence (X', u') is cofinally complete. But then (X,v) is preparacompact as was shown in part (1). Next let V E v. Then V has a uniform refinement U consisting of closed sets. For each U E U put U' = Clx'(U) and let U' = {U' Iu' E U). Then U E u' and hence has a countable subcovering say {V;). Then { V,) covers X. In fact, if P E X then P E X' which implies P E vj for some positive integer j. Hence P E Clx'(Uj)' Let 0 be open in X such that P E O. Then 0 = 0' nX for some 0' that is open in X'. Now P E 0' which implies 0' nVj :j. 0 since P E Clx'(Uj)' Then there is atE 0'nVj. t E Vj implies t E X so we have (O'nX)nVj :j. 0. Hence OnVj :j. 0 so that P E Clx(Uj) = V j . Since {V j ) covers X there exists some {Vj ) C V such that uVj = X. Consequently X is countably bounded.
Conversely assume (X,v) is countably bounded and preparacompact. (X',j.!') is paracompact by part (1) so that (X', u') is coftnally complete by Theorem 4.4. Let U' E u'. Since X' is paracompact there exists a locally finite open refinement ~' = {Vp IBE B). Since X is norrn~l we c~n shrink V' to an open covering W = {W~ IB E B} such that Clx'(W~) C V~ for each BE B. Then W' is also locally finite. Since W' is an open covering in a paracompact space X', it must be a member of the universal uniformity u'. But then W = {W'nX IW' E W'} belongs to v which implies there is a countable subcovering {W~,) c W. Since W' is locally finite in X', {W~,) is locally finite in X'. Hence Ui'=l Clx'(W~) = CIX,(Ui'=1 W~) = Clx'(X) = X' since X is dense in X'. Therefore {Clx'(W~)} covers X and
for each positive integer i. Hence {V~,} covers X'. But since V' refines U' there exists a countable subcovering {V~,) c U' that covers X'. Therefore (X',u') is also countably bounded. We are now in a position to show that X' is LindelOf. We will use the fact that (X', u') is cofinally complete and countably bounded to show that each co-directed net in X' clusters. We then invoke Lemma 4.1 to obtain the desired result. Let \jf:D ~ X' be an co directed net and let U' E u'. Then U' has a countable subcovering say {V;}. Put D j = {d E D I \jf(d) E V;) for each i and suppose D j is not cofinal in D for each i. Then there exists a d j E D for each i such that d ~ d j for each dE D j • Since \jf is co directed there is a do E D such that d j ~ do for each i. Since {V;} covers X', \jf(d o) E vj for some positive integer j which implies do E D j which in turn implies D j is cofinal in D since D = ui'=lD j • Therefore \jf is frequently in V j • Hence \jf is cofinally Cauchy and consequently must cluster. Thus X' is LindelOf by Lemma 4.1. Proof of (3): By Problem 3 of Section 4.4, (X, j.!) has a compact completion (X', j.!') if and only if j.! is precompact. But if v is precompact then j.! c v implies j.! is precompact which in turn implies X' is compact. Conversely, if
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X' is compact then /-l' = u' which implies /-l = v and /-l is precompact which implies v is precompact. -
COROLLARY 4.2 Let u be the universal uniformity for a completely regular T I space X. Then: ( 1) (X, u) has a paracompact completion if and only if it is preparacompact, (2) (X, u) has a Linde16f completion if and only if it is countably bounded and preparacompact. COROLLARY 4.3 The completion of a preparacompact uniform space is cofinally complete. COROLLARY 4.4 A countably bounded cofinally complete uniform space is Linde16j. COROLLARY 4.5 A paracompact space is Linde16f if and only if it is countably bounded with respect to the universal uniformity.
In view of the existence of a unique completion for each uniform space, it is natural to ask when a uniform space has a cofinal completion, and if a cofinal completion exists, when is it unique? THEOREM 4.12. A uniform space has a cofinal completion if and only if it is preparacompact. In this case it is unique and identical to the ordinary completion (i.e., the ordinary completion is cofinally complete). Proof: (Sufficiency) Suppose (X, /-l) is preparacompact and (X', /-l') is its completion. It needs to be shown that (X', /-l') is cofinally complete. In the sufficiency part of the proof of Theorem 4.11.(1), it was shown that if v is the uniformity derived from /-l rather than the uniformity /-l and if (X,v) is preparacompact then (X', /-l') is cofinally complete. The proof that (X', /-l) is cofinally complete based on the assumption that (X, /-l) is preparacompact is similar so it will not be included here. (Necessity) Assume (X, /-l) has a cofinal completion (X', /-l'). Let Ix a I a E D) be a cofinally Cauchy net in (X, /-l). Then {xa} clusters to some p E X' so {xa} has a subnet {y~} that converges to p. But then {y~} is Cauchy in (X',/-l'). Since {Y/3} lies in X it is Cauchy in (X, /-l). Consequently, every cofinally Cauchy net in (X, /-l) has a Cauchy subnet so (X, /-l) is preparacompact. Since (X', /-l') is cofinally complete it is also complete. But then (X', /-l') is a completion of (X, /-l). Since the completion of a uniform space is unique this means the cofinal completion (when it exists) is identical to the completion.-
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EXERCISES 1. Show that a completely regular T 1 space is paracompact if and only if it is complete and preparacompact with respect to the universal uniformity.
2. Show that a uniform space (X, J.!) is countably bounded if and only if each directed net is cofinally Cauchy.
(t)
3. Let J.! be a uniformity that generates the topology of X and let U be an open covering of X. We call P E X a residue point for U with respect to V E J.! if there does not exist aU E U with Star (p, V) c U. Let H v be the set of residue points with respect to V. Put Fv = CI(H v ) and G v = X - Fv· Then W = {G v I V E J.!} is called the residue covering derived from U. Show that W is an open covering of X. 4. Show that a completely regular T 1 space is paracompact if and only if each residue open covering with respect to the universal uniformity has a finite subcovering. 5. Show that a completely regular T 1 space is paracompact if and only if each heavily bound open covering (Exercise 5, Section 4.3) with respect to the universal uniformity has a finite subcovering. 6. Show that in a completely regular T I space the following are equivalent: (a) the LindelOf property, (b) each residue open covering with respect to e has a finite subcovering, (c) each heavily bound open covering with respect to e has a finite subcovering. 7. A topological space is said to be entirely normal if the collection of all neighborhoods of the diagonal in X x X forms an entourage uniformity (Exercise 4, Section 2.1). Show that entire normality and almost-2-fully normal (Exercise 6, Section 3.2) are equivalent. 8. A net in a topological space X will be called cofinally t1 Cauchy if for each open covering U of X there is apE X such that the net is frequently in S(p, U). X is cofinally t1 complete if each cofinally t1 Cauchy net clusters. Show that an entirely normal space is paracompact if and only if it is cofinally t1 complete. 9. Show that a metacompact space is cofinally t1 complete. 10. Show that entire normality implies collection wise normality.
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11. Show that a metacompact space is paracompact if and only if it is collection wise nonnal. 12. RESEARCH PROBLEM We would like to have a characterization of preparacompactness in tenns of coverings, perhaps analogous to the characterization of precompactness in tenns of coverings. The following result is in this direction, but something better is needed. A collection ~ of subsets of X is said to be a directed collection if for each A,B E ~, AuB E ~.
THEOREM A uniform space (X, !!) is preparacompact if and only if each heavily bound collection having no finite subcollection that covers X is contained in a heavy directed collection. See Exercises 3 and 4 of Section 4.3.
Chapter 5
FUNDAMENTAL CONSTRUCTIONS
5.1 Introduction In this chapter we consider some important constructions of uniform spaces from other uniform spaces. Our first concern will be to consider the so called classical constructions that are studied for most spaces and algebraic structures that arise in the study of mathematics, namely subspaces, sums, products and quotients. Our approach will be to derive these constructions from a few fundamental concepts. These fundamental concepts take the form of limits of collections of uniformities. We will make these concepts precise in the next section. The reason for waiting until now to introduce the fundamental constructions is that many of the interesting results about these constructions involve concepts related to completeness. For example, it can be shown that the product of complete uniform spaces is a complete uniform space. Without these other concepts, it is difficult to appreciate the utility of these constructions. We opted instead to first develop the theory of completeness in uniform spaces and then see how it applies to these constructions. After the classical constructions, we proceed to some other constructions (some of which also apply to a variety of other structures such as topological spaces). We should remark that we have already seen one such construction; namely, the completion of a uniform space. Of great interest to us will be the concepts of the hyperspace of a uniform space, the inverse limit of a directed family of uniform spaces, the weak completion of a uniform space and the spectrum of a uniform space. If the hyperspace is complete, the original space is said to be supercomplete. Like cofinal completeness. supercompleteness turns out to be a strong form of completeness. In fact, it was recently discovered that supercompleteness is a property that lies between completeness and cofinal completeness. This result will allow us to strengthen several results from Chapter 4. The concept of supercompleteness was introduced by S. Ginsburg and J. Isbell in 1954 in an abstract titled Rings of convergent functions that appeared in the Bulletin of the American Mathematical Society, Volume 60, page 259. But the definition of supercompleteness in the simple form mentioned above is
5.2 Limit Uniformities
III
due to Isbell in a paper that appeared in the Pacific Journal of Mathematics in 1962 titled Supercomplete spaces (Volume 12, pp. 287-290). The concept of the inverse limit of a family of uniform spaces leads to the concept of the spectrum of a uniform space, and the spectral analysis of uniform spaces whose spectra exist. The study of uniform spaces by means of their spectra has been pursued by B. Pasynkov (1963) and K. Morita (1970). Also, the concept of the inverse limit of a directed family of uniform spaces is needed to express the weak completion of a space with respect to a uniformity introduced by K. Morita in 1970. Finally, we will introduce the locally fine corefiection and the sUbfme corefiection of a given uniform space. These were introduced by Ginsberg and Isbell in 1959 and shown to be equivalent by J. Pelant in 1987. This new construction has a variety of uses. In the last section of this chapter we will introduce the concepts of categories and functors. This material is optional as far as being needed in later chapters. It is included for two reasons. First, much of the literature of uniform spaces after Isbell makes use of the category theoretic vocabulary, and second, category theory, like set theory, is a tool that is often helpful in the study of uniform spaces. The term "coreflection" refers to a category theoretic concept, and it will be shown that the locally fine coreflection satisfies this notion. Thereafter, categories and functors will only be used in the exercises.
5.2 Limit Uniformities Given a collection {'ta } of topologies for a set X, there exists a finest topology that is coarser than each 'ta called the infimum topology on X with respect to {'ta } and denoted infa't a . Similarly, there exists a coarsest topology that is finer than each 'ta called the supremum topology on X with respect to {'t a } and denoted supa't a . To see this, note that n'ta is a topology for X that is coarser than each 'tao If't is another topology for X that is coarser than each 'ta , then 't c n'ta so n'ta is the finest topology for X that is coarser than each 'ta (i.e., infa't a = n'ta)' Also note that if L is the collection of topologies that are finer than each 'ta , then L 7= 0 since the discrete topology belongs to L. Consequently, nL is a topology for X. For each a, 'ta C cr for each cr E L. Hence'ta c nL for each a so nL is finer than each 'tao If't is another topology finer than each 'ta , then 't E L which implies nL C 'to Therefore, nL is the coarsest topology that is finer than each 'ta (i.e., sUPa't a = nL). It can be shown (Exercise 1) that a sub-base for sUPa't a is the set u't a . The concepts of infimum and supremum uniformities are analogous to the topological concepts. If lila} is a collection of uniformities for a set X, there is a finest uniformity that is coarser than each Ila called the infimum uniformity and denoted infa/la. There also exists a coarsest uniformity that is finer than each Ila called the supremum uniformity and denoted supalla. By Theorem
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112
2.1, ul-4x is a sub-basis for a uniformity !-lon X which implies I ni'=1 Ua,
IUa,
E
1-4x, for some i = 1 ... n} is a basis for!-l. If!-l' is another uniformity for X that is finer than each I-4x then !-l' must contain each ni'=1 Ua, so !-l C !-l'. Hence!-l = supa!-la. Similarly, if L = {O'y I'Y E G} is the collection of all uniformities that are coarser than each !-la then L :F 0 since I X} E L. Put v = sUPyO'y. If V E v then by the definition of v, there exists 0'1 .. . O'n ELand Wi E O'i for each i = 1 ... n such that ni'=1 Wi < V. For each ex, O'i C I-4x for each i = 1 ... n so Wi E I-4x for each i = 1 ... n which implies ni'=1 Wi E !-la. Hence V E I-4x so !-l C !-la for each ex. So VEL and by the definition of v, v is finer than any other member of L so v is the finest uniformity that is coarser than each !-la. Therefore v = infal-4x.
PROPOSITION 5.1 If l!-la} is a collection of uniformities for a set X and I 'ta } is the corresponding collection of topologies generated by the !-la's. then the topology generated by Supal-4x is Supa't a . Proof: Let 't be the topology generated by supa!-la. Since supa!-la is finer than each !-la, 't is finer than each 'ta so sUPa't a C 'to If U E 't then for each P E U there exists a basic covering n7=1 Ua, E Supa!-la such that S(p, n7=1 Ua) C U. Now S(P,U a) E sUPa'ta for each i = 1 ... n since sUPa't a is finer than 'ta, for each i = 1 ... fl. But then ni'=1 S(P,U a,) E supa't a . If x E ni'=1 S(P,U a,), then X,P E U a, for some U a, E Ua, for each i = 1 ... n. But then x,P E ni'=1 U a, E ni'=1 Ua, which implies x E S(p, ni'=1 U a,). Therefore
which implies U E supa't a . Hence't
C
sUPa't a which implies 't = sUPa't a .-
Unfortunately, a proposition similar to 5.1 does not hold for infa!-la. Whereas u!-la and u'ta are sub-bases for the uniformity supa!-la and the topology sUPa't a respectively, n!-la is not necessarily even a uniformity while n'ta = infa'ta . The reason for this is explored in the exercises at the end of the section. Letf:X ~ Y be a function from a set X into a topological space Y. Thenf determines a coarsest topology 't/ on X such thatfis continuous. A basis for the open sets in 't/ is the collection B = {f-I (U) I U is open in Y}. If F = {fa:X ~ Y a Iex E A} is a family of functions from X into topological spaces Y a , we define the topology 'tF to be suPa't/rJ.' 'tp is called the projective limit topology for the collection F. It can be shown (Exercise 3) that a sub-basis for 'tF is the set S = lral (U a) I U a is open in Y a for some ex} and that 'tp is the coarsest topology on X such that each fa in F is continuous. Similarly, if g:Y ~ X is a function from a topological space Y onto a set X, there is a finest topology 't g on X such that g is continuoUS. A basis for the
5.2 Limit Uniformities
113
open sets in 't g is the collection B = (U c Xlg-I(U) is open in Yl. If G = {g a: Y a ~ X I ex E A} is a family of functions from topological spaces Ya onto X, we define the topology 'tG to be in!a't gex • 'tG is called the inductive limit topology for the collection G. It can be shown (Exercise 4) that a basis for 'tG is the set B = {U c X Ig"ixl (U) is open in Ya for each ex E A} and that 'tG is the finest topology on X such that each g a EGis continuous. The concepts of projective and inductive limits are also relevant to collections of uniformities. The only difference is that now we are interested in making collections of functions uniformly continuous rather than simply continuous. If !:X ~ Y is a function from a set X into a uniform space (Y, v), then by Exercise 2 of Section 2.2, ! determines a coarsest uniformity f..lf on X such that! is uniformly continuous. A basis for f..lf is l (v) = If-I (V) IV E v}. Let F = {!a:X ~ (Ya,v a )I ex E A} be a collection of functions from X into uniform spaces (Ya,v a ). Define f..lF to be sUPaf..lfex' f..lF is called the projective limit uniformity on X. It can be shown (Exercise 5) that a sub-basis for f..lF is the set S = (fal (Va) I Va E va for some ex} and that f..lF is the coarsest uniformity on X such that each!a in F is uniformly continuous.
r
Similarly, if g:Y ~ X is a function from a uniform space (Y,v) into X, by Exercise 3 of Section 2.2, g determines a finest uniformity f..l g on X such that g is uniformly continuous. A basis for f..l g is the set B = {U I U is a covering of X such that g-l (U) E v}. Let G = {g a: Y a ~ X Iex E A} be a family of functions from uniform spaces (Ya,v a ) onto X. We define the uniformity f..lG to be in!af..l g ex and call f..lG the inductive limit uniformity on X. It can be shown (Exercise 5) that a basis for f..lG is the set {U IU is a covering of X and g"ixl (U) E Va for each ex} and that f..lG is the finest uniformity on X such that each g a in G is uniformly continuous. PROPOSITION 5.2 The topology generated by the projective limit uniformity is the projective limit topology.
The proof follows immediately from Proposition 5.1 and the definitions of projective limit topology and projective limit uniformity.
EXERCISES 1. Let {'ta } be a collection of topologies for a set X. Show that the set S = u'ta is a sub-basis for SUPa't a . 2. Let F = Ifa:X ~ Y a Ia. E A} be a family of functions from a set X into topological spaces Y a and let 'tF be the projective limit topology with respect to F. Show that the set S = {fal (U a) I U a is open in Y a for some a.} is a sub-basis
114
for 1F and that continuous.
5. Fundamental Constructions 1F
is the coarsest topology on X such that each
fa
in F is
3. Let C = {ga:Ya ~ Xla E A} be a family of functions from topological spaces Ya onto a set X and let 1c be the inductive limit topology with respect to G. Show that B = {U c X I g~.} (U) is open in Ya for each a E A} is a basis for 1C and that 1c is the finest topology on X such that each g a in G is continuous. 4. Let F = If a:X ~ (Y a,v a ) I a E A} be a family of functions from a set X into uniform spaces (Ya,v a ) and let /-!F be the projective limit uniformity with respect to F. Show that the set S = {fal (Va) I Va E Va for some a} is a subbasis for /-!F and that /-!F is the coarsest uniformity on X such that each fa in F is uniformly continuous. 5. Let G = {g a: Y a ~ X I a E A} be a family of functions from uniform spaces (Ya,v a ) onto a set X and let /-!c be the inductive limit uniformity with respect to C. Show that the set B = {U I U is a covering of X and g~l (U) E Va for each a} is a basis for /-!c and that /-!c is the finest uniformity on X such that each g a in G is uniformly continuous.
5.3 Subspaces, Sums, Products and Quotients In this section, the classical constructions will be examined as special cases of inductive and projective limit uniformities. It will be seen that subspaces and product spaces are special cases of projective limit uniformities, while sum and quotient spaces are examples of inductive limit uniformities. If (X, /-!) is a uniform space and Y c X, define i:Y ~ X by iCy) = Y for each y E Y. Let /-!i denote the projective limit uniformity on Y determined by the single function i. It is left as an exercise (Exercise 1) to show that the uniformity /-!i is identical to the subspace uniformity introduced in Section 4.4. Given a collection F = {X a Ia E A} of uniform spaces with uniformities /-!a for each a, we define the uniform product space (P,1t) to be the Cartesian product set P = I1X a (introduced in the Forward) together with the projective limit uniformity 1t determined by the collection {p a Ia E A} of all canonical projections p a:P ~ X a' Recall that the members of P are functions fA ~ uX a such that f( a) E X a for each a. The mappings p a are defined by p a (f) = f( a) for each f E P. When dealing with uniform product spaces such as P = I1X a, it is customary to refer to the individual uniform spaces X a from which the product space is constructed as the coordinate (uniform) spaces. PROPOSITION 5.3 The topology associated with the uniform product space is the topology of the topological product of the coordinate spaces (considered as topological spaces).
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115
Prool: Let'ta be the topology associated with Ila for each a and let 't denote the product topology on P = IlX a' It will suffice to show that for each sub-basic U E 't containing x E P, there exists a V E 1t with S(x,v) c U and for each S(x,W) where W E 1t, there exists a basic V E 't with x EVe S(x,W). If U E 't is a sub-basic open set then U = If E pi/(p) E U ~} for some open U ~ c X~. Since x E U, p~(x) = x(P) E U~ which implies there exists some V~ E /..l~ with S(x(P),V~) c U~. Since 1t is the projective limit uniformity determined by the collection {p a} of canonical projection, by Exercise 5 of Section 5.2, P ~l (V~) is a sub-basic memberof1t. Let/E S(X'Pi31(V~)) = (Pi31(v~)ix(p) E V~ E V~}. Then IE Pi3 1 (V~) for some V~ E V~ that contains x(P). Hence I(P) E V~ C S(x(P),v~) c U ~ so lEU. Therefore, Sex, Pi3 1 (V~)) c U so 't is coarser than the uniform product topology on P. Conversely, suppose W is a basic member of 1t. Since 1t is the projective limit uniformity determined by the collection {p a} of canonical projections, by Exercise 5 of Section 5.2, there exists a finite collection W a1 . .. Wan of coverings in 1la1 ••• /..la n respectively such that W = n7=1Pa~1 (Wa). For each i = 1 ... n choose Va, E W a, with x(a;) EVa,. Then x E V = n7=1 Va, = n7=1P a~l (Va,) and V is a basic member of 'to If I E V then both x and Ibelong to n7=1Pa~1 (Va). But S(x,W) = (IE PiX,fE n7=1Pa~1(W a) for some W a , E W a, for each i = 1 ... n}. Hence x EVe S(x,W) so 't is finer than the uniform product topology on P. Therefore, 't is the uniform product topology on P. Quotient uniformities are an important example of inductive limit uniformities. If (X, /..l) is a uniform space and R is an equivalence relation on X, let q:X ~ X/R be the canonical projection of X onto the quotient set X/R (introduced in Chapter 0). If we then give Q = X/R the inductive limit uniformity /..lq determined by the single function q, then (Q, /..lq) is called the quotient uniform space with respect to R. It is left as an exercise (Exercise 5) to show that the topology associated with /..lq is the topology of the quotient (topological) space X/R, also introduced in the Introduction. A word of caution is in order here. The space Q may not be Hausdorff (see Exercise 7). To insure that the quotient of a Hausdorff uniform space is again Hausdorff, we need the concept of a uniform quotient (see Exercise 8). It is of interest to observe that every onto mapping I:X ~ Y from X to a set Y determines an equivalence relation Rf on X by taking (x,y) E Rf if I(x) = I(y). Moreover, the canonical projection q:X ~ X/Rf determines a one-to-one function g:X/Rf ~ Y such that I = g © q. g is the function defined by g(x*) = I(x) where x* is the equivalence class containing x. If Y is also a uniform space with uniformity v, then by Proposition 0.20, I is continuous if and only if g is continuous. It is easily seen that this continuity condition also holds for uniform continuity; i.e., I is uniformly continuous if and only if g is uniformly continuous. In fact. if g is uniformly continuous, then I must be uniformly continuous since it is the composition of the uniformly
116
5. Fundamental Constructions
continuous function g and the canonical projection q (which is uniformly continuous by definition). Conversely, if f is uniformly continuous then for each V E v, 1(V) E f.l implies that q [{~l (V)] E f.l q (the quotient uniformity on X/R r ). We will show that g~I(V) = qW1(V)] thereby showing g to be uniiormly continuous. For this let V E V and suppose x* E g ~l (V). Then g(x*) =f(x) E V. Let y E 1[((x)]. Then fey) = f(x) which implies y E x* so q(y) = x*. Thus x* E q(f~I(V» so g~I(V) C q(f~l(V». Next suppose z* E q(f~I(V». 1(V) which implies fez) E V. But g(z*) = fez) so z* E g ~l (V). Then Z E Therefore q(f~l (V» c g~l (V) which implies g~l (V) = q(f~l (V».
r
r
r
In Section 5 of Chapter 1, we introduced the concept of the metric space associated with a pseudo-metric space (X, d) by defining the equivalence relation - in X by x - y if d(x,y) = 0 and defining a metric p on X/- by p(x*,y*) = d(p ~l (x*), P ~l (y*» where p:X ~ X/- was the canonical projection and x* and y* were the equivalence classes with respect to - containing x and y respectively. A similar construction can be done for uniform spaces; i.e., if (X,f.l) is a non-Hausdorff uniform space, it is possible to construct a Hausdorff uniform space from X. For this, define R 1.1 on X by (x,y) E R 1.1 if x E S(y,U) for each U E f.l. Then let q:X ~ X/R 1.1 be the canonical projection of X onto the quotient set X/R 1.1 and let f.l q be thelluotient uniformity determined by q. Then (X/R 1.1' f.lq) is a uniform space called the Hausdorff uniform space associated with (X, f.l). PROPOSITION 5.4 The Hausdorff uniform space associated with (X.f.l) determines a Hausdorff topology and the canonical projection q:X ~ XIR 1.1 is both an open and closed mapping. Proof: Let x* and y* be distinct members of X/R I.l' Then y does not belong to x* which implies y does not belong to S(x,U) for some U E f.l. By Exercise 6 of Section 5.2, B = {V I V is a covering of X/R 1.1 and q ~l (V) E f.l} is a basis for f.l q. Now q(U) = {q(U) I U E U} covers X/R I.l and U < q~l [q(U)] so q(U) E f.l q. If y* E S(x*, q( U» then both x* and y* E q(U) for some U E U which implies x,y E U so Y E S(x,U) which is a contradiction. Hence y* does not belong to S(x*,q(U». Let WE f.l q such that W* < q(U). Then S(x*,W)nS(y*,W) = 0 so (X/R >" f.lq) determines a Hausdorff topology.
To show that the canonical projection q is both an open and closed mapping, first let U be an open set in X and suppose x* E q(U). Then q(x) = q(y) for some y E U. Therefore, x E y*. Let V E f.l such that S(y,v) cU. Then x E S(y,V) which implies x* E S(y*, q(V» c q(U). Hence q(U) is open in X/R 1.1 so q is an open mapping. If F is closed in X and if x* does not belong to q(F) then x E F which implies there exists aWE f.l such that S(x,W)nStar(F,W) == 0. Let V E f.l with V* < W. Suppose S(x*, q(V»nq(F)"#- 0. Then there exists some y* E X/R>, such that x* ,y* E q(V) for some V E V and such that y* E q(F). y* E q(F) implies that there exists some z E F with q(z) = q(y) which in tum implies y E z*, so y E S(z, V). Moreover, x* ,y* E q(V) implies there are r, s
5.3 Subspaces, Sums, Products and Quotients
117
E V with q(r) = q(x) and q(s) = q(y) so Y E S(s,V) and x E S(r,V). Consequently, there exist V I, V2, and V 3 E V such that x,r E V I, s,y E V2, and y,Z E V 3 . But then there are WI and W 2 E W with V I UV 2 C WI and V2 UV 3 cW 2 • NOWX,SE WI ands,zE W 2 impliesW l nW 2 "#0,W I cS(X,W) and W 2 C Star (F, W). Hence S(x,W)nStar(F,W) "# which is a contradiction. Therefore, S(x*, q(V»nq(F) = 0. Thus q(F) must be closed so q is also a closed mapping. -
°
Our last example of an inductive limit uniformity is the uniform sum of a collection {(X a, /-1a)} of uniform spaces. We define a new uniform space LX a by first defining the points of LX a to be the ordered pairs (x, a) where x E X a' The mappings i a:X a ---7 LX a defined by i a(X) = (x, a) are called the canonical injections. The uniformity a of LX a is defined to be the inductive limit uniformity determined by the collection {i a} of canonical injections. The uniform space (LX a,a) is called the uniform sum of the uniform spaces X a' The proof of the following proposition is left as an exercise (Exercise 6).
PROPOSITION 5.5 The topology associated with the uniform sum of a collection of uniform spaces is the topology of the disjoint topological sum of these spaces considered as topological spaces.
The canonical injections essentially transfer the uniformities /-la from the spaces X a onto the disjoint "pieces" i a(X a) of LX a and since a is the inductive limit uniformity determined by the i a's, a is the finest uniformity on LX a that makes all the i a 's uniformly continuous.
EXERCISES 1. Show that if (X, /-1) is a uniform space and Y c X, that the projective limit uniformity /-1, on Y determined by the function i:Y ---7 X such that i(y) = Y for each y E Y is identical with the subspace uniformity on Y introduced in Section 4.4. 2. Show that a net in a product space converges to a point p if and only if its projection in each coordinate subspace converges to the projection of p. 3. Show that a net in a product uniform space is Cauchy if and only if the projection of the net in each coordinate subspace is Cauchy. 4. Show that the product complete.
nx a
of complete uniform spaces (X a' /-1a) is
5. Let (X, /-1) be a uniform space and let R be an equivalence relation on X. Show that the topology associated with the quotient uniform space with respect
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5. Fundamental Constructions
to R is the topology of the quotient (topological) space X/R introduced in the Introduction. 6. Prove Proposition 5.5. 7. Show that if (X, /.1) is a non-normal uniform space with the finest uniformity u, that there exists an equivalence relation R on X such that X/R is not Hausdorff. UNIFORM QUOTIENTS A uniform relation in a uniform space (X, /.1) is an equivalence relation R on X such that for each pair of non-equivalent points x,y E X, there exists a sequence {Un} C /.1 satisfying: (1) If xRx' and yRy', then x' and y' are in no common member of U I . (2) If Un +1 E U n +l , there exists a Un E Un such that if p E U n +1 and pRq, then Seq, Un+d C Un.
A uniform quotient is a quotient uniform space X/R where the equivalence relation R is a uniform relation. 8. Show that a uniform quotient of a Hausdorff uniform space is Hausdorff. 9. Show that the canonical projection q:X uniformly continuous.
--7
X/R of a uniform quotient is
UNIFORML Y LOCALLY COMPACT SPACES A uniform space is said to be uniformly locally compact if it has a uniform covering consisting of compact sets. For each countable ordinal a put X a = a + 1 and let '!a be the order topology on X a' Then X a is compact, so X a has a unique uniformity /.1a consisting of all open coverings. Let (L, /.1) be the sum of the collection {(X a, /.1a) I a < WI }. Let R be the equivalence relation on L defined by x - y if and only if x and y belong to the same X a' Let Y = W1 with the order topology. 10. [So Ginsburg and J. Isbell, 1959] Y has a unique uniformity with a basis consisting of all finite open coverings. R is a uniform relation on (L, /.1). L/R = Y. Let q:L --7 Y be the canonical projection. There exists a U E Il such that q(U) = {q(U) I U E U} is not a uniform covering in Y. L is complete, but q(L) = Y is not complete. L is uniformly locally compact.
5.4 Hyperspaces
119
5.4 Hyperspaces Let (X, /.1) be a uniform space and let X' denote the set of all non-void closed subsets of X. If H, K E X' with He Star (K, U) and K c Star (H, U) for some U E /.1, then Hand K are said to be U-close, denoted by IH - K I < U. Note that this relationship is reflexive, i.e., I H - K I < U implies I K - HI < u. If U E /.1 and F E X', put B(F,U) = {K E x'il F - KI < U) and let U' = {B(F,U) IFE X'). Then define /.1* = {U' IU E /.1). We want to show that /.1* is the basis for a uniformity /.1' on X' but first we establish some useful lemmas. LEMMA 5.] If u* < V.IH-FI < U.and IK-FI < U.then IH-KI
<
v.
Proof: IH - F I < U implies H c Star (F, U) and F c Star (H, U). IK - F I < U implies K c Star (F, U) and F c Star (K, U). Therefore H c Star (Star (K. U),U) c StarCK. V). Similarly K c Star (H. V) so IH - K! < V.LEMMA 5.2 If u* < V and FE X' then S(F.U') c B(FY) c S(F.V'). Proof" Let H E S(F.U'). Then H E B(K.U) for some K E X' such that F E B(K,U). Therefore, H c Star(K,U), K c Star(H.U), Fe Star(K,U). and K c Star (F, U), so H c Star (Star (F. U). U) c Star (F. V). Similarly, F c Star (H, V) which implies H E B(F,V). Therefore S(F,U') c B(F,V). B(F,v) c S(F,V') follows from the definition of S(F.V').PROPOSITION 5.6 /.1* is a basis for a uniformity /.1' on X'. Proof: We need to establish (1) and (3) of the definition of a uniform space. We can do this simultaneously by showing that if V'.W' E /.1*, there exists a U' E /.1* with U' <* V'nW'. For this let U E /.1 such that U* < VnW. Let U' c U'. Then U' = B(E,U) for some E E X'. Let F E Star (U',U'). Then F E B(K,U) for some K E X' such that K E B(E,U) which implies IF - K I < U and IK - E I < U. By Lemma 5.1, IF - EI < V so F E B(E,V) E V'. Therefore, Star (U',U') c B(E,v) so U' <* V'. Similarly. U' <* W'so U' <* V'nW'. Thus /.1* is a
basis for a uniformity /.1' on X'. The uniform space (X', /.1') is called the hyperspace of (X, /.1). Hyperspaces of topological spaces have also been studied. The original notion of a hyperspace is due to F. Hausdorff (see Mengenlehre, 3rd edition, Springer, Berlin, 1927). Hausdorff defined a metric on the set of non-empty closed and hounded suhsets of a given metric space. L. Vietoris generalized the concept to topological spaces (see Bereiche Zweiter Ordnung, Monatshefte fur Mathematik und Physik, Volume 33, 1923, pp. 49-62). N. Bourbaki introduced the uniformity for the hyperspace of non-void closed subsets of a uniform space in terms of an entourage uniformity (Topologie General. Paris. Hermann. 1940).
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In this section we translate Bourbaki's approach in tenns of covering unifonnities. PROPOSITION 5.7 The hyperspace of a uniform space is Hausdorff. Proof: Let (X, 11) be a unifonn space and (X', 11') its hyperspace. Choose H, K E X' with H :f. K. Then either H contains a point not in K or K contains a point not in H. Without loss of generality we may assume there is an x E H such that x does not belong to K. Theorem 2.6 can be used to show there exists U,V E 11 with U* < V such that S(x,V)nStar(K,v) = 0. Suppose K E 8(H,U). Then K c Star (H, U) and H c Star (K, U) which implies x E Star (K, U) which is a contradiction. Therefore, K does not belong to 8(H,U) so X' is T I. Since X' is regular, it is also Hausdorff. LEMMA 5.3 A net {Fa} in X' converges (clusters) to some F and only lffor each V E 11. {F a} is eventually (frequently) in 8(F,v).
E
X'. if
Proof: Let V E 11 and choose U E 11 with U* < V. If {F a} converges to F then {F a} is eventually (frequently) in S(F,U'). But by Lemma 5.2, S(F,U') c 8(F,V). Conversely, if for each V E 11, {Fa} is eventually (frequently) in 8(F,V), then it is eventually in S(F,v,) by Lemma 5.2, so {F a} converges to F.LEMMA 5.4 {Fa} is Cauchy in X' if and only iffor each V exists a ~ such that for each a ~~. Fa E 8(F J3,v).
E
11, there
Proof: Let U E 11 with U* < V. Since {F a} is Cauchy, there exists a ~ such that for some HEX', Fa E 8(H,U) for each a ~~. Then for each a ~ ~, Fa c Star (H, U) and H c Star (F a,U). Also, F J3 c Star (H, U) and H c Star (F J3,U), Therefore, Fa C Star (Star (FJ3'U),U) c Star (FJ3,V). Similarly, FJ3 c Star (F a, V) so Fa E 8(F J3,v) for each a ~ ~. The converse it obvious since 8(F J3'V) E V'. -
Let {Fa} be a net in X'. A point p E X, each of whose neighborhoods meets cofinally many of the Fa's is said to be a cluster point of {F a} (in X). PROPOSITION 5.8 The set K of cluster points of a net {F a} in X' is closed. If {F a} converges in X'. it converges to K. Proof: Let p be a limit point of K and let U be a neighborhood of p. Then U contains a point k E K. Since k is a cluster point of {F a}, U frequently meets {F a} which implies p E K. Hence K is closed. Next assume {Fa} converges to F E X'. Suppose x does not belong to F. Then, as pointed out in Proposition 5.7, there is a U E 11 with S(x,U)nStar(F,U) = 0. By Lemma 5.3, IF a} is eventually in B(F,U). Therefore, there is a ~ with Fa E B(F,U) for each a ~ ~ which implies Fa c Star (F, U) for each a ~~. Hence x is not a cluster point of
5.4 Hyperspaces
121
F. Consequently, F c K and there can be no points of K that are not in F, so F =
K.Our first undertaking with hyperspaces will be to characterize supercompleteness (see Section 5.1 for the definition) in terms of the behavior of a certain class of nets in the original space X. In 1988. B. Burdick informed the author that he had shown that cofinal completeness implies supercompleteness and wondered if the reverse implication held. The author pointed out that real Hilbert space (Chapter 1) can be shown to be a supercomplete metric space that is not cofinally complete (see Exercise 1). Thereafter, Burdick discovered the following characterization of supercompleteness that yields the implication: cofinal completeness => supercompleteness as a corollary. Burdick defines a net Ix a}, a E A, in a uniform space (X, IJ.) to be almost Cauchy if for each U E IJ. there exists a collection C of cofinal subsets of A such that for each K E C, Ixy I'Y E K} c U for some U E U, and such that uC is residual in A. Clearly, Cauchy nets are almost Cauchy and almost Cauchy nets are cofinally Cauchy. His characterization of supercompleteness is that a uniform space is supercomplete if and only if each almost Cauchy net clusters. Burdick's proof depends on a theorem of Isbell on partially convergent functions. Isbell's development of partially convergent functions will be given in a later section. For now we present an alternate constructive proof directly from the definitions of supercompleteness and almost Cauchy nets. This constructive proof has the advantage of making clear just how the clustering of almost Cauchy nets is related to supercompleteness. The proof is rather long, so we decompose it into two lemmas plus the main constructions. We define a net IF a} in X' to be proper if IF a} is ordered by set inclusion (a ~ ~ if and only if F 13 c F a) and if Fa C F E X' for some a then F ElF a}. LEMMA 5.5 A proper net IF a} in X' is Cauchy ~ with F 13 c Star(F )jar each a.
V E IJ. there is a
a.v
if and only if for
each
Proof: Assume IF a} is a proper Cauchy net in X'. Let V E IJ.. Pick U E IJ. with U** < V. By Lemma 5.4, there is a 'Y such that for each a : : : 'Y, Fa E B(F y,U). Now CI(Star(F y,U)) = F 13 for some ~ since IF a} is proper. Let a::::: 'Y. Then Fa E B(F y.U) which implies F y c Star (F a, U) so Star (F y,U) c Star(F a,U*), which implies F 13 c Star (F a, V). If a < 'Y then F y c Fa so F 13 c Star (F y' V) c Star (F a, V). Consequently, there is a ~ with F 13 c Star (F a, V)
for each a. Conversely, assume that for each V E IJ. there is a ~ with F 13 c Star (F a, V) for each a. If a : : : 13 then F~ c Star(F a,V) and Fa C FI3 c Star (FI3'V), Therefore Fa E B(F ~.V) for each a::::: 13 so IF a} is Cauchy.-
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LEMMA 5.6 (X, J..l) is supercomplete if and only if each proper Cauchy net in X' converges. Proof: Let {Fa} be a Cauchy net in X'. For each a put H a = u{F 131 ~ ~ a}. Let {S y} denote the collection of elements of X' containing some H a where {S y} is directed by set inclusion (y < 8 if and only if S8 C S y). Clearly {S y} is proper. Since {F a} is Cauchy, so is {S y }. To see this, let V E J..l and pick U E J..l with U* < V. By Lemma 5.4, there exists a ~ such that Fa E B(F 13 ,U) for each a ~~. Therefore, Fa c Star (F 13'U) for each a ~ ~ which implies CI(H 13) c Star (Fj3,V). Now CI(Hj3) = S8 for some Sii E {Sy}. For each y~ 8, Sy C 58 which implies S y c Star (F 13, V). Also, for each y ~ 8, there exists a A ~ ~ with HA c Sy which implies FA c Sy and Fj3 c Star(FA,U) so Fj3 c Star (Sy,U) which implies 5 y E B(F 13'V) for each y ~ 8. Therefore, {S y} is a proper Cauchy net. Then by hypothesis, {S y} converges to its set of cluster points K. Let V E J..l. Pick U E J..l with U*** < V. By Lemma 5.3, {S y} is eventually in B(K,U) so there exists a 8 with S y E B(K,U) for each y ~ 8. Now there exists a ~ with CI(H 13) c 58' Since {Fa} is Cauchy, there exists a ~ with Fa E B(F ~ ,U) for each a ~~. Pick A such that A ~ ~ and A ~~. Then CI(H A) c CI(H 13) c S 8 so CI(H A) = S cr for some (j ~ 8. Hence CI(H A) E B(K,U). Since A ~ ~, Fa E B(F~,U) for each a ~ A which implies Fa C Star(F~,U) and F~ c Star(F a,U) for each a ~ A. Thus CI(H A) C Star(F ~ ,U*) and F ~ c Stare CI(H A)'U*), Hence CI(H A) E B(F ~ ,U*). Consequently, we have
Then multiple applications of Lemma 5.1 yields I Fa - K 1 < V for each a ~ A so {F a} converges to K. Hence X is supercomplete. Burdick's theorem appeared in an article titled A note on completeness of hyperspaces published in the Proceedings of the Fifth Northeast Topology Conference (1991) by Marcel-Dekker. The following proof is due to the author. THEOREM 5.1 (B. Burdick, 1991) A uniform space (X, J..l) is supercomplete if and only if each almost Cauchy net clusters. Proof: Assume each almost Cauchy net clusters and suppose (X, J..l) is not supercomplete. Then by Proposition 5.8 and Lemma 5.6, there exists a proper Cauchy net {F a}, a E A, in X' that does not cluster to its set of cluster points K. There are two cases to consider here. First, K might be the empty set, in which case K would not belong to X' and therefore could not be a cluster point of {F a}. On the other hand, K may not be empty. This is the case we will assume first. Later we will see that an argument similar to the one we will use here can be used to show that under the hypothesis of this theorem, K cannot be the empty set.
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5.4 Hyperspaces
If K *- 0, there exists a U E /-1 such that {Fa} is eventually outside B(K,U) which implies there exists a ~ with Fa not contained in B(K,U) for each a ~~. Let P E K. Then there exists a cofinal C c A with P E Star (F y,U) for each Y E C. Pick 8 E C with 8 > ~. Then F Ii c F ~ which implies P E Star(F~,U). Therefore, K c Star(F~,U) which implies F~ is not contained in Star (K, U) or else F ~ E B(K,U) which is a contradiction. Similarly, Fa is not contained in Star (K, U) for each a ~~. Suppose y is not greater than or equal to~. Then there is a 8 such that y ~ 8 and ~ ~ 8 which implies F 8 is not contained in Star (K, U) and F 8 C F l' Hence F y is not contained in Star (K, U). Consequently, Fa is not contained in Star (K, U) for each a E A. Put U = Star (K, U) and for each a E A let H a = Fa - U. Then IHa} is a Cauchy net in X' that is directed by set inclusion. Let IGy}, yE B, be the collection of elements of X', directed by set inclusion, that contain a memher of IH a}. Then I G y} is a proper Cauchy net in X'. Let D = I (x, y) E X x B 1x E G y} and define ~ on D by (x, y) ~ (y, 8) if Y ~ 8. For each (x, y) E D put \jI(x, y) = x. Then \jI:D ~ X is a net which we will show is almost Cauchy. For this let WE /-1. Pick U E /-1 with U* < W. Since IGy} is stable, there exists a A with G y E B(GA,U) for each y ~ A. Pick x E Star(GA,U). Then some W E W contains Star(x,U). Let V = IW E wlw contains Star(x,U) for some x E Gd. Let R = I(y,~) E 01 ~ ~ A}. Then R is residual in D. Let (y,~) E R which implies ~ ~ A which in tum implies G~ c Star(GA,U) so y E Star(GA,U) which implies y E S(x,U) for some x EGA' Then there exists aWE V containing y. Therefore, \jI(y, ~) = yEW. Let C w = l(y,~)E RI\jI(y,~)E WforsomeWE V}. ThenulCwlWE V}=R. So I C w W E V} is a family of suhsets of D whose union is residual in D such that \jI(C w ) eWE V. It remains to show that C w is cofinal in D for each WE V. For this let WoE V and (y,~) E D. Then there exists an Xo E G A with S(xo,U)cW o. Pick(z,8)E Rwith(xo,A)~(z,8)and(y,~)~(z,8). ThenG8 eGA which implies Z EGA' But (z, 8) E R implies 8 ~ A which in tum implies G 8 E B(G A'U), so G A C Star (G 8,U). Therefore, x E S(s,U) for some S E G 8 which implies S E S(xo,U). Then (xo, A) ~ (s, 8), (y,~) ~ (s, 8), and \jI(s, 8) E S(xo,U) cWo. Hence Cwo is cofinal in D so \jI is almost Cauchy. 1
°
By hypothesis, \jI clusters to some P E X. Let V be a neighborhood of p. Then there exists a cofinal C c D with \jI(C) c V. Let a E A. Then Fa = G ~ for some p E B. Pick x E G~. Then there exists (y, 8) E C with (x, ~) ~ (y, 8) and \jI(y, 8) E V. Therefore, ~ ~ 8 which implies G 8 C G ~ and \jI(y, 8) = y E G 8 C G~ c Fa SO F arN *- 0. Hence p is a cluster point of IF a} so P E K. But \jI c X - U which is closed, so p E X - U which is a contradiction. Next we have the case K = 0. In this case, let D in the argument above be the set {(x, y) E X x A Ix E Fa}. Define:5; on D as before and define \jI:D ~ X by \jI(x, a) = x for each (x, a) E D. Then just as in the argument above, we can show that \jI is almost Cauchy and hence clusters to some p E X. But then as
124
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the argument above shows, p E K, so K *' 0. Consequently, (X, /J.) must be supercomplete. Conversely, assume X is supercomplete. Let {xa)' a E A, be an almost Cauchy net in X. For each a put H a = {x~ IB ~ a) and Fa = CI(H a). Then {Fa) isanetinX'. Let VE /J.andpick UE /J.with U*< V. Since {xa) is almost Cauchy, there exists a collection {C y), Y E B, of cofinal subsets of A and a collection {U y) C U such that uyC y is residual in A and {x~ I BE C y} C U y for each y. Pick 0:0 E A such that B E uyC y for each B ~ 0:0. Then H ao C uyC y' For each y pick Vy E V such that Star (Uy,U) C V y' We want to show Fao CUyU y. LetYE Fao' IfYE Hao,clearlYYE UyV y . Supposeyisalimit point of H ao' Let U E U be a neighborhood of y. Then there exists an x ~ E Hao with x~ E U. Now x~ E Hao implies x~ E U y for some y so U C V y. Hence F ao C UyV y. Let a1 E H ao' Then F al C F ao which implies F al C Star (F ao' V). Pick y E B. Let BE C y such that B > a1' Then x~ E U y and x~ E F al which implies x~ E VynF al so uV y C Star(F al ,V). But F ao C UyV y so F ao C Star(F al ,V). Hence F al E 8(F ao,v) for each a1 E H ao so {Fa} is Cauchy. Since X' is complete, {Fa) converges to its set of cluster points K. Let p E K. Let W be a neighborhood of p in X. Then there exists a cofinal C C A with Fan W *' 0 for each a E C. If Y E F anW then y E H a or y is a limit point of H a' In either case, W contains some x~ with B ~ a. For each a E C pick K(a) ~ a such that x K(a) E Fan W. Then {x K(a) Ia E C} c W is cofinal in A. Hence. {x a} clusters top. COROLLARY 5.1 Cofinal completeness implies supercompleteness which in turn implies completeness. COROLLARY 5.2 (1. 1sbell. 1962) If X is paracompact. then it is supercomplete with respect to u. Notice that our proof that supercompleteness implies each almost Cauchy net clusters only relies on the fact that there exists a cluster point of {Fa). Consequently, the existence of a cluster point for each proper Cauchy net in X' is equivalent to supercompleteness. Furthermore, if p is a cluster point for the proper Cauchy net F = {Fa) and U E /J., then for each U E U containing p, UnF ~ *' 0 for cofinally many F ~'s in F. Let y be any index. Then there exists a 8 ~ y with UrlF B *' 0 and since F is directed by inclusion we have UnF y *' 0. Therefore, S(p, U)nF a*,0 for each Fa E F. Thus we can pick an Xu E S(P,U)nF a for each U E /J.. Then {xu} converges to p which implies p is a limit point of Fa for each a. Therefore, p E Fa for each a since each F a is closed. Hence nF a*' 0. Similarly, if nF a*,0 then any p E nF ex is a cluster point of F. We record these observations as
5.4 Hyperspaces
125
PROPOSITION 5.9 If (X. 11) is a uniform space then the following statements are equivalent: (I) (X. 11) is supercomplete, (2) Each proper Cauchy net in X' has a cluster point and (3) For each proper Cauchy net F in X', nF 0.
*'
EXERCISES 1. A function f:X -t Y from a uniform space (X, 11) into a uniform space (Y,v) determines a function j':X' -t Y' where X' and Y' are the hyperspaces of X and Y respectively. j' is defined by j'(A) = Cl y (f1Al) for each A E X'. j' is called the hyperfunction of f. Show that j' is uniformly continuous if and only if f is uniformly continuous. THE HYPERSPACE OF A METRIC SPACE 2. The Hausdorff distance h between two (closed) sets A and B in a metric space (X, d) is defined as the maximum of sup I dCa, B) I a E A) and sup {d(A,b) I b E B) where d(x, S) =in!{ d(x,y) lYE S) for any subset S. Show (a) h is a metric on HX (the hyperspace of X) that generates the topology of HX so that the hyperspace of a metric space is again a metric space. (b) If (X, d) is complete then (HX, h) is complete. THE HYPERSPACE OF A COMPACT SPACE 3. Show that the hyperspace of a compact space is compact. 4. Show that a discrete space of the power of the continuum, with the uniformity determined by all countable coverings, is complete but not supercomplete. 5. Show that real Hilbert space is supercomplete but not cofinally complete. 6. Define the limit inferior of a net {Fa) in X' to be the set inf{F a) = {x E X Ifor each neighborhood U of x, {Fa) eventually meets U}. Similarly, define the limit superior to be the set suplF a) = Ix E X I for each neighborhood U of x, {F a) frequently meets U}. If K is the set of cluster points of {F a} then (a) sup{Fa}=K=inf{F a }
(b) (X, 11) is supercomplete if and only if whenever IF a} is Cauchy then for each U E I..l there is a ~ with Fa C StarCK, U) for each ex ~ ~.
5. Fundamental Constructions
126
7. Use the results of Exercise 8 to show that if X is paracompact, then it is supercomplete with respect to u, without reference to cofinally Cauchy or almost Cauchy nets (i.e., do not appeal to Corollary 5.1).
5.5 Inverse Limits and Spectra We begin our discussion of inverse limits of topological and uniform spaces with a special case that will motivate the concept in general. Let {X n ) be a sequence of topological spaces where n ranges over the non-negative integers and suppose that for each positive integer n, there is a continuous function In:Xn --7 X n- 1 • The sequence of spaces and mappings {Xn,fn) is known as an inverse limit sequence. Inverse limit sequences are often represented by diagrams like the one below:
In
In-!
r::
If m < n, then the composition mapping = fm+1 © fm+2 © ... © fn is a continuous mapping from Xn to X m. Consider the sequence of points lx n ) such that for each n, Xn E Xn and Xn = In+l (x n+!). Then {xn ) can be identified with a point of the product space n;:'=oXn by means of the function g:l --7 uX n, where 1 denotes the non-negative integers, defined by g(n) = Xn- By means of this identification, the set Y of all such sequences can be considered to be a subset of n;:,=ox n. Then Y, equipped with the subspace topology, is called the inverse limit space of the sequence {X n, fn)' Y is denoted by lim.,...Xn or by X~. The functions fn are sometimes called bonding maps.
LEMMA 5.7 If {Xn,fn) is an inverse limit sequence with onto bonding maps and for some counrable set {an) of positive inlegers there is a set {x an } such that xa n E Xa n for each n and such that if m < n, then la~m (x an ) = x am ' then there eXlsrs a point in lim.,...X n whose coordinare in Xa n is xa n for each n. Proof: First assume {an) is infinite. For each positive integer m, there exists a least positive integer an such that an ~ m. If an = m put Xm = x an ' otherwise let Xm = f,:: (x an )· Then IXm) is a point of lim.,...X n such that xa n E Xa n for each fl. Next, assume Ian) is finite. Then there exists a greatest an, say aj. If m < aj put Xm = ;:;. (xa). For each m ~ aj, assume Xm has already been defined. Since fm+l is onto, pick Xm+! E Xm+! such that fm+l (x m+!) = Xm • Then by induction we can complete the sequence IXm) such that Ix m } E lim.,...Xn and for each n, xa n E XOn " Lemma 5.7 illustrates the fundamental property of inverse limit spaces, that for any coordinate Xb all elements of lim"",Xn having that coordinate have
5.5 Inverse Limits and Spectra
127
all other coordinates Xm with m < k determined by the inverse limit sequence, whereas there may be some room for choice of the coordinates Xm with m > k. If the bonding maps are not onto, limf-X n may be the empty set. An example of this occurs when the Xn are all countable discrete spaces, say Xn = {x;:'} and the bonding maps fn:Xn ~ X n -1 are of the form fn(x;:') = x;:,~il' Clearly {Xn> fn} is an inverse limit sequence, but if we begin with a point x~, it is only possible to pick the first m coordinates before we reach xi and there does not exist an X'J'+1 such that fm+l (x'J'+l) = Xl' Consequently, there do not exist any sequences {xnl such that Xn E Xn and fn(x n ) = Xn-l for each n > 1. Therefore, limf-X n = 0. Under certain conditions, we can assure that limf-X n l' 0. For instance. THEOREM 5.2 If each space Xn in the inverse limit sequence /XnJn} is a compact Hausdorff space then limf-X n i= 0. Proof: For each positive integer n let Yn C DXn be defined by {Yi} E Yn if for each j < n, Yj-l =Jj(y). Then limf-Xn = nYn' We will show that for each n, Yn is closed. Suppose p E DXn - Yn- Then for some j < n we have Jj+l (Pi+l) i= Pi' Since Xj is Hausdorff, there exists disjoint open sets Uj and Vj containing Pj and Jj+l (Pj+l) respectively. Put Vj+l = J;ll (Vi) and let Up be a basic open set in DXn containing p and having Uj and Vj+l as its /h and j+ Fi factors respectively. Then no point of Yn lies in Up since if q = {qn} E Up, then qj+l E V j +1 which implies qj E Vj so q does not belong to Up. Therefore, Yn is closed. Since {Yn } is a decreasing chain of closed sets in the compact space DXn, nYn i= 0 so limf-X n i= 0. •
There are many applications of inverse limit spaces of inverse limit sequences in topology. What we are interested in here is extending the concept to uniform spaces, and extending it in a more general setting. For this first notice that by changing the definition of an inverse limit sequence {Xn.fn} so that the Xn are now uniform spaces and the bonding maps are uniformly continuous, we get an inverse limit uniform space limf-Xn that is a uniform subspace of the uniform product space DXno This follows from the parallel between product topological spaces and product uniform spaces [see Section 5.3]. But what we have in mind is something more general than this. Let (D,<) be a directed set and suppose that for each a ED, X a is a topological space. Further suppose that whenever a, ~ E D with a :s; ~, there is a continuous function fl:X i3 ~ X a, and that these functions satisfy the following rules: (1)
(2)
.fl;. is the identity function for each a fl © fr =.ff whenever a < ~ < y.
Let Y= {xala
E
D} andF=
UWI
a, ~
E
E
D
D with a:S; Pl. Then the pair {Y,F}
128
5. Fundamental Constructions
is called an inverse limit system. Since the set of non-negative integers forms a directed set, it is clear that an inverse limit sequence is a special case of an inverse limit system. To define the inverse limit space of an inverse limit system, we let Ix a} denote a net in uX a such that x a E X a for each a E 0 and such that if a < ~ in 0 then .!If (x ~) = X a' Then {x a} can be identified with a point p of nx a having coordinates p a = X a' The collection of all such nets, under this identification, is a subspace of nx a that we denote by lim<-X a and call the inverse limit space of the system {Y, F).
LEMMA 5.8 lim<-X a of the inverse limit system I Y, F) is a closed subspace of the product space nx a. The proofs of Lemma 5.8 and of the following Theorem are left as exercises (Exercises 1 and 2).
THEOREM 5.3 The inverse limit space of an inverse limit system of compact Hausdorff spaces is a compact Hausdorff space, and if each space of the inverse limit system is non-empty, then the inverse limit space is non-empty. Again, if in the definition of the inverse limit system of {Y, F), we require the X a E Y to be uniform spaces and the bonding maps in F to be uniformly continuous functions, we get an inverse limit uniform space lim<-X a that is a uniform subspace of the uniform product space nx a' We now give an important example of an inverse limit uniform space that we will use later on. Let (X,v) be a uniform space and let {
5.5 Inverse Limits and Spectra
129
that U~ < U~, then we put A < 11. Then (A, <) is countably directed. To see this let {An} be a countable set of members of A. For each positive integer k let U~ = U}n ... nU} and put
<1>~
ill
X
~
X)l
.t ·A
II ~
X/
.t <1>~
t)l
<1>~
iA X
-7'
XA
~
X/
where i~ denotes the identity map on X considered as a mapping from X)l onto X A' It is left as an exercise [Exercise 3] to show that <1>~ is uniformly continuous with respect to the metric uniformities on X/
veX) is called the weak completion of X with respect to v. A uniform space is said to be weakly complete if each co-directed Cauchy net clusters. We will show that (X,v) is uniformly homeomorphic to a dense uniform subspace of veX) and that veX) is weakly complete. The weak completion of a uniform space with respect to a uniformity was introduced by K. Morita in 1970 in a paper titled Topological completions and M-spaces published in Sci. Rep. Tokyo Kyoiku Daigaku 10, No. 271, pp. 271-288. To see that veX) is weakly complete, let 'I':D ~ veX) be an co-directed Cauchy net. Then for each A E A,1t A © 'I':D ~ X/
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5. Fundamental Constructions
belong to CI('VA(R a». Let m be the least positive integer such that S(y,2-m)nStar(CI('V)...(R a», 2- m) = 0. Now for each n > m, y E 'V),JR n) which implies 'V".
contradiction. Therefore,
5.5 Inverse Limits and Spectra
and only
THEOREM 5.4 (K. Morita. 1970) The mappinR <1>.X if X is weakly complete with respect to v.
131 ~
veX) is onto
if
Proof: We first show that if X is weakly complete with respect to v then is onto. For this, let y E veX). Put 'I' = {~1 (1t,Jy» IA E A}. Let {An} be a
countable collection of indicies in A. Since (A, <) is countably directed, there exists ayE A such that An < y for each positive integer n. Therefore, for each n
Consequently, <1>;;1 (1t-y(y» c n;;'=1 ~~ (1tAn (y» so 'I' has the countable intersection property. Also, if U E v, let
5. Fundamental Constructions
132
Therefore, the point y = {YA IA E A} lies in veX). We want to show that the net <1> © ",:D ~ veX) converges to Y so that if <1> is onto then", converges in X (by Proposition 4.8) and hence X is weakly complete with respect to v. Let S(y,1t~l (<1>A(U~))) be a basic open neighborhood of y in veX). Now <1>A(",(R(A,n») c S(YA' <1>A(U~» so 1t~1 (<1>A(",(R(A, n»))) c 1t~1 (S(YA' <1>A(U~)))
C
S(y, 1t~1 (<1>A(U~))).
By the definition of <1>:X ~ veX), it is clear that for each x E X, <1>(x) E 1t~l(<1>A(X» so
-
EXERCISES 1. Prove Lemma 5.8.
2. Let (X,v) be a uniform space and let {Il>A IA E A} be the collection of all normal sequences of open uniform coverings. For each A E A let X A. and X/tf>A. be the pseudo-metric space and the metric space derived from X with respect to
5.6 The Locally Fine Coreflection
133
<1>1.. as constructed in this section. For each pair A, f.l E A with A < f.l, let <1>~ be the bonding map from X/ll into X/'A' Show that <1>~ is uniformly continuous
with respect to the metric uniformities on X/J.l and X/'A' 3. Let (X,v) be a uniform space and let veX) be the weak completion with respect to v. Let <1>:X --j veX) be the function defined in this section by <1> (x ) = {<1>A(x) IA E A I for each x E X. Show that <1> is uniformly continuous. 4. Let {Y, F) be an inverse system of uniform spaces where Y = {X a I a E Dl and F = {fll a, ~ E D with a < ~}. Let C be a cofinal subset of D. Put Z = {X a Ia E C) and G = {fll a, ~ E C with a < ~ I. Show that the inverse limit of {Z,G} is homeomorphic with the inverse limit of {Y, Fl.
5.6 The Locally Fine Coreflection A uniform space is said to have some property P uniformly locally if there is a uniform covering such that each member of this covering has property P. For instance, a uniform space (X, f.l) is said to be uniformly locally compact if there exists a U E f.l such that each U E U is compact. There are a variety of such properties P that have been studied as uniformly local properties. Similarly, a collection {D a} of subsets of X is said to be uniformly discrete if there exists a U E f.l such that for each pair a,~, StareD a,U)nStar(D ~,U) = 0. In a paper titled Some operators on uniform spaces (Transactions of the American Mathematical Society, Volume 93, pp. 145-168), published in 1959, S. Ginsburg and J. Isbell introduced the notion of locally fine uniform spaces, and the concept of the locally fine corefiection (a finer uniformity) associated with a given uniformity. To define these concepts, they made use of the notion of a uniformly locally uniform covering of a uniform space (X, f.l). A covering V of X is said to be uniformly locally uniform if there exists a U E f.l such that for each U E U, the covering {Un V IV E V} of U is a uniform covering (in the induced subspace uniformity on U). This means that if U = {Va} then for each fixed a, we may replace V with a uniform covering Va = {vg} such that {VanVgl = Va I Va, the trace of Va on Va (Va restricted to Va). A uniform space (X, f.l) is then defined to be locally fine if each uniformly locally uniform covering belongs to f.l. It is easily seen that (X, f.l) is locally fine if and only if it is closed under what Ginsburg and Isbell call the staggered intersection operation {V anVg} where {Val E f.l and {vg} is a uniform covering of Va for each a. Next, they constructed a new uniformity A(f.l) for X, that is the coarsest uniformity for X finer than f.l, that is locally fine. The uniform space (X, A(f.l» is called the locally fine coreflection of X (the meaning of the term corefiection will be explained in the next section - it is not necessary for understanding the uniformity A(f.l».
134
5. Fundamental Constructions
To define A(j..l) we proceed via transfinite induction using the concept of the derivative of a uniformity. Let j..l0 = J.! and define the derivative j..li of j..l to be the family of all coverings which have refinements of the form (U o.nVg:), where (U a I E j..l and for each fixed a, I Vg:) E j..ll U u' It is easily shown that j..li is a filter of coverings with respect to refinement, and that j..l0 C j..ll. Then, for each ordinal number a, we can define j..lo. by the inductive rules 110.+1 = (j..lU)! for all ordinals a and for limit ordinals B by j..l~ = u (j..lu Ia < B). Clearly, there must exist a last derivative 11K = j..lK+1. We put A(j..l) = 11K. It is easily shown that A(j..l) is a filter of coverings with respect to refinement. It is considerably more difficult to show that A(f.l) is a uniformity for X. The proof given by Ginsburg and Isbell (1959) and the revised proof given by Isbell in his book (1964) are not entirely correct. The proof given below incorporates the idea of the lemma they tried to prove prior to the theorem in order to avoid some of the problems with their proofs.
THEOREM 5.6 For a complete metric space (X, j..l) where j..l is the metric uniformity, A(j..l) = u.
Proof: By Stone's Theorem (Chapter I), u consists of all coverings that have open refinements. Each member of j..l has an open refinement. This property is preserved under the operation of taking the derivative. Consequently, A(j..l) C u. To show the converse, let V = (V ~) be an open covering of X. For each positive integer n, let Un = {U~} be a uniform covering of X consisting of the spheres of radius 2- n • Let P be the set of all indices (n, a) of the sets U~ in the covering Un and < be the partial ordering on P defined by: (n+ 1, B) < (n, a) if Ur- l nU~ *- 0, and (n+k, y) < (n,a) if there exists a chain (n+k, y) < (n+k-I,8) < ... < (n, a). Let S be the set of all (n. a) with Star(U~,Un) C V ~ for some B. Since (X, j..l) is complete and each U~ is of radius 2- n , each infinite descending chain in P converges to some limit point x. Since some £-sphere of x is contained in some V 13, only finitely many of the U~ in the chain can have indices in the chain P - S. We now define an ordinal valued function 8 on P - S as follows: if every predecessor of p is in S then 8(p) = O. If 8(q) is defined for each predecessor q of p in P - S, then 8(p) = a where a is the least ordinal greater than all these 8(q). First. we observe that this defines 8 on all of P - S, for otherwise there would exist some PiE P - S on which 8 is not defined which implies there is some predecessor P 2 < P 1 in P - S for which 8 is not defined, and so on, so we get an infinite descending chain ... < Pn < ... < P 2 < P 1 of elements of P - S for which 8 is not defined which is a contradiction. Next, we show that for each (n,a) E P - S, {U~nU~ I(k,Y) E S} E A(j..l) I U~. For this we induct on the values of 8. To get the induction started (and to prove another result we will need later), let p either belong to S or have o(P) == O. In either case, each predecessor of P is in S. Let Q 0 = {s E sis is an immediate
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predecessor of pI, and for each non-negative integer n let Qn +1 = {S E sis is an immediate predecessor of some q E QIl I. Then uQn is the set of all predecessors of pin S. Now let U~ = {UpnUsls E Qol E ~Iup = ~o on Up. Then Uf = {Up nUs Is E U'~l Q;} E ~l on Up and ug c Uf. Continuing this process by induction, we obtain a sequence of coverings {U~ I = {Up nUs Is E Ui~nQi I with U~ c U~+l and U~ E ~n for each non-negative integer n. Consequently. U~ = {UpnUs Is E uQn I E A(~) IUp. Now P = (n,a) for some non-negative integer n and some index a. Then for each positive integer m such that m < n, if (m, '() does not precede (n. a) then U~nU;' = 0. Hence if s E S with s not in uQn and UDnU S ::F 0 then s = (k,~) for some non-negative integer k with k ~ n and some index~. But then U~ E Uk. Since there are only finitely many Uk with k ~ n we see that {UpnU, Is E SI E A(~) i Up. To keep the induction going, we assume {UpnU,ls E SI E A(~)iUp for each pEP with 0(P) < K for some ordinal K. Let q E P - S with o(q) = K. We maintain that {UqnUs Is E SI E A(~) IUq. To see this, note that {UqnUs Is E SI is refined by {(UqnUr)nUs Ir is an immediate predecessor of q and s E S}, that {UqnU r Ir is an immediate predecessor of q I E ~ IUq and for each r < q, {UrnU, I s E S I E A(~) IUr by our induction assumption. To see this last assertion note that if r < q and rEP - S then oCr) < o(s), whereas if rES, then all predecessors of r are in S, which is one of the cases we have already proved. Consequently, {UqnUsls E SI E A(~)IUq, so {UpnUsls E S} E A(~)IUp for eachp E P - S. Now {Up Ip is a maximal element in PIE ~ since if p is maximal, then p
=
(l,a) for some index a. For each maximal p, either PES or pEP - S. If pEP
-Sthen {UpnUslsE SI E A(~)lup. IfpE S,theneachpredecessorofpisinS so {UpnUsls E SI E A(~)lup. Therefore, U = {Usls E SI E A(~) by definition of A(~). But clearly, U < V, so V E A(~). Hence u c A(~), so A(~) =
u.PROPOSITION 5.10 For each U E 11. where (X,~) is a uniform space, there exists a normal sequence {Un) oj locally filllte open members ojA(~) with U j < U. Proof' If U E ~. there exists a uniformly continuous j:X ~ M where M is a metric space, and a uniform covering V of M with 1 (V) < U (Theorem 2.5). Let M' be the completion of M. Then;:X ~ M' is uniformly continuous and l (V') < U where V' is the Morita extension of V to M' (see Exercise 4, Section 4.4). Since M' is paracompact, there exists a normal sequence {V~ I of locally finite open coverings of M' with V'l < V'. By Theorem 5.6, each V~ E A(V) where v is the uniformity of M'. Suppose V~ E va for some ordinal a. By a straightforward induction argument, we can show that Un = 1 (V~) E ~a. Hence Un E A(~) for each n. Clearly Un is locally finite for each nand {Un I is a normal sequence of open, locally finite members of A(~) with U 1 < U.-
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PROPOSITION 5.11 For a uniform space (X, f..l), A(f..l) is a uniformity for X. E A(f..l), there exists a normal sequence {Un I of open, locally finite members of A(f..l) with U 1 < U. By Proposition 5.10, it is true for members of j..I.. Assume a is the least ordinal for which the assertion fails for some U E j..I.u. Now U = {V~n W~ I where {V p} E j..I. K for some K < a and for each index P, {W~ I E f..l K( ) for some K(P) < a. Hence {V p I has an open locally finite refinement {H ol. E A(f..l) with an open locally finite star refinement (K e) E AC/..t). For each 0 choose a P(o) with Hoe V P(o). Since {H 0 I is locally finite, each K e is contained in only finitely many of the H 0, say H 01 . . . H On' Put We = {w~(81l In ... n {w~(8n) I. Since each f..lu is a filter, it can be seen from the induction hypothesis that We has an open, locally finite, star refinement I that is normal with respect to the collection of open, locally finite members of A(f..l). But then, (K I is an open, locally finite, star refinement of U that can be shown to be normal with respect to the open, locally finite members of A(f..l). Therefore, the assertion does not fail after all, so A(f..l) is a uniformity. -
Proof: We will prove inductively that for each U
{Z!
enZ!
COROLLARY 5.3 The uniformity A{f..l) has a uniformly locally finite basis. THEOREM 5.7 For a uniform space (X, f..l). A(f..l) is the coarsest locally fine uniformity finer than j..I.. Moreover, each locally fine uniformity has a basis of uniformly locally finite coverings. Proof: By construction, A(f..l) is locally fine, and finer than f..l. If v is another locally fine uniformity finer than f..l, then it is finer than f..ll. By a straightforward induction argument, v is finer than j..I.u for each ordinal a and hence finer than A(f..l). Therefore, A(f..l) is the coarsest locally fine uniformity finer than f..l. Finally, if f..l is locally fine, then A(f..l) = f..l, so each U E f..l has an open uniformly
locally finite refinement. Therefore, each locally fine uniformity has a uniformly locally finite basis. Let P be a partially ordered set. ReP is residual in P if for each r E R, s R for each s E P with r < s. Note that the restriction R ::f- 0 has been dropped from the corresponding definition for directed sets. A function ",:P ~ X is called a partial net (denoted p-net) in X. A p-net '" in a uniform space (X,v) is partially Cauchy if for each {U u} E v, there exists a family {R u} of residual subsets of P such that for each a, ",(R u) c U U and uR u is cofinal in P. The definition of a cofinal subset of P is the same as the definition for directed sets. We extend this definition by allowing v to be f..lu for some uniformity f..l and ordinal a. In this case, we say", is partially Cauchy with respect to the a 1h derivative of f..l. '" is said to partially converge to some x E X if for each neighborhood U of x, "'-' (U) contains a non-empty residual subset of P. If Pis E
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a directed set, the definitions of partially Cauchy and partially convergent are the same as Cauchy and convergent respectively. LEMMA 5.9 If'l':P ~ (X, Il) is partially Cauchy, so is 'I'.P ~ (X,A(Il)). Proof: Suppose 'I':P ~ (X, A(Il)) is not partially Cauchy. Then there exists a least ordinal ex for which 'I':P ~ (X, IlU) is not partially Cauchy. ex must be a non-limit ordinal, say ~ + 1, for otherwise Ilu is just the union of its
predecessors, for which 'V is partially Cauchy. Let {U ynYS} E f,.lu. Then {U y } E Il~ and for each y, {YS} E Il~. For PEP, there exists a successor q, all of whose successors are in one of the U y' For this y, q has a successor r, all of whose successors are in one of the YS. Hence there exists a residual R(y,8) 1= 0 with 'I'(R(y,8)) c U ynYS such that p .,::; r for each r E R(y,8). But then there exists a cofinal union of residual sets, each mapped by 'V into one of the U yn YS. For each U ynYS not containing one of the.se residual images of 'V, put R(y,8) = 0. Then 'V is partially Cauchy with respect to Ilu which is a contradiction. Therefore, 'I':P ~ (X, A(Il)) is partially Cauchy.· The proof of the following theorem and the terminology used in the proof is somewhat different from Isbell's original proof, but his central ideas are preserved. This different route frees us from having to introduce additional terminology and the concept of a stable Cauchy filter, and from proving some additional lemmas. THEOREM 5.8 (1. Isbell, 1962) For a uniform space (X, Il), the following are equivalent: (1) (X, Il) is supercomplete. (2) X is para compact and A(Il) is fine. (3) Each partially Cauchy p-net in (X, Il) partially converges. Proof: (1) ~ (2) Condition (2) implies each open covering belongs to A(Il). Suppose this is not the case, i.e., suppose the open covering U is not A(Il)-uniform. Let (X', III be the hyperspace of (X, f,.l). We construct a proper
Cauchy net in X' that has no cluster point. By Proposition 5.10 this is a contradiction. For this let B consist of those closed sets F c X with the trace of U on some uniform neighborhood of X - F being A(Il)-uniform. Let F, K E B. Then FnK is closed. F, K E B implies there exists V, WEll with U being A(Il)-uniform on Star(X - F,V) and U being A(Il)-uniform on Star(X - K,W). Let Z = VnW. Then Z Eiland U is A(f,.l)-uniform on both Star(X - F, Z) and Star(X - K, Z). If FnK = 0 then F c X - K so U is A(Il)-uniform on Star(F, Z). Therefore, for each Z E Z, U IZ E A(Il) Iz so U E A(f,.l) which contradicts our assumption, so FnK 1= 0. Also, since U is A(f,.l)-uniform on both Star(X-F,z) and Star(X-K,z), U is A(Il)-uniform on Star(X-F,z)uStar(X-K,z) which contains Star(X-FnK,z). Consequently, FnK E B.
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Let D be the collection of closed sets in X that contain a member of Band define:-O; on D by F:-O; K if KeF. Then (D, :-0;) is a directed set. For each FED put \jI(F) = F. Then \jI:D -? X' is a proper net in X'. We need to show that \jI is also Cauchy. By Lemma 5.5, we need to show that for each V E fJ. there exists an FED with \jI(F) c Star(\jI(K},v} for each KED. For this let WE fJ. with W* < V and put F = Cl( {W E wi U is not A(fJ.)-uniform on W}). Then FEB since the union of the W E W on which U is A(fJ.)-uniform is a uniform neighborhood of X - F containing A = Star(X - F,W). To see this let W E W with Wn(X - F) = 0. Then W is not a member of W on which U is A(fJ.)-uniform. Hence W c A. For each KED, there exists H E B with H c K. Star(H,W) contains a dense subset of F, for if x E U {W E wi U is not A(fJ.)-uniform on W} then x E WOE W on which U is not A(fJ.)-uniform which implies W 0 is not contained in X - H (since H E B) so Woe Star(H,W) which implies x E Star(H,W). But then F c Star(H,W*) c Star(K,V). Since \jI(F) = F and \jI(K) = K we have \jI(F) c Star(\jI(K},v} for each KED so \jI is Cauchy. So \jI:D -? X' is a proper Cauchy net. By Proposition 5.10, \jI has a cluster point P E X. Then there exists a U E U with P E U. Pick V E fJ. with S(p,V*) c U and put F = X - S(P,V). Then F is closed and Star(X - F,V) c S(p,V*) c U so U is A(fJ.)-uniform on Star(X - F,v) since U IS(p,v*) = X IS(p,v*). Hence F E D and S(p,v)nF = 0. Then for each KED with K?: F, KnS(p,v) = 0 so pis not a cluster point of \jI which is a contradiction. Therefore, (1) -? (2). (2) -? (3). By Lemma 5.9 we need only show that a partially Cauchy partial net in (X, A(fJ.» partially converges. Assume this is not the case, i.e., that there exists a partially Cauchy partial net \jI:P -? (X, A(fJ.» that does not partially converge. Then there exists an open covering U of X such that \jI-l (U) contains no non-empty residual subset of P for each U E U. By (2), A(fJ.) = U and since X is paracompact, U E u. Hence \jI is not partially Cauchy which is a contradiction. Therefore, (2) -? (3). (3) -? (1) Let \jI:D -? X' be a proper Cauchy net in the hyperspace (X', fJ.') and suppose \jI has no cluster point. Let P be the collection of subsets S of X that are uniform neighborhoods of sets that meet each \jI(d) for d E D. Then P is partially ordered by set inclusion. For each PEP pickf(P) E P. Thenf:P -? X is a partial net. Let x E X. Since x is not a cluster point of \jI, there exists a U E fJ. and 8 E D with Sex, U*)n\jl(8) = 0. Then Q = Star(\jI(8) , U) E P and S(x,U)nQ = 0. Suppose fpartially converges to x. Then there exists a residual ReP with feR) c S(x,U). Let R E R. Now R = Star(T,v) for some V E fJ. and some T that meets \jI(d) for each d E D. Put A = \jI(8)nT and let W = UnV E fJ.. Then B = Star(A,W) E P and B c QnR. Therefore, R :-0; B and feB) E Q which implies feB) does not belong to sex, U) which is a contradiction. Hence f cannot partially converge to any x E X.
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Let {U a} E Jl. If S E P, there exists a V E Jl with Star(T,v*) c S and V** < {U a}. By Lemma 5.5, there exists a 8 E D with ",(8) c Star(",(d),V) for each dE D. Let x E ",(8)nT. Some U y contains R = S(x,V*) c Star(T,V*) c S. Since x E ",(8), each ",(d) meets S(x,V). Hence REP. ReS implies S ~ R in P and all successors of R in P are mapped into U y by f. Therefore, f:P ~ X is partially Cauchy and hence must partially converge by (3). But this is a contradiction, so each proper Cauchy net in (X', Jl) clusters. By Proposition 5.10, (X, Jl) is supercomplete.· In the sense expressed in Theorem 5.8, the uniformity '-(Jl) characterizes supercompleteness in paracompact uniform spaces (X, Jl). In a yet unpublished paper, B. Burdick shows that there exists another uniformity S(Jl) that characterizes cofinal completeness in the same way. We now develop a useful alternate construction of '-(Jl). We denote this new construction by f(Jl) for the time being. I(Jl) is called the subfine corejiection of Jl. To define I(Jl), we first need to define what a subfine uniform space is. A uniform space is said to be subfine if it is a subspace of a fine uniform space. Subfine spaces are a generalization of fine spaces whose definition is much more intuitive than the definition of the locally fine spaces. At the time Isbell published his book in 1964, it was an open problem whether locally fine spaces are subfine. In a 1987 paper titled Locally fine uniformities and normal covers (Czechoslovak Math. Jour. 37(112), pp. 181-187), J. Pelant solved the problem with an affirmative answer. Pelant's complex proof, together with other proofs he references would take up more space than we can allot to this subject. The interested reader can find his proof and the necessary references in the citation above. In what follows, it will first be shown that every uniform space (X, Jl) has a coarsest subfine uniformity I(Jl) finer than Jl, and that by Pelant's Theorem, I(Jl) = '-(Jl). For the construction of I(Jl), we need the concept of an injective uniform space. A uniform space Y is injective if whenever A is a uniform subspace of X, every uniformly continuous function f:A ~ Y can be extended to a uniformly continuous function F:X ~ Y. THEOREM 5.9 The closed interval [0,1] is an injective space. Proof: Let i:A ~ X be a uniform imbedding and f:A ~ I = [0,1] a uniformly continuous function. Let Jl be the uniformity of X and Jlp the collection of all coverings of X refined by finite members of Jl. It is left as an exercise to show that Jlp is a uniformity for X that is coarser than Jl (Exercise 6). Then the identity function jx:(X, Jl) ~ (X, Jlp) is uniformly continuous. Let n:X ~ JlpX be the uniform imbedding of (X, Jlp) into its completion JlI3X,
Now Jl131 A is a uniformity on A coarser than Jl so the identity function ~ (A, Jlp) is .uniformly continuous and Cl l1Bx (A) is the completion of (A, Jlp) since completIons are unique. So Jli3A is a closed subspace of Jli3X, Since Jli3 is a precompact uniformity, by Proposition 4.19, Jli3X is compact.
h :(A, Jl)
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Since I is precompact,f:A --+ I is also uniform continuous with respect to I-l~ IA. By Theorem 4.9, f can be extended to a uniformly continuous function f~:I-l~A --+ I. Then since I-l~A is closed in I-l~X and I-l~X is normal by Proposition 0.32, we can apply Tietze's Extension Theorem (0.2) to get a continuous extension g:I-l~X --+ I. Since both I and ~~X are compact, their unique uniformities are fine, so g is uniformly continuous. Define F: X --+ I by F = g © n: © h. Then for each a E A, F(a) = g(n:(jA (a») = g(a) =f~(a) =f(a), so F is an extension of J. F is uniformly continuous since it is the composition of uniformly continuous functions. Hence [0,1] is an injective space.COROLLARY 5.4 Any closed interval! c R is an injective space. PROPOSITION 5.12 Each product of injective spaces is injective. Proof: If Y = nY a is the product of injective uniform spaces Y a and the uniform subspace A c X is mapped uniformly continuously into Y, then the functions fa = p a © j:A --+ Y a (where p a is the a th coordinate projection) is a uniformly continuous function, so it can be extended to a uniformly continuous function g a: X --+ Y a . Then the function g: X --+ Y defined by g(x) = {g a (x) } extends f:A --+ Y, since f(a) = {fa(a)} for each a E A. To see that g is uniformly continuous, let U be a uniform covering of ny a' Then there exists a finite collection al ... an of indices and uniform coverings Val ... Van of Y a ] ... Y an respectively such that Pa~ (Va] )n ... n Pa~ (Va.) < U. Then
which is uniformly continuous in X since ga~ (Va) is uniformly continuous for each i = 1 ... n. Therefore, g -I (U) is a uniform covering of X so g is uniformly continuous. Consequently, Y is an injective uniform space. A uniform covering U of a uniform space X is said to be realized by a uniformly continuous function f:X --+ Y if for some uniform covering V of Y we have 1 (V) < U.
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PROPOSITION 5./3 If a family of uniformly continuous functions f a: X --+ Y a realizes every uniform covering of X, then the mapping f:X --+ nY a defined by f(x) = (f a (x)) for each x E X, is an imbedding. Proof: Clearly fis uniformly continuous by an argument similar to the one used to prove that g is uniformly continuous in Proposition 5.12 above. To show that fis an imbedding, we need to show that for each uniform covering U ofX,/(U) is a uniform covering of ny a . To see this, note that there exists an index y for U and a uniform covering V of Y y such that (V) < U implies (p y © f)-I (V) <
f:/
5.6 The Locally Fine Coreflection
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U which in tum implies J (Prj (V» < U. But Prj (V) is a sub-basic uniform covering of OY a and since f is uniformly continuous J (Prj (V» is a uniform covering of X. Hence feU) = PrJ (V) which is a uniform covering of OY a' -
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THEOREM 5.10 Every uniform space can be uniformly imbedded in a product of metric spaces. Proof: By Theorem 2.5, for each uniform covering U a of a uniform space X, there exists a uniformly continuous function fa that maps X onto a metric space M a such that the inverse image of each set of diameter < 1 is a subset of an element of Ua. Therefore, U a is realized by fa. Then by Proposition 5.13, the mapping f:X ~ OM a defined by f(x) = {fa(x)} for each x E X is an imbedding.
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For a set S, the metric space I ~(S) of all bounded real valued functions on S with the distance function d(f, g) = sup{ If(x) - g(x) II x EX}, is called the I ~ -space of S. THEOREM 5.11 Each metric space can be uniformly imbedded in the unit sphere of an 1~-space(S)for some set S.
Proof: Let (M, d) be a metric space. For each pair x,y E M put 8(x,y) = min {d(x,y),l }. Clearly (M, 8) is uniformly homeomorphic with (M, d). For each x E M, let x* be the function on M defined by x*(y) = 8(x,y). Then d(x*,y*) = sup{ 18(x,z) - 8(y,z)llz EM}. For any z E M, the largest 8(x,z) can be is 8(x, y) + 8(y, z) because of the triangle inequality. Therefore, sup { 18(x, z) - 8(y, z) II Z E M} ~ sup { 18(x,y) + 8(y. z) - 8(y, z) II z E M} = 8(x,y) so for each pair x,y E M, d(x* ,y*) ~ 8(x,y). Conversely, Ix*(Y) - y*(y) I = 18(x,y) - 0 I = 8(x,y) so sup{ Ix*(z) - y*(z)llz E M} = 8(x,y). Hence, for each pair x,y E M, d(x* ,y*) = 8(x,y), so (M, 8) is uniformly homeomorphic with f(M, 8) c I ~(M) where f:M ~ I ~(M) is the uniform imbedding defined by f(x) =x* for each x E M. Since (M, d) is uniformly homeomorphic with (M, 8), we have (M, d) is uniformly homeomorphic with f(M) c I ~ (M). Since Ix*(m) I ~ 1 for each m E M, (M, d) can be uniformly imbedded in the unit sphere of an I ~-space. PROPOSITION 5.14 The unit sphere of an I ~-space is injective. Proof: Let S J (X) denote the unit sphere of the I ~ -space I ~ (X). Let Z be a uniform space andf:Z ~ S J (X) a uniformly continuous function. We first show that f is uniformly continuous if and only if the function g:Z x X ~ [-1,1] is uniformly continuous where g is defined by g(z, x) = [f(z)] (x) and X is considered to be uniformly discrete. For this, suppose f is uniformly continuous and V is a uniform covering of [-I,ll. Then there exists a uniform covering U of Z such that f maps any two points that are U-close in Z into two functions that are V-close in S J (X). Then the product U x W where W = { {x} Ix EX}
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is a uniform covering of Z x X. Moreover, U x W is finer than g(V*). To see this note that g(V*) = {g(Star(V,V» I V E VI. For each V E V, g(Star(V,v» = {(z,x) E Z x xl [f(z)](x) E Star(V,V)}. Now if (ZI ,x) and (z2. x) E U x {xl for some U x {x} E U X W. then z 1 and z 2 are U-close in Z which implies t(z ") andf(z2) are V-close in S 1 (X). so [fez J )l(x) and [f(Z2)](X) are V-close in [-1J I. Hence [f(zd](x), [f(Z2)](X) E Vo for some Vo E V, so g(l! x {x)) c Star(V" , V). Therefore, U x W is finer than g(V*) so g is uniformly continuous. Conversely, suppose g is uniformly continuous. Then there exists a uniform covering U of Z and a uniform covering W of X such that U x W < g(V). Without loss of generality we may assume W = { {x} Ix EX}. Pick x IJ E X. If zl, z2 E U for some U E U. then (z 1. xo). (Z2. xo) E U x {xo} c g(V) for some V E V. But g(V) = {(z, x) E Z X X I [f(z)](x) E V} so [fez 1 )](x), [f(Z2)](X) E V. Since this holds for each Xo in X, f(zd and f(z2) are two functions in S 1 (X) that are V-close. Hencefis uniformly continuous. To show that S 1 (X) is an injective uniform space. let A be a uniform subspace of Y and let f:A ~ S 1 (X) be a uniformly continuous function. Then by the above result, g:A x X ~ [-1,1] is uniformly continuous where g(a, x) = [f(a)](x). g can be extended to a uniformly continuous functIOn G:Y x X ~ [-1,1] (this follows from Theorem 4.9 and Exercise 4 of Section 2.2). But then F:Y ~ S 1 (X) is uniformly continuous where for each Y E Y, F(y) is the function in S 1 (X) defined by [F(y)](x) = G(y. x). Clearly, F is an extension of j, so S 1 (X) is an injective space. • THEOREM 5.12 Each uniform space can be uniformly imbedded in an injective uniform space. Proof: Let X be a uniform space. By Theorem 5.10. X can he imhedded in a product I1M u of metric spaces. By Theorem 5.11, each M u can be imbedded in the unit sphere S 1 (M u) of the I ~-space I ~(M u). Consequently, X can be imbedded in OS 1 (M u). By Proposition 5.14, S 1 (M u) is an injective uniform space for each a. Then by Proposition 5.12, OS 1 (M u) is an injective uniform space, so X can be imbedded in an injective uniform space. •
For a uniform space (X, ~), let (Y. v) be an injective uniform space in which (X, ~) can be imbedded. Let u be the finest uniformity on Y and put l(~) = u I X. Clearly (X, l(~» is a suhfine uniform space. The next lemma shows that l(~) is well defined. l(~) is called the subfine coreflection of ~. LEMMA 5.10 If (X, ~) is uniformly imbedded In an injective uniform space (Z, ~), and if u is the fine Uniformity on Z, then (X, l(~)) and (X, m(/l)) are uniformly homeomorphic where m(/l) = u Ix. Proof: Let f:A ~ (X, ~) be uniformly continuous where A is a subspace of a fine uniform space B. Thenfhas an extension g:B ~ (Y, v) since Y is injective
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and (X, /-1) is imbedded in (Y, v). Since B is a fine space, g is uniformly continuous into (Y. u) where u is the fine uniformity on Y. But then the restriction of g to A is uniformly continuous, so f:A ~ (X, 1(/-1» is uniformly continuous. Therefore, any uniformly continuous function f from a subfine space into (X, /-1) is also uniformly continuous into (X, [(/-1». The identity function i:(X, m(/-1» ~ (X, /-1) is uniformly continuous, so by the above result, i:(X, m(/-1» ~ (X, [(/-1» is uniformly continuous. If (Y, v) were replaced with (Z, 0 in the above argument, we would obtain the result that any uniformly continuous function f from a subfine space into (X, J.1) is .also uniformly continuous into (X, m(/-1». Hence i-I :(X, 1(/-1» ~ (X, m(J.1» is uniformly continuous, so (X, m(/-1» is uniformly homeomorphic with (X. 1(/-1». -
COROLLARY 5.5 The sUbfine corejiection of an injective space or a complete metric space isfine. Proof: The injective case is clear from the proof of Lemma 5.10. If (X, /-1) is a complete metric space, then by Theorem 5.11 and Proposition 5.14, (X, /-1) is a closed subspace of an injective metric space (Y, v). Since X is closed in Y, each open covering of X together with the set Y - X is an open covering of Y. Hence the fine uniformity on Y induces the fine uniformity on X.LEMMA 5.11 If(X. /-1) is sUbfine then it is locally fine. Proof: Assume (X, /-1) is subfine. It suffices to show that 1-(/-1) = /-1. For this let (X, /-1) be uniformly imbedded in some fine space (Y, v). Then /-1 = v I X. Then as shown in the proof of Lemma 5.l0, the identity function i:(X, /-1) ~ (X. [(/-1» is uniformly continuous. Clearly i-I :(X, [(J.1» ~ (X, 11) is uniformly continuous since [(/-1) is finer than J.1. Therefore, (X. /-1) is uniformly homeomorphic with (X,[(/-1» so J.1 = [(/-1).-
LEMMA 5.12 [(/-1) is the coarsest sUbfine uniformity for Xfiner than J.1. In particular. /-1 is sUbfine if and only if [(/-1) = /1. Proof: Suppose there exists a subfine uniformity 8 for X such that /-1 C 8 C [(/-1). Then i:(X, 8) ~ (X, /-1) is uniformly continuous, so by the proof of Lemma 5.l0, i:(X, 8) ~ (X, 1(/-1» is uniformly continuous. Clearly i-I :(X. [(/-1» ~ (X, 8) is uniformly continuous since 1(/-1) is finer than 8. Therefore, (X, [(11» is uniformly homeomorphic with (X, 8), so 1(/-1) is the coarsest subfine uniformity for X finer than /1.LEMMA 5.13 A point finite uniform covering has a uniformly locally finite uniform shrinking. A a-disjoint uniform covering has a a-discrete uniform shrinking.
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Proof: Let {Va} be a point finite unifonn covering. By Exercise 6 of Section 2.2, there exists a unifonn covering {Va} that is a unifonnly strict shrinking of {Va}. Let W be a uniform covering such that for each a, Star(V a , W) eVa. Let p E X and pick W E W with pEW. If Wn V 13 1:- 0 then p E V 13' Consequently, W can meet at most finitely many of the Va's since p can be in at most finitely many of the Va's. Therefore, {Va} is unifonnly locally finite.
If {Va} is a a-disjoint unifonn covering, there exists countably many subcollections {VB} c {Va} such that each {VB I is a disjoint collection. Since for each a, Star(V a , W) eVa, the collection {VB} is unifonnly discrete. Hence {Va} is a a-unifonnly discrete unifonn shrinking of {Va}. LEMMA 5.14 Each uniform covering of a sUbfme uniform space has a locally finite uniform refinement and a a-disjoint uniform refinement. Proof: Let (X, 11) be a subfine uniform space. Then there exists a fine unifonn space (Y, 11*) such that (X, 11) is a subspace of (Y, 11*). Let U E 11. There exists a U* E 11* with U ::: U*nX. By Theorem 2.5, there exists a uniformly continuous f:X ~ (M, d) where d is a metric for M, such that for each x E X, f(S(x, U*)) ::: S(f(x), 1/2). Since M is paracompact, the covering V ::: {SCm, 1/2) I m E M} has a locally finite refinement W that belongs to the finest unifonnity for M. Since f is continuous, l (W) E 11* because 11* is fine. Then l (W) IX is a locally finite member or 11 that refines U.
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Similarly, V has a a-disjoint refinement Z. Hence l (Z) is a a-disjoint open refinement of U* that is itself uniform, so l (Z) IX is a a-disjoint unifonn refinement of U.-
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THEOREM 5.13 Each subfine uniform space has a basis of uniformly locally finite coverings and a basis of a-uniformly discrete coverings.
The proof of Theorem 5.13 follows from the preceding lemmas. A unifonnity is said to be point finite if each member has a unifonn point finite refinement. THEOREM 5.14 ffll is a point finite uniformity, so is the derivative Ill. Proof: Let {V anV~} E Ill. Let U::: {Va) and Va::: {V~} for each a. Without loss of generality we may assume U is point finite. Choose a point finite A ::: {A y I E f.! with A <* U and for each a, a W a <* Va in 11. Each A y is contained in at most finitely many of the Va, say Val . . . Van' Put G y ::: W al n . . . nWan and let zy ::: {ZS) be a point finite unifonn refinement of GY. Then {Aynzs) E III is point finite.
To show that {AynZS} <* {VanV~}, note that for any pair 'Y, 0 we can choose Va containing Star(A y , A) and a V~ containing Star('4, wet). If Aenzt
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meets AynZS then A e c Star(A y, A) c U u which implies Ze < G e < W U , so ~ c Star(Z{>" W U ) c vg. Hence Aen~ c AynZS. We conclude that {AynZSI <* {U unV~ I so f..ll is also a point finite uniformity.COROLLARY 5.6 Each sUbfme uniform space is locally fine. Proof: Let (X, f..l) be a subfine space and (Y, f..l*) a fine space such that (X, f..l) is a unifonn subspace of (Y, f..l*). From the definition it is clear that (X, f..ll) is a unifonn subspace of (X, f..l * 1). By Theorem 5.13 f..l and f..l* are point finite, so by Theorem 5.14, f..ll and f..l* I are also. Since f..l* is fine, f..l* 1 = l..l* which implies f..ll = f..l. A straightforward induction on a yields Jlu = f..l for each a, so A,(f..l) = f..l.
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EXERCISES
DERIVATIVES OF UNIFORMITIES It took over 20 years to settle the question of whether the a th derivative f..lu of a unifonnity f..l is a unifonnity or not. Ginsburg and Isbell announced a proof in 1955 that the answer is affinnative. In their 1959 paper referenced earlier in this section, they retracted their announcement and left the question open. 1. [J. Isbell, 1964] If f..l is a unifonnity that has a basis consisting of point finite coverings, then f..ll (and hence any f..lU) is a unifonnity. 2. [J. Pelant, 1975] A unifonnity f..l for X has a basis consisting of point finite coverings if and only if the first derivative of the product unifonnity I1~1 f..la, where each f..la = f..l, of the product space (X, f..l)K is a unifonnity for each cardinal K. 3. [J. Pelant, 1975] There exists a unifonn space with no basis consisting of point finite coverings. Consequently, by Exercise 2, there exists a unifonnity with a derivative that is not a unifonnity. (Note: The existence of a uniform space without a basis consisting of point finite coverings was also an open problem from Isbell's book.) INJECTIVE SPACES 4. Let X and Y be unifonn spaces and let U(X,Y) denote the collection of all unifonnly continuous functions from X into Y. For an imbedding i:A ~ X, define the function F:U(X,Y) ~ U(A,Y) by F(h) = h © i for each h E U(X,Y). Show that Y is injective if and only if F is onto. 5. Show that if a unifonn space Y has the property that whenever A is a closed
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uniform subspace of X, every unifonnly continuous function J:A ~ Y can be extended to a unifonnly continuous function F:X ~ Y, then Y is complete and hence injective. [Hint: Suppose Y is not complete and Z is its completion. Put A = Y and construct X as follows: Let K be a regular cardinal. K + 1 is a compact Hausdorff space in the order topology. A continuous function Jon K or K + 1 into a metric space is finally (residually) constant. Since Y is imbeddable in a product of metric spaces, K can be chosen large enough to make each continuous function on K finally constant. Also, K should be greater than the cardinality of any subset of Y having a limit point in Z - Y. Let S c Y have a limit point in Z - Y. Put X = [Y X (K + 1) - K]U[S X (K + 1)Ju[P x KJ. Then Xc Z X (K + 1). Identify Y with Y x (K + 1) - K so that Y c X. It can be shown that Y is closed in X and any continuous extension over S x (K + 1) of the identity i: Y ~ Y must agree with coordinate projection on a unifonn neighborhood of S x [(K + 1) - K] and cannot be extended over {p) x K.J 6. Let (X, /-l) be a unifonn space and let p(/-l) be the collection of all coverings of X refined by a finite member of /-lo Show that p(/-l) is a unifonnity for X that is coarser than /-l. 7. Show that a unifonn retract of an injective unifonn space is also injective. A retraction is a mapping r such that r © r = r. A uniform retract (Y, v) of a unifonn space (X, /-l) is a unifonn subspace such that there exists a unifonnly continuous function r:X ~ Y such that r is a retraction.
5.7 Categories and Functors In this section we introduce the elements of category theory. None of the following sections or chapters will depend on a knowledge of this subject, so this section can be skipped without sacrificing the accessibility to subsequent sections. However, there will be exercises at the end of future sections that will involve an elementary understanding of category theory. The approach here is to keep the development of category theory separate from our development of unifonn spaces to allow the reader the choice of pursuing this area or not. Some researchers in the theory of unifonn spaces tend to combine categorical notions with unifonn space concepts in theIr publications. Consequently, a knowledge of category theory is helpful in reading some of the literature in the theory of uniform spaces. On the other hand, the reliance on categorical notions and terminology by these unifonn space researchers has often been criticized as making this field almost inaccessible to anyone but the specialist. In future sections and chapters, all unifonn concepts will be introduced without reliance on categorical terminology or concepts, to allow as immediate access to the ideas as we know how. However, in order to illustrate the current usage of category theory as a tool in the study of unifonn spaces (just as set theory is used as a tool), we will
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present a selection of problems that involve categorical notions that should suffice to prepare the reader to access that branch of uniform space theory that uses category theory. For those who have studied even elementary set theory, it is understood that the class of all sets cannot itself be a set. The assumption that it is leads to a statement that is simultaneously true and false, as first noted by B. Russell. This is the famous Russell Paradox. However, it is still useful to retain the notion of a class of all sets, just it is useful to have the notion of the class of all uniform spaces or the class of all topological spaces. By a concrete category C, we mean a class 0 of sets (referred to as objects of C), and for each ordered pair of objects (X, y), a set of functions f:X -7 Y (denoted by M(X. Y) and called the morphisms of C with domain X and range Y) such that (1) The identity function on each object X is a morphism of C with
domain X and range X. (2) Each composition of morphisms of C is a morphism of C. The class 0 of sets may be larger than any cardinal number, and will be in all examples given in this book. By a covariant functor F:C -7 D of two concrete categories C and D, we mean a pair of functions F 0 (called the object function) and F M (called the morphism function) such that F 0 assigns to each object X in C an object F o (X) in D, and FM assigns to each morphism f:X -7 Y of C a morphism F M(f):F0 (X) -7 F o(Y) such that (3) For each identity morphism Ix of C, F M(1X) = IFo(x). (4) For each composed morphism g © f of C, FM(g © f) = F M(g) © F M(f). Since F 0 only applies to objects and F M only applies to morphisms, it is customary to denote both F 0 and FM simply by F as no confusion is likely to arise. This shorter notation is useful for denoting the composition of covariant functors F:C -7 D and G:D -7 E by G © F(X) = G(F(X) and G © F(f) = G(F(f). An isomorphism is a covariant functor F:C -7 D such that there exists a covariant functor F-1:D -7 C with both F- 1 © F and F © F- 1 being the identity functors on C and D respectively. If the functor F:C -7 D is an isomorphism, the concrete categories C and D are said to be isomorphic and both C and D are said to be instances. of the same abstract category. A categorical property is a property P that if possessed by a concrete category C, is preserved by isomorphisms (i.e., if D is a concrete category isomorphic with C, then D also has property P). Similarly, we can define categorical concepts and definitions.
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For instance, the notion of a mapping f:X ~ Y having an inverse 1 : Y ~ X is a categorical concept in the following sense. If we define f:X ~ Y has an inverse Y ~ X to mean that there exists a mapping y ~ X such that ©f = 1x and f © 1 = 1y, then if X and Y belong to the concrete category C such that F:C ~ D is an isomorphism, then it is easily seen from the definitions that the morphism F(j):F(X) ~ F(Y) also has an inverse.
rl:
rl:
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On the other hand, the notion that the morphism f:X ~ Y is one-to-one is not a categorical notion. However, being one-to-one implies an important categorical notion that is closely related to being one-to-one. In fact, in uniform spaces, being one-to-one is equivalent to this notion. A morphism f:X ~ Y of a concrete category C is said to be a monomorphism if for each pair of morphisms g:Z ~ X and h:Z ~ X we have f © g = f © h implies g = h. Clearly if f is one-to-one it is a monomorphism. On the other hand, if f is not one-toone then f(x) = fCy) for some x yin X. But then the constant functions g:Z ~ {x} and h:Z ~ {y} are not equal, yet f © g = f © h. Hence f is not a monomorphism. Therefore, f is a monomorphism if and only if it is one-to-one. It is the fact that constant functions are morphisms in the category of uniform spaces that makes monomorphisms equivalent to one-to-one mappings.
'*
Similarly to this left cancellation property that defines the monomorphisms, the right cancellation property defines the epimorphisms. f is said to be an epimorphism if for each pair of morphisms g:Y ~ Z and h:Y ~ Z, we have g © f = h © f implies g = h. Epimorphisms are closely related to the concept of being onto. Indeed, an onto morphism is clearly an epimorphism. But being an epimorphism is a more general property, for iff(x) is merely dense in Y, it suffices to be an epimorphism. To see this, note that if g © f = h © f and y E Y, we can pick a net q> inf(X) converging to y. Since g and h are morphisms (uniformly continuous functions in the category of uniform spaces), g © q> converges to gCy) and h © q> converges to hCy). But g © f= h © fand q> cf(X) implies g © q> = h © q>. Since nets converge to unique limits in uniform spaces, we have gCy) = hCy). Therefore,fis an epimorphism. A morphism f:X ~ X is called a retraction if f2 =f © f =f. If a retraction is either an epimorphism or a monomorphism, it is an identity morphism. By a contravariant functor F:C ~ D of two concrete categories C and D, we mean a functor F that assigns to each object X of C an object F(X), but to each morphism f:x ~ Y of C, a morphism in the opposite direction F(j):F(Y) ~ F(X) of D to C that satisfies property (3) above, but instead of satisfying property (4), it satisfies (4') For each composed morphism g © fof C, F M(g © f) = F M(j) © F M(g). Observe that the composition of two contravariant functors is a covariant func-
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tor. Also, functors of different variances can be composed with the resultant functor being a contravariant functor. A duality is a contravariant functor F:C ---* D such that there exists a contravariant functor F-1:D ---* C with both F- 1 © F and F © F- 1 being identity functors. If C and D are concrete categories, and F:C ---* D is a duality, then every concept, definition, or theorem about C gives rise to a dual concept, definition or theorem about D. To illustrate what is meant by dual concepts and definitions, we consider the concepts of sums and products of objects of an arbitrary concrete category C. As we shall see, these concepts are dual concepts. We have already considered the concepts of sums and products of uniform spaces. We now give categorical definitions for sums and products of objects in arbitrary concrete categories in such a way that the ordinary definitions of sums and products of uniform spaces are equivalent to the categorical definitions in the category of uniform spaces. Let {X a} be a collection of objects in the concrete category C and let {i a) be a collection of morphisms such that i a:X a ---* L for some L E C. Then L is called the sum of the objects {X,,} if the following conditions hold: (5) If f:L ---* Y and g:L ---* Yare distinct morphisms, then for some index a, the morphisms f © i a and g © i a are distinct. (6) For each family of morphisms f a:X a ---* Y, there exists a morphism f: L ---* Y such that f © i a = fa for each a. Similarly, an object rr E C is said to be a product of the objects {X a} if there is a collection {p a} of morphisms such that P a:rr ---* X a for each a, if the following conditions hold: (7) If f:Z ---* rr and g:Z ---* rr are distinct morphisms, then for some index a, the morphism P a © f and P a © g are distinct.
(8) For each family of morphisms f a:Z ---* X a, there exists a morphism f:Z ---* rr such that P a © f = fa for each a. To show that the categorical definitions of sum and product are dual, we must show that if L is the sum of {X a} in C, then rr = F(L) is the product of {F(X a)} in D and vice-versa. For this, assume L is the sum of IX a} in C and put rr = F(L). Suppose F(j) and F(g) are distinct morphisms such that F(j):F(Y) ---* rr and F(g):F(Y) ---* rr. Then f:L ---* Y and g:L ---* Y. Since F- 1 © F is the identity, F- 1 © F(j) = f and [F- 1 © F](g) = g. Therefore, f = g implies [F-l © Fl(j) = [F-I © F](g), so F(j) = F(g) which is a contradiction. Therefore, f and g are distinct. Since L is the sum of {X a}, there exists an a with f © i a ::j:. g © i a. Then by an argument similar to the above, F(f © i a) ::j:. F(g © i a) which implies F(i a) © F(j) ::j:. FU a) © F(g). For each a, put p a = F(i a) and let Z = F(Y). Then if F(j):Z ---* rr and F(g):Z ---* n are distinct morphisms, for some index a, the
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morphisms Pu © F(j) and p u © F(g) are distinct Roughly speaking, condition (5) in C forces condition (7) in D. Next, suppose F(f u):Z --? F(X u) is a family of morphisms in D. Then --? Y is a family of morphisms in C, so by (6), there exists a morphismf:L --? Y with f © i u = fa for each a. But then F(j):Z --? D such that p u © F(j) = F(f u) for each a. Again, roughly speaking, condition (6) in C implies condition (8) in D. Hence D = F(L) is the product of {F(X u») in D.
f u: X u
The proof that if D is the product of (X u) in D, then L = r! (D) is the sum of IF- 1 (X u») is similar. This establishes the duality of the concepts of sum and product in arbitrary concrete categories. To illustrate what is meant by dual theorems we will show that the categorical theorem: sums are unique (in C), gives rise to the dual (categorical) theorem: products are unique (in D). To see this, let L be the sum of IXu) in C. Then D = F(L) is the product of I F(X u») in D as we have already seen. If D' is another product of IF(X u»), then L' = F- 1 (il,) is another sum of IX u) so there exists a uniform homeomorphism i:L' --? L (since sums are unique in C). Therefore, F(i):D --? D' and since i-I © i = IE' and i © i-I = IE, we have:
Similarly, F(i- l ) © F(i) = IF(E), so F(i) is a uniform homeomorphism of D = F(L) onto D' = F(L'). Therefore, products are unique in D. The theorem that products are unique in C implies sums are unique in D is proved similarly. This establishes the duality of these two theorems. The principle of duality says that any categorical concept, definition or theorem gives rise to a dual concept, definition or theorem. It is also of interest to give categorical definitions of the concepts of subspace and quotient space. For the concept of a subspace, we recall that for a uniform space X, a subspace can be considered to be a uniform space S such that there exists a one-to-one uniformly continuous function i:S --? X such that the uniformity of X relativized to i(S) is identical with the identification uniformity of i(S). By a uniform imbedding of S into X, we mean a uniform homeomorphism of S onto a subspace S' of X. Clearly, if S is a subspace of X, the mapping i:S --? X above is a uniform imbedding.
Then a one-to-one mapping i:S --? X is a uniform imbedding if and only if whenever i = g © f where g and f are uniformly continuous and f is one-to-one and onto, then f is a uniform homeomorphism. To see this first assume i:S --? X is a uniform imbedding and i = g © f where g and f are uniformly continuous and f is one-to-one and onto. Let v be the uniformity of Sand Il the uniformity of X restricted to i(S). For V E v, i(V) = {i(V) I V E V} E Il since i is a uniform imbedding. Let A be the uniformity of S' = f(S). Since g is uniformly
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continuous, g-I(i(V» E A. Butf= g-I © i because i is one-to-one and g = i © i which implies g is one-to-one. Hence f(V) E A, so f is a uniform homeomorphism.
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Conversely, if the one-to-one mapping i = g © f where g and f are uniformly continuous and f is one-to-one and onto implies f is a uniform homeomorphism, then if V E v, f(V) E A. Assume i(V) does not belong to /.!. Then g(g-I U(V))) is not in /.! since g is one-to-one. But f(V) E A implies g -I U(V» E A since f = g-I © i. Therefore, the identification uniformity /.!* of g is strictly finer than /.!. Let A' be the uniformity induced on 5' by /.! with respect to g. Then A' is strictly coarser than A = gl (/.!*). Hence f is uniformly continUOUS with respect to v and A', but (5,v) and (5', A') are not homeomorphic which is a contradiction. Therefore. dV) E /.! so i is a uniform imbedding. Next. we use the characterization of uniform imbeddings to motivate the categorical definition of imbeddings. A monomorphism i:S --7 X in a concrete category C is said to be an imbedding if whenever i = g © f where g and f are morphisms, and f is one-to-one and onto. then f is an isomorphism. This means that if f:5 --7 5' then 5' has the same structure as 5. so that no structure can be inserted on 5 that is between the structure of 5 and that of i(5). In the case of uniform spaces, this means that the uniformity of 5 is the one induced by the function i. It is also clear that in any concrete category, two imbeddings with the same image have isomorphic domains. The dual categorical concept of an imbedding is that of a quotient morphism. An epimorphism q:X --7 Q in a concrete category D is said to be a quotient if whenever q = d © h where d and h are morphisms and d is one-toone and onto. then d is an isomorphism. To show that in the category of uniform spaces, a uniform quotient mapping satisfies this categorical definition, let (X, j..l) be a uniform space and q:X --7 Ya uniform quotient mapping. Then q is onto and Y has the identification unifonnity with respect to /.! generated by q. Let v be the uniformity of Y, then V E v if and only if l (V) E /.!.
r
Let q = d © h where d and h are uniformly continuous and d is one-to-one and onto. Let A be the uniformity of Q' = heX). If W E A, then h -I (W) E /.!. Since q is a uniform quotient, q(h- I (W» E v. But then d(h(h- I (W))) = deW) E v, so d is a uniform homeomorphism. Consequently, in the category of uniform spaces, unifonn quotient mappings satisfy the categorical definition of a quotient morphism. A subcategory S of a concrete category A is a category such that each object of S is an object of A and each morphism of 5 is a morphism of A. S is said to be a full subcategory of A if every morphism of A whose domain and range are both objects of S is a morphism of 5. A full subcategory R of a category A is said to be a reflection of A if for each object X E A, there is an object X* E R and a morphism r:X --7 X* (called the reflection morphism) such that each morphism f:X --7 Y where Y E R can be factored uniquely over X* by r
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(i.e., f = g © r for a unique g:X* -+ Y). The space X* is called the reflection of X in R or simply the R-reflection of X. There are many interesting reflections of the category of all uniform spaces. We have already encountered some of them unknowingly. For instance, the class of all complete uniform spaces C is a reflection of the category U of all uniform spaces. For each uniform space X E U, the completion X* E C is the C-reflection of X and the natural imbedding i:X -+ X* is the reflection morphism such that any morphism f:X -+ Y E C factors uniquely over X* by i. For any reflection R of a category A, an R-reflection of an object X E A is unique up to isomorphism. To see this, note that if X* and X+ are both R-reflections of X, then there exists reflection morphisms r*:X -+ X* and r+:X -+ X+ such that any morphism f:X -+ Y E R factor uniquely over both X* and X+ by r* and r+ respectively. Therefore, r* = g+ © r+ and r+ = g* © r* for some unique g* and g+. Now g*:X -+ X+ and g+:X+ -+ X*. Moreover, r+ = g* © (g+ © r+) = (g* © g+) © r+ so (g* © g+) = lx+. Similarly, g+ © g* = Ix'. Therefore, g + = g*-l so X* and X+ are isomorphic (in the class of uniform spaces this means uniformly homeomorphic). The fact that an R-reflection is unique allows us to make the simplifying assumption that if X E R, the only R-reflection of X is X itself and the only reflection morphism r:X -+ X is the identity morphism. This allows us to state the following fundamental theorem about reflective categories.
THEOREM 5.15 If R is a reflection of a concrete category A, X, YEA and X*, y* are their R-reflections respectively, then any morphism f:X -+ Y determines a unique morphism f* :X* -+ y* such that r © f = f* © s where r:X -+ X* and s:Y -+ y* are the reflections morphisms respectively. Proof: Since s © f:X -+ Y*, it can be factored uniquely over X* by r. Letf* © r be this factorization. Then s © f = f* © r and since f* is unique, it is the unique morphism that satisfies the theorem. Theorem 5.15 shows that the correspondences X -+ X* and f -+ f* determine a functor P called the reflection functor. The object function of P is Po:A -+ R defined by Po (X) = X* for each X E A and the morphism function PM is defined by PM(f) = f* for each f E A. To see that P is a functor, we need to show that Po and PM satisfy (3) and (4) of the definition of a functor. To demonstrate (3) we note that by Theorem 5.15, the identity Ix has a corresponding unique morphism l~:X* -+ X* such that l~ © r = r © Ix = r. Now Ix> © r = r so Ix> = 1~ since G is unique. Therefore, PM(1X) = Ipo(x), To demonstrate (4), if g © f is a morphism of A, then by Theorem 5.15, there exists unique morphisms g*, f* and (g © j)* such that (g © j)* © r =r © (g © j),f* © r = r © fand g* © r = r © g. But then r © (g © j) = g* © r © i=
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g* © 1* © r. Since (g © j)* is unique, (g © j)* = g* © 1*. Therefore, PM(g) © PM(/)'
The dual concept to the concept of a reflection is the concept of a corejlection defined as follows. A full subcategory K of a concrete category A is said to be a coreflection of A if for each object X E A, there exists an object X* E K and a coreflection morphism k:X* ~ X such that each morphism /:Z ~ X where Z E K can be factored uniquely over X* by k (i.e., / = k © g for a unique morphism g:Z ~ X*). The space X* is called the coreflection of X in K or simply the K-reflection of X. As with reflections, there are many interesting coreflections of the category of uniform spaces. Possibly the simplest and most useful is the fine corejlection that maps each uniform space (X, J..l) onto (X, u) where u is the finest uniformity for X. Here, the coreflection morphisms are the identity mappings. To see that the class F of fine uniform spaces is a coreflection of the category U of uniform spaces, it suffices to show that each morphism j:Z ~ X where Z is a fine space can be factored uniquely over (X, u). But since /:Z ~ X is continuous and Z is fine, j:Z ~ (X, u) is uniformly continuous, so / can be factored over (X, u) by the identity mapping. By the principle of duality, for any coreflection K of a category A, a K-coreflection X* of an object X is unique up to isomorphism. As with reflections, the uniqueness of K-coreflections allows us to make the assumption that if X E K, the only k-coreflection of X is X itself and the only coreflective morphism k:X ~ X is the identity morphism, so we have the dual theorem: THEOREM 5.16 1/ K is a corejlection 0/ a concrete category A, X, Y E A and X*, y* are their K-corejlections respectively, then any morphism/:X ~ Y determines a unique morphism 1* :X* ~ y* such that / © k = j © 1* where k:X* ~ X and j:Y* ~ Yare the corejlective morphisms respectively.
Theorem 5.16 shows that the correspondences X ~ X* and / ~ 1* determine a functor K called the coreflection functor. The object function of K is !Co:A ~ K defined by !Co (X) = X* and the morphism function KM is defined by KM(/) = 1* for each/E A.
EXERCISES 1. Show that the definitions of uniform sums and products are equivalent to the categorical definitions of sums and products in the concrete category of uniform spaces.
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DUAL THEOREMS 2. Show that the categorical theorem P: A retraction f that is a monomorphism is an identity implies the dual theorem Q: A retraction f that is an epimorphism is an identity. REFLECTIONS 3. Let U be the category of uniform spaces and for each (X, Jl) E U let p(Jl) denote the uniformity consisting of all coverings refined by finite members of Jl (see Exercise 6, Section 5.6). Then (X, p(Jl» is a precompact uniform space. Show that the class P of all precompact uniform spaces is a reflection of U with reflection morphisms being the identity functions. Let P denote the reflection functor (called the precompact reflection). Show that for each (X, Jl) E U that Po(X, Jl) = (X, p(Jl» and for each morphism f:(X, Jl) ~ (y, v) that PM(j) =f. 4. For each (X, Jl) E U let e(Jl) be the collection of all coverings of X refined by a countable member of Jl. Show that e(Jl) is a separable uniformity for X. [A uniform space is separable if it has a basis consisting of countable coverings.] Show that the class U w of all separable uniform spaces is a reflection of uniform with reflection morphisms being the identity functions. Let e denote the reflection functor (known as the separable reflection), and show that for each (X, Jl) E U, eo(X, Jl) = (X, e(Jl» and for each morphism j:(X, Jl) ~ (Y, v) that eM(j) = f·
5. Show that the separable uniformities are precisely the countably bounded uniformities. 6. Show that the reflections P and e commute, i.e., for each (X, Jl) p(e(X,Jl» = e(p(X,Jl». 7. Let 1t denote the completion reflection that maps each (X, Jl) Show that the reflections P and 1t commute.
E
E
U,
U into /lX.
8. It was long an unsolved problem (due to Ginsberg and Isbell) whether the reflections e and 1t commute. 1. Pelant (1974) showed that in general they do not. THE SAMUEL COMPACTIFICATION 9. Show that the composition 1t © P of the reflection 1t and P is again a reflection. Since for each X E U, p(X) is precompact, it is clear that 1t(p(X» is compact and p(X) is uniformly homeomorphic with a dense subspace of 1t(p(X». 1t(P(X» is called the Samuel compactification of X, or the uniform compactification of X.
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COREFLECTIONS 10. Let U be the category of unifonn spaces and for each X = (X, ~) E U let denote the locally fine coreftection of~. Put X* = (X, A(~» and let L be the category of locally fine unifonn spaces. Show that A:U ~ L defined by A(X) = X* for each X E U and A(j) = f is a coreflective functor so that the locally fine coreflection is really a coreftection categorically. A(~)
Chapter 6
PARACOMPACTIFICATIONS
6.1 Introduction In the late 1950s and during the 1960s, K. Morita and H. Tamano worked (independently) on a number of problems that involved the completions of uniform spaces. We state some of these problems here even though we have not yet defined some of the terms used in the statements of these problems. (1) (Tamano) Characterize the uniform spaces with paracompact completions" (2) (Morita) Characterize the Tychonoff spaces with paracompact topological completions. (3) (Morita) Characterize the Tychonoff spaces with LindelOf topological completions. (4) (Morita) Characterize the Tychonoff spaces with locally compact topological completions. (5) (Tamano) Is there a paracompactification of a Tychonoff space analogous to the Stone-Cech compactification or the Hewitt realcompactification? By the topological completion of a uniform space, we mean the completion of the space with respect to its finest uniformity. By a paracompactification of a topological space, we mean a paracompact Hausdorff space Y such that X is homeomorphic with a dense subspace of Y. If we identify X with this dense subspace (which is customary), then Y is a paracompact Hausdorff space containing X as a dense subspace. Certain paracompactifications are compact. We call these compactifications. The theory of compactifications is well developed. The theory of paracompactifications is not. Notable among compactifications is the Stone-Cech compactification. We will devote a significant amount of effort in this chapter to its development. We are not yet in a position to define what a rea[compact space is. Like paracompactness, realcompactness is a generalization of compactness. We will show in the next chapter thatfor all practical purposes, realcompactness is also a generalization of paracompactness. We will of course, formalize what we mean by "for all practical purposes," but for now, we only mention that this formalization is a set theoretic matter. It turns out that the statement All
6.1 Introduction
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paracompact spaces are realcompact is known to be consistent with the axioms of ZFC, but it is not known if it is independent of ZFC. Furthermore. it is known that if there exists a Tychonoff space that is paracompact but not realcompact, that its cardinality is larger than any cardinal with which we are familiar.
Once we know what a realcompact space is, we will be able to define a rea/compactification analogously with the way we defined a paracompactification. Like the theory of compactifications, the theory of realcompactifications is well developed. Among the various realcompactifications a space may have, there is one, known as the Hewitt realcompactification that plays a role in the theory of realcompactifications analogous to the role played by the Stone-Cech compactification in the theory of compactifications. Since "for all practical purposes," paracompactifications lie between realcompactifications and compactifications, it is natural to ask if there exists a paracompactification that plays a role analogous to the role played by the Stone-Cech compactification and the Hewitt realcompactification in the theories of compactifications and realcompactifications respectively. This is precisely Problem 5 above that we will refer to as Tamano's Paracompactification Problem. In this chapter we will show that, in general, there does not exist such a paracompactification. We will also characterize those spaces for which such a paracompactification exists and examine some of their properties. In 1937, two papers appeared that characterized the Tychonoff (uniformizable) spaces as those having compactifications. The two papers were: Application of the Theory of Boolean Rings to General Topology that appeared in the Transactions of the American Mathematical Society (Volume 41, pp. 375-481) by M. H. Stone and On bicompact spaces by E. Cech that appeared in Annals of Mathematics (Volume 38, pp. 823-844). Their approaches were very different, but both exhibited a compactification that is the largest possible compactification of a Tychonoff space. Clearly every compactification of a Tychonoff space X is the completion of X with respect to some totally bounded uniformity (since we can relativize a compactification's unique uniformity to X to obtain the desired totally bounded uniformity on X). In 1948, P. Samuel published a study of compactifications obtained as completions of uniform spaces (see Ultrafilters and compactifications, Transactions of the American Mathematical Society, Volume 64, pp 100-132). As a result, compactifications obtained as the completion of a uniform space are sometimes called the Samuel compactification of the uniform space. In this chapter we will employ this approach. The Stone-Cech compactification will be obtained as the completion of a Tychonoff space with respect to its ~ uniformity (introduced in Chapter 3). If X is uniformizable, the
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completion (X',W) of (X,~) is usually denoted by central role in general topology.
~X.
The study of
~X
plays a
Since ~ is the finest totally bounded uniformity, the Stone-Cech compactification is realizable as the completion with respect to the finest uniformity for which a compactification exists. A similar situation occurs with respect to realcompactifications. We will find in the next chapter that realcompactifications are precisely the completions with respect to the countably bounded uniformities and the Hewitt realcompactification is the completion with respect to the finest countably bounded uniformity (namely the e uniformity). Consequently, if Tamano's Paracompactification Problem is to be answered affirmatively for a given space X, the solution needs to be the completion of X with respect to the finest uniformity for X that has a paracompact completion. K. Morita was interested in when the topological completion of a Tychonoff space is paracompact. He showed that if X is an M-space (to be defined later in this chapter), then X has this property. Morita called the topological completion of an M-space X the "paracompactification of X." Consequently, we will call a paracompactification that is a solution to Tamano's Paracompactification Problem the Tamano-Morita paracompactification. In studying Tamano's Paracompactification Problem, several questions come to mind. The first is: Which spaces admit a paracompactification? This question is easily answered (for T J spaces). Since compactifications are paracompactifications, the existence of the Stone-Cech compactification implies that all Tychonoff spaces have paracompactifications. Conversely, if a space has a paracompactification it must be Tychonoff since paracompact Hausdorff spaces are normal (hence uniformizable) and as we saw in Chapter 5, a subspace of a uniform space is also a uniform space. Consequently, the class of spaces is not expanded by considering those with paracompactifications as opposed to those with compactifications. The next question that arises is: Are all paracompactifications obtainable as a completion with respect to some uniformity? This question is also easy to answer. If X is a Tychonoff space and PX some paracompactification of X, then X is a dense subspace of PX. Since PX is paracompact, by Theorem 4.4, it is cofinally complete with respect to its finest uniformity u, and hence complete with respect to u. Now u induces the v (derived) uniformity on X (see Section 4.5). Since completions are unique (Theorem 4.10), (PX,u) is the completion of (X,v). Consequently, all paracompactifications are obtainable as completions with respect to some uniformity. A completion of a uniform space that is cofinally complete will be called a uniform paracompactification. We have already seen (Theorem 4.12) a necessary and sufficient condition for a uniform space to have a uniform paracompactification. Clearly, a Tychonoff space can have many paracompact_ ifications in the same manner that it can have multiple distinct Samuel
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compactifications. We begin our study of paracompactifications by first considering compactifications. We wish to obtain the Stone-Cech compactification of PX of a Tychonoff space X as the completion of X with respect to its P uniformity. Since Phas a basis A consisting of all finite normal coverings it is clearly precompact, so the completion of eX, P) is compact. But what makes the completion of eX, P) the Stone-Cech compactification? E. Cech characterized PX in the following way:
THEOREM 6.1 (E. Cech, 1937) Any Hausdorff compactification BX of X is the image of PX under a unique continuous mapping f that keeps X pointwise fixed and such that f(~X - X) = BX - X. M. H. Stone characterized ~X in another way:
THEOREM 6.2 (M. H. Stone. 1937) Iff is any continuous mapping of a Tychonoff space X into a compact Hausdorff space Y. then f has a unique continuous extension ff'J :~X -7 Y. We propose to show that the completion of (X, ~) satisfies Theorems 6.1 and 6.2 in the sense that if we replace ~X with the completion of (X, ~) in the statement of both these theorems, they both remain true. Then we will show that any compactification satisfying Theorems 6.1 and 6.2 is the completion of eX, P). Thus, among the various compactifications a Tychonoff space X may have, the completion of eX, P) is distinguished as the Stone-Cech compactification of X.
6.2 Compactifications In what follows, we will adopt the following notation: if (X, f..l) is a uniform space, then f..lX will denote its completion. For us then, ~X will denote the completion of (X, ~). We first prove Theorem 6.2 using this interpretation of ~X.
(Proof of Theorem 6.2) Let Wbe the unique uniformity of ~X (Theorem 2.8). Then (BX, W) is the completion of (X, ~). Let B be the unique uniformity on Y. By Lemma 3.8, each Tychonoff space admits a uniformity that has a basis consisting of all finite normal coverings. Let U E B be one of them. For each UE UputVu=Unj(X)andlet V= (VUIUE Ul. Then V is a finite normal covering of I(X) and 1 (V) = {f-l (Vu) IU E U} is a finite normal covering of X. By definition of B, r 1 (V) E B. But then f is uniformly continuous with respect to ~ and B. By Proposition 4.20, Y is complete, so by Theorem 4.9, I has a unique uniformly continuous extension/f'J:BX -7 Y.-
r
(Proof of Theorem 6.1) X can be identified with dense subspaces of both BX and BX. Define f:X c ~X -7 X c BX by I(x) = x for each x E X. Then, as
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in the proof of Theorem 6.2, B' (the unique uniformity on BX) has a basis consisting of all finite normal coverings. If U' E B' is one of these finite normal coverings, we have just seen that r 1 (Unf(X)) is a finite normal covering of X, so f is uniformly continuous. Therefore, we can apply Theorem 6.2 and obtain a unique continuous extension f~:PX ~ BX. Since PX is compact, f(PX) is closed in BX. But X c f(PX) c BX which implies f(PX) = BX since CI(X) = BX. It remains to show that f~(PX - X) c BX - X. For this let x E PX - X and let {x u} C X be the fundamental Cauchy net in X (see Theorem 4.9) that is used to define x' = j'(x). Suppose x' E X. Now {xu} converges to x and {f(x u )} converges to x'. But since f(x) = x for each x E X, {f(x u)} = {x u}. But then {xu} converges to x' E X and {xu} converges to x which is not in X which is a contradiction. Hence x' does not belong to X. Therefore,j~(pX - X) c BX - X. Since f~(X) = X, we have f~(PX - X) = BX - X.Finally, assume BX is a compactification of X that satisfies Theorem 6.1. Then there exists a unique uniformly continuous function f:BX ~ PX such that f(BX) = PX, f(x) = x for each x E X, and f(BX - X) = PX - X. Let U E P be a finite normal covering of X. Then there exists a finite normal covering U' of W such that U' nX = U. Since f is uniformly continuous,j-l (U') is a finite normal covering of BX that belongs to B'. But then r 1(U')nX E B. Since f(x) = x for each x E X and f(BX - X) = PX - X, r1(U')nX = r1(U'nX) = rl(U) = U. Hence B contains all finite normal coverings of X, so pcB. But by Lemma 3.8 and Theorem 2.8, the unique uniformity B' on BX has a basis consisting of all finite normal coverings (on BX). Consequently B has a basis consisting of (perhaps not all) finite normal coverings so B c p. But then B = p, so BX is pX. In order to demonstrate the utility and importance of the Stone-Cech compactification, this section and the next two will be devoted to an analysis of some of the most important topological properties of Tychonoff spaces in terms of subsets of pX. The approach we will follow is due to H. Tamano, and appeared in his 1962 paper titled On compactifications which appeared in the Journal of Mathematics of Kyoto University (Volume I, Number 2). Tamano's approach is usually thought of as characterizing topological properties of a Tychonoff space X in terms of the behavior of subsets of PX - X (which it does). But as we shall see, his approach is deeply involved with certain normal coverings and with extending uniformities of X into pX. In this section we build the tools needed for this development. In the next section we prove Tamano's Completeness Theorem, and in Section 6.4, we present Tamano's Theorem. In Chapter 2, we defined an open set A c X to be regularly open if intx(Clx(A)) = A. If in turn, X is a subspace of Y, we can generalize this concept by taking the closure and the interior in Y rather than in X in order to get what is called an extension of A over Y. In general, if U is open in X and D' is open in Y such that V = V'nX, then V' is said to be an extension ofD over Y. Put VE(y) = Y - Cly(X - V). Then vEry) is an extension of Dover Y since
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UE(Y)nX = (Y - Cly(X - U»nX = X - Cly(X - U)nX = X - (X - U) = U. We call UE(Y) the proper extension ofU over Y. It is clear from this definition that U c V implies UE(Y) e VEly). LEMMA 6.1 Let X be a dense subspace of Y and let U be open in X. Then UElY) is the largest extension of U over Y. Proof: Let U' be an extension of U over Y. If Y does not belong to UE(Y), then y E Cly(X - U). Therefore, every open set V'(y) c Y containing y meets X - U which implies V'(y)nX is not a subset of U = U' nX. Consequently, V'(y) is not contained in U' which implies y does not belong to U' since U' is open. Therefore, U' c UE(y). • LEMMA 6.2 Let X be a dense subspace of Y. Then for any A eX. we have Intx(Clx(A)) = Inty(Cly(A))nX. Proof: Let x E Intx(Clx(A». Then there exists an open set U(x) containing x such that U(x) c Clx(A) c Cly(A). But then there exists an open set U'(x) c Y such that U'(x)nX = U(x). Suppose U'(x) is not a subset of Cly(A). Then there exists ayE U(x) with y E Y - Cly(A) which is open. Since X is dense in Y, U'(x)n[Y - Cly(A)]nX oF 0. Let Z E U'(x)n[Y - Cly(A)]nX which implies Z E U'(x)nX = U(x) c Cly(A). But on the other hand, Z E Y - Cly(A) which implies Z does not belong to Cly(A) which is a contradiction. Therefore, U'(x) c Cly(A) which implies x E Inty(Cly(A»nX so Intx(Clx(A» c Inty(Cly(A»nX.
Conversely, if x E Inty(Cly(A»nX then x E X and there exists an open set U'(x) e Y containing x such that U'(x) c Cly(A) which implies U'(x)nX c Cly(A)nX. Put U(x) = U'(x)nX. Then U(x) is an open set in X containing x with U(x) e Cly(A)nX. Suppose y E Cly(A)nX. Then every open set V'(y) c Y containing y meets A. But then V'(y)nX meets A. Since A c X, every open set V(y) c X containing y meets A which implies y E Clx(A). Therefore, Cly(A)nX e Clx(A). Hence U(x) c Cly(A)nX c Clx(A). Therefore, x E Intx(Clx(A», so Inty(Cly(A»nX e Intx(Clx(A». • LEMMA 6.3 Let X be a dense subspace of Y. If u' is an extension of the open set U eX. then Cly(U') = Cly(U). Proof: Since U = U'nX we have U c U' which implies Cly(U) c Cly(U'). Suppose y E Cly(U'). Then every open set V'(y) c Y containing y contains a point of U'. Since V'(y)nU' oF 0 and since X is dense in Y, there exists an x E X such that x E V'(y)nU' which implies x E U. Therefore, every open set V'(y) c Y containing y meets U which implies y E Cly(U). Hence Cly(U') c Cly(U) so Cly(U,) Cly(U) .•
=
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LEMMA 6.4 Let X be a dense subspace of Y. Then VE(y) is regularly open in Y if and only if V is regularly open in X. Also, if V is regularly open in X, then VE(Y) = Inty(Cly(V)). Proof: If VE(Y) is regularly open, then V = VE(Y)nX = Inty(Cly(VE(Y))nX. Since VE(Y) is an extension of V, by Lemma 6.3 we have V = Inty(Cly(V»nX. Then by Lemma 6.2, V = Intx(Clx(V» so V is regularly open. Conversely, suppose V is regularly open. Then V = Intx(Clx(V» = Inty(Cly(U»nX. Hence V' = Inty(Cly(U» is a regularly open extension of V over Y. Therefore, V' c VE(Y) by Lemma 6.1. We can complete the proof by showing VE(Y) c V'. For this, we suppose on the contrary that VE(Y) is not a subset of V' which implies VE(y) is not a subset of Cly(U'). Since X is dense in Y we have [VE(Y)n(Y Cly(V,)]nX"# 0. Therefore, V = V£(Y)nX is not a subset of Cly(V') which is a contradiction. Hence VE(Y) c V' so VE(Y) = Inty(Cly(U».PROPOSITION 6.1 Let X be a dense subspace of Y and Y a dense subspace of Z. Let V and V be open sets in X and Y respectively. Then the following are valid: (1) VE(Z) nY = VE(Y) , (2) [VE(y)]E(Z) = VE(Z) and (3) VE(Z) c [V nX ]E(Z). If V is regularly open then VE(Z) = [V nX ]E(Z). Proof: To prove (1) we note that VE(Z)nY = [Z - Clz(X - V)]nY = Y [Clz(X-V)nY] = Y - Cly(X - V) = VE(Y). To prove (2) observe that [VE(y)]E(Z) = Z-Clz(Y - VE(Y) = Z - Clz[Y - (Y - Cly(X - V»] = Z - Clz(Cly(X - U» = Z-Clz(X-U) = VE(Z). To prove (3), note that V is an extension of vnX so by Lemma 6.1, V c (V nX)E(Y) and hence VE(Z) c [V nX ]E(Z) by (2) above. If V is regularly open, then vnX is also regularly open since vnX = Inty(Cly(V»nX = Inty(Cly(VnV»nX by Lemma 6.3. But then by Lemma 6.2, vnX = Intx(Clx(VnX». Therefore, V = (V nX)E(Y) by Lemma 6.4. Again by (2) above, VE(Z) = (V nX)E(Z) . _
If we take an open covering U of X and then take the proper extension of each member of U to BX, we get a new covering W(PX) of X in BX called the proper extension of U into BX and we call the set uUE(PX), denoted E u, the extent of U in BX. An open covering V of X is said to be stable if there exists a normal sequence {V n} of open coverings of X such that VI < V and E v = E Vn for each n. The first of Tamano's theorems that we prove shows that an open covering is stable if and only if its extent is paracompact. In general, if X is a dense subspace of Y, we can define the proper extension VE(Y) of V over Y and the extent uVE(Y) of V in Yanalogously.
Letf:X -+ R (reals). We denote by Z(j) the zero set of fdefined by Z(j) = X If(x) = O} and by O(j), the complement of Z(j) in X. We denote by Z(X) the set of all zero sets of X. In Section 1.1, O(j) was called the support of f. Also recall from Section 1.1 that a partition of unity on a space X is a family {x
E
6.2 Compactifications
163
= {A} of continuous functions ,,:X ~ [0,1] such that 11..,,(x) = 1 for each x E X, and such that all but a finite number of members of
THEOREM 6.3 (J. Dieudonne, 1944) For each locally finite open covering {Va} of a normal space, there exists a partition of unity {a} subordinate to {Va} such that O(a) C Vafor each a. Proof: Let X be a normal space and {Va} a locally finite open covering of X. By Lemma 1.3, {Va} is shrinkable to an open covering {Va}. By Theorem 0.3, for each a there exists a real valued continuous function f a:X ~ [0,11 with fa(CI(Va» = 1 andfa(X - Va) = O. Since {Va} is locally finite, so is IVa}. Hence for each p E X,fa(P):t 0 for at most finitely many a. Thus F:X ~ [OJ) defined by F(x) = La!a (x) for each x E X is well defined. To show F is continuous, let p E X. Then there exists an open W(P) containing p such that W(P) meets only finitely many Va, say V aj • • . Van' Therefore, F(x) = L7=da,(x) for each x E W(P). Since each fa, is continuous on W(P), so is F. But then F is continuous at each point of X. For each a, define
PROPOSITION 6.2 A regular Linde16f space has the star-finite property. Proof: Let X be a regular Linde16f space and let U be an open covering of X. For each p E X, there exists an open set V(P) such that CI(V(P» c U for some U E U. Then V = {v(p)lp E X} is an open covering of X and as such has a countable subcovering say {V(pn)}' For each positive integer n let V n E U such that CI(V(pn» c Vn. Since X is normal by Lemmas 1.2 and 3.6, there exist real valued continuous functions fn on X such that fn(CI(V(pn))) = 0 and fn(X-Vn) = 1, so we can apply Theorem 2.16 which yields a star-finite
164
6. Paracompactifications
refinement of U. Consequently, a regular LindelOf space has the star-finite property. PROPOSITION 6.3 If X is a regular space and there exists an open covering U of X such that each member of U is LindelOf and U admits a starfinite refinement V, then X can be decomposed into a sum of disjoint open sets, each of which is Lin de IOf and X isfully normal and has the star-finite property. Proof: If x,y E X and there exists a positive integer n with y E Starn(x, V) then we put x - y. As was seen in Section 2.4, this relation can be used to decompose X into disjoint sets X a, a E A, such that x and y belong to the same X a if and only if x - y [see Theorem 2.15]. Then for each a E A, Starn(x, V) c X a for any x E X a' Since X a N ~ = 0 if a '# ~,X a is both open and closed.
To show a particular X a is LindelOf, we first show that for any open covering W of X and any pair (x,n) such that x E X a and n a positive integer, that there exists a countable subset of W covering Starn(x, V). We induct on n. First note that Star (x, V) c U for some U E U and U is LindelOf. Next assume Starn(x, V) has this property. Then there exists a countable collection {Vi} c V that covers Starn(x, V) which implies Star n+ 1 (x, V) C U;=I Star (Vi, V). But Star (Vi, V) C Vi for some Vi E U, so Starn +1 (x, V) C U;=I Vi' Since each Vi is LindelOf, Star n+ 1 (x, V) can be covered by countably many members of W. But then, for any open covering W of X, X a = U;;'=I Starn(x, V) can be covered by countably many members of W. Since X a is closed, X a is LindelOf. This proves the first part of the proposition. The second part follows from Lemma 3.6 and Proposition 6.2. A space X is said to be locally compact if each p compact neighborhood.
E
X is contained in a
LEMMA 6.5 A locally compact Hausdorff space is completely regular. Proof: Let X be a locally compact space and let p be a point of X contained in some open set W. Since X is locally compact, there exists an open set V containing p such that CI(V) is compact. Put U = VnW. Then CI(U) is also a compact neighborhood of p. Since CI(U) is a compact Hausdorff space, it is normal, so there exists a continuous function g:CI(U) ~ [0,1] such that g(P) = 1 and g(CI(U) - U) = O. Let h be the constant function on X - U defined by h(x) = o for each x E X - U. Define f on X by f(x) = g(x) if x E U or f(x) = h(x) otherwise. Clearly f:X ~ [0,1] is continuous, f(P) = 1 and f(X - W) = O. Consequently X is completely regular.LEMMA 6.6 If U is an open covering of a Hausdorff space X such that the closure of each member of U is compact, and if U admits a star-finite refinement V, then X isfully normal and has the star-finite property.
165
6.2 Compactifications
Proof: Clearly X is locally compact so by Lemma 6.5, X is completely regular. Moreover, any set U of U has the property that for any open covering W of X, U can be covered by countably many members of W. This was the only use made of the Lindelbf property in Proposition 6.3. Hence the same method of proof can be used to prove this lemma. THEOREM 6.4 (K. Morita, 1948) A necessary and sufficient condition for a locally compact Hausdorff space to be paracompact is that it possess the star-finite property. Proof: Let X be a locally compact Hausdorff space. Assume first that X possesses the star-finite property. Then each open covering has a star-finite refinement. Since a star-finite refinement is locally finite, we conclude that X is paracompact. Conversely assume X is paracompact. By Stone's Theorem (l.l), X is fully normal. Since X is locally compact, there exists an open covering U such that the closure of each member of U is compact. Since X is fully normal, U has a star refinement. Then by Lemma 6.6, X has the star-finite property. • E. Cech, in his paper On bicompact spaces referenced in Section 6.1, showed that a T 1 space X is normal if and only if Cll3x(E)nCll3x(F) = 0 for each pair of disjoint closed sets E, F c X. This useful result is needed to characterize the locally compact Hausdorff spaces as those spaces X such that BX - X is compact for any compactification BX of X. This result will also be needed in the proof of Taman 0 ' s characterization of stable coverings.
THEOREM 6.5 (E. Cech, 1937) A T] space X is normal if and only if CI 13x (E)nCI 13x (F) = 0 for each pair of disjoint closed sets E, F eX. Proof: Let X be a T 1 space and assume that for disjoint closed sets E, F c X, Cll3x(E)nCll3x(F) = 0. Since pX is normal, there exist disjoint open sets U,V c pX such that Cll3x(E) c U and Cll3x(F) c V. Then UnX and VnX are disjoint open sets in X containing E and F respectively, so X is normal. Conversely, assume X is normal and that E and F are disjoint closed sets in X. Then there exists a real valued continuous function f:X ~ [0,1] such that feE) = 0 and f(F) = 1. By Theorem 6.2, f can be extended to a continuous functionf 13 :pX ~ [0,1]. Since E is dense in Cll3x(E) andf(E) = O,fI3(Cll3x(E» = O. Similarly f 13 (CI 13x (F» = 1. Then U = {x E pX ifl3 (x) < 1/2} and V = {x E PX ifl3 (x) > 1/2} are disjoint open sets containing Cll3x(E) and Cll3x(F) respectively, so CI 13x (E)nCI 13x (F) = 0 .•
PROPOSITION 6.4 A Tychonojf space X is locally compact if and only
if BX - X is compact for any compactification BX of X. Proof: For each x
E
X, there exists a neighborhood U(x) of x such that
166
6. Paracompactifications
Clx(U(x» is compact, if X is locally compact. Then Clx(U(x» is closed in BX so Clx(U(x)) = Cl ~x(U(x». Therefore. no point of BX - X is contained in Cl ~x(U(x)), so X is open in BX. Hence BX - X is closed and therefore compact. Conversely, if BX - X is compact, then it is closed so for each x E X, there exists a neighhorhood U(x) of y in ~X such that C!(U(x»n[BX - Xl = 0. But then Cl~x(U(x» = Clx(U(x». Hence Clx(U(x» is compact. We conclude that X is locally compact.We are finally in a position to prove the first major theorem from Tamano's 1962 paper. This theorem not only estahlishes the connection among stahle coverings, star-finite partitions of unity and coverings with paracompact extents, hut it is essential in the proof of Tamano's characterization of the LindelOf property in terms of compact subsets of BX - X (where BX is a compactification of X) that we will encounter later.
THEOREM 6.6 (H. Tamano. 1962) The following statements are equivalent for a Tychonojf space x: (1) V is a stable covering ofX. (2) the extent Ev of V in ~X is paracompact and (3) there exists a star-finite partition of unity {
Idx(y) - dAz) I = Id(x,y) - d(x,z) I :;:; Id(x,z) + d(z,y) - d(x,z) I == Id(y,z) I < E.
6.2 Compactifications
167
By Theorem 6.2. the restriction of d x to X can be uniquely extended to a continuous function d~:~X ---7 [0.1]. To show that d~ agrees with d x on Ev we pick Y E Ev - X. and assume drCY) 7: d~(y). Now there exists a net {Ya} eX that converges to Y since d x and d~ are both continuous. {dx(v a )} converges to dAy) and {d~(Ya)} converges to d~(y). But since {Ya} c X, {dxCYa)} = {d~CYa)}. Consequently, {dx(y a)} must converge to two distinct points in a Hausdorff space which is impossible. We conclude that d x can be uniquely extended to ~X. Now d~(r) ~ 1/4 since r E CI~x(U(x». so there exists an open W(r) in ~X containing r such that d~(v) < 1/2 for each Y E W(r)nE v. Therefore, W(r)nE v c sex, 21) = S(x.V)) C Vf for some VE E V'. By Lemma 6.1, W(r) c V' so r E E p. Consequently. Cl ~x(U(x» c E v· Put U = {U(x)lx E Ev}. Since (Ev,'t) is paracompact, there exists a locally finite open refinement {V A} covering E v. Since the coverings V~ consist of open sets in ~X. the set B v = {V~ I V;, E V~ for some n} is a basis for "C. But B v is a subset of the topology on ~X so the topology induced on E v by ~X is finer than 'to Consequently. the covenng {V A} is also open in ~X. An argument similar to the one above can be used to show that Cl ~x(V),J c E v for each A. To show that E v is a paracompact subset of ~X, let {G a} be an open covering of E v. Then for each A, Cl ~x(U A) is covered by finitely many of the Ga's say G I ... Gm • PutH~ = VAnG K for 1(= 1 ... m. Then the family {HO is a locally finite open refinement of {G a}. This completes the proof of (1) ---7 (2). (2) ---7 (3). Assume that E v is paracompact. Since E v is open in ~X, ~X E v is closed and hence compact. Since ~X is also a compactification of E v. by Proposition 6.4, E v is locally compact. Then by Theorem 6.4. E v possesses the star-finite property. Since E v is paracompact. there exists a normal sequence of open coverings {V~} in E v such that VI < Vf • Consider the open covering U = {S(z.V~) IZ E E v}. Since E v has the star-finite property, by Theorem 6.3. there exists a star-finite partition of unity $' = {
It remains to show that E w = E v. For this tirst note that W < V implies Ew c Ev· To show that Ev c Ew, let Z E Ev. Pick x E W'(z)nX, which is possible since X is dense in ~X. Then Ltc I
6. Paracompactifications
168
(3) --7 (1). Let
= Vn
C
V~nX = [V~nX']nX
by Proposition 6.1.(1). Therefore, V~nX' contains V~ so uV~ contains uV~ which implies E vn contains X' for each non-negative integer n. To show that Ev n c X' for each n, let p E Ev n . Then p E V~(x) for some x E X. Let y E
V~(x)nX. Then ~1
L~I I
= L~l
1.
Now ~ I
--7
(1).-
The next step in Tamano's development is to strengthen the hypothesis and conclusion in Theorem 6.6 in the following way: a locally compact Hausdorff space is said to be a-compact if it is the union of countably many compact spaces. In what follows we show that the locally compact paracompact Hausdorff spaces are precisely the topological sums of a-compact spaces. Hence the extents of stable coverings are topological sums of a-compact spaces as can be seen in the proof of (2) --7 (3) in Theorem 6.6. Let us agree to call a
6.2 Compactifications
169
stable covering strongly stable if its extent is (J-compact. Tamano's next theorem establishes the connection between strongly stable coverings and countable star-finite partitions of unity. There are a number of alternate definitions of both local compactness and (J-compactness that are very useful. It will facilitate our development to establish these equivalent deftnitions. The proofs of the next two lemmas are left as exercises (Exercises 1 and 2). LEMMA 6.7 The following four properties are equivalent: (I) X is a locally compact Hausdorff space. (2) For each x E X and neighborhood U of x, there exists a neighborhood V of x with CI(V) c U and CI(V) compact. (3) For each compact K and open U containing K, there exists an open V with K eVe CI(V) c U and CI(V) is compact. (4) X has a basis consisting of open sets with compact closures. LEMMA 6.8 The following three properties are equivalent: (I) X is a (J-compact space. (2) There exists a sequence {Un) of open sets with each CI(Un) compact such that CI(Un) c Un+ 1 for each n, and X = U;;'=l Un. (3) X is a locally compact LindelOf space.
At first glance it might appear that any space that is the union of at most countably many compact spaces is (J-compact, but the rational subset R of R (reals) is a counter example because R is not locally compact. Consequently, local compactness is an essential part of the definition of (J-compactness. Since a (J-compact space is Lindelbf by Lemma 6.8 and completel y regular by Lemma 6.5, then by Lemma 3.6 it is paracompact. Therefore, we get locally compact paracompact spaces by forming topological sums of (J-compact spaces. The next proposition shows that this is the only way to get such spaces. LEMMA 6.9 If a covering {U a}. a E A, has a point (locally) finite refinement {V J3}. ~ E B, then {U a} has a precise point (locally) finite refinement {W a}. If each V J3 is open, then each Wa can be chosen to be open. Proof: Define a function f:B ~ A by assigning to each ~ E B some a E A such that V J3 c U a. For each a let Wa = u{ V J3lf(~) = a}. It is possible that Wa = for some of the a's. Clearly Wac U a for each a, and {W a} is a covering because each V J3 is a subset of some Wa. If {V J3} is point (locally) finite, then each (some neighborhood of) x E X lies in at most finitely many of the V J3 and consequently cannot meet more than finitely many of the W a's. Therefore, {W a } is a point (locally) finite refinement of {U a}. -
o
170
6. Paracompactifications
PROPOSITION 6.5 The locally compact paracompact Hausdorff spaces are precisely the topological sums of a-compact spaces. Proof: In view of the remarks preceding Lemma 6.9, it is clear that we only need to prove that a locally compact, paracompact, Hausdorff space X is a topological sum of a-compact spaces. Let X be covered by open sets U(x) such that x E U(x) and CI(U(x» is compact for each x E X. By Lemma 6.9, there exists a precise locally finite open refinement {V(x) Ix E X I. Since CI(V(x» is compact for each x E X, it meets at most finitely many other V(y)'s. To see this, for each Z E CI(V(x», V(z) meets at most finitely many other V(y)'s and since CI(V(x» is compact, it can be covered by finitely many V(y)'s, say V(z I) ... V(zn). Then W = U?=I V(z;) contains CI(V(x» and meets at most finitely many V(y),s.
For each pair x,y E X put X - Y if there exists a finite family of sets V(z I) .. V(zn) such that x E V(ZI), y E V(Zn) and V(Zi)nV(Zi+d oF 0 for i = 1 ... n-1. Clearly - is an equivalence relation in X. Let {X a I a E A I denote the family of equiValence classes with respect to -. Then the X a cover X and are pairwise disjoint. Each X a is open since it is the union of V(y)'s. Therefore, each X a is a locally compact Hausdorff space. It remains to show that each X a is a-compact. Let K I = CI(V(x» for some x E X a' For each positive integer n, inductively define Kn as the union of all CI(V(y»'s with V(y) meeting K n_l • Since K n - I is compact, it meets at most finitely many V(y)'s (as shown above) so Kn is compact. Since for each y E X a there are finitely many V(z)'s, say V(z d ... V(zn) with x E V(z d, y E V(zn) and V(z;)n V(zi+d oF 0 for each i = I ... n - 1, it is clear that y E K j for some positive integer j. Therefore, X a = U';;=1 KnoB Y Lemma 6.8, X a is a-compact. Therefore, X is a topological sum of a-compact spaces. THEOREM 6.7 (H. Tamano, 1962) The following statements are equivalent for a Tychonoff space x: (I) V is a strongly stable covering of X, (2) there is a countable star-finite partition of unity {
The proof of this theorem can be obtained by modifying the proof of Theorem 6.6 and is left as an exercise (Exercise 3). EXERCISES 1. Prove Lemma 6.7.
2. Prove Lemma 6.8. 3. Prove Theorem 6.7.
6.3 Tamano's Completeness Theorem
171
6.3 Tamano's Completeness Theorem In 1944, in a paper referenced in Section 1.2, J. Dieudonne showed that the product X x Y of a paracompact space X with a compact space Y is paracompact. Hence X x Y is normal. Thereafter, many researchers tried to determine if the reverse implication held; i.e., does the normality of X x Y imply X is paracompact when Y is compact? By 1960, many partial results had been published. Then in 1960, in a paper titled On paracompactness published in the Pacific Journal of Mathematics (Volume 10, pp. 1043-1047), Tam an 0 showed that it does if Y = ~X (more generally if BX is any compactification of X, the normality of X x BX implies that X is paracompact). This result has come to be known as Tamano' s Theorem. In Section 4.5 it was mentioned that H. Tamano and K. Morita had considered the problem: "What is a necessary and sufficient condition for a uniform space to have a paracompact completion?" Tamano published four other papers in 1960. One of them titled Some properties of the Stone-Cech compactijication published in the Journal of the Mathematical Society of Japan (Volume 12, pp. 104-117) details his results in this area on a number of problems having to do with the completion of uniform spaces. One of the theorems in this paper gives a necessary and sufficient condition for a uniform space to be complete in terms of the radical of a uniform space. The concept of the radical is based on the concept of the extent of a covering in ~X. If (X, Jl) is a uniform space, the radical of (X, Il) is defined to be nlEu IU E Ill. i.e., the intersection of all the extents in ~X of uniform coverings. This theorem states that if R is the radical of (X, Jl) then X is complete if and only if R = X. We will refer to this result as Tamano's Completeness Theorem. This section is devoted to its proof. LEMMA 6.10 Let A be a regularly open set in a space X. Let B be an open set which is not contained in A. Then there is an open C c B with CnClx(A) = 0. Proof: If Be Clx(A) then B c Intx(Clx(A)) = A since A is regularly open. But B is not contained in A so B is not contained in Clx(A). Then C = Bn[X - Clx(A)] is the desired subset of B such that CnClx(A) = 0.LEMMA 6.11 Let X be a dense subspace of a space Y. Then (1) If A is regularly open in Y then AnX is regularly open in X, (2) B is regularly open in X if and only if B = Inty(Cly(B))nX, (3) if A is regularly open in Y and B is open in Y, and if AnX contains BnX then B c A and (4) Two regularly open sets A,B in Yare identical if and only if AnX=BnX. Proof: Assume A is regularly open in Y.
AnX c
Intx(Clx (A
nX»
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6. Paracompactifications
Inty(Cly(A nX))nX by Lemma 6.2. Now Inty(Cly(A nX))nX c Inty(Cly(A))nX = A nX since A is regularly open in Y. Therefore, A nX = Intx(Clx(B))nX so A nX is regularly open in X. This proves (1).
To show (2) suppose B = Inty(Cly(B))nX. Since Inty(Cly(B)) is regularly open in Y, by (1) above, B is regularly open in X. Conversely, suppose B is regularly open in X. Then B = Intx(Clx(B)) so by Lemma 6.2, B Inty(Cly(B))nX. Therefore, B is regularly open in X if and only if B = Inty(Cly(B))nX. To show (3) assume A is regularly open in Y and B is open in Y. Suppose B is not contained in A. By Lemma 6.l0 there is an open C c B with C rIA = 0. Since X is dense in Y, there exists apE C such that p E X. But then p E B nX but p is not in A nX which implies B nX is not contained in A nX. Therefore, B nX c A nX which implies Be A. To show (4) assume both A and B are regularly open in Yand A nX = B nX. Then by (3) above A c Band Be A so A = B. Conversely, if A = B then A nX = B nX. Hence two regularly open sets A, B in Yare identical if and only if A nX =BnXTamano worked exclusively with entourage uniformities. His original proof was somewhat different than the proof we give here but the proof given here preserves his approach. The following development is a good example of how working with covering uniformities leads to simpler statements of theorems as opposed to working with entourage uniformities. Tamano's proof was subdivided into a number of propositions. Each of the next five propositions, with the exception of Proposition 6.11, is equivalent to one of his propositions. PROPOSITION 6.6 Let X be a dense subspace of a Tychonoff space Y and let A be a basis for a uniformity ~ consisting of regularly open coverings. For each U E A, Star(VE(y),UE(Y)) c [Star(V,UW(y) so V <* U implies V€(Y) <* u£(Y) for each V E A. Proof: Since each U E A consists of regularly open sets, UE(Y) = {lnty(Cly(V)) IV E U} by Lemma 6.4. Suppose y E Star(VE(Y) , UE(Y)). Then there exists a VIE U such that y E vt(Y) and Vt(Y)nV£(y) "# 0. Therefore, there is a Z E Vl(y) nVE(Y) nX so VIC Star (V, U). Now Y E Vl(y) implies y E Inty(Cly(V d) c Inty(Cly(Star (U, U))) c [Star (V, UWCYl. Next let U, V E A with V <* U. Then Star (V, V) c V for some V E U. By the previous argument, Star (VE(Y) ,VE(Y)) c [Star (V, VW(y) and since Star (V, V) c V, we have [Star (V, VW(Y) c VE(Y). Therefore, Star (VE(Y), VE(Y)) c VE(Y) so VE(y) <* UE(y)._
Let (X, ~) be a uniform space and (X', ~') its completion. Let ~X and ~X' be the Stone-Cech compactifications of X and X' respectively. Then ~X' is also a compactification of X so by Theorem 6.1, there exists a continuous function <1>:
6.3 Tamano's Completeness Theorem ~X ~ ~X' such that <1> is onto, <1>(X) regularly open basis for~.
173
= X and
<1>(~X - X)
= ~X' - X.
Let A he a
PROPOSITION 6.7 For each U E ~, UE(~X') = U'E(~X') Proof: Let U E ~. It suffices to show that UE(~X') = U'E(~X') for each U E U. Since X is dense in X' which is in turn dense in ~X', by Lemma 6.1, U = u'nX which implies U' c UE(X'). Therefore, U'E(~X') c [UE(X')jE(j3X') = UE(~X') hy Proposition 6.1.(2). But U c U' which implies U E(j3x') c U'E(j3X'). Therefore, U E(j3x') = U'E(j3X')._ PROPOSITION 6.8 For each U E A, <1>-1 (U'E(~X') < W(j3X). Proof: Let U E A. It suffices to show that <1>-1 (U'£(j3X') c UE(~X) for each U E U. By Proposition 6.7, U'E(~X') = U£(j3x') so <1>-1 (U'E(~X') = <1>-1 (UE(~X'). Now
is open in ~X and U£(j3X) is regularly open in ~X since U E A. Therefore, by Lemma 6.11.(3) it suffices to show that <1>-1 (U£(j3X')nX c U£(j3X)nX = U. For this let y E <1>-1 (U E(j3X')nX which implies Y E X and <1>(y) E U£(j3x'). Now Y E X implies <1>(Y) = Y E U£(j3x')nX = U. We conclude that <1>-1 (U E(j3x') )nX c U. -
<1>-1 (U£(j3X')
LEMMA 6.12 R
= ndu{S(x,U£(~X)) Ix E
Xl)
= ndu{S(x,Ul(j3X) Ix E
Xl). Proof: From the definition of the radical, R = n A[u {U£(~X) I U E U I) c n A[u{S(x,W(j3X)Ix E Xl) c n A[{uS(x,Ul(j3X) Ix E Xl). Conversely, suppose Y does not belong to ndu{S(x,Ul(j3X) Ix E Xl). Then there exists a V E A such that y does not belong to S(x,V l(j3X) for each x E X which implies y does not belong to U E(j3X) for each U E U which in turn implies y does not belong to R. Hence, R = n A[u{S(x,U E(j3X) Ix E Xl) = n A[u{S(x,W(j3X) Ix E Xl).PROPOSITION 6.9 A point p E ~X is contained in the radical if and onlyif<1>(p)E X'. Proof: If pER then for each U E A, p E Sex, W(~X) for some x E X. Put \jf(U) Hence K u = S(p, U E(j3X)nX '# 0. Let V(P) be an open neighborhood of p in ~X. Then V(p)nS(p,U£(j3X) '# 0 which implies V(p)nK u '# 0. Consequently, p E Clj3x[K u l for each U E A. Since <1> is continuous, <1>(P) E Cl j3x,[K ul. We wish to show that nACl j3x,[K ul is a single point q E X' which will mean <1>(P) EX'.
=x.
For this consider the net \jf:A ~ X. For each U E A, \jf(U) E K u. Let V, A such that W* < V* < U. Then p E S(\jf(W),WE(j3X)nS(\jf(V),V Ec!>X) which implies [S(\jf(W),WW(!3X)n[S(\jf(V),VW(j3X) '# 0 which in turn implies S(\jf(W),W)nS(\jf(V),v) '# 0. Hence \jf(W) E Star (S(\jf(V),v),V) c S(\jf(V),U). W
E
6. Paracompactifications
174
So, 'l' is Cauchy in X and in X'. Therefore, 'l' converges to some q want to show that {q} = nACI~x,[K ul. For this let F u E
U
= {'l'(V) I V
E
A and
V*
E
X'. We
< U). Then F u c K u for each U
A. 'l' converges to q implies q E Clx,(F u) c Cl ~x,(F u) c Cl ~x,(K u) for each E A. Therefore, q E Cl ~x'(K u). Suppose y =F q is another point of ~X'. Then
there exists an open A c ~X' containing q such that y does not belong to Cl ~x'(A). Then AnX' is an open set in X' containing q so there exists a U' E A' with S(q,U') c AnX'. Let V' E A' with V' <* U'. 'l' converges to q implies there exists a w~
E
A' with W;) <* V such that
W' E A' with W' <* W~ which implies 'l'(W) E S(q,V') for each W' E A' with W' <* W~. Now 'l'(W) E S(p,WE(~X) so P E S('l'(W).WE(~X). For each x E K w. P E S(X,WE(~X) so there exists WI, W 2 E W with p E W1(~X)nW~(~XJ and 'l'(W) E W) uW 2. Since W1(~X)nW~(~x) =F 0, W! nW 2 =F 0 which implies there
exists a V) E V with W) uW 2 C V). Since 'l'(W) E S(q,v'). there exists a V~ E V' such that q. 'l'(W) E V~. Then V')nV~ =F 0. Consequently, there exists a U' E U' with V') uV~ c U'. Then q, x E U' which implies K w c Seq, U') c AnX' c A. Hence y does not belong to K w which implies y does not belong to = {q}.
nACl~x'[K uj· Therefore, nACl~x'[K ul
Conversely, if <jl(P) E X' then <jl(P) E U' for some U' E U', for each U' E A'. But then <jl(P) E U'E(~X') C UE(~X) by the proof of Proposition 6.8. Hence p E E u for each U E A which implies pER. We conclude that pER if and only if <jl(P) EX'.-
THEOREM 6.8 if and only if R =X.
(H.
Tamano, 1960) A uniform space (X,I1) is complete
Proof: Assume X is complete. It is evident that X cR. Let pER and for each U E A let 'l'(U), K u and F u be defined as in the proof of Proposition 6.9. Let V E A such that V* < U. By Proposition 6.6, VE(~X) <* UE(~X). Let y E K v = S(p,VE(~X)nX. Then there exists a V) E V with y, p E Vj(~X) and a V2 E V with P,'l'(V) E V~(~X). Then Vj(~X) n V~(~X) =F 0 and since VE(~X) <* UE(~X). there exists aU E U with Vj(~X)uVi(~X) c UE(~XI Therefore, Y E UE(~X)nX = U so Y E S('l'(V),U) for each V E A with V* < U. Hence. F v c K v c S('l'(V). U) for each V E A with V* < U, so 'l' is Cauchy in X. Since X is complete 'l' converges to some q E X. By an argument similar to the proof of Proposition 6.9, n{ Cl ~x[K ull U E A} = {q} and p E Cl ~x[K u 1for each U E A. Therefore, p = q E X so R = X. 0
Conversely, assume R = X and suppose (X, 11) is not complete. Then there exists a point q E X' such that q does not belong to <jl(X). Let p E pX such that <jl(P) = q. Then pER by Proposition 6.9. Since p does not belong to X we have R =F X which is a contradiction. We conclude that (X, 11) is complete. -
6.3 Tamano's Completeness Theorem
175
A Tychonoff space X is said to be topologically complete if there exists a uniformity 11 for X such that (X, 11) is complete. Clearly a Tychonoff space is topologically complete if and only if it is complete with respect to its finest uniformity. THEOREM 6.9 (H. Tamano, 1960) X is topologically complete only if for each p E ~X - X there is a partition of unity $ Cl ~x(O(<1>,,)) does not contain pfor each <1>" E $.
= {<1>,,}
if and
such that
Proof: Assume X is complete with respect to a uniformity 11, and that p E ~X X. By Tamano's Completeness Theorem, X = n{E u I U Ell\. Then, there is a U E 11 such that p E ~X - E u. Let (U,,) be a normal sequence with U 1 <* U.
Let d be the pseudo-metric on X defined (as in the proof of Theorem 2.5) with respect to {U,,}. Then the collection of spheres of radius 2-" coincides with the point-star coverings associated with U" [i.e., Sex, 2") = sex, Un) for each nJ and d:X x X ~ [OJ]. Let 't denote the topology on X induced by d. By Stone's Theorem, (X,'t) is paracompact. Now consider the open covering V = (V(x) Ix EX} of (X,'t) where Vex) = (y E X I d(x,y) < li41 and let {V d be an open locally finite refinement of V. By Theorem 6.3 there exists a partition of unity $ = {<1>,,} subordinate to (V" I such that 0(<1>,,) c V" for each <1>" E $. Since 't is coarser than the original topology, $ is also a partition of unity with respect to the original topology. We now show that p does not belong to Cl ~x(V(x» for each x E X which implies p does not belong to CI~x(O(<1>,,)) for each <1>" E $ which will establish the necessity part of the theorem. We can show this by demonstrating that Cl ~x(V(x» c E u for each x E X. For this let Z E X and for each x E X put dz(x) = d(z, x). Then dz:X ~ [0,1] is uniformly continuous with respect to the pseudo-metric uniformity on X by an argument similar to the one in (1) ~ (2) of Theorem 6.6. By Theorem 6.2, d z can be uniquely extended to a continuous function dr~X ~ [0,1]. Let r E Cl ~x(V(x». Then d~(r) .,; 1/4 so there exists an open W(r) in ~X containing r such that d~(y) < 1/2 for each y E W(r)nX. Therefore, W(r)nX c S(z, 1/2) = S(z, U 1) c U for some U E U. But then W(r) c Cll3x(U) c Eu. Hence CIBx(V(z» c Eu. This completes the necessity part of the proof. To prove the sufficiency part of the theorem, let $ = (<1>,,) be a partition of unity on X such that p does nor belong to Cl BX(O«!>A)) for each <1>" E $ where p E ~X - X. For each x E X and each positi~e integer n, put Vn(x) = ( Y E: X I .q I<1>,Jx) - <1>" (y) I < 2- n I and let Vn = IV n (x) Ix EX}. Then {V,,} is a normal sequence in X, so Vn E U for each positive integer n. Let p E ~X - X and let x E X. Then <1>,Jx) = 0 for all but finitely many (!>A's, say <1>"l ...
176
6. Paracompactifications
m} or else L7'=l I<1>", (x) - "i (y) I = L7'=l I<1>", (x) I = 1. Therefore, Y E 0(<1>,,) which implies Vn (x) C U7'=l 0(<1>,,) which in turn implies CI ~x [Vn (x) 1 c CI~X[U7'=lO(,,)], so p does not belong to CI~x[Vn(x)J. Therefore, p does not belong to R (the radical) with respect to the u uniformity. Hence R = X. By Tamano's Completeness Theorem, (X, u) is complete. Consequently, X is topologically complete. Let X be a Tychonoff space. For each f E C*(X) define If I to be sup{lf(x)lxE X}. Then I I iscalledthesupremumnormonC*(X). We can define a metric don C*(X) by d(j,g) = If - gl = sup{ If(x) - g(x) I Ix E X}. It can be shown (Exercise 1) that (C*(X), d) is a complete metric space. The following lemma appeared in Irving Glicksberg's 1959 paper Stone-Cech Compactijications of Products (Transactions of the American Mathematical Society, Volume 90, pp. 369-382). LEMMA 6.13 (I. Glicksberg, 1959) Let X and Y be Tychonoff spaces and let f E C*(X x Y). If the mapping :Y ~ C*(X), defined by (y) = fy where fix) =j(x,y) for each x E X, is continuous, then f has a continuous extension f~ to ~Xx Y. Proof: For each y E Y, let ft denote the extension of fy from X to ~X, and put ~(Y) =ft· Then ~:Y ~ C*(~X). Now C*(~X) with the metric d~ generated by the supremum norm is isometric with C*(X) with the metric d. To see this, note that there is a natural one-to-one correspondence 8:C*(X) ~ C*(~X) defined by 8(f) = f~. It is easily shown (Exercise 2) that d~(j~, g~) = d(j,g) for any pair f, g E C*(X) so 8 is an isomorphism. Consequently, the continuity of implies the continuity of ~ . Notice that for each (x,y) E X x Y that f(x,y) = fY(x). This motivates our definition of f~ as follows: for each (x,y) E ~X x Y putf~(x,y) = ft(x). Clearly, f~ is an extension of f, Since sup {ft (x) Ix EX} = sup {fy(x) Ix EX} for each y E Y, we have thatf~ is bounded. To show f~ is continuous, let (xo,Yo) E ~X x Y and let £ > O. Since fto E C*(~X), there exists a neighborhood U of x 0 in ~X with Ifto(x) - fto(xo) I < £/2 for each x E U. Since ~ is continuous, there is a neighborhood Y of Yo in Y with d~(~(Y) - ~(Yo» < £/2 for each y E V. Therefore, 1ft - fto I < £/2 for each y E Y, so Ift(x) - fto(x) I < £/2 for each x E ~X. Thus for each (x,y) E U X Y, If~(x,y) - f~(xo,Yo)1 = Ift(x) - fvo(x) I $ Ift(x) - fto (x) I + Ifto (x) - fto (x 0) I < £/2 + £/2 = £, which establishes the continuity of f~ at (xo,Jo). Since (xo,Yo) was chosen arbitrarily,f~ E C*(PX).
-
6.3 Tamano's Completeness Theorem
177
THEOREM 6.10 (H. Tamano, 1962) The following conditions on a Tychonoff space are equivalent: (1) X is topologically complete. (2) For each p E ~X - X there is a normal sequence {Vn} of open coverings of X such thatfor each Va E Va, Cl~x(V 0) does not contain p. (3) If P E ~X - X then (p) x X and M are functionally separated (by a member of C*(X x ~X)). Proof: Assume X is topologically complete and let p E ~X - X. By Theorem 6.9, there exists a partition of unity = {
But then Cl ~x[V o(x)] does not contain p for each x E X. Thus (1)
~
(2).
To show that (2) ~ (I), let {Vn} be a normal sequence of open coverings of X such that for each V 0 EVa, Cl ~x(V 0) does not contain p. Using Corollary 2.2, we can construct a pseudo-metric d on X such that W < Vo where W = {W(X)IXE X} andW(x)= lYE Xld(x,y)< I} foreachxE X. Let-rdenotethe topology of (X, d). Then (X,-r) is paracompact. Clearly W(x) is open with respect to -r and W(x) does not contain p for each x E X. Since (X,-r) is paracompact, we can use Theorem 6.3 to construct a partition of unity = {
178
6. Paracompactifications
defined by (y) = dy for each y E X, is continuous. Therefore, by Lemma 6.13, d has a continuous extension d*(z,y) over pX x X. It is clear that d* = 1 on {p} x X and d* = 0 on L1X. Thus (1) ~ (3). To show (3) ~ (1), let P E PX - X and let {p} x X be functionally separated from L1X by f E C*(PX x X). Without loss of generality, we may assume f(L1X) = 0 and f( {p} x X) = 1. For each x E X let fx be defined by fxCz) = f(z,x) for each z E pX. Then put d(x,y) = SUpzE!3X IfAz) - fy(z) I. Then d is a pseudo-metric on X. Let 't denote the topology of (X, d), and consider the space (X,'t) which is paracompact. For each x E X, let Vex) = {y E Xld(x,y) < 1/2}. Then Vex) is open in (X,'t). Put U = {V(x) x EX}. Then there exists a partition of unity <1> = {A} that is subordinate to U. Since 1: is coarser than the original topology, U is an open covering with respect to the original topology and <1> is a partition of unity with respect to the original topology. 1
Now d(x,y) < 1/2 implies that SUpzE!3X IfAz) - f/z) 1 < 1/2 so Ifx(y) - fy(y) 1 < 1/2. Butfy(Y)=f(Y,y)=Oso Ifx(y) < 1/2. Therefore, fAz)~ 1/2foreachz 1
E Cl 13X[V(x)] since fx is continuous on pX. On the other hand,fx(P) = f(P, x) = I for each x E X. It follows that CI 13x [V(x)] does not contain P for each x E X. Since <1> is subordinate to U, CI 13x [O(A)] does not contain P for each <1>1.. E <1>. By Theorem 6.9, this implies X is topologically complete.EXERCISES 1. Let d be the metric constructed from the supremum norm on C*(X) for a Tychonoff space X. Show that (C*(X), d) is a complete metric space. 2. Show that d 13 (f13,g 13) = d(f, g) in the proof of Lemma 6.13, for any pair f, g E C*(X).
6.4 Points at Infinity and Tamano's Theorem Although Tamano's Theorem is not about uniform spaces, the proofs of Tamano's Completeness Theorem and Tam an 0 ' s characterization of topological completeness (Theorem 6.9) hold the key to its proof. Also, it is an important example of using points at infinity to prove things about the space itself. If we consider the space R of real numbers, we notice that R has a compactification R* that is constructed by adding two additional points that we denote by _00 and +00. R is sometimes denoted by (-00, +(0) while R* is often referred to as the extended real numbers and denoted [-00, +00]. The point -00 is assumed to precede all members of R and +00 is assumed to follow all members of R. In this way, the natural ordering on R is extended to R*. The topology of R* is formed by taking as a basis, all open intervals in R together with all sets in R* of the form [-00, r) and (r, +00] where [-00, r) = {y E R* 1 y < r} and (r, +00] = {y ER*ly>r}.
6.4 Points at Infinity and Tamano's Theorem
179
The points -00 and +00 are called points at infinity. In general, if (X, Jl) is a unifonn space and (X', Jl) is its completion, the set X' - X is called the set of points at infinity of X. Similarly, if P is a paracompactification of X, PX - X is referred to as the set of points at infinity of X. Points at infinity are always relative to some uniformity. Points at infinity with respect to precompact unifonnities (e.g., ~X - X) play an important role in general topology. If Jl is a unifonnity for X and H C JlX - X, then H is called a set at infinity. Compact sets at infinity will be important to us in what follows. Sets at infinity can be very large with respect to the original unifonn space. For example, the space N of positive integers with the usual topology is countable whereas ~N - N is not. We have already seen a characterization of local compactness in tenns of compact sets at infinity, namely, Proposition 6.4. Our next characterization of paracompactness in tenns of compact sets at infinity relies on a well known characterization of paracompactness by E. Michael that appeared in the Proceedings of the American Mathematical Society in 1953 (Volume 4, Number 3, pp. 831-838). LEMMA 6.14 (E. Michael, 1953) The following properties of a regular topological space are equivalent: (1) X is paracompact, (2) every open covering of X has a locally finite (not necessarily open) refinement and (3) every open covering of X has a locally finite closed refinement. Proof: (1) ~ (2) is obvious. To show (2) ~ (3) let U be an open covering of X. Since X is regular, there exists an open covering V of X such that the closures of elements of V is a refinement of U. By assumption, there exists a locally finite refinement W of V. Then the closures of elements of W is the desired locally finite closed refinement of U.
To show that (3) ~ (1), let U be an open covering ofX. Let V be a locally finite refinement of U and let W be a covering of X consisting of closed sets, each one of which intersects only finitely many members of V. Now let Z be a locally finite closed refinement of W. For each V E V, let V' = X - u{Z E Z IVnZ = 0}. Then V' is an open set containing V such that if Z E Z, then Z intersects V' if and only if Z intersects V. For each V E V pick a U v E U such that V C U v· Let U' = {V' nUv I V E V). Then U' is an open refinement of U and since each element of a locally finite covering Z intersects only finitely many elements of U', U, is locally finite. Notice the similarity between statement (2) of the following theorem and the statement of Theorem 6.9. By replacing points at infinity with compact sets at infinity in the hypothesis, the strength of the conclusion is raised from topological completeness to paracompactness.
180
6. Paracompactifications
THEOREM 6.11 (H. Tamano. 1960) Let BX denote any compactification of X. Then the following statements are equivalent: (1) X is paracompact. (2) For each compact set C at infinity. there exists a partition of unity
To show (3) ~ (1), we use Lemma 6.13. Let {V 13} be an open covering of X and for each ~ let V~ denote the proper extension of V 13 over BX. Put C = BX - uV~. Then C is a compact set at infinity. By hypothesis there is a collection {G a.} of compact subsets of BX with X c uG a., CnG a. = 0 for each a, and for each x E X there is an open neighborhood V'(x) of x that meets only finitely many of the Go. 's. Now each G a. is covered by a finite number of the V~'s, say vy ... V~. Put H~ = Go.nV~nX. Then each H~ is contained in some VI3 = V~nX and Go.nX = uZ=IH~. Constructing the H~'s for each Go. in this manner yields a locally finite refinement {H~} of {V 13 }. It follows that X is paracompact. Let C(X) denote the collection of all real valued continuous functions on a space X and let C'(X) be a subset ofC(X). If there is anfE C'(X) such thatf(x) = 1 for each x E F and f(x) = 0 for each x E G, then F and G are said to be functionally separated by a member of C'(X). The following theorem contains Tamano's Theorem, i.e., (1) and (3) are equivalent. THEOREM 6.12 (H. Tamano, 1962) Let BX be any compactification of X. Then the following statements are equivalent: (J) X is paracompact. (2) For each compact C c ~X - X, there is a normal sequence of open coverings {V n I of X such that for each V 0
E
Va. CIBx(V o)nC = 0.
(3) X x BX is normal.
(4) IfG = X x C is closed in X x BX and GnlV( = 0, then G and IV( are functionally separated (by a member of C*(X x BX)).
6.4 Points at Infinity and Tamano's Theorem
181
Proof: The proof of the equivalence of (1) and (2) is almost identical to the proof of the equi valence of (1) and (2) in Theorem 6.10 and therefore will not be included here. It consists of exchanging the point p E ~X - X with the compact set C c ~X - X and adjusting the proof accordingly.
To show that (1) ---+ (3) we use the original proof of Dieudonne (1944). For this let U be an open covering of X x BX where X is assumed to be paracompact. For each point (x,y) E X x BX there are neighborhoods V(x,y) of x in X and W(x,y) of y in BX with (x,y) E V(x,y) X W(x,y) c U for some U E U. For a fixed x E X, the sets V(x,y) x W(x,y) such that y E BX form an open covering of the compact set {x} x BX. Therefore, there exists a finite number of points y 1 ... Yn E BX such that {V(X,Yi) x W(X,Yi) Ii = 1 ... n} covers {x} x BX. Put Vx = n;'=1 V(X,Yi). Then {Vx x W(x,y;) I i = 1 ... n} is a covering of {x} x BX and Vx is a neighborhood of x in X. Let V = {Vx Ix EX}. Then by assumption there exists a locally finite open refinement V' of V. For each V' E V' pick xv' E V'. Then XV' E V' C VXo for some Xo E X. Let Y 1 ... Ym be the finite collection of points in BX corresponding to x 0 such that {Vxo x W(XO,Yi) Ii = 1 ... m} covers {xo} x BX. Then the collection {V' x W(xv"Y;) Ii = 1 ... m} covers {xV'} x BX. Hence U' = {V' x W(xv' ,y;) IV' E V} is an open covering of X x BX that clearly refines U. Furthermore, U' is locally finite since if (a,b) E X x BX, there exists an open set A in X containing a such that A meets only finitely many members of V'. But then A x BX meets only finitely many members of U'. This completes the proof that (1) ~ (3). To prove (3) ~ (4) note that LlBX (the diagonal of BX) is closed in BX x BX so that LlBXn[X x BX] is closed in X x BX. But LlBXn[X x BX] = LlX, so LlX is closed in X x BX. Since X x BX is normal, there exists a continuous function f E C*(X x BX) with f(LlX) = 0 and f(G) = 1, so G and LlX are functionally separated. The proof that (4) ~ (1) is very similar to the proof the (3) ~ (1) of Theorem 6.10 and therefore will not be included here. It consists of exchanging the point p in ~X - X with the compact set C c ~X - X, and adjusting the proof accordingly. •
EXERCISES 1. Show that (1) and (2) of Theorem 6.12 are equivalent.
2. Show that (4) implies (1) in Theorem 6.12. 3. [A. Hohti, 1981] A uniform space (X, Jl) is uniformly paracompact if and only if for each compact K c ~X - X there is a U E Jl with Cll3x(S(x,U» c ~X K for each x E X.
182
6. Paracompactifications
4. lA. Hohti, 1981] A uniform space (X, IJ.) is unifonnly paracompact if and only if (X x ~X, IJ. x ~) is C-nonnal. C-nonnality is defined as follows: (X, IJ.) is C-normal if for each pair A, B of disjoint closed sets, there exists a U E IJ. such that given any x E X, there is a U x E IJ. with Star(A,Ux)nS(x,U)nStar(B,U x ) =
0.
6.5 Paracompactifications
If there is a paracompactification n X of X that plays a role analogous to the role of the Stone-Cech compactification and the Hewitt realcompactification in the theories of compactification and realcompactification respectively, what properties should 11: have in order that it be considered to play an analogous role? First, if X is a paracompact space then we should expect that 1I:X = X and second. 1I:X should be realizable as the completion of X with respect to some uniformity 11: such that 11: is the finest uniformity to have a paracompact completion. Every paracompactification of X is realizable as the completion of X with respect to some uniformity as shown in Section 6.1. We will show that in general, such a paracompactification does not exist. and that a necessary and sufficient condition for 11: to exist is for the finest uniformity u to be preparacompact, in which case u = 11:. Consequently, if X is paracompact then 1I:X exists and 1I:X = X. In the case 11: exists, we call 1I:X the Tamano-Morita paracompactification of X. We begin our discussion by proving a lemma about the relationship of any Tychonoff space Y that contains X as a dense subspace and the subspaces of ~X that contain X. Let X and Y be Tychonoff spaces. A function f:X - j Y is said to be a perfect mapping if f is closed, continuous and the inverse image of each point of Y is compact in X. Let i:X - j Y be an imbedding of X into a dense subspace of Y. Let j: Y -+ PY be the imbedding of Y into its Stone-Cech compactification and let k: PX -+ PY be the unique continuous extension of j © j such that k(~X - X) = ~Y - X. Finally, let Z = k- 1 (Y). LEMMA 6.15 (J) The restriction k z = k I Z is a closed mapping, (2) Z is the largest subspace of ~X to which j © i has a continuous extension to Y, and (3) Z is the only subspace of~X containing Xfor which k z is a peifect mapping.
Proof: Since ~X and ~Y are both compact, the mapping k is closed and the inverse image of each compact set in ~Y is compact in ~X. Then kz has the same properties because it is the restriction of k to the total inverse image of Y. By definition of Z, Z is the largest subset of ~X to which j © i has a continuous extension to Y. Furthermore, we claim that Z is the only subspace of ~X for which k IZ is a closed mapping such that the inverse image of each point is compact. To show this suppose W is another subspace of ~X containing X having the property that k IW is a closed mapping and the inverse image of each
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point is compact. Put S = WnZ. Then XeS c Z and since Z 7' W, S is a proper dense subspace of Y. Moreover, k~l(p)nS is compact for each p E Y. Choose Z E Z ~ S and let q = k (z). Then z has an open neighborhood V such that Clz(V)n[k~l (q)nSj = 0. Thus q does not belong to k(Clz(V)nS). But Clz(Clz(V)nS) = Clz(V) and q E k(Clz(V». Hence q E k(Clz(Clz(V)nS» c Cly(k(Clz(V)nS». Therefore, k(Clz(V)nS) is not closed in Y. Hence kl S takes the closed set Clz(V)nS into a set that is not closed in Y. Therefore, k Is is not a closed mapping which is a contradiction. Consequently Z is the only subspace of BX containing X for which the extension k of j © i is a closed mapping such that the inverse image of each point is compact. The following development of the Tamano~Morita paracompactification appeared in the author's paper titled Paracompactifications. Preparacompact~ ness and Some Problems of K. Morita and H. Tamano that appeared in Questions and Answers in General Topology. Volume 10, pp. 191-204.
THEOREM 6.13 Each completion J.!X of a uniform space (X. J..I.) where J..I. isfmer than B is homeomorphic to a subspace ofBX containing X.
Proof Let i:X ---+ flX be the imbedding of X into its completion and }:flX---+
PflX the imbedding of flX into its Stone-Cech compactification. By the above lemma, there exists a unique continuous extension k of} © i such that kCPX - X) = BJ..I.X - X and such that if Z = k~l (J..I.X) then (1) the restriction kz = k IZ is a closed mapping and (2) the inverse image of each point under kz is compact. Let J..I.' denote the uniformity of J..I.X and let J..I.+ be the coarsest uniformity on Z that makes kz uniformly continuous. Then a basis for J..I.+ is the collection i kZ1 (U') I u' E J..I.' I. Clearly X is dense in Z. To see that (X, J..I.) is a uniform subspace of (Z, J..I.+), let U" E J..I.+ such that U'" = k Z1(U') for some U' E J..I.'. Since kz keeps X pointwise fixed, U+ nX = U' nX = U for some U E J..I.. Conversely, if U E J..I. then there exists a U' E J..I.' such that U = U'nX. But then U = kZI (U')nX and kZI (U') E J..I.+. Consequently (X, J..I.) is a dense uniform subspace of (Z, J..I.+). Clearly J..I.+ generates the same topology on Z as the one induced on Z by BX since J..I.+ is finer than B. Finally we claim that (Z, J..I.... ) is complete and therefore (Z, J..I.+) is the completion of (X, J..I.). Since completions are unique, we can identify J..I.X with Z so that X c J..I.X c BX. To see (Z, J..I.... ) is complete. let \jf:D ~ Z be Cauchy with respect to J..I.+. Since kz is uniformly continuous, kz © \jf is Cauchy in J..I.X and hence converges to some q E J..I.X. Suppose \jf does not cluster to any p E k Z1 (q). Since k Z1 (q) is compact, there exists an open set 0 containing k Z1 (q) such that \jf is eventually in Z - O. Then kz(Z - 0) is closed in J..I.X and does not contain q. Then J..I.X - kz(Z - 0) is an open set containing q such that kz © \jf is eventually in its complement. Therefore'll does not converge to q which is a contradiction. Hence'll must cluster to some p E kZ1 (q). Since'll is Cauchy, 'II converges to p. Therefore, (Z, J..I.+) is complete. But then (Z, J..I.+) is the completion of (X, J..l).-
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PROPOSITION 6.10. The Stone-Cech compactification p~ of a J..lX c ~X of X with respect to some uniformity /-l of X is ~X (i.e., ~J..lX
completion = ~X).
Proof: Let i:/-lX ~ ~X be the identity imbedding of /-lX into ~X and let j be the identity mapping on ~X. By Lemma 6.14, there exists a unique continuous extension k:~/-lX ~ ~X such that k(~/-lX - /-lX) = ~X - /-lX and if Z - k- 1 (~X) = ~/-lX then k IZ = k is a closed mapping such that the inverse image of each point is compact. We claim that by an argument similar to the one in the proof of Theorem 6.13, we can see that ~X is homeomorphic to ~/-lX.
For this let Wdenote the uniformity of ~X and put W= ~~ (the coarsest uniformity on ~/-lX that makes k uniformly continuous). Then (X, ~) is a dense uniform subspace of (~/-lx' W) by an argument similar to the one in the proof of Theorem 6.13 that showed (X, /-l) to be a dense uniform subspace of (Z, /-l). Finally, we claim (~/-lX, W) is complete by an argument similar to the one in the proof of Theorem 6.13 that showed (Z, /-l') to be complete. Since completions are unique, ~/-lX is homeomorphic to ~X (i.e., we can identify ~/-lX with ~X so that ~/-lX = ~X.) -
If X is a subspace of Y such that each f E C(X) can be extended to a continuous function over Y, then X is said to be C-imbedded in Y. If each g E C*(X) can be extended to a continuous function over Y, then X is C*-imbedded in Y. Then every closed subspace of a normal space is C*-imbedded (by Theorem 0.4). The following theorem has many applications. Notice that Proposition 6.10 is a special case of the equivalence of (5) and (6) in this theorem. THEOREM 6.14 Let X be dense in the Tychonoff space Y. Then the following statements are equivalent: (1) Every continuousf:X ~ K, where K is a compact Hausdorff space, has a continuous extension fY'Y ~ K. (2) X is C*-imbedded in Y (3) Any two disjoint zero sets in X have disjoint closures in Y. (4) IfZ, z' E Z(X) then Cly(ZnZ') = Cly(Z)nCly(Z'). (5) ~X = ~Y. (6) X eYe ~x.
Proof: (1) ~ (2) If f E C*(X) then f is a continuous mapping from X into the compact subset C I R (f(X». Hence (2) is a special case of (1), so (1) implies (2).
(2) ~ (3) Let Z, Z' E Z(X) with ZnZ' = 0. Then there exists f, f' E C*(X) such that Z = Z(j) and Z' = Z(f'). Let g = maxI IfI,11 and g' = 1 maxI 1f'1 ,11. Put h' = minlg,g'l and let h = maxlh',ll. Then 0 ~ h ~ 1, h(Z) = o and h(Z') = 1. Let h Y be the continuous extension of h to Y. Let U = {Y E Y Ih(y) < 1/21 and V = lyE Y Ih(y) > 1/21. Then U and V are disjoint open sets
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in Y such that Z c U and Z' c V. Consequently, Z and Z' have disjoint closures in Y. (3) ~ (4) Clearly Cly(ZnZ') c Cly(Z)nCly(Z,). To show the reverse inclusion, let p E Cly(Z)nCly(Z,). Then there exists a zero set neighborhood V of p in Y. P E Cly(V nZ) and p E Cly(V nZ'). Thus (3) implies (V nZ)n(V nZ') i= 0 which in tum implies Vn(ZnZ') i= 0. Therefore, p E Cly(ZnZ'). Consequently, Cly(Z)nCly(Z,) c Cly(ZnZ').
(4) ~ (5) We show ~X = ~Y by showing that (X,~) is a uniform subspace of (Y, ~) and appealing to the uniqueness of the completion (Theorem 4.10). To do this, it suffices to show that every finite normal covering U of X can be extended to a finite normal covering U' of Y such that U = U' nX. If U is a finite normal covering of X, then it can be shown (Exercise 4) that there exists a normal sequence {Un) of finite open coverings of X such that U 1 < U, and for each positive integer nand U E Un' X - U is a zero set of X. Next, notice that we can extend statement (4) of the theorem to any finite collection Z 1 •.. Zn so that Cly(n7=1 Zi) = n7=1 Cly(Zi)' Now UE(Y) is an extension of U into Y for each positive integer n, and U~(Y) is an extension of Un into Y. Furthermore, by the extended version of (4),
we have for each positive integer n, n{Y - VE(y) IV E Un) = n{ Cly(X - V) IV E Un) = Cly(n{ X - V IV E Un)) = Cly(0) = 0. Consequently. U~(Y) is a covering of Y for each positive integer n, so WrY) is a covering of Y by Proposition 6.6. Also, by Proposition 6.6, U n+ 1 <* Un for each positive integer n so U~(r? <* U~(Y) for each n. Hence UE(y) is a finite normal covering of Y and U = UE(y) nX. (5)
~
(6) is obvious.
(6) ~ (I) By Theorem 6.2, a continuous f:X ~ K, where K is a compact Hausdorff space, has a continuous extensionfj3:~X ~ K. But thenfY = fj31 Y is a continuous extension to Y. THEOREM 6.15 The completion of a Tychonoff space X with respect to its finest uniformity u is uX = n {J.!X IJ.! is a uniformity for X and J.!X is a subspace of~X). Proof: To see that nJ.!X is topologically complete. observe that for each point p E ~X - nJ.!X, P must belong to ~X - AX for some uniformity A of X. Since AX is complete, AX is topologically complete. By Theorem 6.9, for each q E ~AX AX, there exists a partition of unity cI> = {<1>d on AX such that C l j3AX (0(<1>,"» does not contain q for each <1>)" E cI>. By Proposition 6.10, ~AX = ~X so CL.j3,Y(O(<1>:.J) does not contain p for each <1>)" E cI>. For each <1>)" E cI> put '1')" = <1>)" InJ.!X.'Fhen 'I' = {'I'),,) is a partition of unity on nJ.!X such that Cl j3x(O('I',..» does not contain·~ p for each '1'", E '1'. Since this holds for each p E ~X - nJ.!X. we have that nJ.!X is topologically complete.
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Let v be the finest unifonnity on nl..lX and let v' be the unifonnity induced on X by v. Since v is the finest uniformity on n~..IX, v is finer than the unifonnity u' induced on n~X by u. But then v' is finer than u which implies that v' = u. Since n~X is topologically complete, n~X is the completion of X with respect to v'. Since completions are unique, n~X = uX. -
LEMMA 6.16 paracompact.
A perfect preimage of a paracompact space is
Proof: Let {Va} be an open covering of X. For each y E Y, let {Va, Ii = 1 ... n(y)} be a finite subcovering for the compact set 1 (y). Then F(y) = u7=C.jl Va, is closed in X so f(F(y» is closed in Y. Put V(y) = Y - f(F(y». Then V(y) is an open neighborhood of y and rl(V(y» c u7=Q)V a ,. Next let {W(y)} be a precise locally finite open refinement of {V(y) lYE Y}. For each i = 1 ... n(y), put W(y,i) = r 1 (W(y»nV a,' Then the collection {W(y,i) lYE Y and i = 1 ... n(y)} is an open covering of X that refines {Va}. Moreover, {W(y,i)} is locally finite since if p E X, there exists a neighborhood V of f(P) that meets at most finitely many of the W(y) and then the neighborhood 1 (V) of p meets at most finitely many of the W(y,i). -
r
r
John Mack sketched an outline of the proof of the following theorem for the author late one night at the 1991 Northeast Topology Conference after a lengthy discussion of paracompact subspaces of BX containing X. It plays a major role in the author's solution to Tamano's Paracompactification Problem.
THEOREM 6.16 (1. Mack, 1991) Let u be the finest uniformity for the space X. Then for each p E BX - uX there is a paracompact Y c BX - {p} that contains uX. Proof: By Theorem 6.13, X c uX c BX so by Theorem 6.14, BuX = BX. Since uX is topologically complete and BuX = BX, by Theorem 6.10, if p E BX - uX, there exists an F E C*(uX x BX) such that F(x, x) = 0 and F(x, p) = 1 for each x E uX. Putf= max{F,l} and for each pair x,Y E uX put 8(x,y)
= sup { If(x, q) - f(Y, q) II q E
BX}.
Then max{f(x, y),f(y, x)} :<::; 8(x,y) :<::; 1. It is easily shown that 8 is a pseudometric on uX. For each pair x,y E uX define x - y if and only if 8(x,y) = 0 and set [x] = lYE uxI8(x,y)=0}. LetM= {[x]lxE uX} and put d([x],[y]) = S(x,y). Then (M, d) is a metric space and the quotient mapping ':uX ~ M defined by '(x) = [xl for each x E uX is continuous. Let (M*, d*) be the completion of (M,d).
Now put = j © i © <1>' where i:M -+ M* is the imbedding of M into its completion and j:M* -+ I3M* is the imbedding of M* into its Stone-tech
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compactification. Then :uX -7 ~M*. Let ~ be the extension of over ~uX = ~X. Since ~X and ~M* are both compact, ~ is closed and the inverse image of each compact set in ~M* is compact in ~X. Put Y = (~tl (M*) and let ~ = ~ I Y. Then ~:Y -7 M* and since Y is the total inverse image of M*, ~ is also closed and the inverse image of each compact set in M* is compact in Y. Therefore, ~ is a perfect mapping. By Lemma 6.15, Y is paracompact. Clearly, uX c Y. By a proof similar to the proof of Lemma 6.12, we can show that d*:M* x [0,1] has a continuous extension d~:M* x ~M* -7 [OJ]. Set g = d~ © (~ x ~). Then g E C*(Y X ~X) and g(Y x ~X) c [0,1]. Let (x,y) E X X X. Then g(x,y) = d~(~(x),~(Y» = d~(x,y) = d*(x,y) = S(x,y) ~f(x,y). Therefore, M*
-7
fl X x X
~ g I X x X which implies
fl X x ~X
~ g IX x ~X.
Hence 1 = f(x, p) ~ g(x, p) for each x E X. Therefore, 1 ~ g(x, p) for each y E Y. Since g(x, x) = Sex, x) = 0 for each x E X, g(y,y) = 0 for each y E Y. Consequently, p does not belong to Y. Hence Y is a paracompact subset of ~X {p) containing uX. -
THEOREM 6.17. lfrr is the finest uniformity for a space X such that the completion rrX is paracompact, then rr = u (the universal uniformity for X). Proof: Put PX = Il{Y c ~xlx c Y and Y is paracompact). Clearly PX *- 0 and since ~X is paracompact, rr is finer than ~. Therefore, rrX c ~X which implies PX c rrX. Now, for each paracompact space Y such that X eYe ~X, it is easily shown that Y is the completion of X with respect to some uniformity 'A that is finer than ~. Then by Theorem 6.15, uX c PX. Therefore, PX = ux.
There are two uniformities on X that bound rr, namely, u (the finest) and sup{'AI'AX c ~X and 'AX is paracompact), which we denote by sup'A. Clearly sup'A c rr c u. Now rrX is paracompact and PX c rrX. We claim that the completion sup'AX of X with respect to sup'A is PX. Since X c PX c 'AX for each uniformity 'A such that 'AX is a paracompact subset of ~X, 'A' (the uniformity of the completion 'AX) induces a uniformity 'A+ on PX and the uniformity 'A on X. Consequently, sup'A+ = sup{'A+) is a uniformity on PX that induces the uniformity sup'A on X. Thus (X, sup'A) is a dense uniform subspace of (PX, sup'A+). To see that (PX, sup'A+) is complete, let \jI be a Cauchy net in PX with respect to sup'A+. Then \jI is Cauchy in PX with respect to each 'A+. But then \jI is Cauchy in 'AX which implies \jI converges to some P A. E 'AX c ~X for each 'A. Since ~X is Hausdorff, p A. = P ~ for each pair of uniformities 'A, ~ such that 'AX and ~X are paracompact subsets of ~X. Therefore, there exists apE ~X such that \jI converges to p and p E PX since p = p A. E AX for each A. Hence (PX, SUPA+) is complete, and since completions are unique, SUpAX = PX.
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Finally, we show that rrX = uX. Since PX c rrX, rr induces a unifonnity rr+ on PX that is finer than sup/...+ since rr is finer than sup/.... Then (X, rr) is a dense uniform subspace of (PX, rr+). Now (PX, rr+) is complete since (PX, sup/...+) is complete and rr+ c sup/...+, and since completions are unique, rrX = PX = uX. But then uX is paracompact, so by hypothesis, rr = u. Analogously to the definition of topologically complete, a space will be called topologically preparacompact if it is preparacompact with respect to its finest unifonnity.
THEOREM 6.18 (N. Howes, 1992) A necessary and sufficient condition for the existence of the Tamano-Morita paracompactification is topological preparacompactness. Proof: Suppose X is topologically preparacompact and let u be the universal unifonnity on X. Then uX is paracompact which implies that rr exists and equals u. Conversely, if rr exists, then by Theorem 6.17, rr = u which implies that uX is paracompact which in turn implies that u' is cofinally complete. Therefore, u is preparacompact which implies X is topologically preparacompact. We now show that some of the results in Chapter 4 can be generalized by replacing cofinally Cauchy nets with almost Cauchy nets. Part (I) of the following theorem is due to J. Isbell and appeared in his paper Supercomplete spaces referenced in Chapter 5.
THEOREM 6.19 (1. Isbell, 1962, N. Howes, 1992) Let X be a Tychonoff space and let u be the finest uniformity on X. Then (1) X is paracompact if and only if (X, u) is supercomplete, (2) X is LindelOf if and only if (X, e) is supercomplete and (3) X is compact if and only if (X, P) is supercomplete. Proof: (1) is essentially the equivalence of (1) and (2) in Theorem 5.8. To prove (2), we need only show that if (X, e) is supercomplete then X is Linde16f since by Theorem 4.4 and Corollary 5.1, a Lindelof space is supercomp1ete with respect to the e unifonnity. By Lemma 4.1 it will suffice to show that each (I)-directed (countably directed) net is almost Cauchy with respect to e. For this let ",:D -7 X be an (I)-directed net. Let U E e. Then there exists a countable nonnal covering {Vii that refines U. For each i put Cj = {d E D I",(d) E Vii and let E = u{ Cj ICj is not cofinal in D}. If E = 0 pick any d E D. Then {Cj } is a collection of cofinal subsets of D such that each d' ;::>: d is in some C j and for each i, ",(C j ) c V j c U for some U E U. If E "# 0, then for each i such that Cj is not cofinal in D, there exists a d j E D with R(dj)nC j = 0 where R(dj) = {d E Dldj ~ d}. Since", is (I)-directed, there exists a d' E D with d j ~ d' for each i such that Cj is not cofinal in D.
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Hence R(d')ilC i = 0 for each i such that Ci is not cofinal in D. Therefore, R(d')ilE = 0. For each j put D j = CjilR(d'). Then {D j } is a collection of cofinal subsets of D such that each d+ ~ d' is in some D; and for each j, \jf(D j ) C Vj C U for some U E U. Since U was chosen arhitrarily, \jf is almost Cauchy. To prove (3), we need only show that if (X, ~) is supercomplcte then X is compact since by Theorem 4.4 and Corollary 5.1, a compact space is supercomplete with respect to~. For this, it will suffice to show that each net in X is almost Cauchy with respect to~. Let \jf:D ~ X be a net in X. Let U E ~. Then there exists a finite normal covering {V I ... V n} that refines U. For each i = 1 ... n put Ci = (d E D I \jf(d) E Vi}' Then an argument similar to the above can be used to show that there exists a d E D such that the collection {C i I Ci is cofinal in D) is a collection of cofinal subsets of D such that each d' ~ d is in some C i and for each i, \jf(C;) C Vi C U for some U E U. Therefore, each net in X is almost Cauchy with respect to ~. If instead of Theorem 4.4, we were to use Theorem 6.18 to motivate our definition of preparacompactness, then we would define preparacompactness to be the property that each almost Cauchy net has a Cauchy suhnet. To distinguish between these two notions we define a uniform space to he almost paracompact if each almost Cauchy net has a Cauchy subnet. Then we can prove: THEOREM 6.20 (N. Howes. 1992) Let (X. fJ.) be a Uniform space and let v be the uniformity on X derived from fJ.. Then (1) (X. fJ.) has a paracompact completion if and only if (X.V) is almost paracompact. (2) (X. fJ.) has a LindeLOf completion if and only if (X.V) is countably bounded and almost paracompact. Proof: To prove (1) let (X', fJ.') be the completion of (X, fJ.) and let u' he the finest uniformity for X'. Assume (X,v) is almost paracompact and that \jf:D ~
X' is an almost Cauchy net with respect to u'. Let E = D X u' and define < on E by (d,U') < (e,V') if d < e and V' <* U'. For each (d, U') E E put Sed, U') = a for some a E X such that a and \jf(d) both belong to some U' E U'. Then the correspondence (d, U') ~ Sed, U') defines a net S:E ~ X. Let U' E u' and pick V' E u' such that V' <* U'. Since \jf IS almost Cauchy, there exists a d E D and a collection {C a} of cofinal subsets of D such that each d' ~ d belongs to some C a and for each index a, \jf(C a) C V~ for some V~ E V'. For each index a put Ea
=
{(e,W')leE CaandW'<*V').
Then Ea is cofinal in E for each index a. Let (e,W') E Ea. Then S(e,W') = y E X such that y and \jf(e) both belong to some W' E W'. Since (e,W') E E a , e E
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C a which puts \jI(e) in V~. Consequently,
y E Star(V~,W') c Star(V~,V') c Va for some V~ E U'. Therefore, 8(£ a) C Va for each index a.. To show 8 is almost Cauchy it remains to show there exists an e E E such that each e' ::::> e belongs to some £ a' For this choose W' E u' such that W' <* V' and put e = (d,W'). Now dEC j3 for some index Pwhich implies e E £ /3' If e' > e then e' = (d', Z') for some d' > d and Z' <* W'. Then d' E C y for some index yand Z' <* V' so e' E £ y' Hence 8 is almost Cauchy with respect to u'. Since 8 lies in X, 8 is almost Cauchy with respect to v and therefore has a Cauchy subnet
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191
PROPOSITION 6.12. A uniform space is countably bounded if and only if each ill-directed net is almost Cauchy By a supercompletion of a uniform space (X, Il), we mean a supercomplete uniform space (y, A) such that X is uniformly homeomorphic to a dense subspace of Y. It is natural to inquire about the existence and uniqueness of a supercompletion for a given uniform space. In fact, A. Hohti, posed this question in a paper titled Uniform hyperspaces (Proc. 1983 Hungarian Topology Colloquium, pp. 333-334). The following theorem provides an answer.
THEOREM 6.21. (N. Howes, 1992) A necessary and sufficient condition for a uniform space to have a supercompletiofl is that it he almost paracompact. In this case, the supercompletion is unique and identical with the ordinary completion. The proof is left as an exercise (Exercise 3).
EXERCISES 1. Prove Proposition 6.11. 2. Prove Proposition 6.12. 3. Prove Theorem 6.21. [Hint: Adapt the proof of Theorem 4.11 using Corollary 6.1] 4. Show that if U is a finite normal covering of a Tychonoff space X that there exists a normal sequence {Un} of finite open coverings of X such that U 1 < U and for each positive integer nand U E Un' X - U is a zero set of X. RELATIVEL Y FINE SPACES In 1993, B. Burdick gave an example of a uniform space that is cofinally complete with respect to transfinite sequences but not cofinally complete (uniformly paracompact) thereby showing the existence of uniform properties that cannot be characterized by transfinite sequences in the same way they are characterized by nets. The class of relatively fine uniform spaces is a class of uniforms spaces in which the use of transfinite sequences is equivalent to the use of nets for studying several uniform properties like uniform paracompactness. Let K be an infinite cardinal. A uniformity f.l is said to be K-bounded if each uniform covering has a subcovering of cardinality less than K. For a uniformizable space X and for each infinite cardinal K there exists a finest
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K-bounded uniformity. It is the supremum of all K-bounded uniformities and is denoted by UK' The uniformities UK where K is an infinite regular cardinal are called the relatively fine uniformities. Note that u, e and ~ are all relatively fine uniformities. 5. [Howes, 1994] For a relatively fine space (X, equivalent:
UK)'
the following are
(1) (X, UK) is uniformly paracompact. (2) (X, UK) is cofinally complete. (3) (X, Uk) is cofinally complete with respect to transfinite sequences. (4) X is paracompact and each transfinite sequence with no subsequence of cardinality less than K clusters. (5) X is paracompact and each K-directed net clusters. (6) Each directed open covering is uniform. 6. [Howes, 1994] Let (X, /J.) be relatively fine where /J. #-~. If there do not exist measurable cardinals then: (1) The following are equivalent: (a) (X, /J.) has a paracompact completion. (b) X is topologically preparacompact. (c) X is topologically almost paracompact. (2) The following are equivalent: (a) (X, /J.) has a LindelOf completion. (b) X is topologically preparacompact and U = e. (c) X is topologically almost paracompact and U = e. 7. [Burdick, Howes, 1994] For a locally fine space (X, /J.) the follwing are equivalent: (1) (X, /J.) is cofinally complete. (2) (X, l-l) is cofinally complete with respect to transfinite sequences. RESEARCH PROBLEM 8. What other uniform properties can be characterized in locally fine spaces by transfinite sequences and are these characterizations analogous to their net characterizations? 6.6 The Spectrum of ~X We denote the weak completion u(X) of X with respect to u, developed in Section 5.5, by uX since u(X) is homeomorphic to uX. To see this, recall that
6.6 The Spectrum of PX
193
(X, u) is unifonnly homeomorphic to a dense subspace of u(X). Hence the unifonnity of u(X) is the finest unifonnity for u(X). Then by Theorem 5.5, u(X) is complete. Since completions are unique, we have that u(X) is unifonnly homeomorphic with uX. In this section {", IA E A I will denote the collection of all nonnal sequences of open coverings of X. Let {'1\ Iy E r} be the collection of all nonnal sequences of open coverings of ~X such that 'I'y begins with a finite covering. If Y E r then 'I'y = {We I for some nonnal sequence of open coverings such that W~ is finite. Then for each n, W n = {W j3 nX IW j3 E We} is an open covering of X and
Proof: The proofs of (1) and (2) are trivial. To see (3) let x E G. Then there exists a positive integer n(x) such that S(x,Wn(x) c G. Put K = u{S(x,We(X) Ix E G}. Clearly KnX = G. Let H be the interior of K in [~Xlr. Clearly G c H c K so HnX = G. This proves (3).-
For any continuous f:X ~ Y where X and Y are both Tychonoff spaces, there exists a continuous extension fj3:~X ~ ~Y. By Lemma 6.16, X/y is a unifonn subspace of ~X/'I'y. Let 11y:X/y ~ ~X/'I'y denote the inclusion mapping. Then we have the following diagram:
II
Clearly ~X/'I'y is compact since it is the quotient of a compact space, so ~(~X/'I'y) = ~X/'I'f Now 9y restricted to X is 11y © <1>y so by the above diagram we see that 9y =11y © <1>~.
194
6. Paracompactifications
Let \fI",\fIfl be two nonnal sequences of open members of W (the unifonnity of ~X) such that A < f.!. Then, just as in Section 5.5, there exists a bonding map (<j>~)~:~X/\f'fl ~ ~X/\f'". Clearly (<j>~)~ restricted to X/<1>fl is just <j>~. Then, just as in Section 5.5, (~X/\f'd<j>~)~} is an inverse limit system of unifonn spaces, and the inverse limit of this system, which we denote by W(~X) is the weak completion of ~X with respect to W. Since ~X is complete, by Theorem 5.4, W(~X) = ~X, so ~X is the inverse limit of {~X/\f'd<j>~)~}. If \fI" is a nonnal sequence of open members of Wthen there exists a \fI fl such that f.! E r with A < f.!. To see this let \fI" = {W~} and let Y E r. Let \fly = {V~}. Then V~ is finite and {W~nV~} is a normal sequence. Put U~ = V~ and for each n > I put U~ = W~nV~. Then let f.! be the element of r such that \fIfl = {U~}. Clearly, A < f.!. Therefore, {\fly lYE r} is cofinal in the collection of all nonnal sequences of open members of W. Consequently, by Exercise 5 of Section 5.5, ~X is the inverse limit space of the inverse limit system {~X/\f'y, (<j>~)~} where YE r. Now since lly:X/<1>y ~ ~X/\f'y is the inclusion mapping, ll~ must be one-toone, so ~(X/<1>y) can be identified with the subset ll~[~(X/<1>y)] of ~X/\f'"y" But then ~X/\f'y is a compactification of X/<1>y containing a copy of ~(X/<1>y) so ~X/\f'r can be identified with ~(X/<1>y). Moreover, for each pair A, f.! E r with A < f.!, <j>~:X/<1>fl ~ X/<1>" has a unifonnly continuous extension ~(<j>~):~(X/<1>fl) ~ ~(X/<1>,,). Since ~(X/<1>fl) and ~(X/<1>,,) can be identified with ~X/\f'fl and ~X/\f'" respectively, we can consider ~(<j>~) as a mapping from ~X/\f'fl into ~X/\f'". Then ~(<j>~) restricted to X/<1>fl is the same as (<j>~)~ restricted to X/<1>w Therefore, ~(<j>~) = (<j>~)~. Consequently, we have the following: THEOREM 6.22 (K. Morita, 1970) {~(X/<1>,,), (<j>~)~} is an inverse limit system whose inverse limit can be identified with ~x. Proof: The proof of this theorem has essentially been proved in the preceding discussion. It remains to observe that since r is cofinal in A, the inverse system in the hypothesis of the theorem has the same inverse limit as the system where A ranges over r, namely, ~X (see Exercise 5, Section 5.5). Consequently, ~X has the spectrum W(X/<1>,,), (<j>")~}. Recall from Section 5.5 that uX = u(X) has the spectrum {X/<1>", <j>t}. Also recall the unifonn imbedding <j>:X ~ uX defined by <j>(x) =( <j>dx)} E uX. It's extension <j>~:~X ~ ~uX = ~X can be seen to be the mapping defined by <j>~(x) = (<j>~(x)} E
~X.
THEOREM 6.23 (K. Morita, 1970) For any uniformizable space X, uX c ~X and uX = n {(<j>~rl (X/<1>,,) IA E A} = n {(f~rl (T) I f:X ~ T is continuous with T metrizable}. Proof: That uX c
pX follows from either Theorem 6.13 or Exercise 1.
To prove
6.6 The Spectrum of PX
195
the first equality, suppose x E uX. Then x E fhX/A such that for each pair A, Il E A with A < Il, ~(xh!) = x A. Hence x E (~rl(X/A) for each A E A. Therefore, uX c n{(~rl(x/A)IA E A}. Conversely, suppose x E n{(~rl(X/A) IA E A}. Then ~(x) E X/A for each A E A. Therefore,
1
To prove the second equality, let f be a continuous mapping of X into a metric space T. For each positive integer n, let Un = {Set, 2- n) t E T}. Then the sequence {Un} forms a basis for the metric uniformity of T. For each n put Vn = 1 (Un)' Then {V n} is a normal sequence of open coverings of X, so {V n} =
r
To see that g is continuous let t E T and let U be an open neighborhood of t in T. Then there exists a positive integer n such that Set, 2- n ) c U in T which implies Set, Un) C U which in tum implies Star(f-l (t),V n) c F- 1 (U). Pick x E 1 (t). Then S(x,V n) is open in X A' Put V = S(x,v n). Then fey) c U, so (g © ~ © i A)(V) c U. Since i A(V) = V, fey) = (g © ~)(V). For any open W c X Awe have:
r
W
=
(XE XAIS(x,V n) c W for some positive integer n}.
From this it is easily seen that (~rl(A(W» = W so that ~ is also an open mapping. Hence Z = ~(V) is open in X/A and g(Z) c (g © ~)(V) =fey) c U. Moreover,
«g
1
LEMMA 6.18 fiX eYe uX then uY = uX.
196
6. Paracompactifications
Proof: If g: Y -7 T is continuous where T is a metric space, then f = g IX: -7 T is continuous. Thenr:uX -7 uT, gU:uY ~ uTand (ix)u:uX -7 uY, where i x :X-7 Y is the inclusion (identity) mapping. Since (ixt = iuX , uX c uY. Since gUlux:ux -7 uT,r = gUlux. Butr =f~lux and gU = g~luY sof~lux = (g~ I uY) IuX = g~ I uX which impliesf~ = g~. Conversely, for any continuousf:X -7 T where T is a metric space,r:uX -7 uT. Since T is metric, T is weakly complete with respect to the metric uniformity which implies T is weakly complete with respect to u so uT = T by Theorem 5.4. Therefore, g = IY is a continuous mapping from Y into T such that g IX =f.
r
Then by Theorem 6.22, uY = ill (g~rl (T) Ig: Y -7 T where g is continuous and T is metrizable) c ill (f~rl (T) If:X -7 T where f is continuous and T is metrizable) = uX. Therefore, uY c uX. Similarly, uX c uY so uY = uX.-
THEOREM 6.24 (K. Morita. 1970) uX is characterized as a space Y with the following properties: (1) Y is topologically complete and contains X as a dense subspace. (2) Any continuous f:X ~ T where T is a metric space can be extended to a continuous mapping from Y into T. Proof: uX is the inverse limit of the inverse system of metric spaces I X/cD", ~} and hence by Theorem 5.5 is topologically complete. Thus uX satisfies (1). To show uX satisfies (2), letf:X -7 T where T is a metric space. Thenr:uX -7 uT. But since T is a metric space, T is weakly complete, so by Theorem 5.4, uT = T. Therefore, is a continuous mapping from uX into T.
r
Conversely, let Y be a space satisfying (1) and (2). Let g E C*(X). Then g(X) is a bounded subset of R and hence metrizable, so g:X -7 g(X) is a mapping of X into a metric space and by (2) g can be extended to Y. But then X is C*-imbedded in Y so by Theorem 6.14, BX = BY and X eYe BX. Again by (2), for any continuous mapping f from X into a metric space T, there is a continuous extension g:Y -7 T. Since BX = BY we have f~ = g~ where f~ and g ~ are the unique continuous extensions from BX and BY into BT given by Theorem 6.2, and g~ I Y = g. Now g~(Y) c T so f~(Y) c T which in turn implies Y c (f~rl (T). Therefore, by Theorem 6.23, Y c uX. Then by Lemma 6.17, uY = uX. But since Y is topologically complete, Y = uX.-
EXERCISES 1. If I Y, F) is an inverse limit system where Y = I Y a Ia E A} and F = {~:Y a -7 Y ~ Ia,B E A with B < a} and if for each a E A, X a C Y a , then the inverse limit of {X,G} is a subset of the inverse limit of {y, F} where X = {X a I a E A} and G
= {~IXal a,B
E
A with B < a}.
6.7 The Tamano-Morita Paracompactification
197
2. Show that uX is characterized as a topologically complete space Y which is the smallest with respect to properties (1) and (2) below: (1) Y contains X as a dense subspace. (2) Every real-valued bounded continuous function on X can be extended to a continuous function over Y. 3. Show that if X is the topological sum of IX A IA topological sum of I uX A I A E A}.
E
A}, then uX is the
6.7 The Tamano-Morita Paracompactification As an example of the Tamano-Morita paracompactification of a space X, we consider the case where X is an M-space. This example was given by K. Morita in his paper Topological completions and M-spaces referenced in Section 5.5, as an example of a space whose topological completion is paracompact. M-spaces were introduced in 1964 by Morita in a paper titled Products of Normal Spaces with Metric Spaces (Math. Annelen, Volume 154, pp. 365-382). Paracompact M-spaces, unlike paracompact spaces, are countably productive. M-spaces are defined as follows: a space X is said to be an M-space if there exists a normal sequence {Un} of open coverings of X satisfying the condition (M) below: (M)
If IKn} is a decreasing sequence of non-empty closed sets in X such that Kn C S(x, Un) for each positive integer n, and for some x E X, then nKn 0.
*
Let X be an M-space and let {$A I A E A'I be the collection of all normal sequences of open coverings of X satisfying condition (M). It is left as an exercise to show that A' is cofinal in A. Hence by Exercise 1 of Section 6.6, uX is the inverse limit of the inverse limit subsystem IXN.lI,., ~ IA E A'I of IX/CPA' ~ IA E AI. Recall that a continuous function f:X ~ Y is said to be a perfect mapping if f is closed and for each p E Y, 1 (P) is a compact subset of X. f is said to be quasi-perfect if f is closed and 1 (p) is countably compact for each pE Y.
r
r
LEMMA 6.19 For each A E A', A:X ~ X/CPA is a quasi-peifect mapping. Proof: Let A E A' and let CPA = {Un I. Then A:X ~ X/CPA is the mapping defined in Section 5.5 by <1>1.. = ~ © i A where i A:X ~ X A is the identity mapping on X viewed as a mapping from X onto X A and ~ is the quotient mapping from X A onto X/CPA' X A is the pseudo-metric space (X, d A) where d A is the pseudo-metric generated by ct>A' From the proof of Theorem 6.23, ~ is an open mapping. To show that <1>1.. is closed, let F be a closed subset of X and
198
let y* E CI«h.(F». Let x we have:
6. Paracompactifications E
<1>~I(y*).
Since Star(5;(x, U n+l ), Un+d c sex, Un),
Hence V = <1>~(/n{XA (S(x. Un))) is open in Xj
r
1(F) is closed in X so (g © /)(/-1 (F» = g(/(/-l (F») = g(F) is closed in Z. Therefore, g is a closed mapping. Next let z E Z. Then (g © I rl (z) is countably compact in X which implies I«g © f)-I (z» is countably compact in Y. But I«g © f)l (z» =1(/-1 (g-l (z») = rl (z) so g is a quasi-perfect mapping. -
Proof: Let F be closed in Y. Then
As a consequence of Lemmas 6.18 and 6.19, if "A., j..l E A' with "A. < j..l, we have <1>A = <1>~ © <1>)1; and since <1>A and <1>)1 are quasi-perfect, we must have that <1>~ is quasi-perfect. Let (<1>~)~ denote the Stone extension of <1>~. Then (<1>~)~(X/ct>)1) = X/
This fact allows us to establish the following theorem: THEOREM 6.25 (K. Morita, 1970) II X is a Tychonojf M-space then uX = (/~)-l (T)for any quasi-perfect mappingffrom X onto a metric space T.
6.7 The Tamano-Morita Paracompactification
199
Proof: The proof consists in showing that a quasi-perfect mapping f from X onto a metric space T coincides with ,,:X ~ X/cD" for some A E N. From the proof of Theorem 6.23, we already know that f = g © <1>" for some A E A and for some continuous g:X/cD" ~ T defined by g(x*) = fez) where z E x*. Let! Un l and! Vn l be as defined in the proof of Theorem 6.23. Clearly g is onto since for l (t) which implies g(x*) = [(x) = t. To show g each t E T we can pick an x E is one-to-one, let x* and y* be distinct elements of X/cD". Then there exists a positive integer n such that S(x,Vn)nS(y,V n) = 0 for some x E x* and y E y*. Therefore, S(f(x) , Un)nS(f(y), Un) = 0. Since g(x*) = j(x) and g(y*) = fey), g(x*) g(y*) so g is one-to-one.
r
"*
Finally, to show g is open, let U be open in X/cD" and let x* E U. Then there exists a positi vc integer n such that S(x*, 2 -n) C U so g(S(x*, 2- n c g(U) and contains g(x*). Therefore, if g(S(x*, 2- n» is open in T then x* is an interior point of g(U). Since x* was chosen arbitrarily, g(U) is open. Let x E x*. Then
»
g(S(x*,2- n»
=
UE TIY*E S(x*,2- n),YE y*,andf(y)=tl =
Now X - SexY n) is closed so f(X - SexY n» = T - S(f(x), Un) is closed in T since f is a closed mapping. Therefore, S(f(x) , Un) is open in T so g is an open mapping. Hence g is a homeomorphism so we can identify T with X/cD" which implies f coincides with <1>". It remains to show that A E N. We do this by showing that {V n} satisfies (M). Let K = {Kn} be a decreasing collection of closed sets in X such that for
some p E X, Kn c S(P,V n) for each positive integer n. Put q =f(P). Thenf(K n) c Seq, Un-I) for each n since Kn c S(p,V n). Now the f(Kn) are closed subsets of T and {f(Kn)} is a decreasing sequence so q E f(Kn) for each positive integer n. Therefore, l (q)nK n 0 for each n. Hence {f-I (q)nKn l is a decreasing 1 (q) sequence of closed sets in X. Since is countably compact, I'In[j-l (q)nKnl 0 so nKn 0. Therefore, {V n l satisfies (M). •
r
"*
"*
r
"*
LEMMA 6.21 Let f:X ~ T where f is continuous and T is metric. If X is an M-space then :uX ~ T and the following assertions hold: ( J ) f is onto if and only if is onto. (2) f is closed if and only if is closed and (3) f is quasi-perfect if and only if is perfect.
r
r
r
r
Proof: From the proof of Theorem 6.23, we know there exists a continuous function g:X/cD" ~ T such that f = g © <1>" for some A E A. The proof can be modified to show that A can be chosen in such a way that A E N. For this let {Un} and {V n} be as in the proof of Theorem 6.23. Then cD" = {V n }. Pick J..1. E N such that A < J..1. and let cDll = {W n ). Then
200
6. Paracompactifications
positive integer n, Y2 E S(y 1,W n). Since A < ~, Y2 E S(y t>V n)for each positive integer n which implies f(Y2) E S(f(y d,U n ) for each n. Therefore, fey 1) = fey 2)' Define h: X/c.l>" ~ T by h(y*) = fey 1) where Y I E y*. Then f = h ©
X
f
T
~
t~ uX
~
t
X/c.l>"
Clearly if f is onto then r is onto. If r is onto then h is onto since r = g u ©
f is closed and let F be closed in
uX. Put H =
FnX. Then f(H) is closed in T which implies h- l (f(H» is closed in X/c.l>" which in tum implies u«
r
To prove (3), assume f is quasi-perfect. Then by (2) above, is closed. If p E uT then p E ~T which implies (f~rl (P) is closed and hence compact in ~X. Since X is an M-space, by Theorem 6.25, uX = (f~rl (T) so (furl (P) = (f~rl (P). Hence (furl (P) is compact so is perfect. Conversely, assume is perfect. By (2) above, f is closed. Hence, as shown in the proof of Theorem 6.25, h is a homeomorphism of X/c.l>" onto f(X) c T. Identifying X/c.l>" with f(X), we see that f agrees with
r
r
THEOREM 6.26 If X is a Tychonoff M-space, then X is topologically preparacompact and the Tamano-Morita paracompactification of X is (f~rl (T) for any quasi-perfect mapping from X onto a metric space T. Proof: Let A E N. Then
6.7 The Tamano-Morita Paracompactification
201
EXERCISES 1. Let {
~
Y be a perfect mapping. Show that if Y is countably compact then
Chapter 7
REALCOMP ACTIFICATIONS
7.1 Introduction From Stone's characterization of ~X (Theorem 6.2), it can be seen that if! is a real valued bounded continuous function on X, then! can be uniquely extended over ~X. Consequently, any continuous function !:X -7 [0,1] has a unique continuous extension over ~X. In fact, ~X can be characterized by this property. To see this we need only establish that if each continuous function f:X -7 [0,1] has a unique continuous extension over ~X, then any continuous function !:X -7 Y where Y is a compact Hausdorff space has a continuous extension over ~X. For this let g:X -7 Y be a continuous function from X into a compact Hausdorff space Y. Let F be the collection of all continuous functions from Y into [0,1] and for each! E F put If = [0,1]. Let P = [lfeF!f and define the function <1>:Y -7 P by <1>Cy) = [lfeFfCy) E P. Then <1> is a continuous function so <1> © g:X -7 P is a continuous function. Now for each! E F,f © g:X -7 [0,1] can be uniquely extended to a continuous function ht=~X -7 [0,1]. Define h:~X -7 Pby h(P)
= ( ... ,ht
Then h is also continuous. Let {xa} be a net in X that converges to p. Then { hex a)} is a net in P that converges to h(P). Also, {g(x a)} is a net in Y that converges to some Yp E Y. Hence {<1> © g(x a)} is a net in P that converges to <1>Cyp)' Since Xa E X for each a, and since hf is an extension of!© g,f© g(x a ) = hfix a ) for each a, so {h(xa)} = {<1> © g(xa)}· Consequently, {<1> © g(x a )} converges to h(P) which means <1>Cyp) = h(P). Define y:~X -7 Y by Y(P) = Yp for eachp E ~X. To show that y is a continuous extension of g, suppose x E X. Then clearly Therefore, y is an extension of g over ~X. To show that y is continuous, first notice that <1>:Y -7 <1>(Y) c P is a homeomorphism. This is because the family F separates points and can distinguish points from closed sets [see Lemma 1.6]. Then Y(P) = Yp = <1>-1 (h(P» for each p E ~X so Y= <1>-1 © h. Therefore, y is continuous. Finally, to show that y is unique, suppose B:~X -7 Y is another continuous extension of g over ~X. By Corollary 2.3, both y
Yx
= g(x) so )'(x) = g(x).
7.2 Realcompact Spaces
203
and 8 are uniformly continuous with respect to W(the uniformity of ~X). Byan argument similar to the proof of Theorem 6.2, g:X ---7 Y is also uniformly continuous with respect to~. Since ~X is the completion of (X, ~), by Theorem 4.9, g has a unique uniformly continuous extension over ~X. Therefore, y = O. We record this result as: THEOREM 7.1 ~X is characterized among compactifications of X by the property that each continuousfunctionf:X ---7 [0,1] has a unique continuous extensionfl3:~X ---7 [0,1].
Since ~X is the only compactification of X for which each real valued bounded continuous function can be continuously extended, it is natural to inquire about the nature of spaces Y such that X is dense in Y and every real valued (not necessarily bounded) continuous function on X has a continuous extension over Y. This investigation leads to the concept of a real compact space. In a 1948 paper titled Rings of real valued continuous functions 1 (Transactions of the American Mathematical Society, Volume 64, pp. 45-99), E. Hewitt introduced the concept of a Q-space. Q-spaces are defined as follows: let Z(X) denote the zero sets of real valued continuous functions on the Tychonoff space X [cf. Section 6.2]. A non-empty subfamily Z of Z(X) is said to be CZ-maximal if Z contains no countable subset with empty intersection and Z is maximal with respect to this property. A Q-space is a Tychonoff space such that the intersection of any CZ-maximal family is non-void. This definition of a Q-space is due to Shirota. It is equivalent to Hewitt's original definition [see Definition 13 and Theorem 50, p. 85 of Hewitt's paper]. Shirota, in his 1952 paper A Class of Topological Spaces (referenced in Section 3.2), showed the equivalence between Q-spaces and Tychonoff spaces that are complete with respect to their e uniformity which he called e-complete spaces.
7.2 Realcompact Spaces Today, Q-spaces and e-complete spaces are known as realcompact spaces, a term introduced by L. Gillman and M. Jerison in their book Rings of Continuous Functions published by Van Nostrand in 1960. Hewitt showed that realcompact spaces are precisely the closed subspaces of Cartesian products of real lines. He also showed that for each Tychonoff space X, there exists a realcompact space uX, called the Hewitt realcompactification of X such that X c uX c ~X and uX has the properties: (1) uX is the largest subspace of ~X containing X such that each real valued continuous function can be extended to uX and (2) uX is the smallest realcompact space between X and ~X (i.e., X is realcompact if and only if X = uX). Further, Hewitt showed that uX is completely determined (up to homeomorphism) by the following properties: (1) uX is realcompact, (2) uX contains X as a dense subspace, and (3) every real valued
204
7. Realcompactifications
continuous function on X can be continuously extended to uX. Hewitt proved parts (2) and (3) of the next theorem. The proof given below is due to Shirota.
THEOREM 7.2 (E. Hewitt, 1948 and T. Shirota, 1952) For a Tychonoff space X, the following conditions are equivalent: (1) X is e-complete, (2) X is realcompact, (3) X is homeomorphic to a closed subset of a Cartesian product of real lines with the usual topology. ~ (2). Let (X, e) be complete and let Z be a CZ-maximal family of X. Let U = {Un} be a countable normal covering of X. We first show that there exists a Z E Z and a U E U such that Z c U. As shown in the proof of Lemma 3.5, we can construct a precise closed refinement {Fn} of {Un}, and then use Corollary 2.2 and Proposition 1.4 to construct real valued continuous functions fn:X ~ [0,1] such thatfn(Fn) = 1 andfn(X - Un) = O. For each n, the set Zn = {x E X Ifn(x) ~ 1/2} can be shown to be a zero set and Fn c Zn C Un. If for each positive integer n, Zn does not belong to Z then for each n there exists a countable collection {Z;:') c Z such that Znn[n;;;,n=IZ;:.J = 0. Then n;;;,n=IZ;:' = [n;;;,n=1 Z;:']nX = [n;;;,n=1 Z;:']n[ U;;'=I Zn] = U;;'=I (Znn[n;;;,n=1 Z;:.]) = 0. But this contradicts the assumption that Z is a CZ-maximal family. Hence for some positive integer n, Zn E Z. Then Zn is the desired Z CUE U.
Proof: (1)
If Z I, Z2 E Z then Z 1 nZ2 "# 0 and if Z 1 is the zero-set of f and Z2 is the zero-set of g, then Z 1 nZ2 is the zero-set of fg. Clearly Z 1 nZ2 E Z. Since Zu{Z 1 nZ2) contains Z and satisfies properties (1) through (3) of the definition of a CZ-maximaI family. Let E = {(Z, x) IZ E Z and x E Z) and define <s: on E by (Z I, x) <S: (Z2,y) if Z2 c Z I. Then (E, <S:) is a directed set. For each (Z, x) E E put \jI(Z, x) = x. Then \jI:E ~ X is a net. If U E e, there is a Z E Z and aU E U with Z c U. Then \jI(Z, x) E U for each x E Z and if (Z',y) E E with (Z, x) <S: (Z',y), then Z' c Z so \jI(Z',y) E U. Therefore, \jI is Cauchy with respect to e and hence converges to some p E X. Let Z E Z and let U(P) be an open set containing p. Since \jI is eventually in U(P), there is a (Z', x) E E with Z' c Z and \jI(Z', x) = x E U(P), Then U(p)nZ "# 0 so P E Clx(Z) = Z, Hence p E n {Z IZ E Z). Therefore, X satisfies condition (2). This completes the proof of (1)
~
(2).
To show that (2) ~ (3), let C denote the family of all real valued continuous functions on X and for each f E C let Rf = R. Set P = TIfEcRf and let h:X ~ P be the product mapping defined in Section 1.3 by [h (x)]f = f(x). Since X is Tychonoff, the family C can distinguish points and can distinguish points from closed sets. Consequently, by Lemma 1.6, h is an imbedding, so heX) is homeomorphic with X. We may therefore assume X c P. Let 1t be the product uniformity on P and let (X, 1t) denote the uniform subspace of (P, 1t). It remains to show that X is closed in P. By Exercise 3 of Section 4.4, it will suffice to show that (X, 1t) is complete.
7.2 Realcompact Spaces
205
For this let {x a } be a Cauchy net in (X, 1t'). For each a put H a = Cl{x~ Ia E 1t, there is an a such that Hac U for some U E U. Moreover, since there exists a continuous function f:X """"* [0,1) with f(H a) = 0 and f(X - U) = 1, we have for the zero-set Z(j) that Hac Z(j) C U. Let Z' be the family of zero-sets of X that contain some H a' Then if Z 1, Z2 E Z', Z1nZ2 E Z'. By Zorn's Lemma (Theorem 0.1(4», it is possible to find a maximal subfamily Z of Z(X) with respect to the finite intersection property that contains Z'. We now show that Z is also CZ-maximal.
< ~}. It is easily shown that for each U
Suppose, on the contrary, there is a countable subfamily {Z;} of Z with ni'=IZi = 0. Then for each positive integer i, there is anj; E C with Zi = Z(fi). In fact, we can assume that for each i, fi:X """"* [0,1]. Now for each positive integer n, define gn:X """"* [0,1] by gn(x) = maxI Ij;(x) II i = 1 ... n} for each x E X. Put g = L;;'=12-ngn' Then since ni'=IZi = 0, we have g(x) > 0 for each x E X. Consequently, g-1 defined by g-I(X) = l/g(x) is continuous. Therefore, g-1 E C. Since 1t is the product uniformity on P, 1t is the coarsest uniformity on P that makes all the canonical projections uniformly continuous. But for each f E C, Pt = f, so g-1 is uniformly continuous. Since for each U E 1t, there is a Z E Z with Z c U for some U E U, and since g -1 is uniformly continuous, then for each £ > 0, there is a Z E Z such that g-1 (Z) is contained in a set of diameter £. Now if x E Z(gn), then g(x) < 2- n+1. Hence there exists a sufficiently large n with g-I(X) > max{g-I(y)ly E Z} for any x E Z(gn)' This implies that Z(gn)nZ = 0. But Z(gn) = n7=1 Z(fi) which implies that Z(gn) E Z. Then Z does not satisfy the finite intersection property which is a contradiction. Therefore, Z is a CZ-maximal family. Consequently, by property (2), Z has a non-empty intersection. Let pEn {Z IZ E Z}. Let U E 1t' and choose V E 1t' such that V* < U. Suppose no Z E Z meets S(P,V). Since there exists a Z(V) E Z and a V E V with Z(V) c V, we see that p does not belong to Z(V) which is a contradiction. Therefore, Z(V)nS(p,V) "# 0 which implies there is an open neighborhood V' of p such that V'nV"# 0 so there is aU E U with V\..;V cU. But then Z(V) c S(p, U) and Z(V) contains H a for some index a. Hence, {x a} converges to p so (X, 1t') is complete. This completes the proof of (2) """"* (3). To show (3) """"* (1), let X be a closed subset of a product P = IlaR a where R a = R for each index a. Since R a is complete for each a, so is P. Let 1t denote the product uniformity on P and let 1t' denote 1t restricted to X. Since X is closed in P, (X, 1t) is complete. It will suffice to show that e is a basis for 1t'. For this let U be a basic member of 1t. Then
for some finite collection of uniform coverings Ua, of R a, where P a, denotes the canonical projection of Ponto Raj for each i = 1 ... n. Now each Uaj has a countable subcovering Vaj since Raj is LindelOf. Furthermore, each Va, is normal since Raj is paracompact. Therefore, P~~ (Val)n ... npu~(Va") is a
206
7. Reaicompactifications
countable normal covering of P that refines U. Hence U is a member of the e uniformity for P. But then e is a basis for the n' uniformity on X. Therefore, e =n', so X is e-complete. Now that we know realcompact spaces are precisely the e-complete spaces, we can use this fact to derive some interesting properties about realcompact spaces.
THEOREM 7.3 Every Lindel6f space is reaicompact. This is because Lindelbf spaces are cofinally complete with respect to the e uniformity by Theorem 4.4 and hence complete with respect to e hy Corollary 4.1.
THEOREM 7.4 A closed subspace of a reaicompact space is reaicompact. This is because if Y is a closed subspace of a realcompact space X, then Y is complete with respect to the e uniformity of X relativized to Y by Proposition 4.18. But the e uniformity of X relativized to Y is precisely the e uniformity of Y.
THEOREM 7.5 A product of reaicompact spaces is reaicompact. Proof: Let X = flaX a where each X a is realcompact and for each a, let e a be the e uniformity of X a' Then for each a, (X a' e a) is complete so X is complete with respect to the product uniformity by Exercise 4 of Section 5.3. Let /-l denote the product uniformity of the e a's. Then a basis element of /-l is of the form
for some finite collection of uniform coverings U a , of e a, where each U a , is countable and where p a, denotes the canonical projection of X onto X a, for each i = 1 ... n. Then U is also countable and therefore belongs to e (the e uniformity of X). Hence /-l c e. Since (X, /-l) is complete and e is finer than /-l, we have that (X, e) is complete so X is realcompact.We have already seen in Sections 2.2 and 5.5 how a normal sequence of open coverings can be used to construct a pseudo-metric. If we consider the family
£
> 0 there exists a d'
E
D
7.2 Realcompact Spaces
207
and a 8 > 0 such that d'(x,y) < 8 implies d(x,y) < e for all x,y E X, then d ED, (3) whenever x -:/- y, there exists a d E D with d(x,y) -:/- O. where dye is defined by dve(x,y) = min {d(x,y), e(x,y)} and is called the join of d and e. Property (3) only holds for Hausdorff uniform spaces and property (2) implies that if dE D, then rd E D for each r> 0; and if dE D and e :::; d then e E
D. To show property (l) holds let d 'J..., deE D and let 'J... and
208
7. Realcompactifications
- (3). In such a case, we call a family D of pseudo-metrics satisfying these properties a pseudo-metric uniformity. It is left as an exercise (Exercise 2) to show that if D is a pseudo-metric uniformity, then the family of coverings f.l = {U~ldE Dand£>O} where U~= {SAx,£)lxE X},isabasisforacovering uniformity on X. It is easily shown (Exercise 3) that the intersection of any collection of pseudo-metric uniformities is again a pseudo-metric uniformity. Since 0 belongs to every pseudo-metric uniformity, the intersection is never empty. Consequently, if S is a non-empty family of pseudo-metrics on X, there exists a smallest pseudo-metric uniformity D containing S. We say S is a subbase for D and that D is generated by S. If B is a subbase for D such that for each dE D and £ > 0, there exists a d' E B and a 8 > 0 with d'(x,y) :-::; 8 implies d(x,y) :-::; £ for each pair x,y E X, then B is called a base for D. It is left as an exercise (Exercise 4) to show that if S is a subbase for D, then the family B of all finite joins d I V ••• vdn such that d; E S for each i = 1 ... n is a base for D. The family C(X) of all real valued continuous functions on X and the family C*(X) of all real valued bounded continuous functions on X can be used to generate two pseudo-metric uniformities C and C* as follows: for each f E C(X) let df be defined as dlx,y)
=
If(x) - fCy) I
for each pair x,y E X. It is an easy exercise (Exercise 5) to show that df is a pseudo-metric on X. Let C = (df If E C(X)} and C* = (df If E C*(X)}. Let c and c* denote the covering uniformities associated with C and C* respectively. Then c and c* have bases of the form b = {U{ I df E C and £ > O}
b* = {U{ Idf E C* and £ > O} respectively, where U{ = (Six, £) I x EX} and where Six, £) is the sphere about x of radius £ with respect to the pseudo-metric df . We now show that c c e and c* c~. For this, let U E c*. Then U = U{ for some f E C*(X) and £ > O. Since f E C*(X) there is some inf a and sup b such that f(X) c [a,b]. f mayor may not assume the end points a and b. But since f is continuous, it assumes each point P such that a < P < b. Let 8 = £/2 and pick PIE X such that f(P I ) = a if f(x) = a for some x E X or else pick PIE X such that f(P I) = a + £. If a + £ > b pick any x E (a,b) and put P I = x. For each positive integer k pick Pk+1 E X such thatf(Pk+l) = f(Pk) + £ if f(Pd + £:-::; b. Otherwise put Pk+1 = b unless f does not assume the value b in which case put PhI =Pk. Let n be the least positive integer such thatf(Pn+l) =f(Pn).
7.2 Realcompact Spaces
209
Clearly {S/'pi, E) Ii = 1 ... n} is a finite subcovering of U{. It is also clear that ufs < U{ and U{; E c. Therefore, every U{ where f E C*(X) has a finite normal refinement and consequently belongs to~. This shows that c* c~. This argument can be modified to show that c c e (see Exercise 6). Consequently, if X is complete with respect to c, then X is real compact and if X is complete with respect to c* then X is compact. In what follows we will show that the completion with respect to c is the Hewitt realcompactification, and the completion with respect to c* is the Stone-Cech compactification, but for now we merely record the following: PROPOSITION 7.1 If a Tychonoff space is complete with respect to the c uniformity, then it is real compact. If it is complete with respect to c*, it is compact. How to translate the concept of uniform continuity into the terminology of pseudo-metric uniformities follows from the next proposition, whose proof we leave as an exercise (Exercise 7). PROPOSITION 7.2 Let (X, M) and (Y, N) be pseudo-metric uniform spaces and let (X, J..l) and (Y,v) be their associated covering uniform spaces. Then f:X -t Y is uniformly continuous with respect to J..l and v if and only if for each e E Nand E >0, there exists a d E M and a 0 > 0 such that if d(x,y) < 0 then e(j(x),f(y)) < E. Proposition 7.2 has the following useful implication:f:X -t R is uniformly continuous with respect to some pseudo-metric uniformity D if and only if df E D. Consequently,fis uniformly continuous with respect to c or c* if and only if df E Cor C* respectively. For C* this means that if f E C*(X) then df E C* so f is uniformly continuous with respect to c* and hence can be continuously extended to the completion of X with respect to c*. By Proposition 7.1, c*X is compact. But then by Theorem 7.1, c*X is the Stone-Cech compactification of X. Since PX has a unique uniformity, we have THEOREM 7.6 c* = ~.
EXERCISES 1. Let d and d' be two pseudo-metrics on X having the same spheres of radius 2- n for each positive integer n. Show that d = d'. PSEUDO-ME1RIC UNIFORMITIES 2. Let D be a pseudo-metrIc uniformity on X and let v = {U~ IdE D and E > 0 I where U~ = {Sd(X, E) Ix E Xl. Show that v is the basis for a covering
210
7. Realcompactifications
uniformity that generates the same topology on X as the pseudo-metric uniformity. 3. Show that the intersection of any collection of pseudo-metric uniformities is again a pseudo-metric uniformity. 4. Let S be a subbase for a pseudo-metric uniformity D and let B be the family of all finite joins d 1 V ... vd n such that d i E S for each i = 1 ... n. Show that B is a base for D. 5. Let f E C(X) and define df by dtCx,y) = If(x) - fey) I for each pair x,y Show thatfis a pseudo-metric on X.
E
X.
6. Show that the e uniformity is finer than the c uniformity.
7.3 Realcom pactifications By a realcompactification of a Tychonoff space X, we mean a realcompact space Y in which X can be imbedded as a dense subspace. From Theorem 7.2 it is easily seen that a realcompactification Y of X can be realized as the completion of X with respect to some countably bounded uniformity. To see this, note that if Y is realcompact, it is e-complete, and the e uniformity of Y relativized to X is countably bounded. The completion of X with respect to this uniformity is Y. Furthermore, e is the finest countably bounded uniformity on X. For this, first notice that the e uniformity is countably bounded since it has a basis consisting of countable normal coverings. Next, if J..l is a countably bounded uniformity of X and U is an open member of J..l, then there is a normal sequence {Un} of open members of J..l such that U 1 < U. Let V be a countable subcovering of U and for each positive integer n put Vn = Unn V. Then {V n} is a normal sequence of open coverings of X such that VI < V C U. Therefore, U E e so e is finer than J..l. Also, this shows that the countable normal coverings V such that V < U for some U E J..l form a basis for J..l, so J..l has a basis consisting of (perhaps not all) countable normal coverings. Therefore, the countably bounded uniformities are precisely the ones that have bases consisting of countable normal coverings (i.e., they are the separable uniformities). Shirota showed that if X is a Tychonoff space, that eX is precisely the Hewitt realcompactification uX. He did this by proving the following:
7.3 Reaicompactifications
211
THEOREM 7.7 (T. Shirota, 1951) Let X be a Tychonoff space. Then eX has the following properties: (1) X is dense in eX, (2) eX is realcompact, (3) eachfE C(X) can be continuously extended to eX. Also, any space satisfying these three properties is homeomorphic with eX. Clearly eX satisfies property (1) of Theorem 7.7. Shirota states in his proof that it is obvious that eX satisfies property (2) also. This would be the case if we knew, for instance, that e' (the e uniformity of eX) is the e uniformity of eX. Although it is not surprising that this should be so, the proof is not what we usually think of as being obvious. Consequently, we first prove this as a lemma before proving Theorem 7.7.
LEMMA 7.1 Let e' be the uniformity of eX. Then e' is the e uniformity of ex' Proof: Each U' E e' has a countable normal refinement. To see this, let V' = {V~} be a closed uniform refinement of U'. Then V = {v~nX} = {V J3} E e and therefore has a countable uniform refinement say {Vi}' For each Vi there is a VJ3, E V with Vi C VJ3 , so
since V' is a closed covering. Thus (CleX(Vi )} refines V' so W' = (lntex(CleX(Vi refines U'. Now W' E e'. To see this, note that {V;} = {V;nX} for some uniform covering {V;} of eX. Pick y E Then
»}
V;.
Therefore, V; c Clex(V;) for each i so (CleX(Vi )} is a uniform covering of eX. But then (Clex(V;)} has an open refinement in e' which implies W' E e'. Hence U' has a countable normal refinement. It remains to show that all countable normal coverings of eX belong to e'. For this let {W;} be a countable normal covering of eX. Then there exists a normal sequence {U~} of open coverings such that CleX(U'I) < {W;}. But then {U~nX} is a normal sequence of open coverings of X such that CleX(U'1 )nX refines {W;nX}. Hence, Clex(U'dnX E e. But for each VEe, CleX(V) = (CleX(V) I V E V} E e'. Therefore
Consequently, {W;}
E
e'. Therefore, e' is the e uniformity of eX. -
212
7. Reaicompactifications
Proof of Theorem 7.7: It remains to show that eX satisfies property (3) and that if Y is another Tychonoff space satisfying properties (1) - (3) then Y and eX are homeomorphic. That eX satisfies property (3) is easily seen from the remarks preceding Proposition 7.1 and those following Proposition 7.2. Iff E C(X) then dJ E C so that f is uniformly continuous with respect to c. Since c c e, f is uniformly continuous with respect to e and consequently can be uniquely extended to a uniformly continuous functionf':eX --7 R. To show that if Y is another Tychonoff space satisfying (1) - (3) then Y and eX are homeomorphic is a little more difficult. For this we will first show that any Tychonoff space Y satisfying properties (1) - (3) also satisfies the following property: (3') If Zn E Z(X) for each positive integer n, then nCly(Zn) = Cly(nZn). For this suppose nCly(Zn) i= Cly(nZn). Then there exists ayE nCly(Zn) Cly(nZn). For eaGh positive integer n, let fn E C(X) such that Zn = Z(fn) and such that Ifn I ::; 1. Since y does not belong to Cly(nZn), there exists a zero set Z c X such that y E Cly(Z) and Zn(nZn) = 0. LetfE C(X) such that Z = Z(f) and put g = L,gn where gn = 2-n( Ifn I + If I). Then g is strictly positive, i.e., g(x) > 0 for each x E X. Let g' be the extension of g over Y and for each positive integer n let g~ be the extension of gn over Y. Now g'l X = g = L,gn = L,g~ IX. By property (1), X is dense in Y so g' = L,g~. Also, for each positive integer n, g~ ::; lin I + 1f'1 where in is the extension of fn to Y and f' is the extension of f to Y. Consequently, g~(y) = 0 for each positive integer n, so g(y) = O. Hence there exists an h E C(X) such that h(x)g(x) = 1 for each x E X. Let h' be the extension of hover Y. Since (hg)' IX = hg = (h'l X)(g' IX) and since X is dense in Y, we have (hg)' = h'g' = 1. Hence g'(y) = g(y) i= 0 which is a contradiction. Therefore, nCly(Zn) = Cly(nZn). We now use property (3') to show that Y and eX are homeomorphic. For this we show that (Y, e) is the completion of (X, e). We need only show that every countable normal covering U of X can be extended to a countable normal covering U' of Y such that U = UP nX. This will show that (X, e) is a uniform subspace of the complete uniform space (Y, e), and since completions are unique, Y is homeomorphic with eX. Now there exists a normal sequence {Un} of countable open coverings of X such that U 1 < U. UE(Y) is an extension of U into Y and for each positive integer n, U~(Y) is an extension of Un into Y. By property (3'), for each positive integer n,
Consequently, U~(Y) is a covering of Y for each positive integer n, so
UE(y)
is a
7.3 Reaicompactifications
213
covering of Y. Furthermore, since Un+! <* Un for each positive integer n, we have that U~rr; <* U~(Y) for each n. Hence W(Y) is a countable normal covering of Y, and clearly U = U£(Y)nX. COROLLARY 7.1 Let X be a Tychonoff space. Then eX has the following properties: (1) X is dense in eX, (2) eX is realcompact, (3) if Zn E Z(X) for each positive integer n then nClex(Zn) = Cl ex (nZn). Also, any space satisfying these three properties is homeomorphic with eX. COROLLARY 7.2 Let X be a Tychonoff space. Then eX = uX.
That eX = uX, the Hewitt realcompactification of X, follows from the fact that in Hewitt's development of uX, he characterized uX by the three properties of Theorem 7.7. It should be noted that Hewitt's development of uX was very different from ours. In fact, it was done in terms of the maximal ideals in the ring C(X) of real valued continuous functions on X, in a manner similar to Stone's construction of ~X. We now prove a theorem that is the analogue of Theorem 6.11 for realcompact spaces. This theorem is very useful for proving things about real compact spaces and realcompactifications. THEOREM 7.8 Let X be a dense subspace of the Tychonoff space Y. Then the following statements are equivalent: (1) Every continuousf:X ~ A, where A is a realcompact Tychonoff space, has a unique continuous extension to Y. (2) X is C-imbedded in Y. (3) If Zn E Z(X) for each positive integer nand nZn = 0 then n(Cly(Zn)) = 0. (4) IfZn E Z(X)for each positive integer n then Cly(nZn) = n(Cly(Zn))· (5) eX = eY. (6)X eYe eX. Proof: (1) ~ (2) By Theorem 7.3, R is realcompact, so (2) is just a special case of (1). Hence (1) implies (2).
(2) ~ (4) This is the same as the proof that (3) implies (3') in Theorem 7.7. (4) ~ (5) We can show eX = eY by showing that (X, e) is a uniform subspace of (Y,e) and appealing to the uniqueness of the completion. But this is similar to the proof that (X, ~) is a uniform subspace of (Y, ~) in (4) ~ (5) of Theorem 6.10. (5)
~
(6) is obvious.
214
7. Realcompactifications
(6) -4 (1) This proof is similar to the proof in Section 7.1 that if every continuous f: X -4 [0,1] has a continuous extension over pX, then each function g:X -4 K where K is a compact Hausdorff space, has a continuous extension over pX. In this case we have Y c eX so each continuous f:X -4 R has a continuous extension r:eX -4 R, so fY = r I Y is a continuous extension of f over Y. If we let F denote the collection of all real valued continuous functions on X and for each f E F we put RI = R, then we can define P = TIl EFRI and <1>: Y -4 P by (y) = TI/Ed(y) E P. The proof in 7.1 then goes through with minor modifications, to show that a function g:X -4 A, where A is a realcompact Hausdorff space, has a unique continuous extension to Y. (4) -4 (3) is obvious.
(3) -4 (4) Suppose Cly(nZn) "# n(Cly(Zn». Then there exists a point p E n(Cly(Zn» - Cly(nZn). Let Z be a zero set neighborhood of p in Y such that ZnCly(nZn) = 0. Then Cly(n(ZnZn» = 0, so by (3), nCly(ZnZn) = 0 which implies ZnCly(nZn) = 0 which is a contradiction since p E ZnCly(nZn).-
COROLLARY 7.3 eX is the largest subspace of C-imbedded.
px
in which X is
Proof: By Theorem 6.l3, eX c pX. Let Y c PX such that X is C-imbedded in Y. Then by (6) of Theorem 7.8, Y c eX.-
px.
COROLLARY 7.4 eX is the smallest realcompact space between X and In particular, X is realcompact if and only if X = eX.
Proof: Clearly X is realcompact if and only if X = eX since real compactness is e-completeness. To show e is the smallest realcompact space between X and pX, let Xc Y c PX and suppose Y is realcompact. By Theorem 6.l1, pY = PX and by Corollary 7.3, eY is the largest subspace of pY = PX in which Y is C-imbedded. Since Y is realcompact, Y = eY. Hence Y is the largest subspace of PX in which Y is C-imbedded. But X c Y implies every real valued continuous function on Y can be extended to YueX since if f E C(Y) then fl X can be extended to eX and since X is dense in YueX, f can be extended to YueX. But then Y is C-imbedded in YueX which implies YueX c Y which in turn implies eX c Y. COROLLARY 7.5 Every continuous function f:X -4 Y, where Y is rea/compact, has a unique continuous extension :uX -4 Y. Moreover, if f~ is the Stone extension off:X -4 Y c pY, then = f~ lux.
r
r
Proof: Since uX = eX, by Theorem 7.8.(1) and (5),f has a unique continuous extension F:uX -4 Y. By Theorem 6.13, Xc uX c pX. Since X is dense in both uX and PX, and bothF andf~ are determined by their values on X, and
215
7.3 Realcompactifications
since both f~lux.-
r
and f~ are unique extensions of f into
py,
it is clear that
PROPOSITION 7.3 An arbitrary intersection subspaces X a of a given space X IS realcompacl.
r
=
of real compact
= nX a' For each a, the identity mapping i:Y ~ X has (by Theorem 7.8) a continuous extension i a:eY ~ X a C X. Since i can have only one continuous extension from eY into X, all the i a must coincide. Hence this common extension maps eY into nX a = Y. But then eY = Y so Y is realcompact. Proof: Put Y
PROPOSITION 7.4 If Y is C-imbedded in X. then Clex(Y)
= eY.
Proof: Since Y is C-imbedded in X and X is C-imbedded in eX, by the transitivity of C-imbedding, Y is C-imbedded in eX. Then Clex(Y) is realcompact by Theorem 7.4, and Y is C-imbedded in Clex(Y). By Theorem 7.8, Clex(Y) c eY. By Corollary 7.4, eY c Clex(Y). Therefore, Clex(Y) = eY.LEMMA 7.2 Iff is a continuous function from a space Y into a space Z. whose restriction to a dense subset X is a homeomorphism. then f(Y - X) = Z f(X).
*'
= f(y) where x E X and y x. Let U be an open neighborhood of X in Y such that Cly(U) c {p i = 0. The homeomorphismfl X maps UnX onto an open neighborhood of f(x) in f(X) c Z. Therefore,f(UnX) = Vnf(X) for some open neighborhood V of f(x) in Z. Since X is dense in Y, every neighborhood of y contains a point of X - U. The homeomorphism fl X maps such a point into Z - V. Therefore. no open neighborhood of y is mapped by f into V. Hence f is not continuous at y which is a contradiction. Consequently, each point p E Y - X cannot be mapped onto f(x) for any x E X. ThereforeJ(Y - X) = Z - f(X). Proof: Suppose f(x)
PROPOSITION 7.5 Let f:X ~ Y be a continuous function from the realcompact space X into the space Y. If K is a realcompact subset of Y then r1(K) is realcompact in X.
r
1 (K) and let i be the identity mapping on H. By Theorem Proof: Let H = 7.8.(1) and (5), i:H ~ HeX has a unique extension i'J:uH ~ X. AlsoJH =fl H has a unique extension fh:uH ~ K. Since H is dense in uH and both of these unique extensions are determined by their values on H, we have fH = f © i which implies fh = (f© i)'J = f© i'J. Now by Lemma 7.2, i'J(uH - H) = X - H so that (f © i)'J(uH - H) c Y - K. Butfh(uH - H) c K. Since fh = (f© i)'J this implies uH - H = 0 so by Corollary 7.4, H = 1 (K) is realcompact in X.-
r
216
7. Realcompactifications
A space is said to have a given property hereditarily if every subspace also has the property. For instance, a space is hereditarily realcompact if every subspace is realcompact.
COROLLARY 7.6 A rea/compact space X, every point of which is a G ii, is hereditarily rea/compact. Proof: For each p E X, there exists an f E C(X) such that Z(j) = {p} since {p} is a G 8 in X. Therefore, X - {p} = X - Z(j) = 1(R - {O)). Since R - {O} is Lindeli:if, it is realcompact, so by Proposition 7.5, X - {p I is realcompact. Now any proper subset of X is an intersection of subsets of the form X - {p} such that p E X. Therefore, by Proposition 7.3, every subspace of X is realcompact. -
r
PROPOSITION 7.6 In any space, the union of a compact subset with a realcompact subset is rea/compact. Proof: Let X = RuK where R is realcompact and K is compact. Suppose X is not realcompact. Then eX X. Pick P E eX - X. Since K is compact, we must have p E Clex(R). The next step will be to show that R is C-imbedded in Ru{p}. For this, letfE C(R). It is possible to construct a function g E C(eX) that is zero on a neighborhood U of K and 1 on a neighborhood V of p. Then (g IR)(j) can be extended to a continuous function h on X by defining h(y) = 0 for each y E K. h can be further extended to Xu{p} since p E eX.
*'
Since f agrees with h on VnR (which is dense in CI(V nR) and which contains p), f can also be extended to p. Consequently, R is C-imbedded in Ru{p}. By Theorem 7.8.(5), eR = e[Ru{p I]. Since R is realcompact, eR = R which implies pER which is a contradiction since p E eX - X. Therefore, X must be realcompact. -
THEOREM 7.9 The following statements about a space Yare equivalent: ( J ) For each space X, if there exists a continuous f:X ---7 Y with 1 (y) compact for each y E Y then X is realcompact. (2) Every space X of which Y is a one-to-one continuous image is realcompact. (3) Y is hereditarily real compact. (4)ForeachYE Y, Y- {y} isrealcompact.
r
Proof: (1)
---7
(2) is obvious.
(2) ---7 (3) Let X be a subspace of Y and let 't be the topology of Y. Define another topology 't' on Y by adding the sets X and Y - X to 'to Then 't' is also completely regular and both 't and 't' induce the same relative topology on X. Since the identity map i:(Y, 't') ---7 (Y, 't) is one-to-one, by (2), (Y, 't') is realcompact. Then X, which is closed in (Y, 't,), is realcompact by Theorem
217
7.4 Realcompact Spaces and Lindell)[ Spaces
7.4. But since 1 and l' induce the same relative topology on X, X is a realcompact subspace of Y. (3)
~
(4) is obvious.
r
(4) ~ (1) Let f:X ~ Y be a continuous mapping such that 1 (y) is compact for each y E Y. By Proposition 7.6, (4) implies Y is realcompact. By Corollary 7.5,fhas a unique continuous extension.f' :uX ~ Y. Now let y E Y. By (4), Y - {y) is reaicompact, so by Proposition 7.5, Z = If''rl(y - {y)) is realcompact. Therefore, by Proposition 7.6, Zur 1(y) is realcompact. Since Zur 1 (y) lies between X and uX, by Corollary 7.4 we have Zur 1 (y) = uX. Therefore, 1(y) = If''rl (y) for each y E Y so.f' maps no point of uX - X onto y. Since this holds for each y E Y, we have uX - X = 0 so X is realcompact.-
r
COROLLARY 7.7 If f:X ~ Y is one-to-one and continuous, and hereditarily realcompact, then X is hereditarily realcompact.
if Y is
EXERCISES 1. Show that a space X is realcompact if and only if whenever X is imbedded in T such that X is dense in T but X "# T, then X is not C-imbedded in T.
2. Show that eX
= ~X if and only if X is pseudocompact.
3. Show that.f' (eX) = f(X) for each f
E
C(X).
4. Let f be a continuous function from X into the realcompact space Y. Let f~ be the Stone extension into ~Y. Show that (f~rl (Y) is realcompact. 5. Show that ifU is open in eX, then XnU is C-imbedded in U. 6. Show that if K is compact in X, then X - K is C-imbedded in eX - K.
7.4 Realcompact Spaces and Lindelof Spaces There is a great deal of similarity between reaicompact spaces and LindelOf spaces. We have already seen that LindelOf spaces are realcompact, that the realcompact spaces are the ones that are complete with respect to their e uniformity and that LindelOf spaces are the ones that are cofinaily complete with respect to their e uniformity. In this section we continue to investigate this similarity. Of particular interest in this regard are two theorems of H. Tamano that appeared in his 1962 paper On compacti/lcations referenced in Chapter 6. Let f E C(X) and let R* = Ru {00) be the one-point compactification of R. Let f* be the Stone extension of !:X ~ R* to ~X. Let XI be the maximal
218
7. Realcompactifications
subspace of ~X to which f:X ~ R can be continuously extended. Then for each p E ~X - XJ we must have f*(P) = or else we could extend f:X ~ R to XjU{p). Consequently XJ = If*tl(R) so XJ is open in ~X. Moreover, by Corollary 7.3, uX = n{XJlf E C(X»). Therefore, by Corollary 7.4, X is realcompact if and only if X = nXJ . We record this as: 00
PROPOSITION 7.7 X is realcompact if and only if X
= nXJ.
Let dJ be the pseudo-metric on X generated by f (i.e., dJx,y) = If(x) f(y) I) and let VJ = (SJx, 1) Ix EX) where SJx,1) is the sphere about x of radius I with respect to df . Let EJ denote the extent of VJ in ~X (cf. Section 6.2).
PROPOSITION 7.8 XJ = EJ.
r
Proof: Let be the extension of f over XJ and let df! be the pseudo-metric on XJ generated by Let Sf!(x,l) be the sphere about x of radius I with respect to df!. Since each member of V£([3X), the proper extension of VJ into ~X, is regularly open in ~X, by Lemma 6.10.(1), each member of V?[3X)nXJ is regularly open in XJ' It is evident that for each p E X,
r.
Sf!(p,I)nX
= SN,1)
c SN, 1)£([3X)nX
= [SN,l)£([3X)nXJlnX
so by Lemma 6.10.(3) Sf!(P,1) c SN, 1)£([3X)nXJ. Now the collection of all spheres Sf! (P,I) where p E X covers XJ' so XJ C EJ. Conversely, if p E ~X is contained in EJ, then PES (x, I )£([3X) for some x E X. Then If(x) - f(y) I < 1 for each y E SJx,1)nX which impliesfis bounded on SJx, l)£([3X) nX. It follows that p E XJ' for if not then f*(P) = which implies that for each positive integer n, there exists a Z E SJx, l)£([3X)nX with fez) > n which is a contradiction. Therefore, EJ C XJ' 00
COROLLARY 7.8 uX = nEJ. THEOREM 7.10 (H. Tamano. 1962) The following statements about a Tychonojf space are equivalent: (1) X is realcompact. (2) For each p E ~X - X. there is a closed G Ii set C in ~X - X containing p. (3) For each p E ~X - X. there is a countable star-finite partition of unity
7.4 Realcompact Spaces and LindelOf Spaces
219
~X - X, since X is realcompact, there exists an f E C(X) that cannot be extended over p. Hence x does not belong to Xf which implies x E Cf' Then Cf is a closed G 6 in ~X containing p.
(2) ~ (1) It is easily shown that each closed G 6 in ~X - X is a Cf for some C(X). Hence uCf = ~X - X which implies X = r-IXf . By Proposition 7.5, this implies X is realcompact.
f
E
(2) ~ (3) If C is a closed G 6 in ~X - X, we can construct a continuous f:X [0,1] such that Z(f*) = C and such that f* assumes all values in [0,1]. For each positive integer n ~ 2 putfn = minlf,l/(n-l)) and gn = max{fn,1/(n+l»). Define hn:X ~ [0,1] by ~
hn(x) = 0 if gn(x) < l/(n-I) and hn(x) = l/(n-I) - I/(n+I) otherwise. Then put k n = gn - l/(n+ 1), In = k n - hn and
THEOREM 7.11 (N. Howes, 1970) The Hewitt realcompactijtcation of X is LindeLOfif and only if X is preparacompact with respect to the e uniformity. Proof: Let (X, e) be preparacompact and let (eX, e') be the completion of (X, e). Then eX is the Hewitt realcompactification of X. By Corollary 4.3, (eX, e') is cofinally complete, so by Corollary 4.4, eX is LindelOf. Conversely, if eX is LindelOf and e' is the e uniformity for eX, then by Theorem 4.4, (eX, e') is cofinally complete which implies (X, e) is preparacompact. Consequently, the proof reduces to showing that e' is the e uniformity for eX. But this has already been proved (Lemma 7.1).-
220
7. Realcompactifications
THEOREM 7.12 (H. Tamano, 1962) Let BX be any compactijzcation of the Tychonoff space X. Then the following statements are equivalent: (1) X is LindelOj. (2) For each compact C c BX - X, there is a countable star-finite partition of unity = {n} on X such that CIBx(O(n))nC = 0 for each n. (3) For each compact C c BX - X, there is a closed G'8 set G of BX such that C c G c BX - X. (4) For each compact C c BX - X, there is a countable family {G n } of compact subsets of BX such that GnnC = 0 for each nand XcuG n · Proof: (1) ~ (2) Assume X is LindelOf and suppose C is a compact subset of BX - X. For each x E X let U(x) be an open neighborhood of x such that CIBx(U(x»nC = 0 and consider the covering U = {U(x) Ix E XI. By Proposition 6.2, U has a countable star-finite refinement V = {Vn I. By Lemma 3.6, X is paracompact and hence normal. By Lemma 1.3, V is shrinkable to a countable star-finite refinement W = {Wnl. Therefore, by Theorem 6.3, there exists a countable star-finite partition of unity
~
(4) is obvious.
(4) ~ (1) Assume statement (4) is valid. Let U = {U a I be an open covering of X. For each a let U~ denote the proper extension of U a to BX. Put C = BX - uU~. Then C is a compact subset of BX - X. By (2), there exists a countable family {G n I of compact subsets of BX such that GnnC = 0 for each n and Xc uG n . Evidently, each Gn is covered by a finite subfamily of {U~I and therefore uG n is covered by a countable subfamily of {U~ I. Since X c uG n , we see that X is covered by a countable subfamily of U. Therefore, X is LindelOf. •
EXERCISE 1. Show that the collection of functions {n} in the proof of (2) ~ (3) of Theorem 7.9 is a countable star-finite partition of unity such that CiI3X(O(n»nZ(f*) = 0.
7.5 Shirota's Theorem
221
7.5 Shirota's Theorem Clearly the discrete space (0 of positive integers is realcompact since (0 is Lindelbf. In fact, if we denote the cardinality I R I of the real numbers by c, it is easily shown that any discrete space X with I X I ~ c is realcompact. To see this note that if I X I ~ c there exists a one-to-one function f:X ---7 R. Since X is discrete, I is continuous and since R is separable, I(X) is Lindelbf and hence realcompact. Then by Proposition 7.5,f-1 (f(X» = X is realcompact. It is natural to ask if all discrete spaces are realcompact. In the light of Shirota's Theorem, this straightforward question takes on unexpected significance. Shirota's Theorem essentially states that for any Tychonoff space X, if all discrete spaces of cardinality ~ 21 x 1 are realcompact, then X is realcompact if and only if it is topologically complete. Consequently, if all discrete spaces are realcompact, topological completeness is the same as realcompactness and eX = uX for any Tychonoff space X.
The problem as to whether all discrete spaces are realcompact reduces to a (still unsolved) set theoretic problem as to whether there exists a certain type of cardinal number called a measurable cardinal. However, it is known that if a measurable cardinal exists, it must be larger than any known cardinal (i.e., any cardinal that can be constructed from the rules of cardinal arithmetic). All known examples of uniform spaces can be constructed from (0 and the rules of cardinal arithmetic, so for all practical purposes the existence of measurable cardinals does not concern us in our study of uniform spaces. The problem of the existence of measurable cardinals has only philosophical interest at the present time. It is known for example that the statement "There do not exist measurable cardinals." is consistent with the axioms of ZFC. Shirota's Theorem is one of the most striking theorems in the theory of uniform spaces. J. Isbell, in his book Uniform Spaces published by the American Mathematical Society in 1964, calls Shirota's Theorem the "first deep theorem of uniform spaces." It is interesting to note that Shirota's proof extends the (now famous) construction in Stone's Theorem of the closed covering {Fn} where each Fn is a discrete collection {Fna I a E Ana} of closed sets. We begin our investigation of this problem by defining what we mean by a measurable cardinal. The concept of a measure will be introduced and investigated in the next chapter, but for now it suffices to introduce a special type of measure called a zero-one measure or {O, I } measure. A measure j.! on a set X is always defined to be a set function on X whose range is a subset of [0,00] or of the complex numbers, i.e., j.! is defined on some collection of subsets of X (called the measurable subsets of X) and the value j.!(E) of a measurable set E c X is called the measure of E. For a discrete space X, the collection of subsets on which a {O,l} measure will be defined is the set 2x of all subsets of X, and the range will be the subset {O,l} of [0,00].
222
7. Realcompactifications
There is more to a measure than just being a set function however. Measures are distinguished from set functions by the property of being additive. A set function WX ~ [0,001 is said to be a-additive if for each pairwise disjoint countable collection {En} of measurable subsets of X,
Consequently, we see that the countable union of measurable subsets must also be measurable. In the case of {O,ll measures, f.!(uE n) can only be 0 or 1, so at most one of the En can have measure 1, i.e., f.!(En) = 1 for at most one n. Hence the series Lnf.!(En) always converges. There are two trivial {O, I} measures on a discrete space X that we can givt: as examples of a {O, I} measure. The first is the zero measure f.!o defined by /JD(E) = 0 for each E c X. The other is the point measure f.!p defined for each p E X by f.!p (E) = 1 if pEE and f.!p (E) = 0 otherwise. For a point measure f.!p' it is clear that f.!p({p}) = 1, i.e., the measure of a single point is 1. The question is: Does there exist a {O,1} measure f.! on the discrete space X that is not one of these two trivial types, i.e., such that f.!( {x}) = 0 for each x E X, but f.!(E) = 1 for some E c X? If there does, X is said to be measurable. If ex is a cardinal number, we can give ex the discrete topology. If ex, with the discrete topology is measurable, then ex is called a measurable cardinal. There is an interesting relationship between the non-zero {0,1} measures on a discrete space X and the CZ-maximal families on X. If Z is a CZ-maximal family on X, let Xz be the characteristic function of Z defined on all subsets of X by Xz(E) = 1 if E E Z and Xz(E) = 0 otherwise. We want to show that Xz is a non-zero (i.e., Xz(A) = 1 for some A c X) {0,1} measure on X. We first show that if E c X then either E E Z or X - E E Z. If E does not belong to Z then En(nEn) = 0 for some sequence {En} C Z, for otherwise we could add E to the collection Z to get a new collection Z' that has the countable intersection property and that contains Z which is a contradiction since Z is maximal. Now nEn E Z for the same reason. Since (nEn)nE = 0, nEn eX - E. But then X E E Z since (X - E)n(nZn) oF 0 for any collection {Zn} c Z because (nEn)n(nZn)
= 0.
Then clearly X E Z so Xz(X) = I which implies Xz is a non-zero set function. Suppose {En} is a pairwise disjoint collection of subsets of X. If m, k is a pair of positive integers such that Xz(Em) = 1 = Xz(Ed then Em' Ek E Z but EmnEk = 0 which is a contradiction. Therefore, at most one of the En can belong to Z. If exactly one of the En, say Ek E Z then LnXz(En) = 1 and Ek c uEn which implies uEn E Z so Xz(uE n ) = 1. Thus
If Xz(En) = 0 for each positive integer n then En does not belong to Z for each n
7.5 Shirota's Theorem
223
which implies X - En E Z for each n. But then IIn(X - En) E Z and (uEn)II[lIn(X - En)] = 0 so uEn eX - IIn(X - En) which does not belong to Z. Hence uEn cannot belong to Z so Xz(uE n) = O. Also. since Xz(En) = 0 for each n, LnXz(En) = O. Thus
Consequently, Xz is a non-zero {O, I} measure on X. Conversely, if 11 is a non-zero {O,!} measure on a discrete space X we can put Zf1 = {E c X IIl(E) = I}. We will show that Zf1 is a CZ-maximal family on X. Let {En} C Zf1 and suppose liEn = 0. For each positive integer n let An = En - Uk=n+1Ek' Then {An} is a sequence of pairwise disjoint subsets of X and since liEn = 0, uAn = uEn. Since 11 is a-additive it is monotone, i.e .. if A c B then Il(A) c Il(B). To see this note that A and B - A are disjoint subsets of X so that: Il(A) + Il(B - A)
= Il(Au[B - AD = Il(B).
Since Il(B - A) = 0 or 1 it is clear that Il(A) ~ Il(B). Since Il(En) = 1 for each n, and En C uEn for each n, we have ll(uE n) = 1 so ll(uA n) = L But ll(uA n) = Lnll(An) which implies exactly one of the An' say Am has Il(Am) = 1. Now AmuEm+l C En and ll(uE n) = I which implies Il(A muEm+1) = I since 11 is monotone. Since AmllEm+1 = 0 and 11 is a-additive, we have IlCA muEm+1) = Il(Am) + Il(En) which is a contradiction since Il(Am) = 1 = Il(E m+1). Therefore, lIEn #-0. Consequently, Zf1 has the countable intersection property. Next we show that Zf1 is maximal with respect to this property, i.e., that Zf1 is a CZ-maximal family. For this, suppose Zf1 is a proper subset of Z and that Z has the countable intersection property. Let Z E Z - Zw Then Il(Z) = 0 which implies Il(X - Z) = I which in tum implies X - Z E Zw But ZII(X - Z) = 0 so Z does not even have the finite intersection property which is a contradiction. Therefore, Zf1 is maximal with respect to the countable intersection property. We record this as: PROPOS/TION 7.9 There is a one-to-one correspondence <1> between the collection of all CZ-maximal families on a discrete space X and the collection of all non-zero {OJ} measures on X. is defined by <1>(Z) = Xz where Z is a CZ-maximalfamily and Xz its characteristic function. THEOREM 7.13 A discrete space X is rea[compact is a non-measurable cardinal.
if and only if Ixi
Proof: Let X be realcompact and suppose 11 is a non-zero {OJ} measure on X. Then Z = {E c X 11l(E) = I} is a CZ-maximal family on X, as shown in the proof of Proposition 7.9. Since X is realcompact, liZ #- 0. Let x E 112, Then
7. Reaicompactifications
224
{x} E Z which implies J.l( {x)) = 1. Therefore, each non-zero {OJ} measure on X is the point measure J.lx for some x E X so 1 X 1 is non-measurable.
Conversely, let 1 X 1 be non-measurable and suppose Z is a CZ-maximal family on X. Then Xz is a non-zero {O,l} measure on X as shown in the proof of Proposition 7.9. Since 1 X 1 is non-measurable, Xz( {p}) = 1 for some p E X which implies {p} E Z which in turn implies p E nZ. Therefore, X is realcompact. We say that a {OJ} measure J.l is M-additive if J.l(uE a) = 0 whenever {E a a E A} is a family of pairwise disjoint sets of measure zero with AI = M. Clearly an M-additive {OJ} measure is N-additive for each cardinal N:<=::: M. It 1
1
is also easily seen that J.l is M-additive if and only if the intersection of any collection of M sets, each having measure 1 is a set of measure 1.
PROPOSITION 7.10 Each {OJ} measure J.l on a non-measurable cardinal M (with the discrete topology) is M-additive. Proof: If J.l is not M-additive, there exists a family of pairwise disjoint sets {E a c MI a E M} such that J.l(E a ) = 0 for each a and J.l(uE a ) = 1. Define the measure A on M by A(E) = J.l(u{Eala E E)). Then A(M) = 1 and A({a)) = J.l(E a) = 0 for each a. Therefore, M is measurable which is a contradiction. Consequently, each {O,1} measure J.l on a non-measurable cardinal M is M-additive. A {O,I} measure J.l is said to be free if J.l({x)) = 0 for each x E X. The intuitive notion behind this definition is that the measure of the whole space is not bound to some p E X in the sense that J.l( {p)) = 1. Clearly, only the free measures are useful in proving that a discrete space is measurable.
1
PROPOSITION 7.11 IfJ.l is a free measure on a discrete space X and E 1 is non-measurable for some E c X, then J.l(E) = O.
Proof: Define the {O,I} measure A on E by A(A) = J.l(A) for each AcE. Clearly A is also free. Since 1 E 1 is non-measurable, we cannot have A(E) = 1. Hence J.l(E) = A(E) = O. We say that a class of cardinal numbers is closed if it is closed under all the rules (of cardinal arithmetic) for forming new cardinals from given ones: addition, multiplication, exponentiation, taking suprema of a collection of cardinals, and the passage from a given cardinal to its successor or any of its predecessors. Clearly the class of all cardinals (this class is not a set) is closed since any cardinal formed from given cardinals using these rules is again a member of the class. The class of all finite cardinals is also a closed class (it is understood that the cardinal number of any index set used in applying these rules is a member of the class in question). Of primary interest to us are the closed classes that contain 1(J) I.
7.S Shirota's Theorem
LEMMA 7.3 A non-empty class of cardinals C is closed whenever ME C, then: (1) N E C for each cardinal N < M. (2) the sum of any M members of C is in C, (3) 2M E C.
225
if and only if.
Proof: The necessity of conditions (1) - (3) is obvious. Conversely, assume C satisfies conditions (1) - (3). We need to show that C contains products, exponentials, suprema and successors. For this let M E C. Then M + 1 <::; 2M so by (3) and (1), M + I E C. Therefore, successors are contained in C. To show that suprema belong to C, let {N a), where a E M, be a collection of cardinals andputN='iN a . By (2),NE C. SincesupNa<::;N,wehavesupNaE Cby(1). Also,
so C contains products. Finally, if N E C, then N M = ON a where N a = N for each a E M. Since products are contained in C, this means N M is contained in
C.THEOREM 7.14 The class of all non-measurable cardinals is a closed class containing Iffi I. Proof: Since ffi with the discrete topology is realcompact, Iffi I is nonmeasurable. We need to show that if C is the class of all non-measurable cardinals, then C satisfies the three properties of Lemma 7.3. To show (1), it will suffice to show that every subspace of a discrete realcompact space is realcompact. But every subspace of a discrete space is closed, so by Theorem 7.4, we have property (1).
To show property (2), let IX I be the sum of M non-measurable cardinals where M E C. Then X is the union of M pairwise disjoint subsets say {E a} such that a E M. Let 11 be a free measure on X. By Proposition 7.ll.ll(E a ) = 0 for each a. Since ME C, Mis M-additive by Proposition 7.10. Hence Il(X) = ll(uE a ) = O. Therefore, 11 is the zero measure so IX I is non-measurable. Hence C contains sums, so property (2) holds. To show property (3), let I X I = M E C and let P be the set of all subsets of X. Then Ipi = 2M. Suppose 11 is a non-zero measure on P. By Proposition 7.10,11 is M-additive. For each x E X let Hx = {A c X Ix E A} and let Kx = {B cXlxdoesnotbelongtoB}. PutZ= {XE XIIl(Hx ) = I}. Then the set E = [nxezHx]n[n{Kx Ix does not belong to Z} 1
is the intersection of no more than M sets of measure 1. Therefore, Il(E) = 1 since 11 is M-additive. But E contains only one element. To see this note that
226
nXEzHx
7. Realcompactifications
= {A c xlz c
A} and n{Kx Ix does not belong to Z}
= {B c xl Be
Z}. Hence
E
=
{AcXIZcA}n{BcXIBcZ} = {Z}.
Therefore, Il is not free which is a contradiction. Thus Ipi = 2M is nonmeasurable, so 2M E C. This shows that property (3) holds.· The smallest closed class containing I co I contains all the cardinals that we know how to construct using the rules of cardinal arithmetic. This class of cardinals is truly immense. A cardinal number is said to be strongly inaccessible if the set of all smaller cardinals is a closed class containing Ico I. Therefore, by Theorem 7.14, the smallest measurable cardinal (if such a cardinal exists) must be strongly inaccessible. It is impossible to prove from the axioms of ZFC that strongly inaccessible cardinals exist. This is because if we have a model for set theory that includes an inaccessible cardinal, then the set of all cardinals less than the first inaccessible cardinal is a model for set theory that does not include strongly inaccessible cardinals and, in which, all the axioms of ZFC hold. Consequently, there is a model for set theory that does not include inaccessible cardinals, so their non-existence is consistent with the axioms ofZFC. It is conceivable that one can prove that no such cardinals exist from the axioms of ZFC, but this is still an open question. A similar situation exists if one tries to prove the existence of infinite cardinals with the axioms of ZFC minus the axiom of infinity. Consequently, most mathematicians believe another axiom must be added to ZFC to prove their non-existence. Clearly their existence requires another axiom. But even if strongly inaccessible cardinals exist, it may still be the case that measurable cardinals do not. Consequently, for all practical purposes, i.e., for working with any space X we can construct from the axioms of ZFC, I X I is non-measurable. Therefore, the question appears to be of philosophical rather than practical interest. LEMMA 7.4 Let Z be a CZ-maximal family on a Tychonoff space X and let f E C(X) such that f '# 0 and Z(f) E Z. IfY = {x If(x) ~ a} for some a > 0, then Z' = {Z( g) E Z(Y) 10 '# Z(f)nZ c Z( g) for some Z E Z} is a CZ-maximal family on Y. Proof: It is easily shown that Z' is a non-empty subfamily of Z(Y) with the countable intersection property. Therefore, it suffices to show that Z' is maximal with respect to the finite intersection property in Z(Y). For this suppose Z' is not maximal, i.e., suppose W is a non-empty subfamily of Z(Y) with the finite intersection property that contains Z' as a proper subset. Let Z(g) E W - Z' for some g E C(Y) and put F 0 = Z(g)nZ(j). Then FoE W so F 0'# 0. NowputF 1 =Y.
227
7.5 Shirota's Theorem
For each rational r E [0, Illet Ur = (xlf(x) < r}, Zr = (xlf(x) ~ r}, U~ = (x I g(x) < r} and Z~ = (x Ig(x) ~ r}. Put G r = UrnU~ and Fr = ZrnZ~. Then G r is open in X and Fr is closed. For any pair of rational numbers r, s E [0, 1] with r < s we have Fr c Gs c Fs. For each x E X put lex) = sup{ r Ix does not belong to Fr}. Then l E C(X) and Z(g') = nFr = nr(ZrnZ~) = Z(j)nZ(g) = Fo. For each r, Fr E Z(X) and Z(j) c Fr. Therefore, each Fr E Z which implies nrFr E Z. Hence FOE Z. Then by the definition of Z', Z(g) E Z' which is a contradiction. Therefore, Z' is measurable after all. THEOREM 7.15 (T. Shirota, 1951) Let X be a Tychonoff space such that IX I is non-measurable. Then X is realcompact if and only if X is complete with respect to u. Proof: Assume X is complete with respect to u. It suffices to show that if Z is a CZ-maximal family of X and U E u, there is aBE Z and a U E U with B cU. The reason for this is that if this is the case, then a proof similar to (1) ~ (2) of Theorem 7.2 will imply that there is apE X with p E nZ. Thus X will be realcompact. For this, observe that U E U implies there is a normal sequence {Un} with U 1 <* U. Let U = {U a I a < y} for some cardinal y. In the proof of Stone's Theorem (Theorem 1.1), it was shown that there is a closed covering H = {Hn " Ia < y and n is a positive integer} of X satisfying the properties: (1) If a of.~, no member of Un meets both H n" and Hn~'
(2) Star(H n" , Un)
C
U a for each a < y and positive integer n.
Also, in Stone's Theorem, the closed covering {Fn} was defined as follows: for each a < yand positive integer n put En" = Star(HnfJ.' U n+3 ) and G n" = Star(Hn" , U n+2 ). Then H n" C En" C CI(En,,) C G n" for each pair a, n and if a of. ~, no member of Un+2 meets both G n" and Gn~' Put Fn = u{ CI(E n,,) Ia < y} for each positive integer n. If for each pair a, n we put Fn" = CI(En,,), we have a closed covering F = {FnfJ.} of X having the following properties: (3) If a of.~, no member of Un+2 meets both F n" and Fn~' (4) Star(Fn", U n+ l ) C U a for each a < yand positive integer n. Then for each positive integer n, Fn = u{Fn" Ia < y}. Suppose that for each n FnnZn = 0 for some Zn E Z. Then nZn = nZnnX = (nZn)n(uFn) = u[Fnn(nZn)] c u(FnnZn) = 0. But this contradicts the assumption that Z is CZ-maximal. Therefore, there exists a positive integer m such that F mnB of. 0 for each B E Z. LetfE C*(X) withf(Fm) = O,j(X - Star(Fm, U4 » = 2 and 0 ~f~ 2. Put Zo = (x If(x) = O} and Z I = (x If(x) ~ I}. Since Fmc Zo and Z is CZ-maximal, Zo E Z. By Lemma 7.4, Z' = (Z(g) E Z(Z 1) 10 of. BnZo c Z(g) for some B E
228
7. Reaicompactifications
Z} is a CZ-maximal family on Z I. Then, just as in the proof (1) ~ (2) of Theorem 7.2, for each countable normal covering V = {Vn } of Z, there is a B' E Z' and a positive integer k with B' c V k • For each ex < 'Y put Za = ZlnStar(Fma' U m+4)' Clearly ZI c Star(Fm,U m+4) and by (3) above, Star(Fma' U m+4)nStar(Fmp' U m+4) = 0 so we can construct a function fa E C*(X) from f (by putting f a (x) = f(x) if x E Star(F ma ,U m+4) and f(x) = 2 otherwise) such that Za = Z(f a). Then Z 1 = uZ a and by (3) above,
Let Y be the discrete space (ex IZa 1= 0) and let h:Z 1 ~ Y be defined by hex) = ex where x E Za. It is easily shown that h is a uniformly continuous mapping from (ZI, e) into (Y,e). Since lyl ~ IAI ~2IXI,byLemma7.3, lyl is nonmeasurable so Y is realcompact by Theorem 7.13. Now for each WEe in Y, h- I (W) E e in Z I. Therefore, there is a B' E Z' with B' c h -I (W) for some W E W. But then h(B,) c W. Hence for each WEe in Y, there exists h(B') E h(Z') with h(B') c W for some W E W. Then just as in the proof of (1) ~ (2) of Theorem 7.2, there exists an ex E Y with ex E h(B,) for each B' E Z'. But this implies B'nZ a 1= 0 for each B' E Z' so Za E Z'. For any B E Z, BnZ 1 E Z' by the definition of Z', so Zan(BnZ I) 1= 0 which implies ZanB 1= 0 which in tum implies Za E Z. But Za c Star(Fma' U m +4) C U a by (4) above, so for each U E u, there exists aBE Z with Be U for some U E U. •
Part II: Analysis
Chapter 8 MEASURE AND INTEGRATION
8.1 Introduction In this chapter and the next, the theory of integration in uniform spaces will be developed. This chapter will only be concerned with those aspects of integration theory that do not depend on the uniform structure of the space. In elementary analysis one encounters the concept of the Riemann integral. Intuitively, the process of Riemann integration in one, two and three dimensional Euclidean space corresponds to calculating lengths, areas and volumes respectively. The formalization of the Riemann integral occurred during the nineteenth century. Briefly, the main idea for one dimensional Euclidean space is that the Riemann integral of a function f over an interval [a,b] can be approximated by sums of the form
where I) ... In are disjoint intervals whose union is [a, b], l'1(1i) denotes the length of Ii and Xi E Ii for each i = 1 ... n. By the end of the nineteenth century and early into the twentieth century, many mathematicians were working on replacements for the Riemann integral that would be more flexible and better suited for dealing with the problems of modern analysis. Some of the more notable researchers were Borel, Jordan, Lebesgue and Young. They realized that a more satisfactory theory of integration could be obtained by replacing the intervals I) ... In in the above approximation with subsets of the real line that belonged to a more general class, e.g., the so-called measurable sets. When this is done, the function f which was usually assumed to be continuous (or nearly so) could be significantly generalized, e.g., the so-called measurable functions. Ultimately, the construction that proved most successful was the one of Lebesgue, so it is customary to call this new integral the Lebesgue integral even though many participated in developing the modern theory of integration. In the theory of Lebesgue integration, associated with the concept of measurable sets is the concept of a measure. In this chapter, a measure is a nonnegative real valued set function (i.e., it maps sets onto non-negative numbers or (0). Intuitively, the measure tells how "big" (in some sense) the sets are. In
8. Measure and Integration
230
this chapter, after a development of the essential properties of measures, we present a development of the Lebesgue integral for topological spaces. In Chapter 9, the theory of Lebesgue integration will be specialized in uniform spaces to invariant integrals.
8.2 Measure Rings and Algebras
Let X be a set and R a collection of subsets of X with the property that if A, B E R then AuB and A - BE R. Then R is said to be a ring in X. It should be noted that this property implies AnB E R since AnB
= [AuB] - [(A - B) u
(B - A)].
If R has the additional property that ACE R whenever A E R then R is called an algebra. Clearly the empty set belongs to every ring R since for any A E R, 0 =A-A.
PROPOSITION 8.1 A ring R in X is an algebra
if and only if X E
R.
Proof: Assume X E R. If A E R then A C = X - A E R. Therefore R is also an algebra. Conversely, assume R is an algebra and let A, B E R. Then 0 = (A B) n (B - A) E R. Since R is an algebra, X = 0 c E R.If R is a ring in X such that whenever {Ai} is a sequence in R then UAi E R then R is said to be a a-ring. If A is both an algebra and a a-ring then A is called a a-algebra.
PROPOSITION 8.2 If a is a collection of subsets of X then there is a smallest ring R containing a. Proof: The class of all subsets of X is clearly a ring containing a. Let R be the intersection of the set of all rings containing a. Clearly R must also be a ring. If M is another ring containing a then ReM since R is the intersection of all rings containing a. Therefore R is the smallest ring containing a. COROLLARY 8.1 If a is a collection of subsets of X then there is a smallest algebra containing a. PROPOSITION 8.3 If a is a collection of subsets of X then there is a smallest a-ring R containing a. Proof: As in Proposition 8.2 we can let R be the intersection of the family of all a-rings containing a since there exists at least one such a-ring. As in the previous proof, R is a ring. To show R is a a-ring let {A} be a sequence in R. This implies {Ai} is a sequence in each a-ring containing a which in turn
8.2 Measure Rings and Algebras
231
implies uA; is contained in each a-ring containing a so uA; E R. Therefore R is a a-ring. If M is any other a-ring containing a then ReM so R is the smallest a-ring containing a. COROLLARY 8.2 If a is a collection of subsets of X such that X then there is a smallest a-algebra A containing a.
E
a
Let (X,'t) be a topological space. Corollary 8.2 implies the existence of a smallest a-algebra B containing 'to The members of B are called the Borel sets of X. In particular, all open and all closed sets belong to B. Furthermore, all F (J 's (countable unions of closed sets) and all G ii'S (countable intersections of open sets) belong to B. Also. Proposition 8.3 implies the existence of a smallest a-ring B 0 containing all compact subsets of X. The members of B 0 are called the Baire sets of X. Let R be a ring and let Jl:R properties: Ml. Jl(0)
~
[0,00]. Let A, BE R. If Jl has the following
= 0,
M2. if A c B then Jl(A) ~ Jl(B), M3. if AnB = 0 then Jl(AuB) = Jl(A) + Jl(B), then Jl is said to be a measure on R and the triple (X, R, Jl) is a measure ring. The members of R are the measurable subsets of X. If R is also an algebra then (X, R, Jl) is a measure algebra. If R is a a-ring and Jl has the additional property M4. Jl(uA;) = LJl(A;) whenever {A;) is a sequence in R with A;nA j = 0 whenever i 'F j, then Jl is said to be a-additive. If R is a a-algebra and Jl is a-additive then (X. R, Jl) is called a measure space. Properties MI and M2 are easily derived from property M3 (Exercise I) but it is customary to list them when defining a measure, probably because properties MI and M2 together with an alteration of M4 are used in defining "outer measures." Outer measures will be introduced in Section 8.4. As will be seen later in this chapter, sets of measure zero are negligible when integrating. Therefore, it is reasonable to expect subsets of these "negligible sets" to be negligible. However, it may happen that for some A E R, Jl(A) = 0 but there exists aBc A with B not belonging to R and consequently B is not measurable. For such subsets we could define Jl(B) = 0, but would this extension of Jl to subsets of measure zero still be a measure on some ring in X? If Jl is a measure on a ring R in X, Jl is said to be complete if whenever A E R with Jl(A) = 0 and Be A then BE R.
232
8. Measure and Integration
THEOREM 8.1 If fl is a measure on a ring R in X and if R* is defined to be the family of subsets H ofXfor which there exists sets A, B E R with A c H c Band fl(B - A) = 0 then R* is also a ring in X. Define fl*(H) = fl(A) in this case and fl*(B) = fl(B) for BE R. Then fl* is a complete measure on R*.
This extended measure fl* is called the completion of fl. With this new measure, all subsets of measure zero with respect to the original measure fl are now measurable. Consequently, we can essentially assume all measures are complete for all practical purposes. Proof of Theorem 8.1.' If H, K E R* then there are sets A I, A 2, B \, B 2 E R such thatA I cHcB I ,A 2 cKcB 2 ,fl(B I -Ad=Oandfl(B2 -A 2 )=O. To show that H - K E R* first observe that A 1 - B 2 C H - K c B I - A 2. It needs to be demonstrated that fl«B I - A 2 ) - (A I - B 2» = O. For this it will first be shown that
Now y E (B I - A 2) - (A I - B 2) implies y E B I and y does not belong to either A 2 or (A I - B 2)' This yield two cases:
Case 1: (y does not belong to A d Suppose y does not belong to (B 2 A 2)nB I' This implies y does not belong to (B 2 - A 2)' But then y does not belong to B 2 since we already know y does not belong to A 2. A I does not contain y implies y E B I - A I. Since y does not belong to B 2 we have y E [(B I -Ad-B2]' Case 2: (yE A l nB 2). ThenYE [(B2 -A2)nBd. In either case we have y E [(B I - A d - B 2]u[(B 2 - A 2)nB d. Sincql is a measure we have
Consequently fl«B I - A 2) - (A I - B 2» = 0 so H - K E R*. It remains to show that HuK E R*. ButA I uA 2 c HuK c B I uB 2 and (B I uB 2 ) - (A I uA 2 ) c (B I - A I )u(B 2 - A 2) since if y does not belong to B I then y E B 2 and y does not belong to A 2 implies y E B 2 - A 2" Then
Thus fl«B I uB 2) - (A I uA 2» = 0 so HuK E R* which implies R* is a ring. Next we show that fl* is well defined on R*. If A I c H c B, and A2 c H B2 with fl(B I - AI) = 0 = fl(B 2 - A 2 ) then A 1- A2 CB2 -A2 so fl(A, -A 2) = O. Similarly fl(A 2 - AI) = O. Therefore fl(A 1) = fl(A 1nA 2) = fl(A 2)' Clearly fl* C
8.2 Measure Rings and Algebras
233
satisfies properties M1 through M3. Consequently 1..1* is a measure on R*. By the definition of J.!* and R*, J.!* is complete. • COROLLARY 8.3 If J.! is a measure on an algebra A in X and if A * is defined to be the family of subsets H of X for which there exists sets A, B E A with A c H c Band J.!(B - A) = 0 then A * is also an algebra in X. Define J.! *(H) :: J.!(A) in this case and J.!*(B):: J.!(B)for BE A. Then J.!* is a complete measure onA*.
It is not surprising that Theorem 8.1 and Corollary 8 .3 can be extended to a-rings and a-algebras respectively where J.! is a a-additive measure. This follows from the fact that if J.! is a a-additive measure on the a-ring R in X and if Hi E R* for each i:: 1,2,3 ... then it is possible to find Ai, Bi E R for each i with J.!(B i - Ai) = 0, such that Ai CHi C B i • Put A:: uA i , B = UBi and H:: uHi . Then A c H c B and since R is a a-ring, A, B E R. Since B-A c u{B i -A i li=1,2,3 ... } we have J.!(B - A) :: 0 if J.!(B i - A;) = 0 for each i since J.! is a-additive. This being the case, we conclude H E R* so that R* is a a-ring. We record this as: THEOREM 8.2 If J.! is a a-additive measure on a a-ring R in X and if R* is defined to be the family of subsets H of X for which there exists sets A, B E R with A c H c Band J.!(B - A) = 0 then R* is also a a-ring in X. Define J.!*(H) = J.!(A) in this case and J.!*(B):: J.!(B)for BE R. Then J.!* is a complete measure on R*. COROLLARY 8.4 If J.! is a a-additive measure on a a-algebra A in X and if A * is defined to be the family of subsets H of X for which there exists sets A, B E A with A c H c Band J.!(B - A) = 0 then A* is also a a-algebra in X. Define J.!*(H) = J.!(A) in this case and J.!*(B) = J.!(B) for B E A. Then J.!* is a complete measure on A *.
Let (X;t) be a topological space and let B denote the Borel sets of X with respect to 'to If J.! is a measure on B then J.! is said to be a Borel measure for (X,'t) or simply a Borel measure if't is understood. By Corollary 8.3, B can be extended to a a-algebra B* consisting of the family of subsets H of X for which there exist sets A, B E B with A c H c Band J.!(B - A) = O. Furthermore, /-1* defined on B* by J.!*(H) = J.!(A) is a complete measure on B*. J.!* is said to be a Lebesgue measure on B*. More generally, a Lebesgue measure is a complete measure on a a-algebra containing the Borel sets of X. Similarly, a measure mo on the a-ring B 0 of Baire sets is said to be a Baire measure for (X, 't) or simply a Baire measure if't is understood. Analogously, B 0 can be extended to a a-ring B~ consisting of subsets K of X for which there
234
8. Measure and Integration
are A, B E B 0 with A eKe B and J..lD(B - A) = O. Then!..l~ defined on B~ by !..l~(K) = J..lD(A) is a complete measure on B~. By a complete Borel or Baire measure is meant a complete measure on B* or B~ respectively. Let L be the collection of members of B that can be covered by countably many compact sets. It is left as an exercise (Exercise 2) to show that L is a a-ring in X, sometimes called the Lebesgue ring in X. This can sometimes lead to confusion since both members of L and B* are said to be Lebesgue measurable sets. In what follows, members of L will be referred to as Lebesgue measurable (L) whereas members of 8* will be referred to as Lebesgue measurable (B*). If neither L or B* is specified, the term Lebesgue measurable set will refer to a member of B*. If!..l is a a-additive measure on L, it is easily shown (Exercise 3) that !..l can be extended to a a-additive measure on B by setting W(A) = !..l(A) if A ELand !..l-(A) = otherwise. Then!..l- is a a-additive measure on the a-algebra B, and as such can be extended to a complete a-additive measure on B*. Therefore, in some sense it can be said that when dealing with Lebesgue measures it suffices only to consider the a-additive measures on L. However, this is not precisely true since there exist finite Lebesgue measures (i.e., !..l(A) < for each A E L) but the extension process discussed above always leads to an infinite measure (i.e., there exists some A E B* for which !..l(A) = 00) if L '# B. 00
00
EXERCISES 1. Show that the defining properties Ml and M2 of a measure can be derived from the property M3. 2. Let X be an uncountable set and let M be the collection of all sets E c X such that E is countable or X - E is countable. Define!..l on M by !..l(E) = 0 if E is countable and !..l(E) = 1 otherwise. Show that M is a a-algebra and that !..l is a measure on M. 3. It is easily shown that if !..l is a measure and A I and A 2 are measurable sets that !..l(A I uA 2) = !..l(A I) + !..l(A 2) - !leA I rIA 2) and if A 3 is also measurable then !..l(A I uA 2 uA 3) = !..l(A 1) + !..l(A 2) + !..l(A 3) - !..l(A I rIA 2) - !..l(A I rIA 3) - !..l(A 2rIA 3) + !..l(A I rIA 2 rIA 3)' Show that in general, if {An} where n = 1 ... N is a finite sequence of measurable sets that
In other words, !..l(u~=IAn) is an alternating finite series of terms such that the term is the sum of all the measures of distinct intersections of n members of
nth
{An}.
8.3 Properties of Measures
235
4. Let L be the Lebesgue ring in a topological space (X;!). Show that L is a a-ring in X.
5. If Jl is a a-additive measure on L show that Jl can be extended to a a-additive measure Jl- on B by setting Jl-(A) = Jl(A) if A ELand Jl-(A) = 00 otherwise.
8.3 Properties of Measures A set function Jl on a collection of sets S is said to be monotone if whenever A, B E S and A c B then Jl(A) ::;; Jl(B). The defining property M2 of a measure can be restated as: Jl is monotone on R. The defining property M3 can be restated as: Jl is additive on R. A set function Jl on a collection S is called subtractive if whenever A, B E S with A c B, B - A E S, and IJl(A) I < 00 then Jl(B - A) = Jl(B) - Jl(A). LEMMA 8.1 IfJl is a measure on a ring R then Jl is subtractive. Proof: If A, B E R with A c B, B - A E R and IJl(A) I < 00, then by M3, Jl(B) =
Jl(A) + Jl(B - A). Since IJl(A) I < 00, it may be subtracted from both sides of this equation which gives Jl(B - A) = Jl(B) - Jl(A).LEMMA 8.2 If Jl is a a-additive measure on a a-ring R and if A with {An} being a sequence in R covering A then Jl(A)::;; LnJl(An).
E
R
Proof: For each positive integer n put Bn = An - Ui
inequalities gives the desired result. PROPOSITION 8.4 IfJl is a measure on a ring Rand {E 1 ... EN} cR covers each point of E E R at least M times, then Jl(E) ::;; (L~=l Jl(En))/M.
= 1 then we must have M = 1 and the proof is trivial. Assume it is true for each positive integer n < N. If M = 1, the result follows from Lemma 8.2, so suppose M > 1. Let {E 1 ••• EN} c R cover each point of E E R at least M times. Put U = {E 1 ••• EN~l } and let V be the set of points of E that are covered at least M times by U. Let p E V and let V be a collection of M members of U that contain p. Then p E nV E R. Moreover n V c V. Clearly, V is the union of the collection of non-void intersections of M Proof: We use induction on N. If N
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8. Measure and Integration
members of U. Since this collection is finite and each of these intersections is measurable, so are V and E - V. By the induction hypothesis we have j..l(V) :s; L~~l j..l(Enf) V)/M and j..l(E - V) :s; L~~l j..l(En - V)/(M - 1). Since E - V C EN we have j..l(E - V):S; j..l(EN)' Then (M - 1)j..l(E - V):S; L~~l j..l(En - V) which implies M j..l(E - V) :s; L~~l j..l(En - V) + j..l(EN)' Therefore, j..l(E) = j..l(V) + j..l(E - V) :s; L~;;l j..l(Enf) V)/M + L~~l j..l(En - V)/M + j..l(EN)/M
=
L~~f j..l(En)/M + j..l(EN)/M = L~=Ij..l(En)/M.PROPOSITION 8.5 Ifj..l is a measure on a ring Rand {E I ... EN} c R is a collection of subsets of E E R such that no point of E is contained in more than M of the En' then j..l(E) ~ (L~=Ij..l(En))/M.
=1 the proof is trivial so assume the proposition holds for each positive integer n < N. If M = 1 the proof is trivial so assume M > 1. Then by the induction hypothesis, j..l(E) ~ L~~f j..l(En)/(M - 1) so (M - 1)j..l(E) ~ L~;;l j..l(En). Therefore, Mj..l(E) ~ L~;;f j..l(En) + j..l(E) ~ L~=l j..l(En) which implies j..l(E) ~ L~=Ij..l(En)/M.-
Proof: Again, we induct on N. If N
PROPOSITION 8.6 If j..l is a a-additive measure on a a-ring Rand {An} is an ascending sequence of sets (i.e., An C An+l for each n) then j..l(uAn) = limn->~j..l(An)'
Proof: Put A 0 = 0. Then j..l(uAn) = j..l(U;;'=1 (An - An-d) = L;;'=l j..l(An - An-d limN->~L~=Ij..l(An - An-d = limN->~j..l(U~=1 (An - An-d) = limN->~j..l(AN)'-
=
PROPOSITION 8.7 If j..l is a a-additive measure on a a-ring Rand {An} is a decreasing sequence (i.e., An+l C An for each n) of members of R of which at least one hasfmite measure then j..l(nA n) = limn->~j..l(An)' Proof: If for some positive integer m, j..l(Arn) < 00 then j..l(An) :s; j..l(Arn) < 00 for each n > m and hence j..l(nA n) < 00. By Lemma 8.1 j..l is subtractive so j..l(Arn) j..l(nA n) = j..l(Arn - nAn) = j..l(u(Am - An». Now the sequence {Arn - An} of members of R is increasing, so by Proposition 8.6 we have: j..l(u(Arn - An» =
Combining these two equations yields:
from which the desired result can be obtained. -
8.3 Properties of Measures
237
A set function J..l:C -7 [0,00] where C is a collection of sets is said to be continuous from below at a set A if for every increasing sequence (An) C C for which uAn = A we have limn ..... ~J..l(An) = J..l(A). Similarly, J..l is continuous from above at A if for each descending sequence I An} C C for which nAn = A and IJ..l(Arn) I < 00 for some positive integer m, we have limn .....~J..l(An) = J..l(A). Propositions 8.7 and 8.8 show that a-additive measures on a-rings are both continuous from above and below. The following proposition shows that the converse also holds under appropriate conditions.
PROPOSITION 8.8 If J..l is a finite, non-negative, additive set function on a a-ring Rand J..l is either continuous from below at each A E R or continuous from above at 0, then J..l is a a-additive measure on R. Proof: Since J..l is additive, property M3 of the definition of a measure holds. By Exercise I of Section 8.2, properties MI and M2 hold so it only remains to show that J..l is a-additive. For this, first observe that the additivity of J..l and the fact that R is a ring implies (by induction) that J..l is finitely additive; i.e., if AI' . An E R with AinA j = 0 for i "# j then J..l(u?=lA;) = L?=lJ..l(A i). Let (An) be a disjoint sequence of members of R and put A = uAn- For each positive integer n put Fn = u?=lA i and Gn = A - Fn. If J..l is continuous from below, then since (Fn) is increasing and A = uFn, by Proposition 8.7 we have:
If J..l is continuous from above at 0, then since (G n) is decreasing and nG n = 0 we have for each positive integer, J..l(A) = L?=lJ..l(A;) + J..l(G n). Taking the limit of both sides of this equation gives:
by Proposition 8.8. Consequently, in either case, J..l is a-additive which is the desired result. -
EXERCISES 1. If J..l is a a-additive measure on a a-ring R and if (An) is a sequence of sets in R, and if we define
show that J..l(lim infn .....~An) $ lim infn .....~J..l(An) and if J..l(ui'=nAi) is finite for at least one value of n, then J..l(lim supn .....~An) ::::>: lim supn .....~J..l(An). 2. Let R denote the real line with the interval topology and let B denote the
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family of all bounded half open intervals of the fonn [a, b) in R. Let R be the collection of all finite disjoint unions of members of B. Define!-l on B by !-l([a,b» = b - a and observe that !-l(0) = !-l([a, a» = a - a =O. (a) Show that if {A 1 ••• An} are disjoint members of B and each Ai is contained in some given set A E B then L?=l!-l(A i ) :s; !-leA). (b) Show that if a closed interval [a 0, b 0] is contained in the union of a finite number of bounded open intervals, say (ai, bi) for i = 1 ... n then b o - ao < L?=l (b i - ai). (c) Show that if {An} is a sequence of sets in B and A E B with A c Ui=lA i then !-leA) :s; Li=l!-l(A i ). (d) Show that R is a a-ring in R.
8. 4 Outer Measures A non-empty collection of sets H in a set X is said to be hereditary if whenever A E H and B c A then B E H. Hereditary collections share the property with rings, a-rings, algebras and a-algebras that the intersection of any set of hereditary collections is again a hereditary collection and consequently, for any collection S of sets there is a smallest hereditary collection Hs containing S. In what follows, we will be interested in those hereditary collections that are also a-rings and a-algebras. Hereditary a-rings are the class of sets upon which we will define outer measures. Outer measures, sometimes referred to a Caratheodory measures (after C. Caratheodory who introduced them in a paper titled Ueber das lineare Mass von Punktmengen - eine Verallgemeinerung des Langenbegriffs (Nachr. Ges. Wiss. Gottingen) in 1914, are an important generalization of measures. Their importance arises from a standard technique employed when attempting to prove a set function is a measure. This technique is to first prove the function is an outer measure on some hereditary a-ring and then use a theorem from the theory of outer measures (to be demonstrated below) to conclude that the function is a measure on a suitably restricted subcollection that is a a-ring. If E is any collection of sets, then H(E) will denote the smallest hereditary L-ring containing E and H(E) will be referred to as the hereditary a-ring generated by E. H(E) is the collection of all sets that can be covered by countably many members of E. That H(E) is indeed a hereditary a-ring is left as an exercise (Exercise 1). Let R be a hereditary a-ring and let !-l*:R ~ [0,00]. If Ii* has the property
M4'. !-l*(UAi) :s; L!-l*(Ai) whenever {Ai} is a sequence in R then !-l* is said to be countably subadditive. If!-l* has the property that whenever A, B E R then 1-1*(AuB) :s; 1-1*(A) + 1-1*(B) then 1-1* is said to be
8.4 Outer Measures
239
subadditive. If!-l* has properties Ml and M2 that define a measure plus property M4' then !-l* is called an outer measure. Measures on rings can be extended to outer measures on the hereditary a-rings that these rings generate.
THEOREM 8.3 If!-l is a a-additive measure on a a-ring R and if for each A E H(R) we define !-l*(A) = inf{L!-l(An) I {An} is a sequence in R covering A} then !-l * is an outer measure on H( R) such that !-l*(A) = !-l(A) for each A E R. Proof: Let A E R and put A J = A and An = 0 for each positive integer n > I. Then {An} cRandAcuA n· By the definition of!-l*, !-l*(A):<:; L!-l(An) =!-l(A) + Ln>l!-l(A n) = !-leA) + O. Conversely, if {An} is a sequence in R covering A, then by Proposition 8.5 !-leA) :<:; L!-l(An) so by the definition of!-l*, !-leA) :<:; !-l*(A). Consequently !-l*(A) = !-leA) for each A E R. Since 0 E R this shows that !-l*(0) = 0 so !-l* satisfies property MI of a measure. To show !-l* satisfies property M2, let A, B E H(R) with A c B. Let {Bn} be a sequence in R covering B. Then {Bn} also covers A so !-l*(A) :<:; L!-l(Bn). But since this is the case for all sequences {Bn} in R covering B, from the definition of!-l* we have !-l*(A):<:; !-l*(B). SO!-l* satisfies M2. To show !-l* satisfies M4' let A E H(R) and let iAn} be a sequence of members of H(R) covering A. Let £ > O. By the definition of !-l*, it is possible to choose a sequence {A;:'} in R covering An such that:
Since {A;:' I n = 1, 2, 3 ... and m = 1, 2, 3 ... } is a countable collection of members of R covering A it follows that:
Since £ was chosen arbitrarily, this shows !-l* satisfies M4'. Thus!-l* is an outer measure on H(R).The outer measure !-l* defined in Theorem 8.3 is said to be an extension of !-l to H(R). Not only is the question of extending measures to outer measures of interest, but of restricting outer measures to smaller classes of sets on which they become measures. Let!-l* be an outer measure on a hereditary a-ring H. A set A E H will be called !-l*-measurable if for every pair of sets P, Q E H with PeA and Q c X - A we have: !-l*(PuQ) = !-l*(P) + !-l*(Q). The concept of !-l*-measurability is of fundamental importance in the theory of outer measures. This is because the !-l*-measurable sets constitute a a-ring in H and !-l* restricted to these sets is a a-additive measure on them. It is often
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difficult to gain an intuitive understanding of the nature of these sets. Outer measures can fail to be measures because they do not satisfy property M3. The definition of Jl*-measurable sets focuses on those sets that in some sense force additivity on all other pairs of sets in H that they "separate." An equivalent way of defining Jl*-measurable sets is to define them to be the sets A E H such that if B E H then Jl*(B) = Jl*(BnA) + Jl*(B - A). This definition may shed some additional light on the nature of these sets. It characterizes them as those members of H that, no matter how they "split" another member B of H, Jl* is additive on the two subsets into which B was split. In view of property M4', this last definition of Jl*-measurable sets can be replaced by Jl*(B)
~
Jl*(BnA) + Jl*(B - A)
in the defining relation. We will denote the class of all Jl*-measurable sets by
M. PROPOSITION 8.9 IfJl* is an outer measure on a hereditary (J-ring H and if M is the class of all Jl*-measurable sets then M is a ring. Proof: If A, B E M and C E H then we have the following three equations: (8.1) (8.2) (8.3)
Jl*(C) = Jl*(CnA) + Jl*(C - A). Jl*(CnA) = Jl*(CnAnB) + Jl*«CnA) - B). Jl(C - A) = Jl*«C - A)nB) + Jl*«C - A) - B).
Substituting (8.2) and (8.3) into (8.1) gives: (8.4)
Jl*(C)
= Jl*(CnAnB) + Jl*«CnA) - B) + Jl*«C - A)nB) + Jl*«C - A) - B).
If we replace C by Cn(AuB) in equation (8.4), the first three terms on the right remain unchanged while the fourth term simply drops out. Therefore, we have
(8.5)
Jl*(Cn(AuB»
= Jl*(CnAnB) + Jl*«CnA) - B) + Jl*«C - A)nB).
Now substituting (8.5) into (8.4) and using the fact that (C - A) - B = C - (AuB) we get: Jl*(C)
= Jl*(Cn(AuB»
+ Jl*(C - (AuB»
which shows that AuB E M. To show A - BE M, we replace C by Cn(X - [AB]) in equation (d) and notice that the second term on the right goes to zero while the other three terms remain unchanged. This gives
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241
(8.6) f.l*(Cn(X - [A - B])
= f.l*(CnAnB) + 1l*«C - A)nB) + f.l*«C - A) - B).
Substituting (8.6) into (8.4) and using the fact that Cn(X - [A - BD we get: f.l*(C)
= f.l*(Cn(A - B»
=C - (A - B)
+ f.l*(C - (A - B»
which shows A - B E M. Consequently M is a ring in H. PROPOSITION 8.10 Iff.l* is an outer measure on a hereditary a-ring H and if M is the class of all f.l*-measurable sets in H. then ifB E H and {An} is a disjoint sequence of sets in M with A = uA n, then f.l*(BnA) = Lnf.l*(BnA n). Proof: Replacing C by B in (8.5) in the proof of Proposition 8.10 and replacing A and B by A 1 and A 2 we get: f.l*(Bn(A 1uA 2» = f.l*(BnA 1nA 2) + f.l*«BnA 1) - A 2) + f.l*«B - A l)nA 2)'
Since AlnA2 = 0, this equation reduces to f.l*(Bn(A 1 uA 2 » = f.l*(BnAd + f.l*(BnA 2)' By induction, we can extend this result to (8.7)
for each positive integer n. Next put Hn = U7=lA i for each n. By Proposition 8.10, M is a ring so Hn E M for each n. Consequently f.l*(B) = f.l*(BnH n) + f.l*(B - Hn). Substituting (8.7) into this equation gives f.l*(B) = L7=1f.l*(BnA i ) + f.l*(B - Hn). Since Hn c A for each n, f.l*(B - A) :<;:; f.l*(B - Hn) for each n. Thus f.l*(B) ~ L?=lf.l*(BnAJ + f.l(B - A) for each n. Hence f.l*(B) ~ Lf.l(BnA n) + f.l(B-A). Since f.l* is an outer measure, f.l*(BnA) = f.l*(Bn[uAnD :<;:; Lf.l*(BnA n). Thus (8.8)
f.l*(B)
~ Lf.l*(BnA n)
+ f.l*(B - A)
~
f.l*(BnA) + f.l*(B - A).
But then A E M by the alternative definition of f.l*-measurability, and therefore f.l*(B) = f.l*(BnA) + f.l*(B - A). But then by (8.8) we have (8.9)
Lf.l*(BnA n) + f.l*(B - A) = f.l*(BnA) + f.l*(B - A).
Since we do not know if f.l*(B - A) is finite we cannot merely subtract f.l*(B - A) from each side of (8.9) to get the desired result. Instead we notice that (8.9) holds for any B E H so we can substitute BnA for B in (8.9). This causes the second term on each side of the equation to vanish, giving the desired result. -
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THEOREM 8.4 If~* is an outer measure on a hereditary a-ring Hand is the set of all ~*-measurable sets in H, then M is a a-ring, every set of outer measure zero belongs to M and the set function ~ defined on M by ~(A) = ~*(A) is a complete a-additive measure on M.
if M
Proof: Let {Mn) be a sequence in M and let BE H. Let A I = M I and let Bn = u?=IM i . Then put An = Mn - B n_l • Since M is a ring (Proposition 8.9), each An E M. Clearly {An) is a disjoint sequence in M. For each positive integer n we
want to show (8.10) Let K be the set of positive integers for which (8.10) holds. Clearly 1 E K. Assume n E K. Since An+1 E M we have ~*(B) = ~*(BnAn+d + ~*(B - An+d. Since An+1 nBn = 0 and B - An+1 E H we have:
Now by Proposition 8.10, Therefore
~*(BnBn)
= ~*(Bn[u?=IAi]) = L?=I~*(BnAi)'
Hence n+ 1 E K so K is the set of all positive integers. Next observe that for each positive integer n, ~*(B - A) ~ ~*(B - Bn) where A = uAn = uMw Thus (8.11 ) holds for each positive integer n. Taking the limit of both sides of (8.11) as n --+ 00 gives 11 *(B) ~ 1: ~rr=lll *(BnA n ) + 11 *(B - A). Again, by Proposition 8.10, ~*(BnA) = L:=I~*(BnAn) so ~*(B);::: ~*(BnA) + ~*(B - A). But then ~*(B) = ~*(BnA) + ~*(B - A) WhICh shows A E M. Consequently, M is a a-ring. To show ~ is a-additive let {An} be a disjoint sequence in M and let A = uA n • Then substituting A for B in equation (8.9) we get:
so L~(An) = ~(A) or in other words L~(An) = ~(A) so ~ is a-additive. Next let ME H such that ~*(M) = O. Then for each B E H we have ~*(B) = ~*(M) + ~*(B) ~ ~*(BnM) + /-l*(B - M)
8.5 Measurable Functions
243
so M E M. Hence every set of outer measure zero belongs to M. Finally, to show that Il is complete let A E M with 1l(A) = 0 and let B cA. Then 1l*(A) = Il(A) = 0 and hence 1l*(B) = O. But then B E M so Il(B) = O. Hence Il is complete. -
EXERCISE 1. Let E be a collection of subsets of a set X. Show that there exists a smallest hereditary a-ring H (E) containing E and that H (E) consists of the collection of all sets that can be covered by countably many members of E.
8.5 Measurable Functions Let X be a set and M a a-algebra on the set X. Then X is said to be a measurable space and the members of M are called measurable sets. If X is a measurable space and Y is a topological space, a function f:X ~ Y is said to be a measurable function if rl(U) E M whenever U is open in Y. The proof of the following proposition is left as an exercise. PROPOSITION 8.11 If f:X ~ Y is a measurable function from a measurable space X into a topological space Y and g:Y ~ Z is continuous where Z is a topological space, then the composition function g © f:X ~ Z is measurable. THEOREM 8.5 If u and v are real measurable functions on a measurable space X and g is a continuous mapping of the plane into a topological space Y then h:X ~ Y defined by h(x) = g(u(x), v(x)) for each x E X is measurable. E X put f(x) = (u(x), vex»~. Then f maps X into the plane. Since g is continuous and h = g © f, by Proposition 8.11, it suffices to demonstrate the measurability of f. For this let I x J be an open rectangle in the plane where I and J are open intervals in R. Then r l (I x J) = u- I (I)rw- I (J) is measurable since both u and v are measurable functions. Every open set U in the plane is a union of these basic open sets I x J of the product topology. Since the plane is separable, U can be constructed from countably many of these basic rectanglesB n • Sincerl(U) =rl(U";=lB n ) = u";=d-1(B n ) we see thatrl(U) is measurable. Hence f is a measurable function. -
Proof: For each x
PROPOSITION 8.12 If X is a measurable space and u and v are real valued measurable functions on X then/:X ~ C defined by f(x) = u(x) + iv(x)/or each x E X is a complex valued measurable function on X.
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Proof: The proof follows from Theorem 8.5 with g(z) = z for each complex number z.PROPOSITION 8.13 If f:X where j(xj = u(x) + iv(x) for each x measurable functions on X.
C is a complex measurable function X then u,v and ifl are all real valued
~
E
Proof: The proof follows from Proposition 8.11 with g(z) I z I respectively for the functions u, v and IfI. -
= Re(z),
Im(z) and
PROPOSITION 8.14 If f and g are complex valued measurable functions on a measurable space X, then so are f + g andfg. Proof: If u and v are real valued measurable functions, and g(u, v) = u + v, then by Theorem 8.5, hex) = g(u(x), vex»~ = u(x) + vex) = (u + v)(x) is measurable. Since f and g are complex measurable functions, f = u I + iv I and g = u 2 + iv 2 for some real valued measurable functions u I,V I and U2,V2' Then f + g = (UI+U2) + i(v) + v2)' As shown above, UI + U2 and VI + V2 are real valued measurable functions, so by Proposition 8.12, f + g is a complex valued measurable function. The proof that fg is measurable follows from the same argument with g(u(x), vex»~ = u(x)v(x).PROPOSITION 8.15 If E is a measurable set in X, then the characteristic function XE of E, defined by Xdx) = 1 if x E E and XE(X) = 0 otherwise, is a measurable function.
r
Proof: If U is an open set in R containing 1 but not 0 then l (U) = E. If U contains 0 but not 1 then l (U) = X-E. If U contains both 1 and 0 then l (U) = X and if U contains neither then 1 (U) = 0. In all of these cases, 1 (U) is a measurable set so XE is a measurable function. -
r
r
r
r
PROPOSITION 8.16 Iff is a complex measurable function on X, there is a complex measurable function a on X such that Ia I = 1 andf = a IfI. Proof: Let Z = {x If(x) = O}. Then Z is measurable since fis measurable. Put Y = C - (O}. Define g:Y ~ C by g(y) = yllyl for each y E Y. Then for each x E X put a(x) = g(j(x) + Xz(x». If x E Z, a(x) = 1. If x does not belong to Z then a(x) = f(x)/ If(x) I, In either case, Ia I = 1. Since g is continuous on Y and Z is measurable, a is measurable on X. Finally, if x E Z then f(x) = 0 = IxO = a If(x) I and if x does not belong to Z then a If(x) I = fJ(x)/ If(x) Illf(x) I = f(x). Let X be a space and B the Borel a-algebra on X. If f:X ~ Y is a continuous function where Y is a topological space, then clearly l (U) is a Borel set whenever U is open in Y. Therefore, every continuous mapping of X
r
245
8.5 Measurable Functions
into any topological space is Borel measurable. If Y is R or C, the Borel measurable functions will simply be called Borel functions. THEOREM 8.6 Let M be a a-algebra in X and letf:X -7 Y where Y is a topological space (l)lf A = {E c ylr l (E) E M} then A is a a-algebra in Y. i (E) E M. (2) Iff is measurable and E is a Borel set in Y then (3) IfY = [-00,00] andr l ((a. 00]) E Mfor each real a, thenfis measurable.
r
Proof: If A, BE A thenrl(B - A) =rl(B) - rl(A). Sinceri(A) andrl(B) E M we have r l (B) - r l (A) E M so r l (B - A) E M which implies B - A E A. If {An} is a disjoint collection of members of A, then r l (uAn) = uri (An) E M since r 1 (An) E M for each nand M is a-additive. Therefore, uAn E A so A is a a-ring. But X = 1 (Y) and X E M since M is an algebra. Therefore, YEA so A is a a-algebra. This proves (1).
r
To prove (2), let A be the a-ring defined in (1). If f is measurable, then A contains all open sets in Y since by the definition of measurability, r 1 (U) E M for each open U c Y and therefore U E A. But then A contains all the Borel sets so r 1 (E) E M by the definition of A. To prove (3), let A be the a-algebra defined in (1). Since A is a a-algebra in [-00,00] and (a, 00] E A for each a E R, then [-00, a) E A since [_00, a)
= u[-oo, a - lin] = u(a - Iln,oo]e
where (a - lin, oor is the complement of (a - lin, 00] in [-00,00]. Consequently, (a, b) = [-00, b)n(a, 00] E A. Since every open set in [-00,00] is a countable union of open intervals (a, b), every open set in [-00, 00] is in A. Therefore,f is measurable. Let {xn I be a sequence in [-00,00] and put Sk = SUP{Xko Xk+l, Xk+2 ... } for each positive integer k. Then put a = mfl Sk I. We say a is the limit superior of {xn I and denote a by lim sup Xn • The limit inferior is defined similarly: for each positive integer k let i k = inflxko Xk+l, Xk+2 ... } and put lim inf Xn = sup{id. Clearly, liminfxn
= -lim sup (-xn).
If {xn} converges to x it is easily shown that
If for each positive integer n,fn:X lim sup fn can be defined on X by:
-7
[-00, 00], then the functions sup fn and
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246
(sup fn)(x) = sup (fn(x» and [lim sup fn](x) = lim sup (fn(x». If the limit f(x) = limn-4~fn (x) exists (i.e., if (fn(x)) converges) for each x then we say that f is the point-wise limit of {fn }.
E
X,
PROPOSITION 8.17 If {fn} is a sequence of measurable functions from (X, M) into [-00,00] where M is a a-algebra, then the functions g = sup fn and h = lim sup fn are both measurable. Proof: By Theorem 8.6.(3), it suffices to show that if a E R, then g-I«a, 00]) and h- 1«a, 00]) are measurable. For this, we first show that g-I«a, 00]) = u]"I«a,oo]). If x E g-I«a, 00]), then g(x) E (a, 00]' so g(x) > a. Therefore, there exists a positive integer k such that a < hex) ~ g(x). Hence hex) E (a, 00] which implies x E .fkl «a, 00]). Therefore, g -I «a, 00]) c U]"I «a, 00]). Conversely, if x E u]"I«a, 00]) then x E .fkl«a, 00]) for some positive integer k which implies hex) E (a, 00] which in tum implies fk(x) > a. Since g(x) = sup{fn(x)}, we have a < hex) ~ g(x) which implies g(x) E (a, 00]. But then x E g-I«a, 00]) so u]"I«a, 00]) c g-I«a, 00]). Hence g-I«a, 00]) = U],,1 «a,oo]). Since each fn is measurable,],,1 «a, 00]) is measurable. Since Mis a a-algebra, u]"I«a, 00]) E M so g-I«a, 00]) is measurable. Therefore, g is a measurable function. A similar argument holds with infreplacing sup. Now since h = lim sup fn' for each x E X, hex) = inA~1 {suPn;::dfn(x)}}. For each positive integer k put
COROLLARY 8.5 The limit of every point-wise convergent sequence of complex measurable functions is measurable. Proof: The proof follows immediately from Proposition 8.17 for the real valued case. Proposition 8.12 can then be used to establish the complex case.COROLLARY 8.6 Iff and g are measurable functions from a space X into [-00,00], then so are max{f, g} and min{f, g).
r r -r·
A special case of Corollary 8.6 is the function = max{f, O}. Another is = -min {f, O}. is called the positive part of f while is called the negative part of f. Clearly If I = + rand f =
r
r
r
r
A function s on a measurable space (X, M) whose range is a finite subset of [0, 00) is called a simple function. If {a 1 ... an} is the range of the simple function s and if Ei = (x
E
X Isex) = ai), then:
8.5 Measurable Functions
247
where XE, is the characteristic function of E; for each i = 1 ... n. Clearly, S is measurable if and only if E; is measurable for each i. In this case, S is called a simple measurable function. THEOREM 8.7 Iff:X -7 [0,00] is a measurable function , there exists a sequence {sn} of simple measurable functions on X such that (l)05, s l 5, s 2 5, .. . f, (2) {sn(x)) converges to f(x) for each x E X. Proof: For each positive integer n and for each i E [1, n2n], put En, = r1([an"b n where an, = (i - 1)/2n and b n, = i/2 n and let Fn = 1 ([n, 00]). Then, for each positive integer n, define Sn by
,»
r
By Theorem 8.6.(2), the En; and Fn are measurable sets in X. Therefore, the Sn are simple measurable functions. Let m, n be positive integers such that m < n and let x E X. To prove (1), there are three cases to consider: Case I: (l(x) sn(x).
E
Fn) Since Fn c Fm,f(x)
E
Fm- Therefore, sm(x) = m < n =
Case 2: (l(x) E Fm butf(x) does not belong to Fn) Then sm(x) = m. Let k be the least positive integer such that x E E nk . Then k > m 2 n which implies a nk > (m2n - 1)/2n = m - l/2 n which in turn implies a nk ~ m. Therefore, sn(x) ~ sm(x). Case 3: (l(x) does not belong to F m) Let k be the least positive integer such that x E E nk . Then x E Emj where j = k mod (2n-m). Hence
Thus sm(x) 5, sn(x). In all of these cases, sm(x) 5, sn(x). To show that for each n, Sn 5, f, first note that if f(x) = 00 for some x E X that sn(x) 5, f(x) for each n. Therefore, suppose x E X and f(x) < 00. Let k be the positive integer such that f(x) E [k, k + I). Then Sk(X) = k 5,f(x). For each positive integer n > k, there are 2n subintervals [(k
+ i - l)/2 n , (k + i)/2n]
where i = I .. . 2n that partition [k, k + I). Let i be the positive integer such that
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248
f(x)
E
[(k + i -1)J2n, (k + i)/2n]. Then sn(x) = (k + i - 1)/2n ~f(x). Consequently, each positive integer n. This proves (1).
Sn ~ffor
To prove (2), first note that if f(x) = 00 then sn(x) = n for each positive integer n so {sn(x)} converges to f(x). Therefore, assume that f(x) < 00. As shown above, if k is the positive integer such that f(x) E [k, k + 1), then for each positive integer n > k, there are 2n subintervals [(k + i - 1)/2n, (k + i)/2n] that partition [k, k + 1) andf(x) belongs to one of them, say [(k + j - 1)J2n, (k + j)/2nj and sn(x) = (k + j - l)/2 n. Hence If(x) - sn(x) I < 2- n. Consequently, the sequence (f(x) - sn(x)} is Cauchy and therefore converges to O. Therefore, (sn(x)} converges to f(x). This proves (2).COROLLARY 8.7 Sums and products of measurable functions in£O [0.00] are measurable.
Proof: Letfand g be measurable functions from X into [0,00]. By Theorem 8.7, there exists sequences {fn} and {g n} of simple measurable functions such that 0 ~ fl ~ h ~ ... ~ f, 0 ~ g I ~ g2 ~ ... ~ g, (fn(x)} converges to f(x) and (gn(x)} converges to g(x) for each x E X. It is an easy exercise (Exercise 1) to show that sums and products of simple measurable functions are simple measurable functions. Consequently, {fn + gn} and {fngn} are sequences of simple measurable functions. We can also show (Exercise 2) that ([fn + gn](x)} converges to [f + g](x) and ([fngn](x)} converges to [fg](x) for each x E X. Hence f + g andfg are measurable.-
EXERCISES
I. Show that sums and products of simple measurable functions into [0, 00] are simple measurable functions. 2. Show that if {an} and {b n } are sequences in [0,00] such that 0 ~ a I ~ a 2 ~ •• . , 0 ~ b 1 ~ b 2 ~ , . . , {an} converges to a and {b n } converges to b, then I an + b n } converges to a + b and {anb n } converges to abo 3. Show that the set of points at which a sequence of measurable real valued functions converges is a measurable set. 4. Show that if f is a real valued function on X such that Ix measurable for each rational number r, then f is measurable.
E
X If(x) ~ r} is
5. Let IAn} be a sequence of measurable sets in a a-algebra M and let J..l be a measure on M. Define the limit inferior denoted lim infAn of the sequence
249
8.6 The Lebesgue Integral
(An 1 by lim infAn = U;;'=1 (nk=nAn) and define the limit superior denoted lim supAn by n;;'=1 (uk=nAn)' Show that (a) Il(iim injAn) ::::: lim injll(An) and (b) if Il(U;;'=1 An) < 00 then lim sUPIl(A n) ::::: Il(lim supAn).
8.6 The Lebesgue Integral In the theory of integration we often encounter the concept of infinity and the symbols 00 and -00. We have already defined a measure 11 to be a set function on a set X into [0, 00 L In order not to have to make special provisions for dealing with these concepts and symbols in some of the following theorems, we define addition (+) and multiplication (x) on [-00,00) as follows:
a + 00 = 00 + a = 00 for each a such that -00 < a, a - 00 = -00 + a = -00 for each a such that a < 00, a x 00 = 00 x a for each a such that < a, x 00 = 00 x = 0, a x 00 = 00 x a = -00 for each a such that a < 0.
°
°
°
With these definitions, it can be shown that the commutative. associative and distributive laws hold for [0, 00). Since -00 + 00 and (-00) x (00) are not defined, we cannot extend these laws to [-00,00], but fortunately, we will not need to. The cancellation laws also hold in [0,00) with the following modifications:
a + b = a + c implies b = c if a < 00, ab = ac implies b = c whenever < a < 00.
°
If s is a simple measurable function on X where (a 1 ••• an 1 c [0, 00) is the range of s and for each i = 1 ... n, Ei = (x E Xls(x) = ad, and if M is a a-algebra on X and 11 is a measure on M, then for E E M. we define the Lebesgue integral of s with respect to 11 as:
IEsdll = L?=1 aill(EnEi). If j:X ~ [0,00) is a measurable function, we define the Lebesgue integral of j with respect to 11 to be the supremum of all simple measurable functions s such that 0::::: s :::::j, i.e.,
I Efdll = sup {IEsdlll
°: :
s ::::: j and s is a simple measurable function I.
Clearly, fEfdll E [0,00] and the two definitions of Lebesgue integral given above for the case where j is a simple measurable function are equivalent. The Lebesgue integral behaves in the same manner as the Riemann integral, as the following theorem shows.
8. Measure and Integration
250
THEOREM 8.8 Let (X, M) be a measurable space and let E. F E M. Let j.! be a measure on M andfand g be measurablefunctionsfromX into [0,00]. Then (1) 1fO ~f~ g then Jtfdj.! ~ fEgdj.!. (2) If E c F then Jtfdj.! ~ JF/dj.!. (3)
If c E [0, (0) then JEcfdj.! = cftfdj.!.
(4)
Ifj(x) = 0 for each x
E
E then Jtfd!J. = O.
= 0 then Jtfd!J. = O. (6) Jtfd!J. = JxXtfdj.!. (5)
Ifj.!( E)
The proof of Theorem 8.8 is straightforward and is left as an exercise (Exercise I). The next proposition reveals an interesting property about certain integrals, namely, that they are also measures. PROPOSITION 8.18 Let M be a a-algebra on X, j.! a measure on M and s a simple measurable function on X. For each E E M put A(E) = JEsdj.!. Then A is a measure on M. Proof: Let the range of s be {a 1 . . . ak} and for each i = 1 ... k let Ai = {x E X Isex) = ai}' Suppose {En} is a sequence of disjoint members of M such that E =uEn • Then since j.! is a-additive we have:
Consequently, A is also a-additive so property M3 of a measure is satisfied. Clearly A(0) = 0 so MI is satisfied. By Theorem 8.8, M2 is satisfied. Therefore, A is a measure on M. PROPOSITION 8.19 If M is a a-algebra on X, and sand t are simple measurable functions on X, then: fx(s + t)dj.!
j.!
is a measure on M
= fxsdj.! = fxtdj.!.
Proof: Let the range of s be {ai ... am} and the range of t be {b 1 ••• bk }. For each i = I ... m and each j = I ... k put Ai = {x E X Isex) = ad and Bj = {x E X It(x) = bj }. Now for each pair i, j put Eij = AinBj • This yields fEi/s + t)dj.!
= (ai + b)j.!(Ei) = aij.!(Ei) + b/E;) = fEi/dj.! + fEij tdj.!.
Thus the conclusion of this proposition holds for each Eij in place of X. Then by Proposition 8.18, we have:
251
8.6 The Lebesgue Integral
fx(s + t)dJ.!
= Lrj~JE'J (s + t)dJ.! = Lrj~dfE'J sdj.l + fE,/dj.l] =
L7'J,~JE
" )
sdj.l + L7'J,~JE tdJ.! " )
= fxsdj.l + fxtdj.l. •
The great success of Lebesgue's definition of the integral is largely due to the ease of passing to the limit of certain sequences of measurable functions. One example of this is the following celebrated theorem.
THEOREM 8.9 (H. Lebesgue, 1904) If M is a a-algebra on X, Il a measure on M and {fn} a sequence of measurable functions on X such that 0 :,; Ji(x) :,; /j(x) :,; 00 for each pair i, j with i < j and for each x E X, then if {fn(x)} converges to f(x) for each x E X,f is measurable and
limn~JxfndJ.!
= fxfdj.l.
Proof: By Theorem 8.8(1), fxfndJ.! :,; fxfn+!dj.l for each positive integer n, so there exists an a E [0,00] such that lfxfndj.l} converges to a. By Proposition 8.17, f is measurable. Since for each x E X, {fn (x)} is a non-decreasing sequence, we have fn(x) :,; f(x) for each positive integer n. Therefore, by Theorem 8.8(1), fxfndj.l :,; fxfdJ.! for each positive integer n. Consequently, a :,; fx/dj.l because the convergence of dxfndj.l} to a implies a is the supremum of lfxfndj.l} since lfxfndj.l} is non-decreasing. Let s be a simple measurable function such that 0 :,; s :,; f and let c E (0,1). For each positive integer n put En = {x E X Ifn(x) ~ cs(x)}. Clearly each En is measurable and En C En+! for each n. Also, X = uEn, for if f(x) = 0 then x E E! whereas if f(x) > 0, cs(x) < f(x) since c < 1, and since {fn(x)} converges to f(x), there exists a positive integer k such that cs(x) < hex). Therefore, x E Ek • Then
for each positive integer n. By Proposition 8.18, A(E) = fEcsdj.l is a measure on M. Since {En} is an ascending sequence with uEn = X, by Proposition 8.7, {A(En)} converges to A(X). Then, since lfxfndJ.!} converges to a, lfEncSdJ.!} converges to fxcsdj.l, and for each n fEncsdj.l:'; fxfndj.l, we have dxsdj.l = fxcsdj.l :,; a. Since this is true for each c < 1, it is clear that fxsdj.l :,; a. Hence, for each simple measurable function s with 0 :,; s :,; f,
ixsdll :,; a :,; fxfdj.l. By definition, fx/dj.l is the supremum of all such simple measurable functions, we have a = Jxfdj.l. Therefore, Ifxfndj.l} converges to Jxfdj.l. •
252
8. Measure and Integration
In what follows, some of the sequences of numhers to which we will have occasion to refer will have rather complex representations such as the sequence of integrals {fxfnd/-ll in the proof of the preceding theorem. Rather than stating that this sequence converges to the numher fxfd/-l (we include DO as a numher) we will adopt a slightly simpler notation by writing
THEOREM 8.10 If Un} is a sequence of measurable functions from X
into [0. DO] andj(x)
= 'Ln!n(x)jor each x E
X, then fxfd/-l
= 'Lnfxfnd/-l.
Proof: By Theorem 8.7, there exist sequences {sn} and I (n} of simple measurable functions such that 0 :<; S I :<;; S2 :<; ... :<; fl' 0 :<;; (1 :<; t2 :<; ... :<; 12, Sn(X) --7 fl (x) and tn(x) --7h(x). For each positive integer n put un(x) = sn(x) + (n(x) for each x E X. It is easily shown that I un} is a sequence of simple measurable functions that converges to (f I + h). By Proposition 8.19, fx(Sn + (n)d!J. = fxSnd/-l + fx(nd/-l
for each positive integer n, and by Theorem 8.9, fxsnd/-l
--7
fXfl d/-l and fx(nd/-l
--7
fxhd/-l.
Consequently, by Execise 2 of Section 8.5, fxund/-l
= fx(sn + (n)d/-l = fxsnd!J. + fx(nd/-l
--7
fxfl d/-l + fxhd/-l.
But it is also easily shown that 0 :<; U 1 :<; U2 :<; ... :<; (fl + h), so another application of Theorem 8.9 shows that (fl T h) is a measurahle function and fxUnd/-l
--7
fx(fl + h)d!J.·
Since limits are unique, this implies
This result can be extended by an induction argument to show that for each positive integer k we have 'L~=dn is a measurable function and
fx('L~=dn)d!J.
=
'L~=1 (fxfnd/-l)'
For each positive integer k put gk = 'L~=dn' Then {gn} is a sequence of measurable functions such that 0:<;; g 1 :<;; g2 :<;; ... :<;; 'Ln!n and gn(x) --7 'LJn(x) for each x E X by the definition of a series 'LJn(x). By yet another application of Theorem 8.9, 'LJn is a measurable function and
8.6 The Lebesgue Integral
253
But then the partial sums ~i=l (fxlid/.!) ---7 fx(~,Jn)dJ..l. By the definition of a series, the partial sums ~i=l (fxlid/.!) ---7 ~n(fxfndJ..l). Consequently, ~n(Jxfnd/.!) = fx(~,Jn)d/.!. -
THEOREM 8.11 integer n. then
If fn:X
---7
[0. 00] is measurable for each positive
Proof: For each positive integer k put gk(X) = inf{fn(x) Ik ~ n} for each x E X. Then for each positive integer n, gn is a measurable function by Proposition 8.17 and gn ~ fn' so by Theorem 8.8(1), fxgnd/.! ~ fxfndJ..l. Then by the definition of limit inferior we have:
and for each x E X, lim inf fn(x) = sup {gn(x) }. Therefore, 0 ~ g 1 ~ g2 ~ ... ~ lim inf fn' so by Theorem 8.9, fxgndJ..l ---7 fx(lim inf fn)dJ..l. Since the sequence {ixgndJ..l} converges, it converges to its limit inferior, i.e., fxgndJ..l ---7 lim inf fxgndJ..l. Therefore, fx(lim inf fn)dJ..l = lim inf fxgndJ..l ~ fxfndJ..l.Theorem 8.11 is known as Fatou's Lemma. It is possible that the inequality in Theorem 8.11 is strict, i.e., that
Jx(lim inf fn)d/.! < lim inf fxfndJ..l. To see this let X = [0, 1], let g(x) = 0 if x ~ 1/2 and 1 otherwise, let hex) = 0 if x > 1/2 and 1 otherwise, and putfn = g if n is even andfn = h if n is odd. Clearly lim inf fn = 0 so fx(lim inf fn)dJ..l = O. If J..l(A) > 0 for each A c X with nonempty interior, then both fxgdJ..l > 0 and fxhd/.! > 0 which implies lim inf fxfndJ..l > O. Hence strict inequality holds. If we define J..l(A) = f6X A dx where XA is the characteristic function of A and f6X A dx is the ordinary Riemann integral, then J..l is such a measure.
THEOREM 8.12 If M is a a-algebra on X and J..l is a measure on M, and iff:X ---7 [0,00] is measurable, then A defined by A(E) = fEfdJ..lfor each E E M is a measure on M and fxgdA = fxgfdJ..lfor each function g:X -7 [0,00]. Proof: Let {En} be a sequence of disjoint measurable sets in X with E = uEn. Then for each x E X, [XEJ](X) = ~n[XEnJ](X) since x can belong to at most one En. Therefore, XEf= ~nXEJ· Consequently, A(E) = ixXddJ..l and A(En) = fXXEJdJ..l.
254
8. Measure and Integration
By Theorem 8.l0 we have A(E) = ixXEfdll = LJXXEJdll = LnA(En).
Since 1l(0) = 0 implies A(0) = 0, we see that A is a measure on M. Let E and put g = XE. Then since XE is a simple measurable function,
E
M
Consequently, the equation in the conclusion of the theorem holds whenever g is the characteristic function of a measurable set. Next let g be the simple measurable function L~=l anXAn where An = {x E X Ig(x) = an}. Then fxgdA =
L~=l anA(An)
=
L~=l aJxMn dA
=
L~=l aJxXAJd!!
=
L~=l ixanXAJdll.
The last equality being attained by Theorem 8.8(3). By Theorem 8.10
L~=JxanXAJdll
=
fxL~=1 anXAJdll
= fxgfdll.
Hence the conclusion of the theorem also holds whenever g is a simple measurable function. If g is a measurable function, then by Theorem 8.7, there exists a sequence {sn} of simple measurable functions on X such that 0 ::; s I ::; S2 ::; ... ::; g and such that sn(x) ~ g(x) for each x E X. Then, from what we have already shown, for each positive integer n, JXSndA = JxsJdll. By Theorem 8.9, JXSndA ~ fxgdA so JxsJdll ~ fxgdA. But since 0::; s I ::; S2 ::; ... ::; g we have 0 ::; s If::; s2!::; ... ::; gf and since sn(x) ~ g(x) for each x E X, we have [s,J](x) = sn(x)f(x) ~ g(x)f(x) = [gj](x) for each x E X. Again by Theorem 8.9, fxsJdll ~ Jxgfd!!. Therefore, JxgdA = Jxgfdll.The equation in the statement of Theorem 8.12 is often written dA = fdll, or even as f = dA!dIl. The latter notation is suggestive of the role f plays but has no meaning as a ratio.
THEOREM 8.13 (H. Lebesgue) If {fn} is a sequence of complex, measurable functions on X with f(x) = limfn(x) for each x E X and if there exists a measurable function g on X with fx Ig Id!! < 00 such that Ifn(x) I ::; g(x) for each x E X, then fx If Idll < 00 and limn ~JX Ifn -fl dll
=0
and limn ~Jxfn dll
= fxfdll·
Proof: Clearly If I ::; g and since f is measurable by Corollary 8.5, Jx If Idll < 00. If - fn I ::; 2g, we can apply Theorem 8.11 to the functions 2g - If - In I to
Since get
8.6 The Lebesgue Integral
255
Now lim inln -4~ [2g - II - In I] = 2g and lim inln -4~Jx(2g - II - In 1)dJ..l = Jx 2gdf..l lim SUPn-4~Jx II - In I df..l. Hence lim SUPn-4~Jx II - In I dJ..l $; O. Therefore, limn-4~Jx II - In I df..l = O. By Exercise 2, IJx(f - In)df..ll $; Jx II - In I df..l which implies IJxldJ..l - JxlndJ..l1 $; Jx II - In I dJ..l for each positive integer n. But since limn -4~Jx II - In I df..l = 0 we have limn -4~ IJxldf..l - JxlndJ..l1 = 0 or in other words,
limn-4~Jxlndf..l = Jxldf..l.-
EXERCISES 1. Define L 1 (f..l) to be the collection of complex measurable functions on X for which Jx III df..l < 00. The members of L 1 (f..l) are called the Lebesgue integrable functions on X with respect to f..l. If IE L 1 (f..l) and I = u + iv where u and v are real measurable functions, define the integral of I (with respect to f..l) over E as
fddf..l = fEU+df..l- fEU-df..l+ ifEv+dll- dEv-dJ..l for each measurable E c X. Each of the four integrals on the right are finite, so the integral on the left is a complex number. Show that if I, gEL j (f..l) and a and b are complex numbers, then al + bg ELI (f..l) and
fx(al + bg)df..l = afxldf..l + bfxgdf..l. 2. Show that if IE L 1 (f..l) then IJxldf..ll
$;
Jx III df..l.
3. Let {fn} be a sequence of complex measurable functions on X such that In (x) -7/(x) for each x E X. Show that if there exists agE L 1 (f..l) such that lin (X) I $; g(x) for each x E X and each positive integer n, then: (a) IE L i (f..l). (b) limn--+~Jx lin - II dJ..l = O. (c) limn-4~Jxlndf..l = JxldJ..l. 4. Suppose In:X -7 [0,00] is measurable for each positive integer n and In+l ? In for each n. Also assume In (x) -7/(x) for each x E X and liE L 1 (f..l). Show that
ixl n dJ..l-7 ixldJ..l. 5. Show that the condition liE L 1 (f..l) is essential in Exercise 4. 6. Let I ELI (f..l). Show that for each < £ whenever f..l(E) < 8.
£
> 0 there exists a 8 > 0 such that JE III df..l
256
8. Measure and Integration
8.7 Negligible Sets If ~ is a measure on a a-algebra M and if E E M such that ~(E) = 0, then E is said to be a negligible set with respect to J..l. Negligible sets are negligible in the theory of integration. If P is a property that a point x may have (e.g., some function I may be continuous at x or differentiable at x) we say x has property P and write P(x). If A is a set and E is a negligible subset of A such that for each x E A - E we have P(x), we say that P holds almost everywhere on A. This is frequently abbreviated P holds a.e. on A. It is common to make statements like "I is continuous almost everywhere on A," in which case it is meant that the measure of the set of points of discontinuity of Ion A has zero measure. If I and g are measurable functions such that ~(E) = 0 where E = Ix I/(x) * g(x)} then X - E and E are disjoint sets whose union is X. Consequently,
fxld~ = fx-dd~+fd* = fX-Egd~+O = fx-Egd~+fEgd~ = fxg*. Hence, functions that are the same almost everywhere have equal integrals. This is what is meant by saying that negligible sets are negligible in the theory of integration. Since functions that are equal almost everywhere behave the same with respect to integration, we can generalize our definition of measurable function in the following way. Let E E M and let I:E ~ R. If ~(X - E) = 0 and 1 (U)nE E M for each open U in R then I is said to be measurable on E. Clearly, if we put/(x) = 0 for each x E X - E, we extend/to a function on X that is measurable with respect to the old definition. Intuitively, it should not matter what values we assign to I on X-E. We would like to be able to assign values to I on X - E in an arbitrary manner and still get a measurable function with respect to the old definition.
r
But here, a problem arises! It may be the case that certain subsets of X - E are not measurable. Fortunately, Corollary 8.4 states that every measure can be completed, i.e., every subset of a set of measure zero is itself a subset of measure zero. It will be to our advantage to just deal with complete measures. More measurable sets just mean more measurable functions. Then we can extend the function I:E ~ R above in any arbitrary manner to X and be assured that the extended function is measurable with respect to the old definition of measurability. This new definition of measurability has many consequences. For example, Theorem 8.10 can be modified to allow the sequence {In} to be a sequence of measurable functions on X that converges almost everywhere on X. With our new definition of measurability it is easily shown that the limit of (fn ) is still a measurable function I and Jxld~ = I.Jxlnd~, without having to restrict ourselves to the set on which the convergence actually takes place.
8.8 Linear Functionals and Integrals
257
EXERCISES 1. Show that if f:X ~ [0,00] is measurable, E almost everywhere on E,
2. Show that if f everywhere on X.
ELI (/1)
and
°
E
M and
Jdd/1 = for each E
E
fdd/1
=
M, then
°
then f =
°
f = 0 almost
3. Show that if f ELI (/1) and IJxfd!! I = Ix If Id!!, then there exists an a such that af= If I almost everywhere on X.
E
R
4. Show that if {fn} is a sequence of complex measurable functions defined almost everywhere on X such that Lnix Ifn i d!! < 00, then the function f(x) = LJn (x) converges almost everywhere on X,f ELI (Jl) and fxfdJl = LnfxfndJl. 5. Show that if Jl(X) < 00, f ELI (Jl), FcC is closed and the averages A£ = l/Jl(E)fddJllie in F for each E E M with Jl(E) > 0, then f(x) E F for almost all x EX. 6. Let {En} c M such that LnJl(En) < 00. Show that almost all x most finitely many of the En.
E
X lie in at
8.8 Linear Functionals and Integrals Recall from linear algebra that a vector space V over the scalar field F is a set V whose elements are called vectors and whose two operations are called addition and scalar multiplication. A linear transformation of V into another vector space W is a mapping A of V into W such that A(ax + ~y) = aA(x) + ~A(y) for all x,y E V and a,~ E F. In the special case W = F (the field of scalars), A is called a linear functional. Exercise 1 of Section 8.6 shows that L I (Jl) is a vector space whose scalar field is C. It is easily seen that the mapping I Jl:L I (Jl) ~ C defined by I Jl (f) = fxfdJl is a linear functional on L 1 (Jl). In the special case where V is the set of all continuous complex valued functions on the unit interval [0, 1], and F = R then the linear functional J:V ~ F defined by J(f) = f6f(x)dx (the ordinary Riemann integral) is clearly a positive linear functional. Since integrals are linear functionals, it is natural to ask: when are linear functionals integrals? In 1909, F. Riesz provided the following remarkable answer for the vector space C of all continuous complex valued functions defined on [0, 1]: for each positive linear functional A on C, there exists a finite positive Borel measure Jl on [0, 1] such that A(f) = f6fdJl. In fact, we now develop the celebrated Riesz Representation Theorem in a setting more general than the vector space C.
8. Measure and Integration
258
LEMMA 8.4 If X is a locally compact Hausdorff space, U is open in X and K c U is compact, then there exists an open V with compact closure such that K eVe CI(V) c U. Proof: Each point of K has an open neighborhood with compact closure, and K is covered by finitely many of them. Therefore, K lies in an open set W with compact closure. If U = X, put V = W. Otherwise, for each p E X - U let Wp be an open set containing K whose closure does not contain p and put Fp = WnWpn[X - U]. Then IFp} is a collection of compact sets such that nFp = 0.
Then IX - Fp} is an open covering of X. Pick Fq E {Fp}. Then some finite collection I (X - Fp I) ... (X - FpJ} covers Fq so FqnFp1 n ... nFpn = 0. Then V = UnWqnWpI n ... nWPn is open and contains K. Furthermore, CI(V) c CI(Wq)nCI(Wpl)n .. . nCl(WpJ since FqnFp1 n ... nFpn = 0.The collection of all complex valued functions f on a space X whose support (denoted O(/) has compact closure is denoted by CK(X). It is easily shown that CK(X) is a vector space under the operations of functional addition [(f + g)(x) = f(x) + g(x)] and scalar multiplication «aj)(X) = a[f(X)]). The notation K
r
A function f will be called lower semi-continuous if 1(a, 00) is open for each a E R and upper semi-continuous if 1 (-00, a) is open for each a E R. The following facts about upper and lower semi-continuous functions follow from the definitions. A real valued continuous function is continuous if and only if it is both upper semi-continuous and lower semi-continuous. Characteristic functions of closed sets are upper semi-continuous and characteristic functions of open sets are lower semi-continuous. The inf of any collection of upper semi-continuous functions is upper semi-continuous and the sup of any collection of lower semi-continuous functions is lower semicontinuous.
r
LEMMA 8.5 If X is a locally compact Hausdorff space, V is open in X and K c V is compact, then there exists an f E CdX) such that K (f ( V, 0 5,f 5, I, andj(x) = 1 for each x E K. Proof: Letr, =Oandr2 = 1. Let Irn},n~3,beawellorderingoftherationals between 0 and 1. By Lemma 8.4, there exists open sets U 0 and U 1 such that CI(U 0) is compact and K CUI c CI(U 1) c U 0 c CI(U 0) c V. We contend that it is possible to find a sequence {Urn} such that for each n, Urn is an open set with compact closure such that if ri < rj then CI(Ur) C Uri' For this, let H be the set of all positive integers such that n E H if it is possible to find n + 1 open
8.8 Linear Functionals and Integrals
259
°...
sets {V~, Ii = n} with compact closures such that if ri < rj (i, j ~ n) then CI(V~) c V~ and for each m < n, V,:, = V~ for i = 1 ... m. }
I
I
I
Clearly, 1 E H. Suppose n E H. Then one of the members rl ... rn' say ri, is the largest one smaller than rn+h and one, say rj' is the smallest one larger than rn+I' By Lemma 8.4, we can find an open V~n+}l with compact closure such that CI(V r)n ) c Vn+1 rn+1
C
CI(V nrn+l +l ) c V '1n .
°...
Also, for each m ~ n put V~:l = V~m' Then {V~,+l Ii = n + I} is the desired collection for n + 1, so n + 1 E H. This completes the induction argument. For each positive integer n put Vrn = V~n' Then {V rn } is the desired sequence.
°
For each rational r E [0, 1] define fr:X -7 [0, 1] by f(x) = r if x E Vr and otherwise. Define gr:X -7 [0, 1] by gr(x) = 1 if x E CI(V r) and r otherwise. Then put f(x) = sup {fr(x)} and g(x) = inf{ gr(x)}, for each x E X. Then f is lower semi-continuous and g is upper semi-continuous. Moreover,f(X) c [0, 1] and f(K) = 1. Finally, since O(fr) c CI(Ur) c V 0 for each rational r E [0, 1) we conclude 0(1) c CI(V 0)' Consequently, the proof will be complete if we show f =g.
Suppose there exists rationals r, s E [0, 1] such thatfr(x) > gAx) for some x X. From the definition of fr and gs we see this implies r > s, x E Vr and x does not belong to Vs. But r> s implies CI(Ur) c Vs which is a contradiction. Hence fr ~ gs for each pair of rationals, r, s E [0, 1). Consequently,f = supfr ~ infgr = g so f ~ g. Conversely, suppose f(x) < g(x) for some x E X. Then there exists rationals r, s such that f(x) < r < s < g(x). Now f(x) < r implies x does not belong to Vr and s < g(s) implies x E Cl(Us). But r < s implies CI(Vs ) c Vr which is a contradiction. Therefore,f(x) = g(x) for each x E X.E
LEMMA 8.6 If V I ... Vn is a finite collection of open sets in a locally compact Hausdorff space and K is a compact subset ofu'/=! Vi. then there exists a partition of unity on K subordinate to {V I ... V n }.
Proof: For each x E K, there exists (by Lemma 8.4) a neighborhood Vx with compact closure such that CI(Vx) c Vi for some i = 1 ... n. Since K is compact, there are finitely many points P! ... Pm with K c nj'=1 Vpj • For each positive integer i = 1 ... n, let Fi = u{ CI(Vp) Ipj E V;}. By Lemma 8.5, there exists an fi E CK(X) with Fi < fi < Vi, ~fi ~ 1 andfi(x) = 1 for each x E F i . Put g! = fl and for each i = 2 ... n put gn = (1 - 11)(1 - fz) ... (l - In-dfn. Clearly gi < Vi for each i = 1 ... n. A simple induction argument can be used to show that for each x E K, L'!=lgi(X) = 1 - (1 - Id(1 - fz) ... (1 - fn). Since K c u'!=IFi , at least one fi has fi(x) = 1 for each x E K. Hence L'!=l gi(X) = 1 for each x E K. Therefore, {g! ... g n} is a partition of unity subordinate to {V I . . . Un}. -
°
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8. Measure and Integration
A Borel (Baire) measure is said to be outer regular on a Borel (Baire) set E if f.!(E) = infl f.!(U) lEe U and U is open}. In case f.! is only a Baire measure, we assume U is an open Baire set in the definition of outer regularity. f.! is said to be inner regular on E if f.!(E) = sup I f.!(K) IKe E and K is compact}. f.! is said to be outer regular if it is outer regular on all Borel (Baire) sets and inner regular if it is inner regular on all Borel (Baire) sets. f.! is regular if it is both outer regular and inner regular. f.! is said to be almost regular if it is outer regular and inner regular on all Borel (Baire) sets E such that either E is open or f.!(E) < The regularity condition asserts that the values of the measure can be calculated from its values on the topologically important open and compact sets. 00.
The original version of the followmg theorem was given in 1909 by F. Riesz for the case where the underlying space X is the interval [0, lJ. The more general version of the theorem discussed below is known as the Riesz Representation Theorem and was first proven for locally compact Hausdorff spaces by S. Kakutani in 1941 in a paper titled Concrete representation of abstract (M)-spaces (Annals of Mathematics, Volume 42, pp. 934-1024). The version of the theorem stated below is due to P. Halmos. Rather than prove this lengthy theorem here, it is developed as a series of exercises at the end of the section. THEOREM 8.14 (Riesz, 1909, Kakutani, 1941. Halmos. 1950) If X is a locally compact Hausdorff space and A is a positive linear functional on CK(X). then there exists a a-algebra M on X containing all Borel sets. and a unique. complete. almost regular measure f.! on M such that f.!(K) < for each compact K e X. that represents A in the sense that A(f) = fxfdf.!for each f E CK(X). 00
THEOREM 8.15 Let X be a locally compact. a-compact Hausdorff space and let A be a positive linear functional on CK(X). Let M and f.! be the a-algebra and measure respectively from Theorem 8.14. Then M and f.! have the following properties: (1) If E E M and £ > O. there exists a closed set F and an open set U such that FeE e U and f.!{U - F) < E. (2)!! IS a regular Borel measure on X. (3) IfE EM. there are sets G. H eX such that H is an F (J. G is a Gil. He E e G and f.!(G - H) = 0 Proof' The proof of this theorem is closely related to the proof of the Riesz Representation Theorem. For that reason it is also left as an exercise at the end of the section. THEOREM 8.16 If X is a locally compact Hausdorff space in which every open set is a-compact and f.! is a Borel measure on X with f.!(K) < 00 for each compact K. then f.! is regular.
8.8 Linear Functionals and Integrals
261
Proof: For each f E CK(X), put A(j) = fxfd/l. Clearly, A is a linear functional. Suppose f E C K(X) such thatf~ O. Then f is real valued and hence bounded by some real numher r on X. Moreover. f(X - K) = 0 for some compact K c X. Since X is open, there exists an ascending sequence {Kn I of compact sets such that X = uKw Since f is real valued, m(E) = fdd/l for each E E M is a measure on M by Theorem 8.12. By Proposition 8.6, m(X) = limn-.=m(K n). Hence Jxfd/l = limn -.=kfd/l. For each fl, kfd/l = fKnnKfd/l ~ fKid/l ~ r/l(K). Since /l(K) < 00. {JKJd/l} is an ascending sequence of non-negative real numbers bounded above by r/l(K). Hence fxfd/l is a non-negative real number. Therefore. A is a positive lineal functional. By Theorem 8.14, there exists a measure v such that for each f E CK(X), Jxfd/l = fxfdv.
Let U be an open set. Then U = U£=lFi where {F;} is a sequence of compact sets. Pick fl E CK(X) such that F 1 < fl < U. Let K 1 be the closure of the support of fl' Assume that for each m ~ fl thatfm E CdX) has been chosen such that [u 7'=1 1 F i ]U[u7'=ll KJ < fm < U. Pickfn+l with [Ui'=l F i]U[Ui'=l K J
EXERCISES THE RIESZ REPRESENTATION THEOREM
1. Let X be a locally compact Hausdorff space and let A be a positive linear functional on CdX). For each open U c X put /leU) = SUp{A(j) If < U}. Then for each E c X let /leE) = inf{ /leU) lEe U and U is open}. Let N be the class of all E c X such that /leE) < 00 and /leE) = sup {/l(K) IKe E and K is compact}. Let M be the class of all E c X such that EnK E N for each compact K. Show that: (a) /l is well defined (on open sets U eX). (b) /leA) ~ /l(B) if A c B. (c) /leE) = 0 implies E E Nand E E M. (e) f~ g implies A(j) ~ A(g). 2. Show that if {En} is a sequence of sets in X, then /l(uE n) ~ Ln/l(En). 3. Show that N contains all compact sets.
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8. Measure and Integration
4. Show that N contains every open set U with
~(U)
< 00.
5. Show that if E = uEn, where {En 1 is a sequence of disjoint members of N, then ~(E) = Ln~(En). 6. Show that if E = uE n, where {En 1 is a sequence of disjoint members of N, and if ~(E) < 00 then E E N. 7. Show that if E E N and E > 0, there exists a compact K and an open U such that K c E c U and ~(U - K) < E. 8. Show that if A, BEN, then A - B, AuB and AnB E N. 9. Show that M is a a-algebra on X that contains all Borel sets. 10. Show that N consists of those E E M such that ~(E) < 00. 11. Show that ~ is a measure on M. 12. Show that for each IE CK(X) that 'A(j) = Jxld~. [Hint: It suffices to prove this for I real, so it is enough to show 'A(j) ::; Jxld~ for each real I E CK(X). Let E be the support of some real I E CK(X), and let [a, b] contain the range of f. Let E > and choose a finite setYl ... Yn with Yi - Yi-l < E andyo < a < Yl < ... < Yn = b. For each i = 1 ... n put Ei = {x E X IYi-l < I(x) ::; Yi InE. Show there exists open sets VI' .. Vn with Ei c Vi for each i such that ~(Vi) < ~(EJ + Eln for each i and such that I(x) < Yi + E for each x E Vi. Then show there are functions hi < Vi with Lhi = 1 on E so I = LhJ. Use this to finish the proof.]
°
13. Complete the proof of Theorem 8.14. 14. Prove Theorem 8.15 LEBESGUE MEASURE Euclidean n-space R n is a real vector space with respect to coordinate-wise addition and scalar multiplication. If x, Y E R n are given by (x 1 •.• x n ) and (y 1 ... Yn) respectively, there is an inner product defined on R n by (x, Y) = L!=lXiYi and Ix I is defined as (x, X)1/2. The norm I I satisfies the triangle inequality; i.e., Ix - yl ::; Ix - zl + Iz - yl for any z ERn. This should be familiar to the reader from linear algebra. The function d(x, y) = Ix - Y I is a metric on Rn. IfE c R n and x E R n we define the translate of E by x as the set E + x = {y + x lYE E I. A set of vectors in R n of the form C = {x Iai < Xi < hi} or a set obtained by replacing any or all of the < with ::; in the n inequalities defining C will be called an n-cell. The volume or measure of an n-cell is defined to be v(C) = TI!=l (hi - a;). If a E R n and e > 0 we call the set B(a, e) = {xl ai ::; Xi < ai + e}
8.8 Linear Functionals and Integrals
263
the box at a with side £. We will also refer to such a box as an £-box. For each positive integer m let Am be the set of all vectors in R" whose coordinates are integral multiples of 2-" and let Bm be the collection of all 2-" boxes at vector& XE A". 15. Show the following: (1) For each m, each x E R" lies in one and only one member of Bm. (2) If U, E Bm and U 2 E Bk where m < k then either U leU 2 or
U,nU 2 =0.
(3) If U E Bm then v(U) = 2- mn (4) If m < k and U E Bm then A" has precisely 2(k~m)" vectors in U. (5) Each non-void open set in R" is a countable union of disjoint boxes, each belonging to some Bm. 16. Show that there exists a positive, complete measure m defined on a a-algebra Min R" having the following properties: (1) m(U) = v(U) for each £-box U. (2) M contains all the Borel sets in R".
(3) E E M if and only if there exists an F cr set F and a G Ii set G with FeE c G and meG - F) =o. (4) m is regular. (5) For each x E R" and E E M, m(E + x) = m(E). (6) The property (5) is called translation invariance. If)l is a positive, translation invariant Borel measure on R" with )l(K) < for each ompact set K, then there exists a real number r with )leE) = rm(E) for each Borel set E in R". 00
Chapter 9 HAAR MEASURE IN UNIFORM SPACES
9.1 Introduction In 1933, in a paper titled Die Massbegrifj der Theorie der Kontinuierlichen Gruppen published in the Annals of Mathematics (Volume 34, Number 2), A. Haar established the existence of a translation invariant measure in compact, separable, topological groups. Translation invariance of a measure f.l in a topological group G means that if E is a measurable set then f.l(E + x) = f.l(E) for each x E G. Here, E + x = {Y E G Iy = a + x for some a E G). E + x is called the x-translate of E. The transformation Tx defined on G by Tx(y) = Y + x is called the x-translation or simply a translation. Topological groups will be defined later in the chapter and these concepts will be developed formally. In 1934, in a paper titled Zum Haarschen Mass in topologischen Gruppen (Comp. Math., Volume 1), J. von Neumann showed the uniqueness of Haar's measure, and in 1940, A. Weil published L'integration dans les groupes topologiques et ses applications (Hermann Cie, Paris) where Haar's results were extended to locally compact topological groups. In 1949, I. Segal extended Haar's results to certain uniformly locally compact uniform spaces that we will call isogeneous uniform spaces (Journal of the Indian Mathematical Society, Volume 13). In 1958, Y. Mibu, evidently unaware of Segal's work, independently established similar results for this same class of spaces (Mathematical Society of Japan, Volume 10). The Haar measures of both Segal and Mibu were Baire measures rather than Borel measures. Recall from Chapter 8 that Baire measures are defined on a smaller class of sets than Borel measures, namely, on the smallest a-ring containing all the compact sets. In 1972, G. Itzkowitz extended Haar's results to the Borel sets of a class of locally compact uniform spaces (Pacific Journal of Mathematics, Volume 41) that he called equi-homogeneous uniform spaces. That equi-homogeneous uniform spaces are equivalent to isogeneous uniform spaces is the subject of Exercise 2 at the end of this section. Itzkowitz showed the existence of a Haar integral (translation invariant, linear functional on the set of real valued continuous functions with compact support) for locally compact equihomogeneous uniform spaces. His approach was to show that a locally compact equi-homogeneous uniform space (X, f.l) is homeomorphic to a quotient G/H of topological groups, where H is a stability subgroup of G, and then apply Weil's
9.1 Introduction
265
theory of invariant measures on these quotients. as recorded in Chapter 3 of L. Nachbin's book The Haar Integral. (Van Nostrand, New York, 1965) to obtain a unique Haar measure. Itzkowitz's approach involved showing the modular function on H is constant and then appealing to theorems in Weil's theory to show this implies the existence and uniqueness of a Haar measure. In Section 4 of his paper. he also presented an alternate proof of the existence part of his development of a Haar measure on the Borel subsets of a locally compact equi-homogeneous uniform space. His proof that locally compact equi-homogeneous uniform spaces are quotients of topological groups contains an error as pointed out by the author in a paper titled On Haar Measure in Uniform Spaces (Mathematica Japonica, 1995) with a counterexample to his proof of Lemma 2.1 which is used in an essential manner to prove his Theorem 2.2. However. his alternate existence proof of a Haar measure on the Borel subsets of a locally compact equi-homogeneous uniform space is valid. This leaves his extended theory of Haar measure (on the Borel subsets of a locally compact isogeneous uniform space) incomplete in the sense that the uniqueness part of the proof has not been established. In 1992, the author, unaware of Itzkowitz's work, showed the existence and uniqueness of a Haar measure on the Borel subsets of locally compact isogeneous uniform spaces and presented that development in a series of lectures at the 1992 Topology Workshop at the University of Salerno. The existence part of the author's development is essentially the same as Itzkowitz's alternate existence proof. But the author's uniqueness proof is a uniform space argument rather than an appeal to Weil's theory of invariant measures on quotients of topological groups. The author's development, as presented at the Salerno Workshop is given in this chapter. It turns out that Itzkowitz's theorem that locally compact isogeneous uniform spaces are quotients of topological groups is true. We will prove this in Section 3 using a modification of Itzkowitz's approach that allows us to avoid the use of his erroneous Lemma 2.1. We will nor show that the rest of ItZkowitz's approach can be corrected because the topology we get with our new proof is finer than Itzkowitz's topology on the group G and this necessitates additional work to straighten out his approach which is beyond the Scope of this chapter. What we will show is that the converse of this result is true, i.e .. that quotients of topological groups are isogeneous uniform spaces. This characterizes the locally compact isogeneous uniform spaces as locally compact quotients of topological groups and leads to necessary and sufficient conditions for locally compact uniform spaces to have a topological group structure that generates the uniformity or to have an abelian topological group structure that generates the uniformity. The Segal-Mibu approach uses a generalization of K. Kodaira's construction given in a paper titled Uber die Beziehung zwischen den Massen und den
266
9. Haar Measure in Uniform Spaces
Topologien in einer Gruppe, (Proc. Phys. Math. Soc. Japan, Volume 23, No.3,
1941, pp. 67-119) whereas the Itzkowitz-Howes approach uses a generalization of A. WeiI's technique published in his paper referenced earlier in this section. At the present moment it may appear that the Weil technique is more powerful in isogeneous uniform spaces in that it can be used to obtain a measure on a larger class of sets. However, it is probable that the two methods are equivalent. If this is the case, we would have a way of constructing the measure directly on the Borel sets using a simple combinatoric method. A uniform space (X, /.1) is said to be isogeneous if there exists a basis v for /.1 and a collection H of uniform homeomorphisms of X onto itself such that: H, and each pair of points x, y E X, Y E S(x, U) if and only if i(y) E S(i(x), U) for each U E v. (2) For each pair x, y E X, there exists an i xy E H that carries x onto y.
(1) For each i
E
The members of H are called isomorphisms with respect to v or simply isomorphisms. v is called an isomorphic basis for /.1. Clearly, if i is an isomorphism and U E v, then i(S(x, U» = S(i(x), U) for each x E X. Also, it is easily seen that compositions and inverses of isomorphisms are again isomorphisms. A topological space is said to be homogeneous if for each pair of points p, q E X there exists a homeomorphism of X onto itself that carries p onto q. Clearly isogeneous uniform spaces are homogeneous topological spaces. There are various types of isomorphisms. An isomorphism t:X ~ X is called a translation if t has no fixed points. If the isomorphism r:X ~ X has a proper subset F '" 0 of fixed points and F does not separate X - F, then r is called a rotation. If F separates X - F, r is called a reflection. In what follows, we will show that locally compact isogeneous uniform spaces have a unique integral that is not only translation invariant, but also invariant under rotations and reflections. All topological groups are isogeneous uniform spaces with respect to the classical group uniformities and the classical group translation Tx defined by Tx(y) =Y + x for each x '" 0 in a topological group satisfies the above definition of translation. Let C(X) denote the ring of real valued continuous functions on X and CK(X)' the members of C(X) whose support have compact closures. For any f E CK(X) and isomorphism i:X ~ X, denote f © i by /; E CK(X). By a Haar integral for X, we mean a positive linear functional! on CK(X) such that Iif;) = I(f) for each isomorphism i:X ~ X. A Haar measure for X is an almost regular, Borel measure m satisfying m(i(E» = m(E) for each Borel set E and each isomorphism i:X ~ X. The following lemma is left as an exercise.
E
LEMMA 9.1. If (X, /.1) is a locally compact uniform space, then eachf CK(X) is uniformly continuous.
9.2 Haar Integrals and Measures
267
Isogeneous unifonn spaces were introduced in a series of lectures by the author at the 1992 Topology Workshop, held at the University of Salerno, Italy. The remaining material in this chapter is from the Workshop lecture series.
EXERCISES 1. Prove Lemma 9.1. EQUI-HOMOGENEOUS UNIFORM SPACES
Let (X, U) be an entourage unifonn space. A function f:X ~ X is said to be nonexpansive with respect to a base B for U if for each U E B and (x, y) E U, the relation (f(x), f(y» E U also holds. By a B-nonexpansive homeomorphism f of a unifonn space (X, U) onto itself, we mean a homeomorphism f of X onto itself such that f is nonexpansive with respect to a base B for the unifonnity U. A unifonn space (X, U) will be called an equi-homogeneous space if there is a group G of homeomorphisms acting on X such that (i) G is transitive (i.e., given p, q E X, there is agE G such that g(P) = q, and (ii) there is a base B for U such that G is a group of B-nonexpansive homeomorphisms of the unifonn space. 2. Show that the equi-homogeneous unifonn spaces are precisely the isogeneous unifonn spaces. 3. A collection G of functions from a unifonn space (X, U) to a unifonn space (Y,V) is said to be equi-continuous if for each V E V, there is a U E U such that for each g E G, [g x g](U) c V. Show that if G is a group of homeomorphisms acting on a unifonn space (X,U), then the following are equivalent: (i) there is a base B for the unifonnity such that G is a group of B-nonexpansive homeomorphisms of (X, U), and (ii)
G is an equi-continuous group of unifonn homeomorphisms on the unifonn space (X, U). 4. Show that if (X, U) is a locally compact isogeneous unifonn space, then (X,U) is unifonnly locally compact.
9.2 Haar Integrals and Measures In this section, (X, J.I.) is assumed to be a locally compact isogeneous unifonn space. Let v be an isomorphic basis for J.I. and H a collection of isomorphisms with respect to v that satisfy condition (2) in the definition of an isogeneous unifonn space. Let g ~ 0 be in CK(X) such that g(x) ~ b for each x in some U-sphere S(y, U) where U E v. If f ~ 0 is in CK(X) with f(X - K) = 0 for some
268
9. Haar Measure in Uniform Spaces
compact K c X then there is a finite subset I x I . . . xn} of K such that {S(x I ' U) ... S(x n , U)} covers K. If Iflj = sup {f(x)Ix E S(Xj' U)} and aj ~ iflJib for each j = 1 ... n, then f(x) ~ L7=1 ajgxi(x) for each x E X where gXj = g © LXI} for some isomorphism ixjy:X -7 X that carries xi onto y. The finite collection {a I .. an} is said to dominate fwith respect to g. Put [fl g 1= inf{ Liaj I {aj} dominates fwith respect to g}.
THEOREM 9.1. The number Ifl g] is non-negative,finite and satisfies: [Ji Ig] = Ifl g]for each isomorphism i:X -7 X. (2) Ifl + 12 I g] ~ Ifl Ig] + 1f21 glfor each pair fl.h E CdX). (3) [afl g] = alfl glfor each a > O. (4)fl ~h implies Ifllg] ~ 1f2 lg]foreachfl ~ 12 E CK(X). (5) Ifl h] ~ [fl g][g I h]for each h E CdX) with h O. (6) [h Ij]-I ~ [fl g]/[h Ig] ~ [fl hlfor each h E CK(X) with h O. (1)
'*
'*
Proof: We prove only (1). (2) through (6) are left as an exercise. Now Ji(x) = [f©i](x) ~ L}=laj[gxj © n(x) for each {ail that dominatesfwith respect to g. Also, gxj © i = g © [ix j y © i] where [i xj y © i] is an isomorphism so {a;} , dominatesfi with respect to g. Hence [Ji Ig] ~ [fl g]. Now fi,-1 = f© i © i-I = f and from what has just been proved, we have [fi g] = [Ji,-J i g] ~ [Ji! gJ. Therefore, [fi Ig] = [fl g] for each isomorphism i.· Let CK(X) denote the non-negative members of CK(X) and choose some k E CK(X). For each g E CK(X), define 1~:CK(X) -7 [0,00) by Ig(f) = [flg]/[klg] for eachfE CK(X).
COROLLARY 9.1. For each g E C-;rXJ. (l)lg~O.
(2) [klj]-I ~/g(f)~ [flk]. (3) Ig(fi) Ig(f)for each isomorphism i:X -7 X. (4) Ig(rtf) = aIg(f)jor each a > O.
=
(5) Ig(fl + h) ~ 19(fl) + l~(h)·
Since X is locally compact, for each U E g(X - K) = 0 for some U-small compact set K.
LEMMA 9.2. For each 11,12
~
there exists agE C;(X) with
'* '*
CKrX) with II 0 12 and lor each e ~ Ig(11 + h) + elor each g E 0 and such that the support of g is U-small. E
> 0, there exists a U E v such that Ig(fl) + Ig(hJ ct;(X) with g
'*
Proof: Let e > O. Assume [fl + h](X - K) = 0 for some compact K and letfE CHX) such that/(K) = l. For any 8 > 0 put <1> = II + fz + 8fand for i = 1,2 put hi = /;/<1> for <)lex) "" 0 and 0 otherwise. Clearly <jl, hi, h 2 E Ct: and hi + h 2 ~ 1. By Lemma 9.1, for each e' > 0 there exists a U E v with Ihi(X) - hi(y) I < e' for each x, y E X that are U-c1ose. Consider any g E Ct: with g 0 such that the
'*
9.2 Haar Integrals and Measures
269
support of g is U-small. Suppose <1> ~ Ljajgxj • Since gXj(x) 7:- 0 implies x and Xj are U-close, we have Ih;(xj) - h;(x) I < £' for i = 1,2. Therefore, fi(x)
=
[<1>h;](x) ~ [Ljajgx/x)]h;(x) ~ Ljajgx/x)[h,(xj ) + £'l
so Iii Ig] ~ Ljaj[h;(x) + £'l which implies Iii I g) + [f21 g] ~ LjajO + 2£'). Consequently, [f] Ig] + [12 Ig] ~ [<1> Ig](l + 2£'). Division by [k Ig] yields Ig(1]) + Ig (12) ~ Ig(1] + h) + 2£'l g(11 + h) + 0(1 + 2£')/ g (j). Since both Ig(j) and Ig(1] + h) are finite, the last two terms on the right hand side of this inequality can be made less than £ for sufficiently small 0 and £'. THEOREM 9.2. (G. Itzkowitz, 1972) Each locally compact, isogeneous uniform space has a non-zero Haar integral. Proof: Let D be the set of all f E CHX) with f 7:- O. For each fED put Hf = [[k 1.1]-] ,[fl kll c R. Then Y = W f is a compact Hausdorff space. For each U E vputG u = {lglgE Dand the supportofgis U-small}. LetFu =CI(G u )· Then {F u} has the finite intersectIOn property so rtF u #. 0. Let I E nF u· Then 1 can be interpreted as a functional on D where I(f) = pJ/) and Pf is the projection of Y onto the lh factor space H f . We now show I satisfies the following: (1) I(j) > 0 for each fED.
(2) 1(Ji) = I(j) for each isomorphism i and eachfE D. (3) I(af) = aJ(j) for each a> 0 and fED. (4) 1(1] + h) = l(1d + 1(12) for eachfl J2 E D. First note that for each neighborhood U(I) = {\jI E YII\jI(fj) -1(fj)1 < £} for some finite collection If] ... fn} c D, and each U E v, there exists some Ig E U(I)nG u · . (1) follows from the fact that for each fED. l(t) = pJ/) E Hf = [[k 1.1]-1 ,[fl k]] and hence I(j) ~ [k 1.1]-1 > 0 since k #. O. (2) follows from observing that for each £ > 0, there exists a g € E D with 1g F E {\jI E YII\jI(fj)-I(fj)1 < £} for j= 1,2 wheref] =fandh =fi. Then 1/(Ji) -1(j)1 ~ 1/(Ji)-lgE(Ji) I + 1/8£(Ji)-1(j)1 andsinceIg£(Ji)=lg£(j) we have 1/(Ji)-I(j)1 < 2£. (3) follows by using the inequality I/(af) - aJ(j)1 ~ I/(af) -lgE(af)1 + I/ g£ (af) - aJ(j) I where 18£ E {\jI E yll \jI(fj) - l(fj) I < £} for j = 1,2 where fl = fandh = afand obtaining I/(af) - aJ(j) I < £ + ac.
To prove (4), let £ > 0 and put h = f] + h. By Lemma 9.2. there exists a U E v such that Ig(1l) + Ig(12) ~ Ig(1] + h) + £ for each g E D whose support is U-small. Let/gEE {\jIE yll\jl(fj)-I(fj)1 <£}nG u forj=I,2,3. Then
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9. Haar Measure in Uniform Spaces
and since I g£(fl + h) = Ig£(fd + I g£(f2) we have I/(fl) + 1(f2) -I(fl + 12)1 ~ I/(fd- l g£(fl)1 + 1/(f2)-lg£(f2)1 +£<3£. We can extend I to a linear functional on Cl()0 by defining 1(0) = O. To extend I to a linear functional J on CK(X) note that each f E CK(X) can be written as for some E CK<X). We define J(j) to be l(f+) - l(f-). For each ~ = -a (where a > 0) we have J(~.I) = J(a(-.I) = I(a(-.I)+) -/(a(-.I)-) = a[/«-.I)+) -1«-.1)-)] = a[l(f-) -Ier)] = -a[l(f+) -I(f-)] = ~J(j). Clearly J(fd + J(f2) = J(f) + h) for each h'/2 E CK(X). Finally,fl = © i andfi = ©i so J(f) = l(fl) - 1(Ji) = ler) - l(f-) = J(j) for each isomorphism i:X ~ X. Therefore, J is a uniform integral for X. -
r -r
r, r
r
r
By the Riesz Representation Theorem (Chapter 8), there exists a a-algebra M in X containing all Borel sets, and a unique, positive, regular measure m on M which represents J in the sense that J(j) = Jxfdm for each f E Cdx) and such that m(K) < 00 for each compact K eX. THEOREM 9.3 m is a Haar measure for X. Proof: We first show that if K is compact then m(i(K) = m(K) for each isomorphism i:X ~ X. Let U be an open set containing K. Then there is an fu E CK<x) withfu(K) = 1 andfu(X - U) = O. Therefore, XK ~fu ~ Xu so m(K)
= fxXKdm ~
fxfudm
~
fxXudm
= m(U).
Since m is outer regular, this shows that m(K) = infl Jxfudm 1 K c U and U is open}. But then m(K) = inftJx[fu © i-I ]dmlK c U and U is open}. Thus, m(i(K)
~
fx[fu © i- 1 ]dm
~
fXXi(U)dm = m(i(U».
Now m(i(K) = Im(V) Ii(K) c V and V is open}. For each open V containing i(K), it is possible to find a V E v such that Star(i(K),V) c V. Since Star(i(K),V) = i(Star(K,V», m(i(K) = infl m(i(U) 1K c U and U is open}, so m(i(K) = m(K). Next suppose E is a Borel set with m(E) < and i is an isomorphism on X. Then there is an ascending sequence {K j } of compact subsets of E with m(E) = limj-t~m(Kj). For each positive integer j let XKj be the characteristic function of K)" Then for each j, XKJ ~ XKj+l' Since, the sequence {XKj } converges almost everywhere to XE, {XKj © r 1 } converges almost everywhere to XE © i-I. By Theorem 8.9, {fxXK.dm} converges to JxxEdm and {fX[XKj © rl ]dm} converges J to Ix [XE © i-I ]dm. Therefore, 00
m(i(E» = ixXi(E)dm = ix[XE © i-I]dm = limj-tJx[XKj © i-I]dm
=
9.3 Topological Groups and Uniqueness ofHaar Measures
271
Consequently, m(i(E» = m(E) for each Borel set E with m(E) < 00, and each isomorphism i. Finally, suppose m(E) = 00. If m(i(E» < 00, then by what has just been shown, m(E) = mW' (i(E))) = m(i(E», so m(E) < 00 which is a contradiction. Therefore, for each Borel set E, m(E) = m(i(E» for each isomorphism i.We will defer the uniqueness proof for m until Theorem 10.5.
EXERCISES 1. Prove Theorem 9.1. 2. Let (X, 11) be an isogeneous uniform space and let i be an isomorphism of X onto itself. Let m be a regular, Borel measure on X. For each Borel set E put miCE) = m(i(E». Show that mi is an almost regular, Borel measure on X. 3. Show that the results of this section still hold if CK(X) is replaced by Cc(X) U:X ~ C Itis continuous and CI(O(/) is compact}.
=
9.3 Topological Groups and Uniqueness of Haar Measures In this section we will introduce the concept of a topological group and see that all topological groups are uniformizable and that locally compact topological groups are cofinally complete (uniformly paracompact) and hence complete. We will also see how the classical existence and uniqueness proofs of the Haar integral in locally compact topological groups follow from the results in the previous section and what form they take in topological groups. This section also contains a solution to the problem of characterizing which uniform spaces have a compatible topological group structure that generates the uniformity for the case when the uniform space is locally compact. The concept of a topological group is simply the combination of the concepts of a group and of a topology on the same underlying set in such a way that the group operation is in some sense compatible with the topology. Precisely, by a topological group, we mean a topological space (X, 't) together with a binary operation +:X2 ~ X satisfying the following conditions: GI. (X, +) is a group. G2. + is continuous with respect to the product topology on X2. G3. The inversion function i:X ~ X defined by i(x) = -x for each x E X is continuous. It is easy to find interesting examples of topological groups. Probably the most
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9. Haar Measure in Uniform Spaces
obvious is the additive group of real numbers (R, +) with the usual top- ology for R. Similarly, coordinatewise addition in R n makes R n a topological group with respect to the Euclidean n-space topology. Another important example is the circle group T c C. The unit circle T is defined as the subset of complex numbers consisting of members z such that Iz I = 1. The group operation is usual multiplication of complex numbers and the topology of T is the subspace topology of T with respect to the usual topology of R2 . PROPOSITION 9.i For a group (X, +) with topology 'T, the conditions G2 and G3 hold if and only If the function f:X 2 ~ X defined by j(x, y) = x - Y IS continuous. Proof: If G2 and G3 hold then fis continuous since f(x, y) = g(x) + i(y) for each pair x, y E X where g is the identity function on X because g x i is continuous on X2 by G3 and addition is continuous by G2 and f is the composition of g x i
and the addition function. Conversely, if f is continuous then inversion is continuous since i(x) = E X, where 0 is the group identity. Also, since x + y = f(x, i(y)) for each pair (x, y) E X2 it follows that addition is continuous.-
f(O,x) for each x
PROPOSITION 9.2 For a topological group (X, 'T, +), the following functions are homeomorphisms of X onto itself: (1) inversion i, (2) right translation ra by some a E X defined by ra(x) = x + a for each x E X, (3) left translation la by some a E X defined by la(x) = a + x for each x E X. Proof: The functions i, ra and la are clearly continuous for each a E X. Since the compositions i © i, ra © r -a' and La © La are all the identity mapping for each a E X, it is also clear that i, ra and fa are all homeomorphisms for each a E
x-
For a set SeX, the notation S-l is used to denote the set consisting of all inverse elements of members of S. S is said to be symmetric in X if S = S-l. If T c X, the notation S + T is used to denote the set consisting of all elements of the form s + t such that s E Sand t E T. PROPOSiTION 9.3 If A is open and B is closed in the topological group (X, +), and ifC, D c X then: (1) CI(D 1 ) = [CI(D)]-1 (2) Cl(x+ D + y) = x + [CI(D)] + yfor each pair x,y EX. (3)D+A andA+Dareopen. (4) B + Y and y + Bare closedfor each y E X. (5) CI(C) + CI(D) c CI(C + D).
9.3 Topological Groups and Uniqueness ofHaar Measures
273
Proof: Since i is a homeomorphism, CI(D- I ) = CIU(D» = i(CI(D» = [CI(D)]I. Similarly, since for each pair x, y E X, Lx and rx are homeomorphisms, Cl[x + D + y]
= ClOAD + y» = CI(lx(ry(D))) =
IxCry(CI(D») = x + CI(D) + y.
This establishes (I) and (2). (3) follows from Proposition 9.2 since for each x E X, X + A = IAA) and D + A = UxED(X + A). Similarly A + D is open. (4) also clearly follows from Proposition 9.2. (5) follows from the continuity of + on X x X, i.e., CI(C) + CI(D) = +(CI(C) x CI(D» c CI(+(C x D» = CI(C + D).PROPOSITION 9.4 If h:X ~ Y is a group homomorphism between topological groups X and Y then: (1) h(A + B) = h(A) + h(B) c CI(h(A)) + CI(h(B)) c h(CI(A + B)) for each A, Be X. (2) h- l (C) + h-I(D) c h-1(C + D)for each C, DeY. (3) CI(h-I(C)) + CI(h-1(D)) c CI(h-1(C + D))jor each C, DeY. (4) If A is symmetric in X then h(A) and CI(h(A)) are symmetric in Y. (5) If C is symmetric in Y then h- l (C) and Cl(h- l (C)) are symmetric in X.
The proof of Proposition 9.4 is left as an exercise, A subgroup H of a group G is said to be normal or invariant if for each x E H, a + x + a-I E H for each a E G. G is said to be Abelian if a + b = b + a for each pair a, bEG. PROPOSITION 9.5 Let H be a subgroup of a topological group G. Then: (1) Hand CI(H) are topological groups. (2) If H
is normal then CI(H) is normal. is Hausdorff and H is Abelian (hen Cl( H) is Abelian. (4) If H is open then H = CI(H).
(3) If G
Proof: H is clearly a topological group since the continuous function f:G 2 ~ G defined in Proposition 9.1 by f(x, y) = x - y has a continuous restriction to H2 c G2 • That CI(H) is also a topological group follows from the fact that since H is a subgroup of G, H + H = H = WI and so CI(H) + [CI(H)r1 = CI(H) + CI(W I ) c CI(H + Wi) = CI(H) by Proposition 9.3. Hence CI(H) is also a subgroup of
G. To show (2), assume H is nonnal. Then x + H - x = H for each x E G, so x + CI(H) - x = Cl(x + H - x) = CI(H) for each x E G by Proposition 9.3 which implies CI(H) is nonnal. To show (3), assume G is Hausdorff and H is Abelian. Define g:G 2 ~ G by g(x, y) = x + y - x - Y for each pair x, y E G. Then g is continuous since + and i (inversion) are continuous. Since G is Hausdorff, {O}
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9. HaarMeasure in Uniform Spaces
is closed where 0 is the group identity element. Since g is continuous, g-I (0) is also closed in G. Now H2 C h- I (0) since H is Abelian, so el(H) x el(H) = el(H x H) c h -I (0) is closed. Hence el(H) is also Abelian. To show (4), assume H is open in G. Then we have G - H = (G - H) + H = + H) which is open since x + H is open for each x E G - H by Proposition 9.3. Therefore, H is closed in G so H = el(H).UxEG-H(X
Let X and Y be topological groups having neighborhood systems N and B at 0 and e respectively, where 0 is the identity element of X and e is the identity element of Y. A homomorphism h:X 4 Y is open at 0 if for each U E N, there is a V E B with V c h(U). PROPOSITION 9.6 A homomorphism h:X 4 Y is continuous (open) and only if it is continuous at 0 (respectively open at 0).
if
Proof: Clearly if h is continuous or open, then it is continuous or open respectively at O. Assume first that h is continuous at 0 and let N be the neighborhood system at O. For each x E X and open V containing y = hex), e (the identity of Y) is in V - y and V - y is open by Proposition 9.3. Since h is continuous at 0, there is a U E N with h(U) c V - y. Since x E U + x and h(U+x) = h(U) + hex) c (V - y) + y = V, we have that h is continuous.
Next assume h is open at 0 and let B be the neighborhood system at e. For each x E X and open U containing x, U - x is an open set containing 0, so there exists a V E B with V c h(U - x) = h(U) - hex) which implies V + y c h(U) where y = hex). Since e E V, Y E V + y which is open. Therefore, hex) is an interior point of h(U). Consequently, h is an open mapping.A key feature of topological groups is that knowing the behavior of the neighborhood system of the identity element is equivalent to knowing the behavior of all neighborhood systems at each point of the group. This derives from the fact that if g is any element of the group and U is a neighborhood of the group identity, then both g + U and U + g are neighborhoods of g and if V is a neighborhood of g, then g-I + V and V + g-1 are both neighborhoods of the identity. Consequently, the following results are especially useful in topological groups. PROPOSITION 9.7 If (X, +) is a topological group and N the neighborhood system of the identity O. then: (I) For each V E N there is a V E N with V + V C u. (2) For each V E N, V-I E N. (3) ForeachVE NandxE X,thereisaVE Nwith[x+V-x]cu. (4) Eachfilter N of subsets of X containing 0 and satisfying (J) - (3) determines a unique topology that makes (X, +) a topological group and has N as its neighborhood system at O.
9.3 Topological Groups and Uniqueness ofRaar Measures
275
Proof: (1) follows from the continuity of + and (2) from the continuity of inversion. Since for each x E X, Lx and rx are homeomorphisms, Lx © rx(U) = [(-x) + U + xl E N for each U E N. Rence V c [(-x) + U + xl for some V E N which implies [x +V + (-x)] cU. Now (1) implies the existence of such a V, so (3) follows. To show (4), let. = {U c X I for each p E U, there exists V E N with V + P c U} .• is easily shown to be a topology on X. Clearly X, 0 E •. If {Va} c . then uVa E • since for each x E uV a , X belongs to some V ~ which implies there exists a V~ E N with V ~ + x c V ~ c uV a . Also, if V I ... Vn E • then n?=1 Vi E • since for each x E n?=1 Vi, there exists a Vi E N for each i with Vi + x C Vi' Since N is a filter, n?=1 Vi E N andx E (n?=1 Vi) + x c n?=1 Vi'
°
Clearly • has N as its neighborhood system at and if cr is another topology for X that makes (X, +) a topological group and has N as its neighborhood system at 0, it is easily shown that cr = ., for if U E • and p E U, there exists a V E N with V + P c U, so U is a neighborhood of p in the topology cr, i.e., • c cr. Similarly cr c •. It remains to show that. itself makes (X, +) into a topological group. For this it suffices to show that f:X 2 ~ X as defined in Proposition 9.1 (f(x, y) = x y) is continuous. For this let (p, q) E X2 and let U be a neighborhood of f(P, q) = p - q. Then W = [U - x + yl E N. By (1), there exists a V E N with V + V C W. By (2), V-I E N so A = VnV- I E N since N is a filter. Then A = A-I and A + A c V, so A + A-I c W. Now A + p is a neighborhood of p. Also, there exists aBE N with [p + B - p] c A by (3). Therefore, [p + B - p] + q is a neighborhood of q, so [A + p] x [(P + B - p) + y] is a neighborhood of (p, q). Moreover,f([A + p] x [(P + B - p) + q]) = {f(a T p,p + b - P + q) Ia E A, bE B) c {a I + P - P - a 2 + P - q Ia I ' a 2 E A) = [A + A-Ij + (x - y) c W + (x - y) = U. Consequently,f is continuous, so the topology. makes (X, +) into a topological group. -
THEOREM 9.4 Let (X, +) be a topological group with topology. and neighborhood system Nat O. For each V E N put U v = {V + x Ix E X} and let A =[U v IV E N}. Then A is a basis for a uniformity f.l that is compatible with •. Consequently, every topological group is uniformizable. Proof: To show A is a basis for a uniformity, it suffices to show that for each pair U v , U w E A, there exists some U u E Athat ,1-refines UvnU w. If U v , U w E A, then V, WEN. As shown in the proof of Proposition 9.7, there exists an A E N with A + A-I c VnW. Then U = AnA- 1 EN so S(x, U u ) = {z E xix, Z E U + yforsomeYE X} = {z+Xlx-y,Z-YE UforsomeYE XI c {ZE XI(z-y)(x - y) E U + U- I } C {z E X I Z - X E V} = V + X E U v . Similarly S(x, U u ) c W + X E U w, so S(x, U u ) c [V + x]n[W + x] E UvnU w. Therefore, U u is a ,1-refinement of UvnU w .
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To show that J..l is compatible with 1:, by Proposition 9.7 it suffices to show that 15(0, U) IU E J..l I =N. For this, first note that if V E N, then V c 5(0, U v ) so 5(0, U v ) E N for each U v E A. Since A is a basis for 11, we see that I S(O,U) IU E J..l I c N. Conversely, if V E N we can choose WEN such that W + W-] C V so S(O,U w ) = Iy E X I 0, YEW + x for some x E XI = (y E xly - x E Wand x E W-] ) = Iy E X lYE W + W-] ) c V. Therefore, N c {S(O, u) Iu E J..l I which establishes 15(0, U) IU E J..ll = N. By now, it has probably occurred to the reader that the uniformity J..l of Theorem 9.4 could just as well have been defined by the basis v = I Wv IV E N I where Wv = Ix + V Ix E X I, and the argument, with minor notational changes, would stilI be valid. The uniformity J..l is called the right uniformity on X whereas the uniformity J..l' generated by v is the left uniformity on X. It is left as an exercise (Exercise 3) to show that the uniform spaces (X, J..l) and (X, J..l') are isogeneous uniform spaces where the collections of homeomorphisms are R = I rx Ix E X I and L = {Ix Ix E X I respectively and the isomorphic bases are A and v (as defined in Theorem 9.4 and above) respectively. Consequently, if (X,+) is a locally compact topological group, then there exist unique Haar measures m and m' for (X, J..l) and (X, J..l') respectively. By definition of U v and Wv we then have m(V + x) = m(V) = m(V + y) for each pair x, y E X and each V E N (neighborhood system at 0) and similarly, m'(x + V) = m'(V) = m'(y + V). From Theorem 9.4 it is straightforward to derive a number of useful consequences that are listed in the following proposition and left as an exercise.
PROPOSITION 9.8 If (x. +) is a topological group with neighborhood systemN at 0, then: (I) X has a neighborhood base consisting of symmetric open (closed) sets. (2) Int(Y) = lYE Ylv+ycYforsomeVE NlforeachYcX. (3) CI(Y) = nlV + yl V E Nlfor each Y cx. (4)XlsT] ifandonlyifO=nlVIVE NI. (5) X is locally compact if and only if there exists a compact V E N. Let (X, +) be a topological group and let Z be a subgroup of X. For each x E X, the set Z + x is called a right coset of Z in X. Similarly x + Z is called a left coset. Both the right cosets and the left cosets form partitions of X. To show this for the right cosets, define the relation p - q in X by p - q E Z. It is easily verified that - is an equivalence relation on X. In fact, p - q E Z if and only if p E Z + q so that the equivalence class [p] containing P with respect tois Z + q for each q E [Pl. Since p E [P] we see that [P] = Z + p. Let Q = X/- be the quotient uniform space defined in Chapter 5. In this section the notation for X/- will be X/Z. It should be noted that X/Z may fail to be a topological group if Z is not normal. Never-the-Iess, X/Z will still be referred to as a quotient of topological groups, the reference being to the quotient topological and uniform
9.3 Topological Groups and Uniqueness ofHaar Measures
277
structures rather than a quotient group structure. The natural projection n:X X/Z is given by n(x) = Z + x for each x E X.
---7
Clearly Z + x = rAZ) and x + Z = IAZ) for each x E X. Also, it is easily seen that the above argument could have been carried out had we chosen to define p - q by -p + q E Z. In this case, the partition would consist of the left cosets (x + Z Ix EX} and we could construct a "left quotient" -IX uniform space. If Z is Abelian, clearly -IX = X/-. PROPOSITION 9.9 The quotient Q = XIZ of a topological group (X. with respect to a subgroup Z is T 1 if and only ifZ is closed in X.
+)
Proof: Assume Q is T 1. Then the singleton set {Z} in Q is closed. Since the canonical projection n:X ---7 Q is continuous it follows that n-1({Z}) = Z is closed in X. Conversely. let Z be closed in X. Let q E Q. By definition, q is some right coset Z + p of Z for some p E X. Now Z + P = rp(Z) and by Proposition 9.2.(2), r" is a homeomorphism, so Z + p is closed in X. Since n- 1(q) = Z + p and n IS an identification mapping, it follows that {q I is a closed subset of Q. Hence Q is T 1. -
If H is a nonnal subgroup of a group G it can be shown (see Exercise 7) that G/H is also a group with respect to the operation + defined on G/H by (H + x) + (H + y) = H + (x + y). In the case G is also a topological group, it is easily verified (see Exercise 8) that Q = G/H is a topological group with respect to the quotient topology, and as such will be referred to as the quotient topological group of G with respect to H. THEOREM 9.5 (N Howes, 1992) Let G be a topological group and H a subgroup of G. Then the quotient GIH with the quotient topology is an isogeneous uniform space. Proof: Let n:G ---7 X = G/H be the natural projection defined by neg) = g + H for each g E G. Then n is continuous and onto X. For each g E G we can define a function g':X ---7 X by g'(f + H) = (g + j) + H for each (f + H) E X. It is easily seen that for each g E G, g' is one-to-one and onto. Furthermore, if x, y E X, there exists agE G with g'(x) = y.
We next show that if V is a neighborhood of e (the identity) in G, then for each g E G, if x, y E n(f + V) for some f E G, then g'(x), g'(y) E n(h + V) for some h E G. Now, x, Y E n(f + V) implies g'(x), g'(y)
E
g'(n(f +
V» =
{g'(n(f + v» I v E
VI
{g+!+v+HivE
VI.
Put h = g + f. Then, for some VI, V2 E V, g'(x) = h + VI + Hand g'(y) = h + V2 + H. But then g'(x), g'(y) E n(h + V). Let G' be the collectIOn of all g' such that g E G and let v be the collection of all coverings n(V) where V = If + Vi! E G
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9. Haar Measure in Uniform Spaces
and V is a neighborhood of e). Then each g' E G' is uniformly continuous since x,y E 1t(j + V) implies g'(x), g'(y) E 1t(h + V) for some h E G. Furthermore, since G is a group, it is easily shown that each g' EGis a uniform homeomorphism. Finally, we show that each g' E G' is isogeneous with respect to v. For this, let x E X. By what we have already shown, g'(S(x, 1t(V» c S(g'(x), 1t(V» for each 1t(V) E v. Since G is a group, we also have that (g-l )'(S(g'(x), 1t(V))) c S«g-l )'(g'(x», 1t(V» = Sex, 1t(V» which implies S(g'(x), 1t(V» c g'(S(x, 1t(V». Hence g'(S(x, 1t(V» = S(g'(x),1t(V» for each g' E G'. Therefore, g' is an isomorphism with respect to v for each g E Go Consequently, v is an isomorphic basis for a uniformity 11 on X. Therefore, (X, 11) is an isogeneous uniform space. • THEOREM 9.6 (N. Howes, 1992) Each locally compact isogeneous uniform space is homeomorphic to a quotient of a topological group. Proof: Let (X, /l) be an isogeneous uniform space and let v be an isomorphic basis for /l. Let H be the collection of isomorphisms on X with respect to v. If 1 are also isomorphisms. f, g E H, it is easily shown that f © g and Consequently, H can be extended to a group G with respect to the operation of functional composition. Without loss of generality, we may assume G to be the group of all isomorphisms on X with respect to v. For each compact K c X and U E v, put [K, U] =
r
{g E G Ithere exists a V E v with S(g(x), V) c Sex, U) for each x E K).
Let p = {[K, U]I K c X is compact and U E v). If P is taken as a system of neighborhoods of the identity (mapping) i E G, then G is a Hausdorff topological group with respect to the operation of functional composition. To see this, first note that every member of Pcontains i. Also, if [K, U], [F, V] E P, then there exists aWE v with W < U and W < V, so [KuF, W] c [K,U]n[F,V]. Consequently, it will suffice to prove:
P,
P P
0) For each [K, U] E there exists [F, V] E with [F,vr 1 c [K, U]. (2) For each [K, U] E ~,there exists [F,VJ E with [F.vf c [K, U]. (3) For each [K, U) E Pand g E G. there exists [F,v] E Pwith g © [F,v] © g-l c [K, UJ. (4) For each [K. U] E ~ and g E [K. U], there exists [F,v) E with g © [F,v] c [K. U].
B
To prove (1), let [K, U] E Band pick W E v with W <* U. Let g E [K, W). Then there is a V E v with S(g(x),V) c S(x,W) for each x E K. Hence, g-l (x) E S(x,W), so there is a neighborhood U(g-l (x» with co~pact ~:osur~ such that CI(U(g-1 (x») c Wx for some Wx E W containing x. Smce g (K) IS compact,
9.3 Topological Groups and Uniqueness ofHaar Measures
279
there exists a finite collection {x 1 ... x n } c K such that U = u7=1 U(g -I (xJ) is a neighborhood of g-I (K). For each i put G j = U(g-I (Xj»' Also for each i:S: n 1, put F j = g-I(K) - U7=j+IG i and put Fn = g-I(K). Then each F j is compact. Choose ZI E v such that for each y E F I, S(y, ZdnStar(X - G I, ZI) = 0. Let k be a positive integer with k < n and assume that for each m:S: k, Zm has been chosen such that for each y E F m' S(y, Zm)nStar(X - G j , Zm) = 0 for some positive integer i:S: m. Let Hm+1 = Fm+1nStar(X - U7'=IG i , Zm). Now Fm+1 Hm+1 is compact so there is a Z~+I E v with
Pick Zm+1 E v such that Zm+1 <* Z~+I and Zm+1 <* Zm. Then for each y E F m+I' S(y, Zm+1 )nStar(X - G j , Zm+d = 0 for some positive integer i :s: m + 1. Consequently, there exists a Zn such that for each Y E g-I (K), S(y, Zn)nStar(X - G j , Zn) = 0 for some positive integer i:S: n. Finally, choose Z E v such that Z < Zn and Z < W. Let x E K. Then S(g-I (x), Z) c Gj c WXj for some positive integer i :s: n. Hence, there is aWE W with x, Xj E W so WnW Xj :F: 0. Therefore, S(g-I(X), Z) c Sex, U) for each x E K so g-I E [K, U]. Thus [K,Wrl c [K, U] so (1) is proved. To show (2), let [K, U] E ~. Pick Y E X and let C(y) be a compact neighborhood of y. Let W E v with S(y,W) c C(y). For each x E K, there exists an isomorphism ix that maps y onto x, so S(x,W) is a neighborhood of x with compact closure. Then there exists a finite collection {x 1 ••• xn } with K c F = u7=1 CI(S(xi,W», Hence there is a V E v with V < U and Star(K,V) c F. If f, g E [F,V], there exists ZI, Z2 E v with S(f(x), ZI) c S(x,V) and S(g(x), Z2) c S(x,V) for each x E F, Then for each x E K, g(x) E Star(K,V) c F. Therefore, S(f(g(x», ZI) c S(x'v) for each x E K, so f© g E [K, U]. Consequently, [F,Vf c [K, U]. To prove (3), let [K, U] E ~ and let g E G. Put F = g(K). Then F is compact. Moreover, g-I © [F, U] © g C {g-I © f© g Ithere is a V E v with S(f(x),V) c Sex, U) for each x E F} C {g-l © f© glthere is a V E v with S(g -I (f(g(y))),V) c S(y, U) for each y E K} c {h E G Ithere exists a V E v with S(h(y),V) c S(y, U) for each y E K} = [K, U]. Consequently, g-1 © [F, U] © g c [K, U]. To prove (4), first note that if g E [K, U] E ~ then there exists a V E v with S(g(x),V) c Sex, U) for each x E K. Let W E v with W <* V. If f E [K,W], then there exists a Z E v with S(f(x), Z) c S(x,W) for each x E K. Therefore, f(x) E S(x,W) for each x E K so that g(f(x» E S(g(x),W) which implies S(g(f(x»,W) c S(g(x),V) c Sex, U). Therefore, g © f E [K, U]. Hence g © [K,W] c [K, U]. We next construct a quotient space from the group G and show that it is homeomorphic with X. For this let Z E X and put H = {g E G Ig(z) = z}. Then H is a subgroup of G. Let Y = GIH be the quotient space of G modulo H. Let g ©
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9. Haar Measure in Uniform Spaces
=
= =
HEY. Then [g © h](z) g(z) for each h E H, so for each f E g © H,f(z) g(z). Suppose Y E X. Then there exists agE G with g(z) y and hence fez) y for each f E g © H. Consequently, there exists a one-to-one correspondence <1>:Y -t X defined by <1>(g © H) g(z) for each g © HEY.
=
=
Define e:G -t X by e(g) = g(z) for each g E G. Then e = <1> © 1t where 1t is the quotient mapping 1t:G -t G/H. To show <1> is a homeomorphism of G/H onto X, we must show that <1> is both continuous and open. Since 1t is a quotient mapping, it will suffice to show that e is continuous and open. To show e is continuous, let g E G. For each U E Y, S(g(z), U) is a basic neighborhood of e(g) = g(z) in X. Also, [{ z}, U] is a basic neighborhood of the identity i E G, so g © [{ z}, U] is a basic neighborhood of g in G. Now e(g © [{ z}, UJ) = {[g©j](z) If E [{ z}, U]} = {g(f(z» Ithere exists a V E Y with S(f(z)'v) c S(z,U)}. S(f(z),V) c S(z, U) implies g(f(z» E S(g(z», U). Therefore, e(g © [{ z} ,UJ) c S(g(z), U) so e is continuous. To show e is open, first note that (4) implies each [K, U] E ~ is open since if g E [K, U], there is an [F,V] E ~ with g © [F,V] c [K, Uj so that each g E [K, U] has a basic neighborhood g © [F,V] c [K, U]. Next, observe that for each x E X and U E Y that [{x}, U](x) = Sex, U). To see this, let y E Sex, U) and let g E G such that g(x) = y. Pick V E Y with S(y,V) c Sex, U). Then S(x,U) c [{x}, U](x) so [{x}, U](x) = Sex, U). Finally, observe that if V eGis open and Y c X, then V(y) = {u(y) IU E V and y E Y} is homeomorphic with [g©V © g-I ](Y) for each g E G. To see this define F:G(X) -t [g © G © g-I](X) by F(h(x» = [g © h © g-I ](x) for each h E G and x E X. Clearly, F is one-toone and onto. If hex) E G(X) and U E Y, then F(S(h(x), U» = S(F(h(x», U), so F is a homeomorphism. Let f = Flu (Y). Then f: V(y) -t [g © V © g -I l(y)' Consequently, V(y) and [g © V © g-I ](Y) are homeomorphic so that if V(y) is open in X, so is [g © U © g-I J(Y). Now we show that if x,y E X and U E Y that [{x}, U](y) is open in X. For this, let g E G such that g(x) = y. It is easily shown that [{x}, U] = {fE Glf(x) E Sex, U)} so g © [{x}, U] © g-I = {g © hlh(y) E sex, U)} = {k E Glk(y) E S(g(x), U)} = [{y}, U]. But then [g © [{x}, U] © g-I](y) = [{y}, U](y) = S(y ,U) which is open in X. From the proof in the previous paragraph, rIx} ,U](y) is homeomorphic with [g © [{x}, U) © g -I](y). Hence [{x}, U](y) is open in X. From this we can show that if [K, Uj E ~ and x E X then [K, U](x) is open in X. For this, pick V,W E Y with W <* V <* U. Then the covering {S(x,W)lx E K} has a finite subcovering S(XI'W) . . . S(xn,W). Let g E ni'=!l {x;} ,W]. Then if x E K, x E S(Xj'W) for some j, so g(x) E S(g(Xj),W). Hence, there are WI, W 2 E W with x, Xl E WI and g(x), g(Xj) E W 2· Since g E [{Xj},WJ, there exists a W3 E W with Xj' g(Xj) E W 3 • Therefore, S(g(x),V) c Sex, U). Consequently, g E [K, U], so ni'=l [{Xi} ,W] c [K,U]. Moreover, ni'=l [{Xi} ,W](x) is open so [K, U](x) is a neighborhood of x.
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Let y E [K, U]. Then y = g(x) for some g E [K, U] which implies there exists an [F,VJ E ~ with g © [F,v] c [K, U]. Now g(x) E [g © [F,v]](x), since the identity i E [F,V], and [g © [F,V]](x) c [K, U](x). [g © [F,V]](x) is a neighborhood of g(x) since [F,V](x) is a neighborhood of x and g is a homeomorphism of X onto itself. Therefore, y = g(x) E [g © [F,v]](x) c [K,U](x) so [K, U] is a neighborhood of each of its points. Hence [K, Uj is open. Finally, for each [K, U] E ~, 8([K, U]) = [K, U](z) is open. Moreover, if g G, then 8(g © [K, U]) = g([K, U](z» which is open in X since g is a homeomorphism. Therefore, 8 is an open mapping. But then <1> is a homeomorphism, so (X, ~) is homeomorphic to the quotient G/H of the topological group G.· E
COROLLARY 9.2 The class of locally compact isogeneous uniform spaces is the same as the class of locally compact quotients of topological groups. Define an isogeneous uniform space (X, ~) to be strictly isogeneous if there exists an isometric basis v for ~ such that the group G of all isomophisms with respect to v contains a subgroup H such that for each x,y E X, there exists only one i E H with i(x) = y. Define (X, ~) to be perfectly isogeneous if G contains a commutative subgroup H such that for each x,y E X, there exists an i E H with i(x) =y.
COROLLARY 9.3 The class of locally compact perfectly isogeneous uniform spaces is the same as the class of locally compact abelian topological groups. Proof: Clearly, a locally compact abelian topological group G is a locally
compact perfectly isogeneous uniform space with respect to the basis v consisting of all coverings of the form {g + V Ig E G I where V is an open neighborhood of the identity e E G, and the family H of all translations Tg defined by Tg(x) = g + x for all g E G. To show the converse, let (X, ~) be a locally compact isogeneous uniform space and let v be an isometric basis for ~ such that the group G of all isomorphisms contains a commutative subgroup H such that for each x,y E X, there exists a j E H with j(x) = y. We can now replace G by H in the proof of Theorem 9.6 so that H becomes not only a topological group, but Abelian. Then, just as in the proof of Theorem 9.6, let z be a fixed element of X and put K = (h E HI h(z) = z I, so that H/K is homeomorphic with X. Let k E K and let x E X. Pick j E H such that j(z) = x. Then k(x) = kU(z» = j(k(z» = )(z) = X so that k = i, the identity mapping on X. Therefore, K consists of the single element i so that H/K = H. Consequently, X is homeomorphic with the abelian topological group H. Since X is locally compact, so is H .•
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COROLLARY 9.4 The class of locally compact strictly isogeneous uniform spaces is the same as the class of locally compact topological groups.
The proof is similar to the proof of the corollary above.
EXERCISES 1. Prove Proposition 9.4.
2. Let (X, +) be a topological group with topology "C and neighborhood system N at O. For each V E N put Uv = {(x, y) E x2 1 y - x E V} and let B = {U v I V E N}. Show that B is the basis for an entourage uniformity U that generates the topology of"C and that for each W E ~ (defined in Theorem 9.4), U {WxW IW E W} is a member of U and for each V E U, Vv = {V[x] Ix EX} is a member of ~.
3. Let (X, +) be a topological group and let ~ and ~' be the right and left uniformities on X respectively. Show that (X, ~) and (X, ~') are isogeneous unifonn spaces with respect to the collections R = {rx Ix E X} and L = {Ix Ix E X} of right and left translations respecti vel y and bases A and v as defined in Theorem 9.4 and following. 4. Prove Proposition 9.8. 5. Show that a continuous homomorphism h:X ~ Y where (X, +) and (Y, +) are topological groups is uniformly continuous with respect to either the right or left uniformities on X and Y respectively. 6. Show that a topological group is pseudo-metrizable if and only if it is first countable and metrizable if and only if it is both first countable and Hausdorff. 7. Show that if H is a normal subgroup of a group G then the quotient G/H is a group with respect to the operation + defined on G/H by (H + x) + (H + y) = H + (x + y) for each pair (H + x), (H + y) E G/H. 8. Show that if the group G in Exercise 7 is a topological group that G/H is a topological group with respect to the quotient topology. 9. Show that the canonical projection 1t from a topological group X onto its quotient Q = X/Z over a subgroup Z of X is an open mapping. 10. Show that the quotient topological group R/Z of the additive topological group R of real numbers over its subgroup Z of integers is unifonnly homeomorphic to the circle group T.
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RESEARCH PROBLEMS
11. Are isogeneous unifonn spaces equivalent to quotients of topological groups without the assumption of local compactness? 12. What is a necessary and sufficient condition for a locally compact unifonn space to have a Haar measure? 13. Find a necessary and sufficient condition for a unifonn space (not necessarily locally compact) to have a topological group structure that generates the unifonnity.
Chapter 10
UNIFORM MEASURES
10.1 Introduction In 1945, L. Loomis introduced an interesting generalization of the Haar measure to locally compact metric spaces that satisfy a property he called the congruence axiom in a paper titled Abstract congruence and the uniqueness of Haar measure (Annals of Mathematics, Volume 46). Loomis' concept of invariance in this paper did not involve a translation (transformation), but rather the property that for a measure I-l. I-l(S 1 ) = I-l(S 2) whenever S 1. and S 2 are compact spheres with the same radius. Because of this distinction, we will refer to these measures as uniform measures rather than Haar measures. As we shall see, Loomis' concept, when applied to uniform spaces, generalizes the concept of Haar measure as presented in the last chapter. In this paper, Loomis showed there exists a unique, regular, invariant (uniform) measure on the Lebesgue ring (Borel sets that can be covered by countably many compact sets) on locally compact, metric spaces that satisfy the congruence axiom. On page 354 of this paper, Loomis mistakenly states that the proofs of the theorems that establish the existence and uniqueness of this uniform measure in locally compact, metric spaces that satisfy the congruence axiom "hold without modification" in locally compact uniform spaces that satisfy the congruence axiom. In 1949, in a paper titled Haar measure in uniform structures (Duke Math Journal, Volume 16), Loomis attempted to extend his generalization of Haar measure to locally compact uniform spaces satisfying the congruence axiom using a different approach than in the 1945 paper. Loomis' 1949 proof was not entirely correct either and, strictly speaking, this is still an open problem. However, we will show that a slightly weakened version of Loomis' claimed results still holds on the Baire ring. In the 1945 paper, Loomis first constructed an outer measure m, on the collection of sets that could be covered by countably many compact sets, that had the invariance (uniform) property and then referred to the classical result that in a locally compact, metric space, the members of the Lebesgue ring are all m*-measllrable (we will not prove that result here as our interest is in general proofs for uniform spaces). Thus, by restricting m* to the Lebesgue ring, he obtained a measure m on the Lebesgue ring with the desired properties. In the 1949 paper, Loomis tries to build the measure up from a more elementary
10.2 Prerings and Loomis Contents
285
set function instead. In the first eight sections of the 1949 paper, Loomis constructs (without the aid of local compactness) a set function A that IS unique (in a sense Loomis defines) on the collection of totally bounded open sets such that whenever S I and S 2 are two totally bounded spheres of the "same radIUs" (generated by the same uniform covering) then A(S d is "almost always" equal to A(S 2) in a sense he defines and calls "invariance." Loomis' development contains some errors. The reader can judge their seriousness but should not confuse these gaps in logic with an unsound approach. Loomis claims A can be extended to a legitimate measure A* on the "Borel field generated by the compact sets." But because of the errors and inconsistencies of definitions (e.g., measure and content) with the classical ones, the classical results to which Loomis appears to be referring (as no proof was given) do not apply. However, Loomis' insight was keen. About half of the results are correct and the other half can be salvaged with better definitions and modified theorems and proofs to obtain a somewhat weaker result (but more nearly like the result in his original metric space paper). The notion of an almost uniform measure is introduced and it is shown that such a measure exists (on the Baire sets) and is unique. Consequently, it is still an open problem whether a uniform measure exists on a locally compact uniform space under Loomis' original assumptions and if so, is it unique? In the next section Loomis' development of these invariant set functions (that we will refer to as Loomis contents) is presented. We will introduce new terminology to clarify the difference between Loomis' definitions and axioms and the classical ones, and in some cases change Loomis' definitions and axioms so they work.
10.2 Prerings and Loomis Contents
In this section, we will find it notationally convenient to use a E Il to mean that the covering a belongs to the uniformity Il rather than our usual notation U E Il. This is because members of the uniformity Il will frequently be used to index various other collections. If (X, Il) is a uniform space and A and B are subsets of X, Loomis defined A < B to mean there exists a u E Il with Star(A, u) c B. If a, bEll, we define a < b to mean there exists aCE Il with Star(S(p,a),c) c S(p, b) for each p E X. Since < is defined on f..!, this could cause confUSIon with the refinement ordering. Therefore, in this chapter, if we wish to denote the refinement ordering, we will use the "upper case" notation of U < V rather than a < b. PROPOSITION fO.1 If (X, Il) is a uniform space then (Il,<) is a directed set. Proof: To show < is transitive suppose a, b, CEil with a < b and b < c. Then
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there exists a', b' E j..l with Star(S(p, a), a' ) c S(p, b) and Star(S(p, b), b') c S(P,c) for each P E X. But then a < c. To show j..l is directed by < note that there exists a Z E j..l with z <* a and z <* b. Then Star(S(p, z), z) c S(p, a) and Star(S(p, z), z) c S(p, b) so z < a and z < b.PROPOSITION 10.2 If a, b
E j..l
with a
< b then there is aCE
j..l
with a
< C < b. Proof: If a < b, there exists a z E j..l with Star(S(p, a), z) c S(p, b) for each P E X. Let x E j..l such that x <* z. Put c = {Star(U, x) IU E a}. Clearly c E j..l. Let qE X. ThenStar(S(p,a),x)=u{VE xIVnS(p,a)700}=u{WE clWnU70 0forsomeUE awithqE U} =U{WE clqE W} = S(p, c). Hencea
To show c < b suppose r E Star(S(q, c), x). Then there exists V,W E x and U E a with rEV, VnW * 0* VnU and q E U. Next, there exists a Z E z with VuW c Z which implies r E Z and ZnU * O, so r E Star(S(q, a), z) c Seq, b). We conclude that Star(S(q, c), x) c Seq, b) for each q E X so C < b.Loomis begins by assuming a "uniform structure" j..l on a space X that is defined as a collection of coverings j..l such that (j..l, <) satisfies Propositions 1O.l and 10.2, i.e., (j..l, <) is directed and satisfies the axiom AI. If x, y E
j..l
with x < y, there is a z E
In addition to axiom AI, Loomis assumes axioms:
j..l
j..l
with x < z < y.
satisfies the following additional
A2. If P E Seq, x) then q E S(p, x). A3. Every y-sphere can be covered by a finite number of x-spheres for each pair x, y E j..l. A4. The smallest number of x-spheres required to cover a y-sphere is the same for all y-spheres for each pair x, y E j..l. It is easily shown that every uniformity j..l satisfies AI, and A2 is part of the definition of a uniformity. So Loomis' "uniform structure" is a generalization of a uniformity. Axioms A3 and A4 are his uniform space version of the congruence axiom for metric spaces. In this chapter we will restrict ourselves to uniform spaces rather than these more general spaces. Loomis' problems begin with his axioms. Axioms A3 and A4 rule out most interesting examples of uniform spaces. Either one of them, for instance, rules out the real numbers R with the usual metric uniformity. Since in a uniform space, the space itself is an x-sphere for x = {X}, A3 implies the space is precompact which R clearly is not because if it were, its completeness would then imply compactness. Similarly R cannot satisfy axiom A4. To see this, consider the uniform covering x consisting of the open interval (-10, 10) and the
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287
open intervals (n-I, n+ 1) for each integer n with In I 2: 10. Let y be the uniform covering {(r-I, r+ 1) IrE R}. The x-sphere about 0 is (-10, 10) while the x-sphere about 12 is (11, 13). Now (11, 13) can be covered by a single y-sphere while (-10, 10) cannot. Surely Loomis did not mean to rule out the most well known uniform space with the archetypical Haar measure when he undertook to generalize the concept of Haar measure to uniform spaces. Consequently, we will assume instead that f..l generates the uniformity rather than being the uniform structure itself, i.e., we assume that f..l is a basis for a uniformity v. We also need to make adjustments to axioms A3 and A4. We assume they only hold for the basis f..l rather than the uniformity v, i.e., that the x and y mentioned in A3 and A4 are members of f..l. Furthermore, this change in assumptions requires a change in the underlying definition of the ordering <. We define A < B to mean that Star(A, u) c B for some U E f..l rather than U E V. A similar change is needed in the definition of a < b where a, b E f..l. In this section and the next, the part of Loomis' development that is essentially correct will be presented. The order in which this material is presented has been rearranged somewhat from Loomis' original paper and the terminology has been translated into the terminology that has just been introduced. Also, some of the details omitted in Loomis' original proofs have been included so the reader may be convinced it is all here. It is often in these areas where Loomis omitted proofs altogether that the difficulties are hidden. PROPOSITION 10.3 For each pair of comparable sets A 0 < A 1 there is an interval {A a Ia E [0, I]} of sets with CI(A a) < Int(A 13) whenever a < ~. Proof: Since A 0 < AI, there exists x, y E f.l with Star(A 0, x) CAl and y <* x. Put A 112 = Star(A 0, y). Then A 0 < A 112 < AI' Let R denote the rationals in [0,11. Well order R as a sequence {rn} such that rl = 0, r2 = 1/2 and r3 = 1. Then ArJ < Ar2 < A r3 . Define Ar. recursively so that Arm < Arn whenever r m < rn. This can be done by letting an = max{rk Ik < n and rk < rn} and b n = min{rk Ik < n and rn < rd. Then an < rn < b n so Aan < A b•. As shown above, it is possible to pick Ar. with Aa n < Arn < Abn' Thus for {r 1 ••• r n } we have Arm < Ar. whenever r m < rn' Clearly, this recursive definition produces a sequence {Ar.} of sets indexed by the members of R such that Ar < As whenever r, s E R
with r < s. Now for each irrational a E [0, 11 put Aa = u{Ar IrE Rand r < a}. Let s R such that a < s. Then for each t E R with t < a, AI < As so At cAs. Consequently, A a C n {As Is E R and a < s}. To show that A a < A 13 whenever a < ~ are in [0, 1], note that there exists a pair of rationals r, s E [0, 1] with a < r < s <~. Now Aa c n{Atl t E R and a < t} cAr and As c u{At I t E Rand t < ~} c A 13' Since r and s are rationals, there exists an x E f.l with Star(A" x) cAs· Hence Star(A a , x) c A 13, so Aa < A 13' Therefore, there exists an interval {A a} E
288
10. Uniform Measures
of sets with Au < A[3 whenever a <~. But Au < A[3 implies CI(Au) < Int(A[3) by the definition of <. Loomis' definition of a measure ring fails to be a ring in the standard sense. For this reason, we call such an object a prering instead. In fact, we adopt the following definitions in order to distingUish Loomis' concepts from the standard ones. Let C be a class of sets and A a collection of members of C. A is said to be a prering in C if A is closed under finite unions. A is hereditary in C if whenever A E A and B E C with B c A then B E A. A hereditarily open prering is a hereditary prering in the class of open sets. A Loomis content is a set function A on a hereditarily open prering P that satisfies the following four properties: Ll. IfU, V E P with U c V then A(U) s; A(V). L2. A(UUV) s; A(Star(U, x» + A(Star(V, x» for each x E /-l. L3. A(UUV) = A(U) + A(V) if Star(U, z)nStar(V, z) = 0 for some z E /-l. Sets U, V c X for which Star(U, z)nStar(V, z) = 0 for some z E /-l are said to be uniformly separated. Note that L3 implies A(0) = O. A is said to be left continuous if it satisfies L4. A(U)
= sup{A(V)lv
A subset P' of P is said to be dense in P with respect to < if for each U, V E P with U < V, there is aWE P' with U < W < V THEOREM 10.1 Let P' be dense with respect to < in a hereditarily open pre ring P and suppose that if U and Vare uniformly separated in P' then UuV E P. Let A be a set function defined on P' that satisfies Ll - L4. Then A has a unique extension to P that is a Loomis content on P. Proof: For each U E P put A*(U) = sup {A(V) I V < U and V E P'}. A* is easily seen to be an extension of A to P. The proof that A* satisfies Ll is trivial. To show A* satisfies L4, note that if WE P with W < U, then there exists a V E P' with W < V < U and by Ll, A*(W) < A*(V), so A*(U) = sup {A*(W) IW < U and W E P}. Therefore A* satisfies L4. To prove L2, let U, V E P and let x E /-l.
Then (10.1 )
A*(UUV)
= sup{A(W)IW
If W < UuV, there exists an a E /-l such that Star(Star(W, a), a) c UuV. Put = W - Star(X - U, a) and Wv = W - Star(X - V, a). Then Wu and Wv are uniformly separated, Star(W u , a) c U, Star(Wv , a) c V,
Wu
10.2 Prerings and Loomis Contents
289
Star(W u , a)uStar(W v , a) c Star(W, a)
and both W u and W v are in P'. Hence [W u u W v] E P'. Therefore, A(W) ::; A(Star(W, a» = A(Star(W u , a)uStar(W v , a» ::; A(Star(Star(W u, a), x» + A(Star(Star(W v , a), x» ::; A*(Star(V,x» + A*(Star(V.x»
for some x E fl. so A* satisfies L2. To show L3, let U, V E P such that for some Z E fl, Star(U, z)nStar(V, x) = = WnU and Wv = WnV. Then Wu < U and Wv < V are unifonnly separated and belong to P'. Hence A(W) = A(WuUWv ) = A(W u ) + A(Wv ), Conversely, for each pair W u , Wv E p' with Wu < V and Wv < V we have WuuWv < UuV and Wu and Wv are unifonnly separated. Therefore, by
0. If W < UuV put Wu
(10.1)
A*(VUV)
= SUp{A(W u ) + A(Wv ) I Wu < U, Wv < V and W u , Wv
E P'}
=
sup{A(A) I A < V and A E P'} + sup{A(B) IB < V and BE P'} = A*(U) + A*(V).
To show A* is unique, let I be another Loomis content on P that is an extension of A. Then for V E P, I(V) = SUp{ACW) IW < U and W E P}. For W E P with W < V, there exists a V E P' with W < V < U and since I is an extension of A, leV) = A(V). Hence I(V) = SUp{A(V) IV < U and V E P'} = A*(V). Therefore I =A* so A* is a unique Loomis content on P.Loomis defined the content A to be invariant with respect to x E fl if A(S(P, x» = A(S(q, x» for each pair p, q E X. It is simply invariant if it is invariant with respect to all members x E /.!. In this case, we write A(y) for the common value A(S(P, y» for any y E fl. We will also use the tenninology that some property P holds for sufficiently small x E fl. By this we mean that there exists ayE fl for which P holds, and for all x E /.! with x < y, the property continues to hold. For this section and the next we will restrict our discussion to a Loomis content A. defined on the hereditarily open prering T of totally bounded open sets. A set A is said to have zero-boundary (with respect to A) if for each £ > 0 there exists a closed set F and an open set VET with F < A < U and A(U - F) < E. Let Z denote the zero-boundary sets. To see that there exist many members of Z, let A and B be sets with A < B. By Proposition 10.3 it is possible to construct an interval {A u} of sets with A 0 = A and Al = B such that for each a < ~ in [0, 1], Au < Aj3. If {Au} c T then A(A u) is an increasing function of a. Hence, from elementary analysis we
10. Uniform Measures
290
know that A(A a) is continuous except (at most) on an increasing sequence {xn} C [0, 1]. If ~ is a continuity point of A(A a), then A p is a zero-boundary member of 1'. To see this, let E > 0. Then there exists a 8 > with IA(A p) - A(A a) I < E whenever a E (~- 8, ~ + 8) and such that A(A a) is continuous in (~ - 8, ~ + 8). Choose r, s E [0, 1] such that ~ - 8 < r < ~ < S < ~ + 8. Then CI(Ar) < A P < As and A(As) - A(Ar) < E. To show that A(As - CI(Ar» < E, note that for each t E [0,1] with ~ - 8 < t < r we have (by L3):
°
since AI and As - CI(Ar) are uniformly separated because AI < Ar < As. Hence
But A(A a) continuous in (~ - 8, ~ + 8) implies that limHrA(A,) = A(Ar) so A(As Cl(Ar» + A(Ar) ~ A(As). Therefore, A(As - CI(Ar» ~ A(As) - A(A r ) < E, so Ap is a zero-boundary member of T. Conversely, if A is a zero-boundary set with respect to A, then its interior is a member of an interval of open sets I A a} C T at which A(A a) is continuous. To see this, we can choose a closed set F 1 and an open set U 1 E T such that F 1 < Int(A) c CI(A) < Uland A(U 1 - F d < 1. Put 0 1 = Int(F d. Next choose a closed set K 2 and an open set V 2 E T with K 2 < Int(A) c CI(A) < V 2 and A(V 2 - K 2) < 1/2. Put F 2 = F 1 uK 2 and U 2 = U 1 (IV 2. Clearly F i < F 2 < Int(A) c CI(A) < U 2 < U 1. Also, let O 2 = Int(F 2)' Continuing this process by induction, we get two sequences {On} and {Un} of members of T such that On < On+l < Int(A) c CI(A) < Un+ 1 < Un for each positive integer n and A(Un CI(On» < 1/n. Now we can use Proposition 10.3 to fill in intervals of open sets between the On and On+l and between the Un+ 1 and Un for each n. Thus the interior of A is a member of an interval of open sets {A a} such that (by construction) each A a E T and such that A a < A P whenever a <~. Moreover, by construction, it is clear that for each E > 0, there exists a pair a, ~ with A a < Int(A) c CI(A) <Ap and A(Ap - Cl(Aa» < E, so A(Aa) is continuousatlnt(A). COROLLARY 10.1 The members of Z are totally bounded (but not necessarily open). A is uniquely determined on T by its values on the zeroboundary members ofT. Proof· The zero-boundary members of T have the properties of P' in the hypothesis of Theorem 10.1 when we put P = T, i.e., they are dense in T, so that A is uniquely determined on T by its values on the zero-boundary members of T.
-
There is a natural way to extend A to all members of Z (not necessarily in We have seen from the construction above that if A E Z then Int(A) is a continuity point of A(A a) for some interval {A a} C T where A a < A p whenever
n.
291
10.2 Prerings and Loomis Contents
P
a < in [0, 1]. Hence if Int(A) = Ar then limHrA(Ar) = A(Ar) so sup{ACO)llnt(A»OE T) A(Ar) inj{A(O)llnt(A) < OE T). IfAr
=
=
=
= =
=
For zero-boundary open sets U and V we have A(UUV) = ACU) + A(V). To see this, first note that if U and V are zero-boundary open sets then A(U) = inj{A(Star(U, x»lx E III and A(V) = inj{A(Star(V. x»lx E Ill. Then by L2, A(UUV) ~ A(Star(U, x» + A(Star(V, x» for each x E 11. Therefore, A(UUV) = injxACUuV) ~ injxA(Star(U, x» + injxA(Star(V, x» = A(U) + A(V). THEOREM 10.2 Z is a ring and A is afinitely additive measure on Z. Prooj: Let A, B E Z. As shown above, there exists intervals {A a I and (B i3 I of members of T with Int(A) = Ar and Int(B) = Bs for some r, s E [0,1] and Ar and Bs continuity points of A(A a) and A(B~) respectively. For each £ > 0, there exists a, y, bE [0,1] with A a < Ar < A ~ and By < Bs < B 8 such that both A(A ~ - C/(Aa» < £/2 and ACB8 - C/(By» < £/2. Now C/(Aa)uC/(B y) < AruBs < A~uB8 and (A~uB8) - (C/(Aa)uC/(B y» C (Ai3 - C/(A a »u(B 8 - C/(By». Clearly both A ~ - C/(A a) and B 8 - C/(B y) are zero-boundary open sets so by the remarks preceding this theorem, A[(A ~uB 8) - (C/(A a)uC/(B y»] ~ A(A ~ C/(A a» + A(B 8 - C/(B y» < £. It follows that AruBs E Z which implies AuB E
p,
Z.
Similarly, C/(Aa) - C/(B 8) < Ar - Bs < A~ - C/(By) and (A 13 - C/(B y»(C/(A a ) - B 8) C (A~ - C/(A a »u(B 8 - C/(By», so A[(A~ - C/(By» - (C/(Aa) B 8)] < £. Hence Ar - Bs E Z which implies A - B E Z. Thus Z is a ring. Now if AnB = 0, then for each A a < Ar and B 13 < B" C/(A a) and C/(B 13) are uniformly separated, so by L3, A(C/(A a)uC/(B ~» = A(CI(A a» + A(C/(B ~». Let <1>:[0, 1] ---7 [0, 11 be strictly increasing, onto and have (r) = s. For each a E [0,1], put C a =B q,(a). Then (A aUC a I is an interval of members of T with A aUC a < A l3uC 13 whenever a < p. Moreover, for each e > 0, there exists a, p, y, b E [0,1] with C/(Aa) < Ar < A i3 , C/(C y) < Bs < C 8 and A(A i3 - C/(Aa» < e/2 and A(C 8 - CI(C y» < £/2. An argument similar to the one above yields A[(Ai3uC8) - (CI(Aa)uCI(C y))] < e, so AruBs is a continuity point of A(A aUC a) which implies sUPa
=
For each a E [0,1], C a Bq,(a), so if a < r then Aa < A and C a < B so C/(A a) and CI(C a) are uniformly separated which implies A(CI(A a)uCI(C a» = A(C/(A a » + A(CI(C a»· Hence sUPa
»
10. Uniform Measures
292
We can generalize Propositions 8.4 and follows:
~(5
to Loomis contents on
r
as
PROPOSI710N 10.4 Each Loomis content A on T has the following properties: L5. If each point of 0 E T lies in at least N of the sets {O 1 •.. OM} c T then A(O) :; r/:!=l A(Star(On' x))/N for each x E ~. L6. If (O 1 ... OM) are subsets of 0 E T and no point of 0 lies in more than N of the On' then A(O)?:. 'L~=lA(On)/N. The proof is left as an exercise.
EXERCISE 1. Prove Proposition lOA Hint: Use Proposition 8.4 and 8.5 on Z and approximate the members 0 and On with memhers of Z.
10.3 The Haar Functions We define the Haar covering function h(A. x), for totally bounded sets A E X and x E ~, to be the smallest number of x-spheres required to cover A. By axiom A4 we have that h(S(P, y), x) = h(S(q, y), x) for each pair p. q E X and x, y E ~. We denote this common value by h(y, x). The Haar function H(A, x) is the inf of the fractions MIN such that for some y < x, there exists a collection of My-spheres covering each point of A at least N times. If y < x, clearly H(A. x) :; H(A, y).
PROPOSITION 10.5 H(A, x) = infIH(A, y) I y < x}. Proof: Suppose H(A, x) "# infIH(A,y) I y < x). Then clearly H(A, x) < infIH(A,y) I y < x}. By the definition of H(A, x). there exists positive integers M and N with H(A. x) < MIN < infIH(A,y) Iy < x} and M w-spheres covering each point of A at least N times for some w < x. By Proposition 10.2, there exists a y E I-l with w < Y < x. Then by the definition of H(A,y) we have H(A,y) :; MIN. which is a contradiction. Therefore, H(A. x) = inf{H(A.y) Iy < x 1.LEMMA 10.1 If a collection U of N open r-spheres covers no point more than n times, and if each point covered by U is covered by at least m spheres from a collection V of M open r-spheres then Nln:; Mlm. Proof: Let U* be a set consisting of m duplicates of each sphere in U and let V* be a set consisting of n duplicates of each sphere in V. If p is the center of a sphere in U* then p is the center of a sphere in U and therefore lies in at least m members of V. Since V* was fonned by duplicating each sphere in V n times,
10.3 The Haar Functions
293
p lies in at least mn formally distinct members of V*. Since U* was formed by duplicating each member of U m times, there are Nm formally distinct centers of spheres in U*. Hence there are at least Nm 2 n formally distinct members of
H
= 1(p, q) I SV),r) E
U* and p
E
S(q,r) E V* l.
On the other hand, if a sphere in V* contained more than mn formally distinct members of U* then its center, say q. would lie in more than mn formally distinct members of U*. Since each member of U was duplicated m times to form U*. this would imply q was contained in more than n members of U. But this is not possible since U covers no point more than n times. Therefore, a sphere in V* can contain at most mn formally distinct centers of spheres in U*. Since there are Mn formally distinct members of H, there can be at most Mmn 2 formally distinct members of H. Thus Nm 2 n ~ Mmn 2 or N In ~M/m.-
LEMMA 10.2 If A E Tand r E !l with Star(A, r) c AT, and Loomis content on T, invariant with respect to r, then for r < r,
if A is
a
O
-
(10.2)
H(A,r) ~ h(A+,x)lhV,x) and A(A)IA(r) ~ h(A+,x)lh(r-,x)
for all sufficiently small x
E
!l.
Proof: Let L be a family of h(A +, x) x-spheres covering A +. Then L covers S(q,r-) for each q E A so at least h(r-, x) members of L meet Seq, r-). If x is sufficiently small that Star(S{jJ, r-), x) c S{jJ, r) for each p E X, then at least h(r-, x) of the centers of members of L lie in Seq, r-). But then q lies in at least h(r-, x) of the r-spheres concentnc with the spheres in L. Let L* be the family
of r-spheres whose centers are concentric with members of L. Then L* is a collection of h(A +, x) r-spheres covering each point of A at least h(r-, x) times so the first inequality in (10.2) follows from the definition of H(A, r). To prove the second inequality in (10.2), assume A is covered by M r- -spheres, say S I . . . SM that cover each point of A at least N times. Then by L5 of Proposition 10.4, A(A) ~ L':!=l A(Star(Sn, x»/N. Since for each p E X, Star(S{jJ, r-» c S{jJ, r), we have A(Star(Sn, x» ~ A(r) for each n = I ... M because A is invariant with respect to r. Therefore, A(A) ~ MA(r)N which implies A(A)/A(r) ~ M/N. By the definition of H(A, r) we get A(A)/A(r) ~ H(A,r). Then the first inequality in (10.2) implies the second.-
10. Uniform Measures
294
LEMMA 10.3 If A E T, and r E J.l with Star(A -. r) c A, and I-.. is a Loomis content on T, invariant with respect to some r- < r, then, (10.3)
H(A,r-) ?: h(A-.x)lh(r,x) and I-..(A)II-..(r-) ?: h(A-,x)lh(r,x)
for all sufficiently small x E J.l. Proof: Let L be a family of h(A -, x) x-spheres covering A -. If q E A, then at most her, x) of the members of L can lie in Seq, r), for otherwise these spheres could be replaced by her, x) x-spheres covering Seq, r), yielding a covering of A - by fewer than h(A -, x) x-spheres which is a contradiction. Thus if x is sufficiently small that Star(S(p, r-), x) c S(p, r) for each p E X, at most her, x) of the members of L have centers in Seq, r-) which implies q lies in at most h(r,x) of the r- -spheres concentric with the x-spheres in L. Then the first inequality in (10.3) follows from Lemma 10.1 and the definition of H(A, r-).
To prove the second inequality in (10.3), let L* be the r- -spheres concentric with the x-spheres in L. Then L* is a h(A -,x) r- -spheres that covers each point of A at most her, x) Star(A -, r) c A we have Star(A -, r-) c A so each member of Then by L6 of Proposition lOA, we have
collection of collection of times. Since L* lies in A.
I-..(A) ?: LS E };*I-..(S)/h(r, x) = h(A -, x)l-..(r-)/h(r, x)
since I-.. is invariant with respect to r-. Let M = h(A - , x) and N = her, x). Then I-..(A)/I-..(r-) ?: M/N so by the definition of H(A, r-), we get I-..(A)/I-..(r-) ?: H(A, r-). Hence the first inequality in (10.3) implies the second.LEMMA 10.4
If A < A + and B- < B where both A + and BET, then
(10.4)
· H (A,r) < t· ' .• h (A + ,x) and t1m sUPr - 1m In)x H(B,r) h(B-,X)
(10.5 )
h(B-,x) < t· ' •• H(B,r) · t1m supx - 1m In) r (A ). h(A +,x) H ,r
If A and B- have non-void interiors, then the right members of (10.4) and (10.5) are finite and positive. Proof: Let r E J.l such that Star(A, r) c A + and Star(B-, r) c B. Then for,- < r, Star(B-, r-) c B. Let r-- < r-. Then substituting B for A in the first inequality of (10.3) and B- for A -, r- for r and r-- for r- yields:
(10.6)
10.3 The Haar Functions
295
Dividing the first inequality in (10.2) by (10.6) gives: (10.7)
H (A,r) H (8,r--)
for all sufficiently small x. Since (10.7) holds for all r-- < r, by Proposition 10.5 we get: (10.8)
H(A,r) H(8,r)
for all sufficiently small x. Taking the limit inferior of the right hand side of (10.8) yields: (10.9)
H(A,r)
Since (10.9) holds for all sufficiently small r, (1004) follows. Now, the reciprocal inequality to (10.8) is: (10.10)
~
H(8,r) H(A,r)
so by an argument similar to the above we get (10.5). In order to use Lemma lOA to construct a measure on Z we will need the following result: LEMMA 10.5 Let Ra, ~) be a real valued function of two real variables a, ~ E [0, 1] that increases as a and ~ increase. Then the discontinuities off lie on a countable family of decreasing curves (allowing vertical and horizontal segments as curve arcs). Proof: Since f is increasing in both a and ~, a point (a, ~) at which f is not continuous has the property that if a- < a < a+ and ~- < ~ < Wthen
and the difference !::.f(a, ~) between the right and left sides of this inequality is referred to as the jump offat (a, ~). Let E£ be the set of points (a,~) at which !::.f(a, ~) ~ e. We will call a finite set of points (ai, ~J for i = 1 ... n an increasing chain if ai < ai+! and Pi < ~i+! for each i = 1 ... n - 1. If the points (ai, ~i) are chosen from EE then clearly n ~ [f(I, 1) - f(O, O)]/e. Let C! be the set of points of E E which are first points of increasing chains in E E but not second points of such chains. Then no pair of points in C! can determine a positive slope for otherwise one of them would be an increasing chain in E E' A
296
10. Uniform Measures
set of points in [0, 1] x [0, 1], no pair of which determine a positive slope, lies on a decreasing curve (where vertical and horizontal segments are allowed as curve arcs - see Exercise 1). Consequently, C 1 can be imbedded in a decreasing curve. Similarly, let C 2 be the set of points in E E which are second points of increasing chains in E E but not third points. Clearly, no pair of points in C 2 determine a positive slope so C 2 can be imbedded in a decreasing curve. Continuing in this manner, we see that E € can be imbedded in n decreasing curves for some n ::; [[(1,1)-f(0,0)]/E. Now if we choose a decreasing sequence {En} converging to zero, the corresponding sets {E En } include all the points of discontinuities of f and the corresponding finite sets of decreasing curves form the countable collection of such curves in the conclusion of the lemma. LEMMA 10.6 Let {Bi3} c T be an interval of sets with non-void interiors. Then there exists an index Po E (0, 1) and a unique increasing function g(P) that is continuous and equal to 1 at Po, such thatfor each interval {A u} c T of sets with non-void interiors, there exists a unique, increasing functionf(a) with
where the limits exist at any point (a, continuous.
P) for
which the functions f and g are
Proof: If {A u} and {B i3} are intervals of totally bounded open sets with non-
void interiors (as obtained by Proposition 10.4). then by Lemma lOA, for sufficiently small E > 0, (10.12)
h(Au_E'x) H(Au,r) H(Au,r) lim supx---- ::; lim infr ( )::; lim sUPr ::; h(Bi3+€,x) H Bi3,r H(B i3 ,r)
(10.13)
lim infr
H(Au+€,r) ( ) . H B~€,r
The four functions of (10.12) and (10.13) are all increasing functions as a increases and P decreases. Since (10.12) and (10.13) hold for any SUfficiently small E > 0, they must be identical for any point where anyone of them is
10.3 The Haar Functions
297
continuous (and hence identical with respect to their corresponding limits as opposed to their limits superior). For each (a, ~) E [0, 1] x [0, 1], put g'(a, ~) = lim supxh(A a, x)/h(B 13, x). Then the function defined by f(a, ~) = g'(a, ~) satisfies the hypothesis of Lemma 10.5, so its set of discontinuities lies on a countable family of decreasing curves which implies the set of discontinuities of g' lie on a countable family F of increasing curves. Then at all points (a, ~) of continuity, g'(a, ~) is the common limit of all four functions in (10.12) and (10.13). Pick ~o so that none of the curves of F contains a horizontal segment with ordinate ~o. Then (a, ~o) is a continuity point for g' for all a except a countable exceptional set that we denote by E 130' Then
is a function only of ~, say g(~), and is therefore independent of the interval of sets {Aa}, except for the original choice of ~o. Definefby f(a) = g(a, ~o). Then, since ~o is a continuity point of g, f(a)
=
g (~)
Now f depends only on the interval {A a }. Furthermore, for any index a, the set A a can be replaced by either its interior or closure without altering its position in the interval (due to the method of construction in Proposition 10.3), the convergence of the ratios, or the value of f(a). Consequently, any other continuity point of g could have been chosen (which would multiply both f and g by a common factor, leaving the ratio unchanged), and the proof would remain valid. Let ~o be the index in the hypothesis of Lemma 10.6 and let P' be the collection of sets A that belong to an interval {A a} C T such that if A = A r then y is a continuity point of the function f(a)/g(~o) of Lemma 10.6. Put A(A) = limxh(A, x)/h(B 130' x) = lim r H(A,r)/H(Bf3o' r). THEOREM 10.3 (L. Loomis. 1949) A is a Loomis content on T and hence can be extended to afinitely additive measure on Z. Moreover P' = ZnT.
Proof: We first show that P' is dense in T with respect to < and that if U, V E P' are uniformly separated then UuV E P'. If U, V E P', there exist intervals {Aa), {Bf3} c Twith Aa < Af3 and Ba < Bf3 whenever a < ~, and such that U = A e and V = B ~ for some 8, ~ that are continuity points of A(A a) and A(B 13) respectively. Let <1>[0, 1] ----+ [0, 1] be a strictly increasing, onto function with
298
10. Uniform Measures
<1>(8) = ~ and for each a E [0, 1] put C a = B <)I(a)' Then as shown in the proof of Theorem 10.2, {A aUC a} C T is an interval with A aUC a < A f3uC f3 whenever a < ~. To show UuV E P', let £ > 0 and pick a,~, y, 8 E [0,1] with Aa < U < A f3 and By < V < B'6 such that A(A f3 - CI(A a» < £/2 and A(B'6 - CI(B y» < £/2. Now U = Ae and V = C e , so there exist r, s E [0,1] withA r < U < As, C r < V < Cs, A(As - CI(Ar» < £/2 and A(Cs - CI(C r » < £/2. Clearly AruC r < UuV < AsuCs and AsuCs - CI(AruC r ) c [As - CI(Ar)]u[C - CI(C r )] so A(AsuCs AruCr ) < A([As - CI(Ar)]u[Cs - CI(C r )]). Hence it only remains to show that A is sub-additive on p' to show that A(AsuCs - CI(AruC r » < £. For this note that for any A, B E p' that h(AuB, x) :S: h(A, x) + h(B, x) for each x E /-!, so A(AuB) :S: A(A) + A(B). Consequently, UuV is a continuity point of A(A aUC a) so UuV E P'. To show p' is dense in T, let A, BET with A < B. By Proposition 10.3, there is an interval {A a} C T with A 0 = A and A I = B and A a < A f3 whenever a < ~. By Lemma 10.6 we can choose a y such that A < A y < Band y is a continuity point of A(A a). Therefore A 'Y E pi so p' is dense in T. That A satisfies L1 on p' is clear from the definition of A. L2 and L3 follow from the fact already shown above that for A, B E pI, h(AuB, x) :S: h(A, x) + h(B, x) for each x E /-! and from the fact that h(AuB) = h(A, x) + h(B, x) if Star(A, x)nStar(B, x) = 0. L4 follows from the fact that if A E P' then H(A, x) = sup {H(A, y) Iy < x} (Proposition 10.6) and the fact that p' is dense in T. Therefore p' satisfies the conditions of Theorem 10.1 if we set P = T, so A can be uniquely extended to a Loomis content on T. Hence by Theorem 10.2, A is a finitely additive measure on the ring Z of zero-boundary sets with respect to A. Since A E p' implies A is a continuity point of A(A a) for some {A a} C T with A E {A a}, A is a zero-boundary set with respect to A by the remarks preceding Corollary 10.1. Therefore A E ZnT. Conversely, if A E Zn T then A is a continuity point of A with respect to some interval {A a} C T that contains A (also by the remarks preceding Corollary 10.1). Hence A E P'. Therefore, P'
=ZnT.This concludes the correct part of Loomis' development. In the final two sections, a development of an only marginally weaker version of Loomis' claims will be presented that is closer in spirit to the approach of Loomis' original construction of a uniform measure on a metric space.
EXERCISE 1. Show that a set of points in [0, 1] x [0, 1], no pair of which determines a positive slope, lie on a decreasing curve (allowing vertical and horizontal segments as curve arcs).
lOA Invariance and Uniqueness of Loomis Contents and Haar Measures
299
10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures In this section we first prove the in variance and uniqueness of a Loomis content on its ring Z of zero-boundary sets. We will then use this result to prove the uniqueness of the Haar measure developed in the previous chapter. We have been using 11 not only as a basis for the uniformity v, but also as an index set for defining limits as in the definition of A in Theorem 10.3. We now want to expand our index set 11 to an index set 11* in such a way that we can have intervals of indicies from 11* analogous to the intervals of sets {Aa} such that CI(A a) < Int(A 13) whenever a < P in Proposition 10.3. For any x, yEll with x < Y, we use axiom A 1 to construct a collection {y a} C 11 where a ranges over the rationals in [0, 1], Yo = x, Y I = Y and Ya < Y 13 whenever a < p. Then for an irrational y E [0, 1] we let Yy be the uniform covering defined by Yy = {Yy(P) Ip E X} where Yy(p) = ua
300
10. Uniform Measures
The function A(A) = limxh(A. x)/h(Bo. x) defined by Theorem 10.3 is invariant with respect to all but countably many indicies from any interval of indicies. If m is any other Loomis content on T which is invariant with respect to all except countably many indices from every interval of indicies. then m is a constant mUltiple of A on z.
The way we will prove the "invariance" part of this theorem is by relaxing the "amount" of in variance in the hypothesis to only include those intervals {A a} C 11* such that Y'Y E 11 whenever y is rational while still proving the "uniqueness." Moreover, as we will see in the next section, even this amount of relaxation is not enough (using our approach) to prove the existence of a genuine "invariant" measure on the Baire ring. We will further have to relax our in variance requirement to only include "spheres with compact closures." The restriction to spheres with compact closures (as opposed to totally bounded spheres) was an assumption in Loomis' 1945 paper. Of course it is possible that the above stated "theorem" can still be shown to be true with another argument, but as we will see in the next section we still need the restriction to spheres with compact closures in our approach to extending A to the Baire sets. Loomis did not include a proof of this in his 1949 paper. Furthermore, to prove the uniqueness of the genuine measure A* that will be constructed in the next section, what is needed is a statement about a smaller ring Z(G) c Z consisting of the zero-boundary members of the hereditarily open prering G of open sets with compact closures. For this we need the following terminology. Let G denote the hereditarily open prering of open sets with compact closures. Then GeT. The zero-boundary sets, denoted Z(G), with respect to A (now considered as a Loomis content on G c 7) are those sets A such that for each E > 0 there exists a closed set F and an open set U E G such that F < A < U and A(U - F) < E. It should be intuitively clear that all of Loomis' results documented in the previous two sections hold for A defined on Z(G). However, to be absolutely sure, one needs to work through all those results replacing T with G and Z with Z( G). We leave this as an exercise for the serious reader. Next we relax Loomis' definition of in variance as follows: a Loomis content l on G is invariant on compact spheres with respect to x E f.l if I(S(P,x» = [(Seq. x» for each pair p, q E X with both S(p, x) and Seq, x) having compact closures. I is simply invariant on compact spheres if it is invariant on compact spheres with respect to all x E j..l. In this case, we do not necessarily have a common value lex) = [(S(P, x» for each p E X unless we know that S(p,x) has compact closure for each p E X. Again, it should be intuitively clear that Loomis' development recorded in the previous two sections can stilI be accomplished with this restriction, but to be absolutely sure. one needs to verify it for themselves. For this we note that it suffices to prove Lemmas 10.2 and 10.3 using this definition of in variance on compact spheres with respect to an r
10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures
301
E ~,
since these are the only places where invarlance is used in the proofs and the rest of the development follows from these two lemmas.
THEOREM 10.4 The Loomis content A of Theorem 10.3 (where T is replaced by G) is invariant on compact spheres with respect to all bur countably many indices of any interval {Ya} c ~ * with Ya < Y 13 whenever a < Pand such that if y is rational then Yy E ~. If I is any other Loomis content on G that is invariant on compact spheres with respect to all but countably many indices of any interval {Ya} C ~*, with Ya < Yj3 whenever a < P and whose members with rational indices are in /-l, then I is a constant multiple of A on Z(G).
Proof: Let {y a} C ~ * be an interval of indicies with Ya < Y 13 whenever a < P such that Yy E ~ whenever y is rational. Let P E X. The function f(a)/g(Po) given by
in Lemma 10.6, is increasing on [0, 1] so it is continuous at all but an increasing sequence {r n} C [0, 1]. Let 8 E [0, 1] such that 8 is not in {r n }. If Y c E ~ then h(y c(P), x) = h(y c(q), x) for each q in X by A4 so A(y c(P» = A(y c(q)) for each q in X. Thus A is invariant with respect to Y c. If Y c is not in ~, 8 is still a continuity point of f(a)/g(po) so there exists an increasing sequence {sn} of rationals with Sn ~ 8 andf(a)/g<po) is continuous at each Sn. Then
because, as shown above, A(ySn (P» = A(Ysn(q» for each q in X since each Sn is rational and because 8 is a continuity point of f(a)/g(Po). Hence A is invariant with respect to all but countably many of the Ya's. Let I be another Loomis content on G that is invariant on compact spheres with respect to all but countably many indices of any interval {Ya} C ~* such that Ya < Y i3 whenever a < Pand such that y y E ~ whenever y is rational. Let Z/(G) be the ring of zero-boundary sets WIth respect to I and Z?c(G) the ring of zero-boundary sets with respect to A. If r is a member of I-l* - I-l, then r- in Lemma 10.2 can still be chosen from I-l and similarly in Lemma 10.3. Therefore, both of these lemmas can be generalized to include the possibility that r E I-l*. Since I satisfies L5 and L6 of Proposition lOA, I satisfies the second inequalities in (10.2) and (10.3), i.e.,
l(A)/I(r) ~ h(A +, x)/h(r-) and I(A)/I(r-) ~ h(A -, x)/h(r, x) for all sufficiently small x
E
~,if I
is invariant on compact spheres with respect
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10. Uniform Measures
to r. By the first inequalities in (10.2) and (10.3), H(A,r) $; h(A ~, x)/h(r-, x) and H(A, r-) ~ h(A -, x)/h(r, x)
for all sufficiently small x E Il. Since I(A)/I(r) and H(A, r) and also I(A)/I(r-) and H(A, r-) play the same roles in these fundamental inequalities from which Lemma 10.6 is eventually proved, it is an easy exercise to show that in Lemma 1O.6,f(a)/g(~o) can be replaced by I (A a)11 (r)
limr
I(B~o)ll(r)
I (A a) =
I(B~o)
since I is invariant on compact spheres with respect to all but countably many indicies from each interval {Ya} C Il* with Ya < Y~ whenever a < ~ and such that Yy E Il whenever y is rational. But then (10.14) whenever Aa E Z1c(G)nZ/(G). Consequently, I is a constant multiple of A on Z1c(G)nZ/(G). It only remains to show that Z(G) = Z1c(G) = Z/(G). For this let A
Z 1c (G). Then A E {A a} for some interval of sets with A a and if A = A y then y is a continuity point of A(A a). Now Z(G) = Z1c(G)nZ/(G) contains all but an increasing sequence {Arn} C {Aa} (by the remarks preceding Corollary 10.1 as applied to both Z1c(G) and Z/(G» and these are the points of discontinuity of either A(A a) or I(A a). Let k > 0 such that I(A a) = kA(A a) for A a E Z( G). Let E > O. Then there is a 0 > 0 such that (y-o,'(+o) c Z( G) and for each pair of indicies a, ~ with y - 0 < a < y < ~ < y + 0, IkA(A~) - kA(A a) I < E since kA(A a) is continuous at y. But then I/(A~) - I(A a) I < E. By the discussion preceding Corollary 10.1, I(A~ - C/(Aa» $; I(A~) -/(A~) < E. Hence Ay is a zero-boundary set with respect to I so A E Z/(G). By a similar argument, if A E Z/(G) then A E Z1c(G). Hence Z(G) = Z1c(G) =Z/(G).E
< A ~ whenever a <
~
Theorem lOA can now be used to prove uniqueness of the Haar measure of Theorem 9.3. THEOREM 10.5 Let (X, v) be a locally compact, isogeneous, uniform space and let Il be an isometric basis for v. Let m be the Haar measure of Theorem 9.3 and suppose h is another Haar measure on (X, v). Then h is a constant multiple of m on (X,V). Proof: Both m and h are defined on G because all open sets are Borel sets. Since m and h are measures they satisfy Ll - L3. Let A E G. Since m and h are inner regular on open sets, meA) = sup{m(K) IKe A and K is compact} and h(A) = sup{h(K) IKe A and K is compact}. Clearly sup{m(V) IV < A and V E
10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures
303
G} ~ m(A) and sup { h(V) IV < A and V E G} ~ h(A), so we can show m and h to satisfy L4 if we show that for a compact K c A, there exists a V E G with K c V
A exists in this space. Also, both m and h satisfy the hypothesis of Theorem lOA. To see this, first note that both m and h are invariant with respect to the members of 11. When we construct 11* from 11 as described in the discussion preceding Theorem lOA, we do so by adding uniform coverings of the form y y where Y y is a member of an interval {y a} C v such that Ya < Y J3 whenever a < ~, Ya E 11 whenever a is rational, and if y is irrational then Y y = {y y(p) Ip EX} where Yy(P) = ua
for each pair p, q E X. Since m is invariant with respect to each rn we have m(yy(p» = m(yy(q». for each pair p, q E X. By a similar argument, h(yy(P» = h(yy(q» for each parr p, q E X. Therefore, m and h are invariant with respect to all members of 11*, so they satisfy the hypothesis of Theorem lOA. Thus both are constant multiples of A on Z. But then they are constant multiples of each
304
10. Uniform Measures
other on Z(G), i.e., there exists some k Z(G).
E
R such that m(Z) = kh(Z) for each Z
E
By Theorem 10.3, Z(G)"G is dense in G with respect to <. Also, as shown above, both m and h are left continuous (L4). Hence for each A E G, meA) = sup{m(V)IV < A and V E G) = sup{m(W)IW < A and W E Z(G)"G} = sup{kh(W)lw < A and W E Z(G)"G} = ksup{h(V)IV < A and V E G) = khCA). Therefore m is a constant multiple of han G. Now let K c X be compact. Then K c U~=l Vn for some finite collection of open sets Vn with compact closures. Let U = U~=l Vn- Then U E G, so m(U) = kh(U). Since K is closed, U - KEG so m(U - K) = kh(U - K). Since m is a measure, m(U) = m(K) + m(U - K) so m(K) = kh(U) - kh(U - K) = kh(K). Hence m(K) = kh(K) for each compact set K. Next, let U c X be open. Since m and h are inner regular on open sets, there exist ascending sequences {Kn) and {C n ) of compact subsets of U such that m(U) = sup{m(Kn») and h(U) = sup{h(C n)}. For each positive integer n put Hn = KnuCn. Then {Hn} is an ascending sequence of compact subsets of U such that m(U) = sup{m(H n)} and h(U) = sup{hCHn)}. Consequently, m(U)
= sup{m(Hn») = sup{kh(Hn») = ksup{hCHn)} = kh(U).
Finally, let E be any Borel set. Since m and h are both outer regular, there exist descending sequences {Un) and {Vn ) of open sets containing E such that m(E) =in/{m(Un») and h(E) = in/{hcYn»)' For each n put Wn = Un"Vn. Then {W n } is a descending sequence of open sets containing E such that m(E) = in/{ m(Wn)} and h(E) = in/{ hCWn)}. Hence m(E)
= in/{m(W n») = illf{kh(W n)} = kill/{h(Wn») = kh(E).
Therefore mCE)
= kh(E)
for each Borel set E, so h is a constant multiple of m on
(X, v),-
EXERCISE 1. Show that in Lemma 10.6,f(a.)/g(~o) can be replaced with I(Aa)/I(B~o)'
10.5 Local Compactness and Uniform Measures In the last paragraph of p. 206 in his 1949 paper, Loomis states (without proof) that if we assume our space to be locally compact, the Loomis content A. "has a unique additive extension to the Borel field generated by the ring of totally bounded open sets and M6 (our L6) holds for any sets from this field." The term Borel field that Loomis uses should not be confused with our term Borel
10.5 Local Compactness and Uniform Measures
305
algebra. Although Loomis does not define the term field in his paper, it is clear
from his proofs of Theorems 2 and 3 that by a field he means a ring (and not a a-ring) in our terminology. It is also clear that by additive he means finitely additive rather than a-additive. Presumably, this "Borel field" is the smallest ring containing the hereditarily open prering T of totally bounded open sets since Loomis says this "Borel field" is generated by the totally bounded open sets. How Loomis intended to extend A to a genuine measure on this "Borel field" is unknown since no proof or reference was given. Had A been a Jordan content (on the compact subsets), one might presume the approach he had in mind was the standard method of Kodaira and von Neumann, referenced in the introduction of this chapter, of extending a Jordan content to a a-additive measure on the Baire ring B o. Even though A is not a Jordan content, it is possible to extend A to a Jordan content A+ (on the compact subsets) in a natural way. However, using Kodaira and von Neumann's method on A+ does not solve the problem because the measure m one obtains is an extension of A+ on the compact subsets and not necessarily an extension of A on the non-compact subsets on which the invariance of A is defined. Consequently, m may not preserve the invariance property. Loomis may have been aware, or partially aware, of this problem because he states on p. 202 that if our underlying space S is locally totally bounded, then its completion T is locally compact and the Loomis content A "can be extended to the completely additive measure of Radon on the Borel field .... generated by the compact subsets of T. It is not clear at this point in what sense the extended measure is a Haar measure; this question will be considered in § 12." Again, Loomis does not prove this assertion or give a reference. In Section 12 of his paper, he does not deal with the question as promised. Instead, he considers the extension provided by his Theorem 9 that extends a Loomis content into the completion T. He does not deal with the problem that when extending the Loomis content A to a genuine measure m on the "Borel field," that the extended measure may not be invariant. Also on p. 202, Loomis uses the term completely additive to describe the extended measure. This may indicate that he uses the term field inconsistently in various places in the paper, and that what he means by the "Borel field" generated by the compact sets of the completion T, is actually the Baire ring B 0 on T. Also, it should be noted that it is not difficult to show that the smallest ring R, containing the hereditarily open prering T of totally bounded open sets, contains all the compact sets when the space is locally compact. To see this let K be a compact set and for each p E K, choose an open neighborhood Vp of p with compact closure. Then it is possible to choose a finite collection of points Pl' ... Pn in K such that V == U7=1 VPi belongs to T, so V - K E T. Since T c R and R is a ring, K == V - [V - K] E R.
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10. Uniform Measures
There is a way around the problem of preserving the in variance of')... (such as it is, given by Theorem 10.4) in the Baire measure m if we are willing to restrict our definition of in variance to the j.!*-spheres of X with compact closure. We will show how this can be done in this section. It seems surprising that Loomis did not do this because he did make this assumption in his earlier paper on extending the concept of Haar measure to locally compact metric spaces in 1945. We have already seen that Loomis abandoned the compact spheres of his 1945 metric space paper in favor of the totally bounded j.!*-spheres in order to be able to construct the measure')... on the ring Z of its zero-boundary sets. However, it seems that returning to the compact j.!*-spheres for the definition of in variance of the extended Baire measure m is the thing to do because it works (i.e., it preserves the result of Theorem 10.4 in the new measure m for the compact j.!*-spheres). Had Loomis not attempted to include j.!*-spheres that do not have compact closures in his definition of in variance in the 1949 paper, he would not have had to go to the trouble of introducing the concept of the extension of ')... to the completion and introducing the weakened axiom A4' (p. 205) to try to achieve invariance of an extension in the completion (which he does not accomplish anyway). This is because it turns out that m is already invariant on B 0 with respect to all but countably many members in any interval {Ya} C j.!* with Ya < Y ~ whenever a < ~ and Ya E j.! whenever a is rational, if we rule out members of j.!* whose j.!*-spheres are not compact. This will be shown in what follows. Based on the previous discussion, we define an almost uniform measure to be a measure m on a uniform space (X, v) such that there exists a basis j.! for v that satisfies axioms Al - A4 and such that for each pair of points p, q in X, and any interval {x a} C j.!* with x a < X ~ whenever a < ~,x 'Y E j.! whenever y is rational, and Cl(S(P, x d) and Cl(S(q, x d) are compact, we have m(S(p, xa» = m(S(q, xa» for all but countably many a. We say m is a uniform measure if m(S(p, x» = m(S(q, x» for each x in j.!* such that S(p, x) and Seq, x) have compact closures. Let K be the collection of compact subsets of X and let 9 be a non-negative real valued function on K with the properties: 11. 9(CuD) ~ 9(C) + 9(D) and 12. if CnD = 0 then 9(CuD) = 9(C) + 9(D) where C, D
E
K.
Then 9 is said to be a Jordan content on K. Let')... be as defined in Theorem 10.4. We can extend')... to KuG by putting ')...+(A) = ').,,(/nt(A». Note that if Int(A) = 0 that ').,,+(A) = 0 since ').,,(0) = O. That ').,,+ is well defined on KuG follows from the fact that Int(A) E G whenever A E K. That ').,,+ is an extension follows from the fact that Int(U) = U if U E G. PROPOSITION 10.6 ').,,+ is a Jordan content on K.
10.5 Local Compactness and Uniform Measures
307
Proof: If either C or D has empty interior, the proof is trivial so assume Int(C) ;tc. ;tc. Int(D). Since Int(CuD) c Int(C)u!nt(D), by the definition of h, h(/nt(CuD», x) ~ h(/nt(C), x) + h(Int(D), x) for each x in!-t. Then
o
. h(/nt(D),x) · h(/nt(CuD),x) < t· h(/nt(C),x) I lm lm x + I lm x x h(B~o'x) h(B~o'x) h(B~o'x)
if Int(CuD), Int(C), Int(D) E Z(G). If any of these sets do not belong to Z(G), we still have
sup{limxh(Z,x)/h(B~o'x) IZ E Z(G) and Z < CuD} ~
sup{limxh(Z,x)/h(B~o' x) I Z E Z(G) and Z < D),
since Z( G) is a ring, H is left continuous on Z( G) and limxh(Z, x)/h(B ~o' x)
=
limxH(Z, x)/H(B ~o' x).
Then by the definition of A on G, we have A+(CuD) ~ A"'(C) + A"'(D). Now suppose CnD = 0. It is easily shown (see Exercise 1) that C and D are uniformly separated. Consequently, Int(C) and Int(D) are uniformly separated so A(/nt(C)u!nt(D» = A(lnt(C» + A(/nt(D». But then A+(CuD)
= A+(CI(lnt(C)u/nt(D») = A(/nt(C)u/nt(D»
A(/nt(C»
+ A(/nt(D»
= A'" (C) + A"'(D).
=
-
Next we construct an outer measure A* from A+ in the following manner. Let 't denote the topology of (X, v) and let U E 'to Put A_(U) = SUpIA+(C) ICc U and C E K}. Then for an arbitrary subset A of X put A*(A) = infl A_ (U) I A c
UE
't}.
PROPOSITION 10.7 A* is an outer measure on X Proof: First notice that if A and B are arbitrary subsets of X with A c B and if
Be U E 't then A CUE 'to Hence A*(A) ~ A*(B). Also notice that if V E 't, then by definition, A*(V) ~ A.(U) for each U E 't containing V. Then since V c V E 't, A*(V) ~ A.(V), Conversely, if V CUE 't then A.(V) ~ A_(U) so A.(V) ~ infl A_ (U) I V cUE 't} = A*(V). Hence A*(V) = A. (V) for each open sel V. Next we show that if C is a compact subset of the union UuV of two open sets, then there exists compact subsets D and E of U and V respectively such that C c DuE. For this let W(P) be a compact neighborhood of p with W(P) c
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U for each p E CnU. Otherwise, for each p E C - U let W(P) c V. Then there is a finite collection x I . . . XN such that C c U~=l W(xn). Put D = u{ W(x n) I Xn E U} and E = u{W(xn ) IXn E V - U}. Clearly C c DuE, D and E are in K, Dc U and E c V. We are now in a position to show that 1.* is finitely subadditive on open sets. We proceed inductively by first considering two open sets V 1 and V 2. First suppose A*(V 1uV 2) is finite. Let £ > 0 and let C be a compact subset of V I uV 2 such that 1.+ (C) > A*(U 1uV 2) - £. As we just showed in the preceding paragraph, there exist compact subsets C 1 and C 2 of V 1 and V 2 respectively such that C c C 1 UC 2 • Now A*(Vd + A*(V 2 ) ~ A+(Cd + A+(C 2 ) ~ A+(C 1UC 2 ) ~ A+(C) > A*(U1UV 2 ) - £. Since £ was chosen arbitrarily, A*(V)uV 2 ):S: A*(Vd + A*(V 2 ). On the other hand, ifA*(V 1UV 2 ) is infinite, then for any a E R. there is aCE K with C c VI uV 2 and A+(C) > a. As before, there exist compact subsets C 1 and C 2 of V 1 and V 2 respectively such that C c C 1uC 2. Then A+(C d + A+(C 2) > a so A*(V d + A*(U 2) > a. Hence A*(V 1) + A*(V 2 ) is infinite. Therefore, in either case, A*(V J uV 2 ):S: A*(V d + A*(V 2 ).
For any positive integer n such that A*(u'!;l Vi) :s: L'!;lA*(Vi ), we can use the proof in the preceding paragraph to show that A* (u'!;l Vi U Vn) :s: L'!;lA*(Vi ) + A*(Un) so we conclude that 1.* is finitely subadditive on open sets by induction. To show 1.* is countably subadditive on open sets, let {V n } be a sequence of open sets and suppose first that A*(uVn ) < 00. Let £ > O. Choose a compact C c uVn with A+(C) > A*(uV,,) - £. Then there is a finite subsequence {Vm,,} of {V n } that covers C. Then A*(uVn) - £ < A+(C) :s: A*(uVmJ :s: LA*(VmJ :s: LA*(Un). Since £ was chosen arbitrarily, A*(uVn ) :s: LA*(Un) whenever A*(uVn ) is finite. Now suppose A*(uV,,) is infinite. Then for each a> 0 there is a compact C c uV" with A+(C) > a. Just as before, there is a finite subsequence {Vm,,} of{Vn } thatcoversCsoLA*(UmJ>a. Hence LA*(Vn ) is also infinite. Therefore 1.* is countably subadditive on open sets. To conclude the proof we must show that 1.* is countably subadditive. For this let {En} be a sequence of subsets of X. Note that it is sufficient to show A*(uEn) :s: LA*(En) where A*(En) < 00 for each n, since if some A*(Em) = 00 the proof is trivial. Let £ > O. For each positive integer n let V" be an open set containing En such that A*(En) > A*(Vn ) - £/2n. Then uEn c uVn and A*(uEn) :s: A*(uVn):S: LA*(V,,) < Ln[A*(En) + £/2n] = LnA*(En) + £. Since £ was chosen arbitrarily, this implies A*(uE,,):s: LA*(En).PROPOSITION 10.8 If W is an open set with compact closure then A*(W)
= A(W).
Proof: A*(W) = A.(W) = sup{A+(C)lc c Wand C E K} = SUp{A(lnt(C»IC c Wand C E K} = sup{A(U) ICI(U) c W and U E 't} since CI(U) E K for each U
309
10.5 Local Compactness and Uniform Measures E 't with CI(U) c W. Now for each U E 't with CI(U) c W there is a V CI(U) eVe CI(V) c W so A*(W) = sup{A(U) I U c Wand U E 'tJ Therefore A*(W) = A(W). -
COROLLARY 10.2 A+(C) ~ A*(C)jor each C
E
E 't
with
= A(W).
K.
COROLLARY 10.3 If x E /..1* and 5i(p, x) and S(q, x) have compacc closures then A*(S(p, x)) A*(S(q, x)) if and only ifA(S(p, x)) A(S(q, x))
=
=
PROPOSITION 10.9 The open subsets of X are A*-measurable. Proof: We need to show that for an open set 0 that A*(S) = A*(OnS) + A*(S-O) for any subset S of X. We start by showing that if U, V E 't and Un V = 0 then A*(UuV) = A*(U) + A*(V). For this it is clearly sufficient to consider the case
where both A*(U) and A*(V) are finite. Let £ > 0 and let C and D be compact subsets ofU and V respectively such that A*(U) - A"'(C) < £ and A*(V) - A'" (D) < £. That this is possible can be seen from A*(U) = A. (U) = sup (1..+ (K) IKe U and K E K) ~ A+(C), for any compact C c U, and similarly for a compact D c V. Then CnD = 0 and A*(U) + A*(V) ~ A+(C) + £ + 1..+(0) + £ = A+(CuD) + 2£ ~ A*(CuD) + 2£ (by Corollary 10.2) ~ A*(UuV) + 2£. Since £ was chosen arbitrarily, A*(U) + A*(V) ~ A*(UuV). But then A*(UuV) = A*(U) + A*(V). Next we show that if U and V are open sets then A*(V) = A*(Un V) + A*(V-U). For this let £ > O. Let C be a compact subset of UnV with A+(C) > A.(UnV) - £ = A*(UnV) - £. Let W be an open set such that C eWe CI(W) c UnV. Then V - U c V - CI(W), so A*(UnV) + A*(V - U) ~ A*(UnV) + A*(V-CI(W» < A+(C) + £ + A*(V - CI(W» ~ A*(W) + £ + A*(V - CI(W». Now W and V - CI(W) are disjoint open sets so as we just showed in the preceding paragraph. A*(W) + A*(V - CI(W» = A*(Wu[V - CI(W)]). Also, since CI(W) c V we have Wu[V - CI(W)] c V so A*(Wu[V - CI(W)]) ~ A*(V). Hence A*(UnV) + A*(V - CI(W» ~ A*(V) + £. Since £ was chosen arbitrarily. A*(UnV) + A*(V - CI(W» :s; A*(V). But then A*(UnV) + A*(V - U) :s; A(V). Conversely, since 1..* was shown to be subadditive in the proof of Lemma 10.7, A*(V) ~ A*(UnV) + A*(V - U). Hence A*(V) = A*(UnV) + A*(V - U). Finally we show that for an open set 0 that A*(S) = A*(OnS) + A*(S - 0) for any SeX. If A*(S) is infinite then A*(OnS) + A*(S - 0) :s; A*(S) so assume A*(S) is finite. Let £ > 0 and choose an open set U containing S such that A*(S) > A*(U) - £. Then OnS c OnU and S - 0 c U - O. so A*(O"S) + A*(S - 0) ~ A*(OnU) + A*(U - 0) = A*(U) as we just showed in the preceding paragraph. Therefore, A*(OnS) + A.*(S - 0) ~ A*(S) + £. Since £ was chosen arbitrarily. A.*(OnS) + A.*(S - 0) ~ A.*(S). Conversely, since 1.* is subadditive, A.*(5) ~ A.*(OnS) + A.*(S - 0), so A.*(S) = A.*(On5) + A.*(S - 0). _ We have now shown that 1.* is a Borel (and hence Baire) measure on X. This follows from Propositions 10.7 and 10.9 and Theorem 8.4. Next we want
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10. Uniform Measures
to show that A* is regular on B 0 (the Baire sets). For this we need a number of results that will eventually enable us to conclude that the mere outer regularity of A* on compact sets is sufficient to imply the regularity of A* on all Baire sets. This is important in our development because the outer regularity of A* follows from the definition of A* using the result in the first paragraph of the proof of Proposition 10.7. Consequently, this will demonstrate the regularity of A* on B o. The remaining results in this section, up to Theorems 10.10 and 10.11 date back to the 1940s or earlier. PROPOSITION 10.10 Every Baire set in X is a-bounded (can be covered by countably many compact sets) and every a-bounded open set is a Baire set. Proof: It is easily shown that the class of all a-bounded sets is a a-ring. Since each compact set is a-bounded, this a-ring contains all compact sets and hence the smallest a-ring (Proposition 8.3) containing the compact sets, namely, B o. Hence every Baire set is a-bounded. The remainder of the proof is left as an exercise. PROPOSITION 10.11 If ~ is a Baire measure that is outer regular on compact sets, then ~ is outer regular on the difference C - D of any pair of compact sets C and D such that DeC. Proof: Since ~ is outer regular on C, for each E > 0 there exists an open Baire set U with C c U and ~(U) < ~(C) + E. Put V = U - D. Then V is open and C D c V. Moreover, ~(V)
- ~(C - D)
= ~(V - [C - D]) = ~(U - C) = ~(U) - ~(C) < E.
Hence for each E > 0 there exists an open V E B 0 that contains C - D such that < ~(C - D) + E, so C - D is outer regular.-
~(V)
We say that an open set is bounded if it is contained in a compact set. PROPOSITION 10.12 If~ is inner regular on each bounded open set, then ~ is inner regular on the difference C - D of any pair of compact sets C and C such that DeC.
The proof of this proposition is similar to the proof of Proposition 10.11 (remember we are assuming X is locally compact) and will be left as an exercise. Moreover, the proof of the next proposition will be left as an exercise for the same reason. PROPOSITION 10.13 If ~ is inner regular on a finite collection of disjoint sets offinite measure, then ~ is inner regular on their union.
10.5 Local Compactness and Uniform Measures
311
THEOREM 10.6 Let /-t be a Baire measure and {En} a sequence of Baire sets. Then the following hold: (1) If/-t is outer regular on each En. then /-t is outer regular on uEn. (2) If/-t is inner regular on each En and {En} is an ascending sequence. then /-t is inner regular on uEn. (3) If/-t is inner regular on En and /-teEn) < 00 for each n. then /-t is inner regular on nEn. (4) If/-t is outer regular on En and /-teEn) < 00 for each n. and if {En} is descending, then /-t is outer regular on nEn. Proof: To show (1) let 10 > O. For each n there exists an open Baire set Un with En C Un and /-t(Un) < /-teEn) + e/2 n. Put V = uUn and E = uEn. If /-teE) = 00 then clearly /-t is outer regular on E. If /-teE) < 00, then
Since V is also an open Baire set, this shows that /-t is outer regular on uEn
=E.
To show (2) put E = uEn. By Proposition 8.6, /-teE) = limn/-t(En). Let 10 > O. Choose n such that /-teE) - 10 < /-teEn). Since /-t is inner regular on En there exists a compact C c En with f.l(E) - 10 < /-t(C). But then /-t is inner regular on uEn = E.
> O. For each n there exists a compact en c En with /-teEn) < /-t(e n) + e/2n. Put C = nen and E = nEn. Then C c E and C is To show (3), let
10
compact. Moreover,
Since C is compact, this shows /-t is inner regular on nEn • To show (4), put E = nEn. By Proposition 8.7, /-teE) = limn/-t(En). Let 10 > O. Choose n such that /-teEn) < /-teE) + £. Since /-t is outer regular on En there exists an open Baire set V such that f.l(En) :5: f.l(V) < /-teE) + e. But then /-teE) :5: /-t(V) < f.l(E) + 10 so /-t is outer regular on nEn = E. -
THEOREM 10.7 A necessary and sufficient condition for a Baire measure /-t to be outer regular on compact sets is that it be inner regular on bounded open sets. Proof: Suppose /-t is outer regular on each compact set and let V be a bounded open set. Let 10 > O. Let C be a compact set such that V c C. Then C - V is compact and hence /-t is outer regular on C - V. Therefore, there exists an open Baire set V such that C - V c V and J.!(V) < J.!(C - U) + e. Since C - V c C (C-V) = V we have: J.!(U) - J.!(C - V) = J.!(U - [C - V]) = J.!(Un V) :5: J.!(V - [C - U]).
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10. Uniform Measures
Since the last term on the right is equal to /.l(V) - /.l(C - U) < e we have that /.l(U) - e < /.l(C - V). Since C - V is compact we have shown that /.l is inner regular on U. Conversely, suppose /.l is inner regular on each bounded open set and let C be compact. Let e > O. Since X is locally compact we can find a bounded open set U containing Co Then, since U - C is a bounded open set, there exists a compact set D c U - C with /.l(U - C) < /.l(D) + e. Since C = U - (U - C) c U - D we have: /.l(U - D) - /.l(C)
= /.l([U - C] - D) = /.l(U - C) - /.l(D)
<
£.
Therefore, /.l(U - D) < /.l(C) + £ and since U - D is a bounded open set, we have shown that /.l is outer regular on C. • THEOREM 10.8 The collection R of all finite. disjoint unions of sets of the form C - D where C and D are compact and DeC is a ring and the Baire ring Bois the smallest a-ring containing it. Proof: We first show that Bois the smallest a-ring containing R. Let L be the smallest a-ring containing R (Proposition 8.3). Then L contains all compact sets, so by the definition of B 0, B 0 c L. But since Bois a ring, B 0 contains R so L c B o. Hence B 0 = L.
To show R is a ring let A, B E R. Then A = U~=l [An - Cn] for some disjoint collection {A 1 . . . AN} of compact sets and collection {C 1 .•• CN } of compact sets such that Cn C An for each n. Similarly, B = Uf=l [Bk - Dd for some disjoint collection {B 1 ••• B K } of compact sets and collection {D 1 D K } of compact sets such that Dk c Bk for each k. For each n = I ... N,
for each k ::F j since the sets Bk are disjoint. Consequently, the collection of sets of the form [An - Cn]n[B k - Dd for each k = 1 ... K is a disjoint collection of the form C - D where C and D are compact and DeC. But since all the An's are disjoint, the collection [An - Cn]n[B k - Dd for each n = I ... Nand k = I ... K is also disjoint. Now AnB =
so AnB E R. Clearly if AnB = 0 then AuB intersections and finite, disjoint unions.
E
R so R is closed under finite
10.5 Local Compactness and Uniform Measures
313
Next we show that if E, FER with E c F then F - E E R. To do this we start with the simplest case, namely, where F = C I - D I and E = C 2 - D2 for compact sets C I , D I , C 2 and D 2 where DIe C I and D 2 C C 2. Now E c F implies [C 2 - D 2J C [C I - DI1 which in turn implies C 2 c C I and [C 2 D2JnDi =0. FirstsupposexE [C I -DI1-[C 2 -D 2J. ThenxE [CI-Dd and x is not in [C 2 - D 2J which implies x E C I , x is not in D I and (x E C 2 implies x E D 2 ). Suppose x is not in [C I - C 2uD Ilu[D 2 - D I J. Then x is not in [C I - C 2uD21 which implies x E C 2 since x is not in D I and x E C j . But then xED 2 which implies xED 2 - D I which is a contradiction. Hence
Conversely, assume x E [C I - C 2uDtlu[D 2 - Dd. Then x E [C I -C 2uDil orXE [D 2 -Dd. IfxE [C I -C 2uDd thenxE C I and X is not in C 2uD I which implies x is not in C 2uD I. Then x E [C I - D I J and x is not in [C 2 - D 2J because D 2 C C 2. Therefore, x E [C I - D d - [C 2 - D 2J. On the other hand, if XED 2 - D I then xED 2 c C 2 and x is not in D I' But since C 2 c C I we have x E C I which implies x E [C I - D d. Moreover, x E D2 which implies X is not in [C 2 -D2J. Therefore,xE [C I -Dd - [C 2 -D2J. Hence
Since D2 c C 2 we see that [C I - C 2uDdn[D 2 - Dd = 0 so [C I C 2uD I Ju[D 2 - D I J is a finite, disjoint union of sets of the form C - D where C and D are compact and DeC so [C I - D I J - [C 2 - D 2 ] E R. Next consider the case where E = H - K for compact sets Hand K with K cHand F = U~=I [Cn - DnJ for compact sets C n and Dn with Dn c Cn for each n and [Cm - DmJn[Cn - DnJ = 0 whenever m '# n. Now E c F implies E c U~=I [Cn - DnJ which in turn implies E c [C k - DkJ for some k since the [Cn-Dn)'s are disjoint. But then as we just showed, [C k - Dd - E E R. Since
it is clear that F - E E R. Finally, we consider the most general case where E = U~I [Hi - KJ and F = U~=I [Cn - Dnl. Since E c F, each [Hi - Kil c F. As just shown, F - [Hi - KJ E R for each i. But F - E = nf'!:l (F - [Hi - KJ) and since R is closed under finite intersections, we have F - E E R. We are now ready to show that R is a ring. Let A, BE R. Now A - B = A (AnB). Since R is closed under intersections, AnB E R and AnB c A, so by what has been shown above, A - BE R. Moreover, AuB = [A - B]u[AnB] is a disjoint union of members of R and since R is closed under disjoint unions, we have that AuB E R. HenceR is a ring.-
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10. Uniform Measures
To prove the theorem showing that the outer regularity of A* on Bois sufficient to imply the regularity of A* on B 0, we need a set theoretic lemma that we now develop. A sequence {En} is said to be monotone if it is either ascending or descending. The limit of {En} is defined to be uEn if {En} is ascending or (lEn if {En) is descending. The limit of {En} is denoted by limnEn. A collection of sets M is said to be a monotone class if for each monotone sequence {En} c M we have limn En E M. Clearly the class of all subsets of X is a monotone class and it is easily shown that the intersection of any collection of monotone classes is a monotone class, so there exists a smallest monotone class containing a given collection A of sets. We denote this smallest monotone class containing A by M(A). LEMMA 10.7 If R is a ring and L(R) is the smallest a-ring containing R. then L(R) = M(R). Proof: A a-ring is a monotone class so M(R) C L(R). Conversely, we will show that M(R) is a a-ring which will imply L(R) c M(R). For this let SeX and let M(S) be the collection of all sets T such that S - T, T - S and SuT all belong to M(R). Then if {En} is a monotone sequence in M(S) we have S - limnEn = limn(S - En) E M(R), limnEn - S = limn (En - S) E M(R) and SU(limnEn) = limn (SuE n) E M(R) so M(S) is also a monotone class if it is not empty.
Now if S, T E R, then since R is a ring, T E M(S). Since this holds for each T E R it follows that R c M(S). Since M(R) is the smallest monotone class containing R, we must have M(R) c M(S). Therefore, if E E M(R) and S E R then E E M(S) and hence S E M(E). Since this holds for each S E R we have R c M(E) so we also must have M(R) c M(E). But M(R) c M(E) for each E E M(R) implies M(R) is a ring. To show M(R) is a a-ring, let {Sn} c M(R). For each positive integer m put Em = U::'=l Sn. Since M(R) is a ring, each Em is in M(R). Now {Em} is an ascending sequence and M(R) is a monotone class so USn = uEm = limmEm E M(R). Hence M(R) is a a-ring which establishes the lemma.THEOREM 10.9 Let ~ be a Baire measure such that ~(C) < for each compact C. Then either the outer regularity of ~ on each compact set or the inner regularity of ~ on each bounded open set is a necessary and sufficient condition for ~ to be regular. 00
Proof: The necessity of either of these conditions is obvious. To demonstrate the sufficiency of either of these conditions, it will suffice to show that ~ is regular on each bounded Baire set since each Baire set is a-bounded by Proposition 10.10 and hence is the union of an ascending sequence of bounded Baire sets. Then by Theorem 10.6, the regularity of ~ on the members of this sequence will imply the regularity of ~ on the union.
10.5 Local Compactness and Uniform Measures
315
Since both the conditions of the hypothesis are equivalent by Theorem 10.7, to prove the sufficiency, we assume them both. Let E be a bounded Baire set and C a compact set such that E c C. It is left as an exercise (Exercise 5) to show that B onC = {BnC IB E B a} is the smallest a-ring containing all the compact subsets of C. Hence B anC is the Baire ring on the compact space C. By the previous theorem applied to the space C, we see that B onC is the smallest a-ring containing the ring R of all finite, disjoint unions of sets of the form A - B where A and B are compact subsets of C with B c A. By Propositions 10.11 and 10.13 we see that f.l is outer regular on every set in R. By Propositions 10.12 and 10.13 we see that f.l is inner regular on every set in R. Let M be the collection of sets S in B onC such that f.l is both outer regular and inner regular on S. By Theorem 10.6 and the fact that f.l(E) < 00 for each S c C we see that M is a monotone class. Also, ReM, so M(R) eM. By Lemma 10.7, M(R) = '1:.(R). But by Theorem 10.8, '1:.(R) = BonC. Hence BonC eM which implies f.l is both outer regular and inner regular on each S E B onS. In particular, f.l is regular on E. Consequently, f.l is regular. THEOREM 10.10 A* is a regular. almost uniform Baire measure on X.
Proof: That A* is a Borel (and hence Baire) measure on X follows from Propositions 10.7 and 10.9. That A* is regular on B 0 follows from the fact that A* is outer regular on all Baire sets (see first paragraph of Proposition 10.7), the fact A*(C) < 00 for compact sets C, and Theorem 10.9. That A* is almost uniform follows from Theorem 10.4, Corollary 10.3 and the definition of an almost uniform measure. THEOREM 10.11 If m is another regular almost uniform Baire measure on X then m is a constant multiple of A* on the Baire sets of X.
Proof: If U E G then U E B a so m is defined on G. Clearly m satisfies L 1 - L3. To show m satisfies L4 let U E G. Since m is regular there exists an ascending sequence {C n } of compact sets with m(U) = limnm(C n). Now for each n there exists an open Vn E G with C n c Vn C U. But then Vn E G for each n. Moreover, m(U) = limnm(Vn) so m satisfies L4. Hence m is a Loomis content on
G. Now by the definition of an almost uniform measure, it is clear that m, considered as a Loomis content on G, is invariant on compact spheres, so by Theorem lOA, m is a constant multiple of A on Z(G). Then an argument similar to one of the arguments in Theorem 10.5 can be used to show that m is a constant multiple of A on G. Since the members of G have compact closures, by Proposition 10.8, A*(U) = A(U) for each U E G. Therefore, m is a constant multiple of A* on G.
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10. Uniform Measures
Consequentl y, an argument similar to the last part of the proof of Theorem 10.5 can be used to show that m is a constant multiple of A* on the Baire sets of X.-
EXERCISES
1. Show that if C and D are compact subsets of a locally compact uniform space then C and D are uniformly separated. 2. Complete the proof of Proposition 10.10. 3. Prove Proposition 10.12. 4. Prove Proposition 10.13. 5. Show that the a-ring B onC = {BnC I B E B 0 } for some compact set C is the smallest a-ring containing all the compact subsets of C. Remember B 0 denotes the Baire ring. 6. Show that A*(C) < 00 for each compact set C. RESEARCH PROBLEMS 7. If a locally compact uniform space satisfies axioms Al - A4, does there exist a regular uniform measure on the Baire ring (as opposed to an almost uniform measure)? 8. What is a necessary and sufficient condition for a locally compact uniform space to have a regular Baire (almost regular Borel) uniform measure? 9. If a locally compact uniform space has a regular Baire (almost regular Borel) uniform measure, is it unique? If not, what is a necessary and sufficient condition for it to be unique? 10. Are there interesting or useful necessary and sufficient conditions for arbitrary (not necessarily locally compact) uniform spaces to have Baire (Borel) uniform measures? 11. Can the construction in this chapter be made to yield a regular Baire (almost regular Borel) uniform measure?
Chapter 11 SPACES OF FUNCTIONS
11.1 U -spaces In this chapter, we will be investigating several types of spaces constructed from functions on a given space X to some other space Y. The spaces X and Y will usually be uniform spaces, but the theory of function spaces is more general than that. Much of this chapter, is independent of uniform spaces. The spaces we will study first will be needed in the next chapter to develop the concept of uniform differentiation. In fact, much of the content of this chapter and the next is needed simply to develop the machinery from classical analysis so that we can discuss uniform differentiation in isogeneous uniform spaces. We will begin by allowing X and Y to be topological spaces and add special constraints as we proceed. Only when we get to the end of this chapter will we consider abstract uniform function spaces. Some of the function spaces we will consider are special cases of more general classes of spaces, but much of the motivation for studying these spaces seems to have been the function spaces themselves. Examples of this are the Hilbert spaces and Banach spaces that we will introduce later in this chapter. The first function spaces we will consider are the so-called U -spaces. To facilitate their study we introduce some results about convex functions. A real valued function 1 defined on an open interval (a, b), where -00:S; a < b :S; 00, is said to be convex if the inequality 1«(1 - c)x + cy)
:S;
(1 - c)l(x) + cl(Y)
holds whenever x, y E (a, b) and 0 :S; c :S; 1. The intuitive idea behind this inequality is that whenever x < z < y, the point (z,f(z» lies on or below the line segment connecting (x, I(x» and (y, l(y» in the plane. This inequality is equivalent to another, namely, (11.1)
1 (z)-I (x)
z-x
:S;
10)-1 (z)
y-z
Whenever a < x < z < y < b. Applying the Mean Value Theorem from Calculus to (11.1) shows that a real valued differentiable function 1 is convex in (a, b) if and only if a < x < y < b impliesf'(x) :S;f'(y) or, in other words, if the derivative f' is a monotonically increasing function.
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11. Spaces of Functions
PROPOSITION 11.1 A convex function on (a, b) is continuous on (ab). The proof of Proposition 11.1 is left as an exercise. If p and q are positive real numbers such that pq = p + q we say p and q are conjugate exponents. Note that this is equivalent to l/p + l/q = 1. Also note that as p approaches I this forces q to approach 00. For this reason we consider I and 00 to be conjugate exponents. In the next theorem, the first inequality is known as Holder's Inequality and the second as Minkowski' s Inequality.
THEOREM 11.1 Let p and q be conjugafe exponents with 1 < P < 00. Let j..t be a positive measure on X and let f and g be non-negative extended real valued measurable functions on X. Then fxfgdj..t :s; IfxfPdj..t}l!PlIxgqdj..t}l!q and
Proof:Puta= IJxfPdj..t}l!P and ~= Uxgqdj..tlliq. Ifa=Othefg=Oa.e. onX by Exercise I in Section 8.7. Consequently. HOlder's Inequality would hold. Also, if a > 0 but ~ = 00, Holder's Inequality would hold. Therefore, if suffices to consider the case where 0 < a < 00 and 0 < ~ < 00. Define the functions hand k by h = fla and k = g/~. Then
fx hP dj..t =
1
f
xfP dj..t = aP
IxfPdj..t = 1. xfPdj..t
J'
Similarly, Ixkqdj..t = 1. If x E X such that 0 < hex) < 00 and 0 < k(x) < 00, there exist real numbers a and b such that hex) = e a1p and k(x) = e b1q . Then since lip + Ilq = I, the convexity of the exponential function implies:
To see this use the substitution (1 - A.) = lip and A. = l/q. Then
But then h(x)k(x) :s; p -1 h(x)P + q -1 k(x)q for each x E X with 0 < hex) < 00 and 0 < k(x) < 00. Now if hex) = 0 or hex) > 0 and k(x) = 00, it is easily seen that the inequality still holds. Consequently, it holds for each x E X. Integration of both sides then yields
11.1 LP -spaces
319
But then (1Ia~)Jxfgdf..t Inequality.
:<::;
I which implies Jxfgdf..t
:<::;
a~ which yields Holder's
To prove Minkowski's Inequality we first observe that it suffices to prove it for the case where the left hand side is greater than zero and the right hand side is less than 00. For this we note that
(f + gY' fx(f+gY'df..t
= f(f + gY'-i
+ g(f + gY'-1 so
= fxf(f+gY'-ldJ.!
+ fxg(f+gY'-Idf..t.
Holder's Inequality implies that
fxf(f + gY' -I dll
:<::;
fxg(f + gY'-ldll
{fxfPdf..t} lip dx(f + g)(p -l)q dJ.! I llq and :<::;
{fxgPdlllllP dx(f + g)(P-l)qdll}llq.
Hence, since (p - l)q = P we have
fx(f+ gY'df..t
:<::;
[{fxfPdlll liP + {fxgPdlllliPlIfx(f+ gY'dll}li q.
Dividing both sides of this inequality by
dx (f + gY' dll I Ilq yields
{fx(f+gY'dIl11IP:<::; dxfPdf..tl liP + {fxgPdf..tl liP since I - l/q = lip. This is Minkowski's Inequality.If Il is a positive measure on X, 0 < p < 00 and if fis a complex measurable function on X, we define
and let U (Il) denote the collection of all such functions f such that If IP < 00. We say If IP is the U -norm of f. If X is a locally compact isogeneous uniform space and Il is Haar measure on X then we write U(X) instead of U(Il). If Il is a positive measure on X andf:X --7 [0,00] is a measurable function, put A = {a E RIIl(f-l«a, 00])) = 01. If A = 0 let b = 00. Otherwise put b = infA. We say that b is the essential supremum of f. If f is a complex measurable function on X, we define If I~ to be the essential supremum of f. We denote the collection of all complex measurable functions f such that If I~ < 00 by L ~(Il)· The members of L ~(Il) are said to be essentially bounded functions. Clearly If(x) I :<::; c for almost every x E X if and only if c ~ If I~.
320
11. Spaces of Functions
PROPOSITION J J.2 If J..t is a positive measure on X, p and q are conjugate exponents, J :S: P :S: 00, and iff E U (J..t) and gEL q (J..t), thenfg ELI (J..t) and Ifg II :S: If Ipig Iq. Proof: If 1 < p < 00 then the desired inequality is just HOlder's Inequality applied to f and g. If p = 00, then If(x)g(x) I :S: IfI~ Ig(x) I for almost all x E X by the definition of Ifl ~. If p = 00 then q = 1, so integrating both sides of this inequality yields:
Finally, if p = 1 then q = 00, so an argument similar to the above again produces the desired inequality. -
PROPOSITION J J.3 If J..t is a positive measure on X, J :S: P :S: U(J..t) and g E U(J..t), thenf + g E U(J..t) and
00,
andf E
The proof of this proposition is left as an exercise. Proposition 11.3 together with the fact that if 1 :S: P :S: 00 and c is a complex number, implies that Icfl p = Ie Ilfl p so cf E U (J..t) which implies U (J..t) is a complex vector space. In fact, U (J..t) is also a pseudo-metric space. If we define the distance between f and g by d(f, g) = g Ip for any pair of functions f, g E U (J..t), then it is easily shown that d is a pseudo-metric on U(J..t). As shown in Chapter 1, a metric space can be constructed from this pseudo-metric space as follows. Define f - g if and only if d(f, g) = O. Clearly - is an equivalence relation on U(J..t). If [f] and [g] are two equivalence classes containingfand g respectively and iff E [f] and g' E [g] then d(f, g) = dC!, g') so that d([f], [g]) is well defined on the equivalence classes of LP(J..t) with respect to -. Hence d is a metric space when considered as being defined on the equivalence classes of U(J..t) with respect to -.
It -
Furthermore, this metric space is also a complex vector space since if f E [f] and g' E [g] then f + g' E [f + g] and cf E [cf]. It is customary to regard U(J..t) as a metric space even though it is the quotient space U(J..t)/- that is reall y the metric space. When this is done, it is with the understanding that f E LP(J..t) really means that f is a representative of the equivalence class (f] to which it belongs. The following theorem is especially significant as we will soon discover.
THEOREM J J.2 For a positive measure J..t and for 1 :S: P :S: metric space LP (J..t) is complete. Proof: First assume that p <
00.
00,
the
Let {fn) be a Cauchy sequence in LP(J..t). Then
11.1 LP-spaces
321
there exists a subsequence {fn,} such that Ifn'+l - fn Ip < 2-; for each positive integer i. Define the functions g and gk for each positive integer k as follows: Put
Then for each positive integer k,
Igklp =
rfxl~r=llfn'+l-fnillpdlllIlP
=
rfx(~r=llfni+l-fn,IYdJl]lIp.
Applying Minkowski's Inequality inductively to the term on the right produces:
Igklp
~ ~f=lrfxlfn'+l-fnYdlllIlP = ~~=llfni+l-fnilp
<
~~=121<1.
For each x E X we have:
so {g~} is an ascending sequence of real valued functions. Moreover, gP(x) = lim infkg~ (x) for each x E X, so by Theorem 8.11,
fx(lim infgDdJl
~
lim inJfxg~dll
which implies:
Therefore,
[fx(~i=l Ifni+l
- fn,
IY' dill lip ~
[lim inAx(I.7=1 Ifni+l - fn, IY'dlll ilp
so Ig Ip ~ lim infl gk Ip ~ 1. But then g(x) <
00
a.e. so the series
converges absolutely a.e. on X. Put f(x) = fn, (x) + ~i=l (fn.+l (x) - fn, (x)) for each x E X where the series converges absolutely and put f(x) = 0 otherwise. Since fni + ~f;;l (fni+l - In) = Ink for each positive integer k, it is clear that f(x) = lim; ~~fni (x) a.e. on X. Even though I is the pointwise limit a.e. of {fn}, it is conceivable that I is not the limit of {In} with respect to the LP -norm I I or that I is not in LP (J.l). It remains to show that this is not the case. For this l:t e > O. Then there exists a positive integer k such that 11m - In I < t whenever m, n > k. For each m > k,
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11. Spaces of Functions
Theorem 8.11 can be used to show that: JXliminfi-.=lfnj(x)-fm(x)lpdJ.!:S; liminfi-.Jxlfn,(x)-fm(x)lpdJ.!:S; fP.
But lim infi-.= Ifn,(x) - fm(x) Ip = If(x) - fm(x) Ip so Ix If - fm Ipd!! :s; £P. Hence If-fm Ip :s; £P so f - fm E U(!!) which implies f E U(!!) since f = fm + (f - fm)' Moreover, we have shown that for each £ > 0, there exists a positive integer k such that if m > k then If - fm Ip :s; £P so f is the limit of {fn} with respect to the LP-norm. The proof that the theorem holds when p left as an exercise. -
=
00
is considerably easier and is
COROLLARY 11.1 If {fn} is a Cauchy sequence in U(!!), where I :s; p :s; 00, and iff is the limit of {fn}, then {fn} has a subsequence that converges pointwise almost everywhere to f
The proof of the corollary is contained in the proof of the theorem. Let SM denote the class of complex valued, simple measurable functions on X whose support has finite measure. Then the members of U (!!) can be approximated by members of SM for 1 :s; P < 00. In fact: PROPOSITION 11.4 SM is dense in U (!!) for 1 :s; P < 00. Proof: Clearly SM C U(!!). First suppose f ~ 0 is in U(!!) and let {sn} be a sequence of simple measurable functions with 0 :s; s 1 :s; S 2 :s; ... :s; f and such that {sn(x)} converges to f(x) for each x E X. That such a sequence exists follows from Theorem 8.7. Since O:s; Sn :S;ffor each positive integer n, we have that Sn E LP(!!) for each n and hence Sn E SM for each n.
Now Ix If Ipd!! < 00 since fEU (!!). But f~ 0 implies If IP = fP = IJP I so Ix IJP Id!! < 00 which implies JP ELI (!!). Furthermore, If - Sn IP :s; JP for each n and {(f - snY} converges to 0 for each x E X. Then by Theorem 8.13, limn-.Jx I (f - snY - 0 Id!! = 0
which implies limn -.=Jx If - Sn IPd!! = 0 which in turn implies limn -.= If - Sn Ip = Consequently, SM is dense in LP (!!). -
o so {sn} converges to f in the LP -norm.
Now let us consider a more specialized class of measure spaces that includes the locally compact isogeneous uniform spaces equipped with Haar measure. Specifically, for the remainder of this section we assume X is locally compact and that!! is a positive, complete, almost regular Borel measure that is finite on compact sets (i.e., the type of measure obtained by applying the Reisz Representation Theorem to a positive linear functional on CK(X)). In such measure spaces we can approximate members of LP(!!), for 1 :s; p < 00 with
323
11.1 LP-spaces
sequences of continuous functions in CK(X). In fact, for such spaces, we will show that CK(X) is dense in U (J..L) for 1 :s; P < 00, with respect to the U -nonn. To do this, we first need a result known in the literature as Lusin's Theorem. It is this result that requires the additional assumptions. Fortunately the Haar measures on locally compact unifonn spaces satisfy most of the restrictions of classical analysis necessary to build the theory of real analysis on locally compact metric spaces. This is why, once we have Haar measure on this class of unifonn spaces we can move the theory of differentiation of a measure to locally compact, isogeneous unifonn spaces.
THEOREM 11.3 If f is a complex measurable function on a locally compact space X and J..L is a positive, complete, almost regular, Borel measure that is finite on compact sets, and ifJ..L(E) < 00, andfis zero outside of E, thenfor each £ > 0 there exists agE CK(X) such that the measure of the set on whichf and g are not equal is less than £. Moreover, we can pick g such that sup 1 g(x) 1 :s; sup Ij(x) I. Proof: First assume E is compact and 0 :s; f(x) < 1 for each x E X. From the proof of Theorem 8.7, there exists a sequence {sn} of simple measurable functions with O:s; SI :s; S2 :s; ... f such that limsn(x) = f(x) for each x E X. Moreover, from the construction in the proof, the s" have the fonn
where E"j = {x 1 (i - 1)/2" :s; f(x) < i/2"} for each positive integer i = 1 ... n2" and F" = {x 1 n :s; f(x) :s; oo}. Since we assumedf < 1 we have F" = 0 for each n and E"j = 0 for i > 2" so s" simplifies to
for each n. Put Then
01
=
SI
and for each positive integer n let
0,,+1
=
S,,+1 -
s".
f(x) = 1:::=10,,(X) for each x E X. By inspection, it is easily seen that for each n > 1, s" - Sn~1 is a simple measurable function that equals 2~" on the set of points where f - S,,_1 ~ 2" and o otherwise. Consequently, 2"0" is the characteristic function of this set which we denote by E". Then E" c E for each n. Let U be an open set containing E such that CI(U) is compact. Then for each n, there exists a compact set K" and an open set Un with K" c En C Un C U and J..L(Un - K,,) < £/2". By Theorem 8.5, there exist functions gn E CK(X) with 0:5 g" :5 1 such that Kn < g" < U", Let
324
1l. Spaces of Functions
Clearly this series converges unifonnly on X. An easy modification to Theorem 1.12 implies g is continuous. Since an = 2- nhn on Kn we have g = f except on u(Un - Kn) which is a set of measure less than £. Thus we have proved the existence part of the theorem for the case where E is compact and 0 s: f < 1. Consequently, it is easily shown that the existence part holds if E is compact andfis merely a bounded complex measurable function. Furthennore, we can remove our compactness assumption, for if Jl(E) < 00, there exists a compact K c E with Jl(E - K) < £ for any £ > O. To prove the general case, let f be a complex measurable function and for each positive integer n put Fn = (x Ilf(x) I > n). Then each Fn is measurable and N n = 0. Consequently, limnJl(Fn) = 0 by Proposition 8.7. For each positive integer n put En = X - Fn. Then f is a bounded function on En. For each n define fn by fn(x) = f(x) if x E En and fn (x) = 0 otherwise. Then f is a bounded measurable function on X and Jl«(xlf(x) =t- fn(x))) s: Jl(Fn). Now let £ > 0 and pick k with Jl(Fk ) < £/2. Then/k is a bounded function on X and Jl( (x [f(x) =t- fn(x»)) < £/2. Then there exists a g E CK(X) with Jl( (x Ig(x) =t- fn(x»)) < £/2 so Jl( (x Ig(x) =t- f(x))) < £. This proves the existence part of the proof. To conclude the proof we first observe that if supx If(x) I = 00 then supx I g(x) [ s: supx If(x) I so assume supx If(x) I < 00. Put b = supx If(x) I and let hex) = x if Ix I s: b and bx/ Ix I otherwise. Then h is a continuous function from C onto CI(S(O, b». If g' E CdX) such that f(x) = g'(x) except on a set of measure less than £ and g = h © l then f(x) = g(x) except on a set of measure less than £ and supx Ig(x) I s: supx If(x) I. This concludes the proof.THEOREM 11.4 If X is a locally compact space and Jl is a positive, complete, almost regular Borel measure on X then CK(X) is dense in LP(Jl)for I S:p < 00 0
Proof: Let SM be the class of complex valued, simple measurable functions on X whose support has finite measure. If s E SM and £ > 0, then by Theorem 11.3, there exists agE CK(X) with g(x) = sex) except on a set of measure less than £ and Igl S:suPxls(x)1 = Isl~. LetE= (xlg(x)=t-s(x»). Then Ig-slp
= [fElg-sIPdJll llP s: rfEl2lsl~lpdJllllP = 2Isl~£lIp.
Therefore, it is possible to find a sequence (gn) c CK(X) such that (gn) converges to s with respect to the LP -nonn metric. But then since SM is dense in U(Jl) by Proposition 11.3, so is CK(X).Theorems 11.2 and 11.4 together say that for 1 s: p < 00, U(Jl) is the completion of CK(X) with respect to the LP -nonn metric. The case where p =00 is different because the definition of L ~(Jl) is essentially different than the
11.1 LP-spaces
325
definition of LP(f..l) for p < DO. Now CK(X) is a metric space with respect to the L ~ -norm metric. To characterize the completion of C K(X) with respect to this metric, we need the following definition: A complex valued function f on X is said to vanish at infinity if for each £ > 0 there exists a compact set K such that If(x) I < £ for each x E X - K. The class of all continuous, complex valued functions that vanish at infinity is denoted by C ~(X). On CK(X) the L ~ -norm coincides with another norm called the supremum norm that is defined by If I
=supx If(x) I. THEOREM 11.5 If X is a locally compact space then C ~(X) is the completion of CdX) with respect to the L = -norm and the supremum norm.
Proof: It is left as an exercise to show that C =(X) is, indeed, a metric space with respect to the L = -norm. We need to show that CK(X) is dense in C =(X) and that C =(X) is complete. That CK(X) is dense in C =(X) follows from the fact that if f E C =(X) and £ > 0, then there exists a compact set K with if(x) I < £ outside K and an h E C K(X) with 0 ~ h ~ 1 and h(K) = 1. If we put g = fh then clearly g E CK(X) and If - g I = = supx If(x) - g(x) I < £. To show C ~(X) is complete let Ifn} be a Cauchy sequence in C =(X). For each £ > 0, there exists a positive integer N such that if m, n > N then Ifm - fn I < £ so supx Ifm(x) - fn(x) I < £. Then for each x E X, the sequence Ifn(x») is a Cauchy sequence in C and hence converges to some point f(x) E C. This pointwise limit function defined by f(x) = limnin(X) is well defined. Moreover, Ifn) converges uniformly to f since for each £ > 0 there exists a positive integer N such that if m, n > N then supx Ifm(x) - fn(x) I < £/2. Also, for each x E X, there exists a k > N with Ih(x) - f(x) I < £/2. But then if m > N, Ifm(x) - hex) I < £/2 so Ifm(x) - f(x) I < £ for each x E X. Now an easy modification of Theorem 8.12 shows thatfis continuous. It remains to show that f E C =(X). For this let £ > O. Then there exists an 1= < £/2 and there exists a compact set K such that Ifn(x) I < £/2 outside K. But then If(x) I < £ outside K so f E C ~(X).-
n with If - fn
EXERCISES 1. Prove Proposition 11.1. 2. Prove Proposition 11.3. 3. Show that d defined by d(J, g) =
If - g Ip
4. Prove Theorem 11.2 for the case p =
is a pseudo-metric on LP (f..l).
DO.
5. (Jensen's Inequality) If f..l is a positive measure on a a-algebra M in a space
326
11. Spaces of Functions
X such that ~(X) = 1 and if g is a real valued function in L I (~) with a < g(x) < b for each x E X, and iffis convex on (a, b), show that:
6. If ai > 0 for each i = 1 ... n such that Lai = 1 and if Xi is a real number for each i = 1 . . . n, show that e(LaixJ :s;; Laiex, where e is the exponential function. 7. Show that if X is a locally compact space and ~ is a positive, complete, almost regular Borel measure on X and if the distance d(f, g) between two functions f, g E CK(X) is defined by d(f, g) = Jx If - g Id~ then (CK(X), d) is a metric space and the completion of (CK(X), d) is precisely the class of Lebesgue integrable functions on X. [Recall that the Lebesgue integrable functions were defined in Exercise 1 of Section 8.6.] 8. Show that if p < q that for some measures ~, U (~) c L q (~) whereas for other measures L q (~) c U (~), and that there are some measures such that neither U (~) or L q (~) contains the other. What are the conditions on ~ for which these situations occur?
11.2 The Space L2(~) and Hilbert Spaces The space L 2 (~) is known as the space of square integrable functions. It plays a major role in modem physics and in many other mathematical applications. In fact, it is the mathematical model that underlies the wave interpretation of quantum mechanics, when X is Euclidean space and ~ is Lebesgue measure on X. L 2(~) is a special case of a more general class of spaces called the Hilbert spaces. Other Hilbert spaces also play important roles in quantum mechanics, in fact, they are the mathematical models behind all quantum phenomenon. We will need some results about Hilbert spaces for our development of uniform differentiation in the next chapter. Hilbert spaces derive their name from David Hilbert who published a series of six papers between 1904 and 1910 titled Grundzuge einer allgemeinen Theorie der linearen Integralgleichungen I - VI in Nachr. Akad. Wiss. Gottingen Math.- Phys. Kl. that involved these spaces. They were republished in book form by Teubner Verlagsgesellschaft, Leipzig in 1912 and reprinted by Chelsea Publishing Co., New York in 1952. We now define these spaces. A complex vector space H is called an inner product space if for each pair of vectors u, v E H there is a complex number (u, v) called the inner product or sometimes the scalar product or dot product that satisfies the following axioms:
11.2 The Space L2()l) and Hilbert Spaces
327
(1) (u, v) = (v, u)* (the * representing complex conjugation). (2) (u + v, w) = (u, w) + (v, w) for u, v, w E H. (3) (cu, v) = c(u, v) for u, v E Hand c E C. (4) (u, u) ~ for each u E H. (5) (u, u) = if and only if u is the zero vector in H.
°
°
°
There are a number of observations we can make about these axioms. First, (3) implies (0, x) = for each x E H and (I) and (3) together imply (x, cy) = c*(x, y) for each pair x, y E H and c E C. Next we observe that (2) and (3) together imply that for each y E H, the mapping defined by A(X) = (x, y) for each x E His a linear functional on H. (1) and (2) can be combined to show that (x, y + z) = (x, y) + (x, z) for x, y, z E H. Finally, by (4) we can define a norm Ix I for each x E H by Ix I = (x, x)1!2 so that Ix 12 = (x, x). PROPOSITION 11.5 (Schwarz Inequality) For each x, y E H, I(x, y) I :s; Ix II y I where the norm on the left is the modulus of the complex number (x,y).
°
Proof: Ifx=Oory=Othen l(x,y)l:s; Ixllyl so assumext:O t:y. Let abean arbitrary complex number. Then (x + ay, x + ay) ~ and (x + ay, x + ay) = Ix 12 + I al 2 1y 12 + a(y, x) + a*(x, y) and a(y, x) + a*(x, y) = 2Re(a(y, x» so
Ix1 2 + laI2IyI2+2Re(a(y,x»~0.
°
Now each complex number a can be represented by a = re il for some real number r ~ and some complex number e il for some real number t. Recall that I eill = 1. Similarly, (y, x) = I(y, x) Ie is for some real number s. Hence Re(a(y,x» = Re(re it I(y, x) leis) = Re(rl (y, x) I ei(s+I) = rl (y, x) IRe(ei(s+I) and Re(e i (S+I) :s; 1. Consequently,
°
so Ixl2 + Irl21yl2 + 2rl(y, x)1 ~ for each real number r ~ 0. Put r = -I x I/ Iy I. Then substituting this value for r in the previous inequality yields Ix II y I :s; 1(x, y) 1.An immediate result of the Schwarz inequality is the so called triangle inequality: 1x + y 1 :s; 1x 1 + 1 y I. It follows from the observation that Ix + y 12 = (x + y,x + y) = (x, x) + (x,y) + (y,x) + (y,y):s; Ixl 2 + 2Re«x, y» + lyl2:s; Ixl2 +21(x,y)1 + lyl2:s; Ixl 2 +2lxllyl + lyI2=(lxl + lyl)2. Consequently, if we define the distance d(u, v) between two vectors u, v E H to be 1 u - vi, it is easily shown that d satisfies the axioms of a metric space. In particular, if u, v, w E H we see that d(u, v) :s; d(u, w) + dew, v) follows from 1 u - vi :s; 1 u - wi + 1 w - v I. If the metric space (H, d) is complete, we call H a Hilbert space. We Now observe that if J.l is a positive measure, then L 2(J.l) is a Hilbert space.
328
11. Spaces of Functions
For this we define an inner product on L 2(Il) by (f, g) = fxfg*dll. Since g E L 2(Il) implies g* E L 2(Il) and the exponents p = 2 = q are conjugate exponents, Proposition 11.2 implies fg* ELI (Il) so (f, g) is well defined on H. Now we observe that if we define Ifl2 = (f, f) or equivalently, If I = (f, f)l!2 then we have
so the L2-norm I 12 on L 2 (Il) is equivalent to the inner product norm I I on L 2(Il). Since Il was assumed to be positive, by Theorem 11.2 we know L 2(Il) is complete with respect to the L 2-norm, so L 2(Il) is complete with respect to I I and therefore is a Hilbert space.
PROPOSITION 11.6 For a given y E H. the mappings defined by f(x) g(x) = (y. x), h(x) = Ix I for each x E X are uniformly continuous functions on H.
= (x,y),
Proof: To show f is uniformly continuous let E > 0 and put 0 = Ix I - x 21 < 0 we have by Proposition 11.5 that:
E/ I y I.
Then if
Therefore, f is uniformly continuous on H. A similar argument shows that g is also uniformly continuous on H. To show h is uniformly continuous, let E > 0 and put 0 = E. By the triangle inequality, if Ix I - x 21 < 0, then Ix I I :0:; Ix I - x 21 + IX21, so IXII - IX21:o:; IXI -x21. Similarly, IX21 - IXII = IX2 -xII = IXI-x21,so IlxI -lx211:o:; IXI -x21. Butthen
so that h is also uniformly continuous on H. Recall that a subset S of a vector space V is called a subspace of V if S is a vector space with respect to the addition and scalar multiplication operations defined on V. A necessary and sufficient condition for S c V to be a subspace is that u + v E S and cu E S for each u, v E S and c E C. If H is an inner product space, a closed subspace of H is a subspace that is closed with respect to the metric topology generated by the inner product norm. A set E in a vector space is said to be convex if for each u, vEE and 0 :0:; x:O:; 1, the vector (1 - x)u + xv is contained in E. One can visualize this property of convexity by imagining a "line segment" being traced out between u and v as x goes from 0 to 1, and that all the vectors on this "line segment" are contained in E. If (u, v) = 0 for some u, v E H we say u is orthogonal to v and denote this by u ..L v. Let u 1- be the collection of v E H which are orthogonal to u. If v, W E u1- then (u, v + w) = (u, v) + (u, w) = 0 and if c E C then (u, cv) = c(u, v) =0 so v
329
11.2 The Space L 2(1l) and Hilbert Spaces
+ W E u1. and cv E u1.. Hence u1. is a subspace of H. Now u1. is the set of vectors x E H where the continuous function g(x) = (u, x) = 0, so u1. is a closed set in H. If S is any subspace of H, let S1. denote the collection of all v E H that are orthogonal to every u E S. Clearly S1. = n{u1.1 u E SI. Since each u1. is a closed subspace of H, so is S1.. PROPOSITION I1.7 Each nonempty, closed, convex set in a Hilbert space has a smallest element with respect to the inner product norm. Proof: Let E be a nonempty, closed, convex subset of the Hilbert space H. Put b = inf{ Iv II VEE I. Then there exists a sequence {vn IcE such that { Ivn I I
converges to b. For any pair x, y E H, Ix + yl2 + Ix - yl2 = (x + y, x + y) + (x-y,.x-y) = 21xl2 + 21y12. If we apply this identity to x/2 and y/2 we get (1/4)lx-yI2= IxI 2/2+ lyI2/2-I(x+y)/212. SinceEisconvex,(x+y)/2E E which implies I(x + y)/21 ~ b, so (11.2) for each pair x, y E H. If we replace x and y in this inequality by Vm and Vn we see that as m, n ~ 00, the right side of (11.2) tends to zero. Hence IVm - Vn I ~ o as m, n ~ 00 which implies {v n I is a Cauchy sequence in E c H. Since H is complete, {v n I converges to some v E H, and since E is closed, vEE. Also, since the norm function h(x) = Ix I is continuous on H by Proposition 11.6, we have I v I = limn IVn I = b. Consequently, there exists a vEE of smallest norm. It remains to show that v is unique. If u is another member of H such that lui = b = lvi, then the inequality (11.2) implies lu - vI2 ::; 0 so u = v. Therefore, E has a smallest element with respect to the inner product norm. Let S be a closed subspace of the Hilbert space H and let u E H. Then the set u + S = {u + v Iv E S I is closed and convex. To see that u + S is closed, let w be a limit point of u + S. Then there exists a sequence {u + Vn I in u + S that converges to w, so {u + Vn I is Cauchy. If E > 0, there exists a positive integer N such that if m, n > N then I u + Vm - U - Vn I < E which implies IVm - Vn I < E, so {v n I is Cauchy in S. Since S is a closed subspace of the complete space H, S is complete which implies {v n I converges to some v E S. Then if E > 0, there exists a positive integer N such that if n > N then Iv - Vn I < E which implies I(u+v) - (u + vn) I < E, so {u + Vn I converges to u + V E U + S. But since limits are unique in metric spaces, u + v = W so W E U + S. Hence u + S is closed. To see u + S is convex, let u + v and u + W E
U
+ S and let 0 ::; A::; 1. Then
(1 - A)(U + v) + A(U + w) = U + (1 - A)V + AW E U + S. Therefore, u + S is convex. Consequently, we can apply Proposition 11.7 to u + S and get a
smallest element Ps.L(u) in u + S with respect to the inner product norm. Next, putps(u) = u - Ps.L(u). Thenps andps.L are functions on H. Since Ps.L(u) E u +
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5, Ps(u) E 5. The function Ps is called the orthogonal projection of H onto 5. The function ps.l is called the orthogonal projection of H onto 5.1. For this later definition to make sense, we need to show thatps.l(u) E 5.1. For this let x = Ps.l(u). Then by the definition of the orthogonal projection onto 5.1 we have, for each y E 5 with Iy I = 1, Ixl2 ~ Ix-cyl2 = (x-cy,x-cy) = IxI 2 -c(y,x)-c*(x,y)+ Icl 2 for each scalar c. If we substitute c = (x, y) into this inequality, we get 0 ~ -I (x,y) 12 which implies (x, y) = 0 for each y E 5 with Iy I = 1. But then (x, y) = OforanYYE 5. Therefore,x=ps.l(u)E 5.1.
THEOREM 11.6 If 5 is a closed subspace of the Hilbert space H, then the orthogonal projections Ps and Ps.l of H onto 5 and 5.1 respectively have the following properties: (1) u = ps(u) + Ps.l(u)for each u E H. (2) The orthogonal projections are unique. (3) The orthogonal projections are linear. (4)lfu E 5 thenps(u) = u andps.l(u) = o. (5)lfu E 5.1 thenps(u) =0 andps.l(u) = u. (6) I u - Ps(u) I = inf{ Iu - v II v E 5j for each u E H. (7) Iu 12 = Ips(u) 12 + Ips.l(u) 12 for each u E H. Proof: (1) follows immediately from the definition of Ps(u). To show (2), first note that 5n5.1 = {OJ. This is because if x E 5n5.1 then (x, x) = 0 which implies x = O. Next let u = v + w where v E 5 and w E 5.1. Thenps(u) + Ps.l(u) = U = v + w which implies Ps(u) - v = w - Ps.l(u). Since the left side of this equation is in 5 while the right side is in 5.1, we conclude that both sides are the zero vector. Therefore, v = Ps(u) and w = Ps.l(u), so the orthogonal projections are unique. To show (3), let u, v E Hand c, dEC. Then by (l),ps(cu + dv) + Ps.l(cu + dv) = cu + dv = c[ps(u) + Ps.l(u)] + d[Ps(v) + Ps.l(v)] so Ps(cu + dv) - cps(u) dps(v) = cps 1. (u) + dps.l(v) - Ps.l(cu + dv). Again, the left side of this equation is in 5 while the right side is in 5.1 so we conclude both sides are the zero vector. Therefore, Ps(cu + dv) = cps(u) + dps(v) and Psl.(cu + dv) = Cps.l(u) + dps 1. (v) so the orthogonal projections are linear. To show (4) and (5), note that if u E 5, then (1) implies u - Ps(u) = Ps.l(u) so the left hand side of this equation is in 5 while the right hand side is in 5.1. Again, we conclude both sides are the zero vector, so Ps(u) = u and Ps.l(u) = O. This proves (4). A similar argument proves (5). To show (6), note that by definition of Psl.(u) we have Iu - Ps(u) I = Ips.l(u)1 =inf{lu+vllvE S} =inf{lu-vllvE S}. This proves (6). To show (7), observe that 1u 12 = (u, u) = (Ps(u) + Psl.(u), Ps(u) + Psl.(u» = Ips(u) 12 +
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(Ps(u), Psl.(u)) + (Psl.(u), Ps(u)) + Ipsl.(u)12. Since (Ps(u), Psl.(u)) = 0 = (Psl.(u), Ps(u)) we have that 1u 12 = Ips(u) 12 + Ipsl.(u) 12 which proves (7).COROLLARY J J.2 If S :t- H then there exists a u E H such that u 1- S and u:t- O. Proof: Pick v E H - S. Put u = Psl.(v). Then v:t- Ps(v) so u:t- O. But u 1- S since
UES.l.In Proposition 11.6 we saw that the function f(x) = (x, y) for a fixed y E H is unifonnly continuous. Since (x, y) E C and since the definition of the inner product causes f to be linear, we see that f is a continuous linear functional. The Riesz Representation Theorem (Chapter 8) showed that positive linear functionals on CK(X) could be represented as positive measures on X. It is therefore natural to ask if continuous linear functionals on a Hilbert space can be represented as inner product functions with respect to a given vector. The answer is affinnative as the next theorem shows. THEOREM J J.7 If A is a continuous linear functional on a Hilbert space H, then there exists a unique v E H such that A(U) = (u, v)for each u E H. Proof: If A(U) = 0 for each u
E H put v = O. Otherwise put K = {u 1 A(U) = OJ. Since A is linear, K is a subspace of H and since A is continuous, K is closed. Since A is not identically zero, Theorem 11.6 shows that K.l :t- {O j. Choose w E K.l such that w :t- O. Then w is not in K so A(W) :t- O. Put c = A(w)/I W12 and let v = c*w. Then v E K.l, v :t- 0 and A(V) = A(C*W) = C*A(W) = c*c 1W12 = cc*(w, w) = (c*w, c*w) = (v, v). For any u E H put x = u - A(U)V/A(v). Then A(X) = A(U) A(U)A(V)/A(V) = 0 so X E K which implies (x, v) = O. Now (u, v) = (X+A(U)V/A(V),V) = (x, v) + (A(U)V/A(V), v) = 0 + [A(U)/A(V)](V, v) = [A(U)/A(V)]A(V) = A(U). Consequently, A(U) = (u, v) for each U E H.
To show that v is unique, let w be another vector in H such that A(U) = (u,w) for each U E H. Put z = v - w. Then (u, z) = (u, v - w) = (u, v) - (u, w) = A(U) - A(U) = 0 for each U E H. But then (z, z) = 0 which implies z = 0 so v = w.
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Recall how a basis is defined in a vector space V. First we define a linear combination of vectors VI ..• Vn E V to be a vector sum of the fonn c 1 v 1 + ... + Cn Vn for some c 1 .•• Cn E C. We define the vectors VI' .• Vn to be linearly independentifclvl +",+cnvn=Oimpliesci=Oforanycj ... CnE C. A subset S of V is said to be linearly independent if every finite subset of S is linearly independent. The set V(S) of all linear combinations of finite subsets of S is clearly a vector space. In fact, it is easily seen to be the smallest subspace of V containing S. V(S) is called the span of S or the subspace spanned by S. If V(S) = V then S is called a spanning subset of V or we say S spans V. Finally,
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we define a basis of a vector space V to be a linearly independent subset that spans V. PROPOSITION 11.8 A linearly independent subset of a vector space is a basis if and only if it is maximal. Proof: Let S be a maximal linearly independent subset of a vector space V. Assume V(S) '# V for otherwise S would be a basis for V. Then 0 E V(S) so there exists a U E V - V(S) with U '# O. Let v I . . . Vn E S and suppose C I v I + ... + CnVn + Cn+1 U = 0 for C I ••• Cn+1 E C. If Cn+1 '# 0 then U = Li'=1 CjVJCn+1 which implies U E V(S) which is a contradiction. Therefore, Cn+l = 0 which implies C I VI + ... + CnV n = 0 so Cj = 0 for each i = 1 ... n. But then Su{u} is linearly independent which implies S is not maximal which is a contradiction. Therefore, S is a basis for V.
Conversely, assume {u a} is a basis for V and suppose i U a I is not maximal. Then there exists a U E H with U '# 0 and {u a }u{ u} linearly independent. Since {u a} spans V, there exists a finite subcollection UI ... Un of {u a } and a finite collection CI ••• Cn E C with U = ClUJ + ... + CnU n. Then CIUI + ... + CnU n + (-I)u = 0 but (-1) '# 0, so {ua}u{u} is not linearly independent after all. • A set of vectors {u a} in a Hilbert space H is said to be orthogonal if (ua,u~)=Oifa'#~. {u a } is said to be normalized if Iual = 1 foreacha. If {u a} is both orthogonal and normalized it is said to be orthonormal. Clearly, {u a} is orthonormal if and only if (u a, u~) = 1 if ex = ~ and 0 otherwise. PROPOSITION 11.9 If UI ... Un is an orthonormal set and v = C I UI + ... + CnU n. then Ci = (v, ui)for each i = 1 ... n and Ivl 2 = Li'=I!c; 12. Proof: For each i = 1 ... n, (v, u;) = (c I UI + ... + CnU n' u;) = Ci(Ui, Ui) = Ci since UI ... Un is an orthonormal set. Also, Ivl 2 = (v, v) = (CIUI + ... + CnU n' C lUI + ... + CnU n) = C ICI *(u" ud + ... + CnC n*(u n, un) = Li'=,1 Ci 12 .• COROLLARY 1J.3 An orthonormal set is linearly independent. Proof: Let {u a} be an orthonormal set and let U! ... Un E {u a}. Suppose C 1 U1 + ... + CnU n = 0 for C I ... Cn E C. By Proposition 11.9, Ci = (0, u;) = 0 for each i = 1 ... n. Consequently, {u a} is linearly independent. • THEOREM 1 J.8 Each Hilbert space has an orthonormal basis. Proof: If H is a non-trivial Hilbert space, then there exists a U E H with IUI = 1. Then {u} is an orthonormal subset of H. Let P be the collection of all orthonormal subsets of H containing (u), partially ordered by set inclusion. Since {u} E P, P '# 0 so by the Hausdorff Maximal Principle (an equiValent
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fonn of the Axiom of Choice), there exists a maximal linearly ordered subcollection Q of P. Clearly {u} E Q, so Q =I- 0. Let R = uQ. If v, W E R then v E A E A and wEB E Q for some A, BE Q. Since Q is linearly ordered by inclusion, either v, W E A or v, wEB. Since both A and B are orthononnal subsets of H, (v, w) = 1 if v = wand 0 otherwise. Therefore, R is orthononnal. Suppose R is not a maximal orthononnal set. Then there exists an orthononnal set 5 containing R with 5 - R =I- 0. Now 5 is not in Q and 5 contains each member of Q so we can adjoin 5 to Q and still have a linearly ordered set with respect to inclusion which implies Q is not maximal which is a contradiction. Therefore, R is a maximal orthononnal set. By Corollary 11.3, R is linearly independent. Suppose R is not a maximal linearly independent set. Then there exists a linearly independent set T with ReT and T - R =I- 0. Let x E T - R. Let V be the subspace of H spanned by R. Then x is not in V which implies y = Pv.i(x) =I- 0 by Theorem 11.7, so (y, v) = 0 for each v E R. Put z = y/I y I which implies Izl = 1 and (z, v) = 0 for each v E R. Therefore, z can be adjoined to R to obtain an orthononnal set in H containing R as a proper subset which is a contradiction. Consequently, R is a maximal linearly independent set so R is a basis. Since R is orthononnal, R is an orthononnal basis. -
E
LEMMA 11.1 If V is a closed subspace of the Hilbert space H and H - V, then the subspace W spanned by Vu {u) is closed.
if u
Proof: Suppose v is a limit point of W. Then v = limn(v n + cnu) where {v n } C V and {c n } C C. Consequently, there exists a b < 00 such that IVn = Cnu I < b for each positive integer n. Assume {c n } has no convergent subsequence. Since closed and bounded subsets in C are compact, this implies limn I Cn I = 00. But then IV n + cnul/lcnl < b/lcnl for each n and limnbllcnl = O. Therefore, limn Ivn!cn + u I = 0, so limn(vn!c n} = -u which implies -u E V since V is closed which is a contradiction.
Hence we may assume {c n } has a convergent subsequence {c mn } that converges to some C E C. Now v = limn(v mn + cmn u) = limn vmn + cu which implies {v mn } converges to v - cu. Since {v mn } c V and V is closed, [v - cul E V and v = [v - cu + cul E W. Therefore, W IS c1osed.THEOREM 11.9 If u 1 . Un is an orthonormal set of vectors In the Hilbert space H and v E H, then 1 v - L;=! (v, Ui)Ui I :s; 1 v - Li'=! CiUi I for all c! .. Cn E C and equality holds if and only if Ci = (v, Ui) for each i = 1 ... n. The vector L;=! (v, Ui)Ui is the orthogonal projection of v onto the subspace W generated by {u! ... un} and if d is the distance from v to W then 1v 12 = d 2 + L;=! I (v, Ui) 12. Proof: The subspace {O} of H is obviously closed. By Lemma 11.1, the subspace of H spanned by {u I} is closed since it is the subspace spanned by {O} u {u 1 }. Proceeding inductively, it is clear that W is closed. Then by
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Theorem 11.6, Pw(v) is an element of W such that Iv - Pw(v) I ~ Iv - w I for each W E W and since the mappings Pw and pw:c are unique, Pw(v) is unique. Consequently, Pw(v) has the property that I v - Pw(v) I ~ I v - L?=1 CiUi I for each collection C 1 ... Cn in C. Let Pw(v) = L?=1 aiUi for some a 1 ... an E C. By Proposition 11.9, ai = (Pw(v), Ui) for each i = 1 ... n. Now v =Pw(v) + Ps.:c(v) so (v, Ui) = (Pw(v), uJ + (Pw:c(v), Ui) = ai + 0 since Pw:c(v) is orthogonal to all the Ui. Hence Pw(v) = L?=1 (v, UJUi' Therefore, Iv - L?=1 (v, UJUi I ~ Iv - L?=1 CiUi I for all C 1 ... cn E C and equality holds if and only if Ci = (v, Ui) for each i = 1 ... n. Finally, the distance d from v to W is the minimum value of Iv - w I such W E W. Therefore, d = I v - Pw(v) I so d 2 = (v - Pw(v), v - Pw(v» = (v-Pw(v),v) - (v - Pw(v), Pw(v». Now (v - Pw(v), Pw(v» = 0 so d 2 = (v - Pw(v), v). To see this, note that for each i = 1 ... n, (v - Pw(v), Ui) = (v - L?=1 (v, UJUi, Ui) = (v, Ui) - (v, Ui)(Ui, uJ = O. Therefore, (v - Pw(v), Pw(v» = (v - Pw(v), L?=1 (V,Ui)Ui) = L?=1 (v, Ui)(V - Pw(v), Ui) = O. Consequently, d 2 = (v - Pw(v), v) = (V-L'?=I(V,Ui)Ui, v) = Ivl 2 - L?=I(V, Ui)(Ui, v) = Ivl 2 + L?=II(v, UJI2 which implies Iv 12 = d 2 + L?=1 I(v, Ui) 12. that
Let {U a} be an orthonormal set in the Hilbert space H. For each vector v Va = (v, ua) for each a. By the symbol Lalval2 is meant the supremum of the set of finite sums of the form I VI 12 + ... + IVn 12 where Vi = (v, Ui) for each finite collection U1 ... Un E {U a}. With this notation we can state and prove the following classical result: E
H put
THEOREM 11.10 (Bessel's Inequality) La Iva 12 ~ IV 12. Proof: For any finite collection U1 . . . Un E {u a}, Theorem 11.9 gives LI=1 I(V,Ui) 12 = IV12 - d 2 where d is the distance from v to the subspace W spanned by the vectors U1 ... Un' Since d ~ 0, this means LI=1 I(v, Ui) 12 ~ IV12 and hence the supremum of such finite sums is less than or equal to Iv 12. -
Sums of the form LaC a where 0 ~ C a ~ for each a and where the summation is defined as the supremum of finite sums of the form C 1 + ... + Cn where C 1 ... Cn E {C a} are especially interesting in Hilbert spaces because they are used in the characterization of the structure of Hilbert spaces. We will now develop this characterization. Let X be a set. For each E E X put Il(E) = 00 if E is infinite and Il(E) = card(E) if E is finite. It is easily seen that 11 is a measure on the a-algebra of all subsets of X. The measure 11 is called the counting measure on X. Let f:X ~ C. Then it is easily seen that LxeX If(x) I, where the summation is the supremum of the finite sums of the form If(x 1) I + ... + If(Xn) I where Xl' .. Xn E X, is the Lebesgue integral of If I with respect to the counting measure on X. We use the notation [2(X) to denote the L2-space L 2(11) where 11 is the counting measure on X. 00
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In particular, if {u a I a E A} is an orthononnal basis in H and for some v E H, v':A ---+ C is defined by v'(a) == Va == (v, ua) for each a E A, then it is a consequence of Bessel's Inequality that v' E [2(A) since:
The importance of the Riesz-Fischer Theorem that we will prove next is that if { U a I a E A} is an orthononnal set of vectors in H then f E [2 (A) implies that f is of the fonn v':A ---+ C for some vector v E H. Before we prove the RieszFischer Theorem, we first observe that Bessel's Inequality implies an even stronger statement about a function v':A ---+ C defined by v'(a) == (v, ua) for each a E A. Bessel's Inequality implies the set of a E A for which v'(a) "# 0 is at most countable. To see this, suppose the set of a E A for which v'(a) "# 0 is uncountable. For each positive integer n put En == {a E AI I v'(a) I > lin}. Then for some positive integer m, Em must be uncountable. But then there exists a finite subset S of Em with LaeS I v'(a) 12 > IV12 which is a contradiction.
THEOREM 11.11 (Riesz-Fischer) If {Ua Ia E A} is an orthonormal set in a Hilbert space H and iff E [2(A), thenf == v' for some v E H. Proof: For each positive integer n put En == {a E A Ilf(a) I > lin}. Then each En must be finite, for otherwise there would exist a finite collection al ... an E A such that Li'=1 If(a;) 12 > If12' For each positive integer n let Vn == LaeEn If(a) IUa. Then each Vn is in H. For each n define v~:A ---+ C by v~(a) == (vn,ua)foreachaE A. ThenforagivenBE A,v~(B)==(LaeEnlf(a)lua,u~) == LaeEn If(a) I(u a , u~) == If(B) I if BEEn and 0 otherwise. Therefore, v~ == If I XE n' Consequently, If - v~ I ~ Ifl2 which implies If - v~ I ~ If I. Since Em C En if m < n, it is clear that limn v~ == f, so by Theorem 8.13, limn If - v~ I == 0 which implies lim" If - v~ I == 0 since we can choose an N large enough so that n > N implies [LaeEn If - v~ 12]112 < LaeEn If - v~ I and therefore If - V~ 12 == [LaeEn If - V~ 12]112 < LaeEn If - v~ I == If - v~ II for n > N. Then since lim" If - v~ 12 == 0, {v~} is a Cauchy sequence in [2(A). Now for each n, Vn == LaeEn If(a) IUa and since E" is finite we can apply Proposition 11.9 to get If(a) I == (v", ua) for each a E E". If m and n are positive integers with m < n, then Em C En. Put E(n, m) - En - Em. Then Iv~ - v~ I == [LaeE(",m) I(v", Ua) - (vm, Ua) 12]1/2 == [:EaeE(n.m) Ilf(a) I - 01 2]112 ==
(LaeE(n,m) If(a) IUa, :EaeE(n,m) Ifla) IUa )1!2 == (v n - vm, v" - vm)1I2 == Iv" - Vm
I,
so 1Vn - Vm I == I V~ - v~ 12 which implies {v n ) is Cauchy in H and therefore converges to some v E H.
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Then v':A ~ C defined by v'(a) = (v, ua) for each a is the desired function. To see this, for a fixed ~ E A, the function g(x) = (x, u~) for each x E H is uniformly continuous (Proposition 11.6). Therefore, since {v,,} converges to v, {g(v,,)} converges to g(v). Hence lim"(v,,, u~) = (v, u~). Then for each a E A, v'(a) = (v, Ua) = lim"(v,,, Ua) = lim" v~(a) =f(a) since {v~} converges to f. Consequently,! = v'. PROPOSITION 11.10 Let {u a} be an orthonormal set in the Hilbert space H. Then {u a} is a basis for H if and only if the set S of finite linear combinations of members of {u a} is dense in H. Proof: Assume {u a} is a basis for H. By Proposition 11.8, {u a} is a maximal linear independent subset of H. Suppose {u a} is not a maximal orthonormal set in H. Then there exists a U E H with IU I = 1 and {u a} U {u} orthonormal. By Corollary 11.3, {u a} U {u} is linearly independent which is a contradiction. Hence {u a} is a maximal orthonormal set in H. Now assume S is not dense in H. Then there exists a v E H - CI(S) which implies there exists ad> 0 such that Iv - U I > d for each U E CI(S). Now v = Ps(v) + Ps~(v) and by the remarks preceding Theorem 11.6, W = Ps~(v)/IPs~(v)1 is in S-L. Since Iwl = 1, { U a} U {w} is orthonormal which is a contradiction. Therefore, S is dense in H.
Conversely, assume S is dense in H and suppose {u a} is not a basis for H. Then {u a} is not a maximal orthonormal set in H, so there exists a U E H with IU I = 1 and U 1.. Ua for each a. Now there exists a sequence {u,,} c S such that lim"u" = u. Clearly U 1.. u" for each positive integer n. Since the function f:H ~ C defined by f(v) = (u, v) for each v E H is uniformly continuous, {f(u,,)} must converge to feu) = (u, u) = 1. But this is impossible since (u, u,,) = 0 for each n. Hence {u a} is a basis for H. The property that S is dense in H has some very interesting ramifications. In fact, it leads to a representation of all Hilbert spaces H as 12(A) where card(A) = card( {II a}) where {u a} is a basis for H. To show this is the case, we need the following lemma. LEMMA 11.2 Let {u a Ia E A} be an orthonormal set in the Hilbert space H and let S be the collection of finite linear combinations of members of {ua}. For each pair x, y E H let x', / E [2(A) be defined by x'(a) = (x, ua) and ira) = (y, ua)for each a E A. Then S is dense in H if and only if (x, y) = (x', y') for each pair x, y E H. Proof: First assume S is dense in H. Let v E H. Now choose e > O. Then there exists a finite col~ec~ion U1 . . . U" E {u a } and c 1 ••. C" E C such thalt w c 1 U1 + ... + C"U n IS wlthm e of v. By Theorem 11.9, Iv - L7;1 (v, uJu; I ~ v - wi < e. Putz=L7;I(V,U;)Uj. Then Iv-zl <e which implies Ivl < Izl +ewhichinturn implies (Ivl - e)2 < Izl2 = I(v, uI)12 + ... + I(v, U,,)12 ~ La Iv'(a)12 by Proposition 11.9. But La Iv'(a) 12 = Lav'(a)[v'(a)]* = (v', V'). Hence, for each v
=
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E H, (v, v) = IV 12 ~ (v', v'). By Bessel's Inequality, (v', v') ~ (v, v) so (v, v) (v', v') for each v E H.
=
Now let u, v E H. Then for each c E C we have (u + cv, u + cv) = (u' + cv' , U' + cv') which implies (u, cv) + (cv, u) = (u ' , cv') + (cv ' , u') or c*(u, v) + c(v, u) = c*(u', v') + c(v', u'). Since this holds for each c E C, it holds for c = 1 and c = i. When c = 1 we have (u, v) + (v, u) = (u', v') + (v', u'). When c = i we have (u,v) - (v, u) = (u', v') - (v', u'). Adding these two equations yields: 2(u, v) = 2(u',v') which implies (u, v) = (u', v'). Conversely, assume (x, y) = (x', y') for each pair x, y E H. Suppose S is not dense in H. Pick u E H - CI(S) and let W = Ps.l(u). Clearly W =I- 0 and (w, u a ) = o for each a. Put x = W = y. Then (x, y) = IW 12 > 0 but (x', y') = La(W,Ua)(w,u a )* = 0 so (x, y) =I- (x', y') which is a contradiction. Hence S must be dense in H.· Recall from Section 1.5 that two metric spaces X and M are isomorphic if there exists a uniform homeomorphism f:X ~ M that preserves distance. Two Hilbert spaces HI and H 2 are said to be isomorphic if there is an isomorphism f:H 1 ~ H 2 that is also a linear transformation, i.e., one that preserves sums and scalar products. Such a mapping is called a Hilbert space isomorphism. THEOREM 11.12 If {U a Ia E A} is an orthonormal basis for the Hilbert space Hand iffor each x E H, x' is the element of 12(A) defined by x'(a) = (x, u a ) for each a, then the mapping A:H ~ 12(A) defined by A(X) = x' for each x E H is a Hilbert space isomorphism of H onto [2(A). Proof: Let u, v E Hand c, dEC. Then A(CU + dv) = (cu + dv)'. Then for each a E A, (cu + dv)'(a) = (cu + dv, u a ) = c(u, u a ) + d(v, Ua ) = cu'(a) + dv'(a) = CA(V) + dA(V) so A is a linear transformation from H into 12(A). That A is onto follows from the Riesz-Fischer Theorem.
Suppose u =I- v but u' = v'. Then for each a E A, (u, u a ) = (v, u a ). Now (u-v)' E 12(A) and (u - v)'(a) = (u - v, u a ) = (u, u a ) - (v, u a ) = 0 for each a so (u-v), is the zero element of 12(A). Since u =I- v, Iu - vi > 0 which implies W = (u-v)/I u - vi is a unit vector in H such that w.1 Ua for each a, so {ua}u{w} is an orthonormal set in H. But then {u a} is not maximal which is a contradiction. Hence u - v = 0 so u = v. Therefore, A is one-to-one. Since (u, u) = (u', u') for each u E H, A preserves inner products and hence distance. Therefore, A is a metric space isomorphism between Hand [2(A). Since Ais a linear transformation, it is a Hilbert space isomorphism. •
EXERCISES
1. Show that the vector space C*(X) of all real valued continuous functions on
11. Spaces of Functions
338
x = [0, 1] is an inner product space with respect to (j, g) = Jxfg*dx (where dx denotes integration with respect to Lebesgue measure) but not a Hilbert space. 2. Show that if S is a closed subspace of the Hilbert space H then (S-L)-L
= S.
3. Show that a Hilbert space is separable if and only if it contains a countable orthonormal basis. 4. Let {un) be an orthonormal sequence in the Hilbert space H. Let S be the set of all v E H with v = L;;'=l CnU n where ICn I ~ lin. Then S is isomomorphic with the Hilbert cube and is an example of a closed and bounded subset of H that is not compact. 5. Show that no Hilbert space containing an orthonormal sequence is locally compact. 6. Show that for each pair of Hilbert spaces, one of them is isomorphic to a subspace of the other.
7. Let U be a member of the Hilbert space H and let S be a closed linear subspace of H. Show that min { IU - v II v E S) = maxI I(u, w) II W E S -L and Iwi = 1).
THE TRIGONOMETRIC SYSTEM Let T be the unit circle in the complex plane. If F:T ~ C is any function on T then the function f defined by f(x) = F(e ix) for each x E R is a periodic function of R of period 2n, i.e., f(x + 2n) = f(x) for each x E R. Conversely, if f:R ~ C is periodic with period 2n, it is easily seen that there is a function F:T ~ C such that f(x) = F(e ix ) for each x E R. Therefore, we can identify the complex valued functions on T with the complex valued 2n periodic functions on R. Define U (D, where 1 ~ p ~ 00, to be the class of all complex valued, Lebesgue measurable. 2n periodic functions on R for which
Iflp =
[1/2nfr-Jr.Jr) If(x) Ipdx] lip <
00.
In other words, U (D = U (IA) where IA is Lebesgue measure on [-n, n) divided by l/2n. The factor 1/2n is not essential here but it simplifies the following development. For instance, with this definition, the U -norm I Ip of the constant function 1 is 1. 8. For each pair f, g E L2(D define (j, g) = l/2nJ1tJ(x)g(x)*dx for each x E [-n,n). Show that (j, g) defines an inner product on L2(D and that If I~ = (j,!) for each f E L 2(D.
11.2 The Space L 2(11) and Hilbert Spaces
339
9. A trigonometric polynomial is a finite sum of the fonn p(x) = a 0 + L~=l [ancosnx + bnsinnx] for each x E R where the an's and bn's are complex numbers. As is well known from calculus, for each x E R we have e ix = cosx + isinx. Show this implies that p(x) can be written as p(x) = L~~NCneinx for some suitable cn's E C for n = -N . .. N. 10. For each integer n (not necessarily positive) put un(x) = e inx for each x E R. Then for each n, Un E L 2 (n. Show that for each pair of integers m, n that (um,u n) = 1 if m = n and 0 otherwise. Consequently, {un In E Z} is an orthononnal set in L 2(n. II. Show that for each positive integer n it is possible to choose Cn such that if = 2- n cn(1 + cosxt for each x E R then the sequence of functions {Pn} converges unifonnly on [-n, -e]u[e, n) for each e > O.
Pn(x)
12. Show that {Pn} is a sequence of trigonometric polynomials such that (1) Pn(x) ~ 0 for each x E R. (2) l/2nJ~llPn(x)dx = 1. (3) If 8 n(e) = SUp {Pn(x) Ix E [-n, -e]u[e, n)} then lim n8n(e) = 0 for each e> O. 13. Let C(n denote the class of all continuous, complex valued, 2n-periodic functions on R with nonn III = = Supx I/(x) I. For each I E C(n we can associate a sequence of functions {P n} defined by Pn(y)
= l/2nfr-lt,lt)l(Y - x)Pn(x)dx.
Show that {P n } is a sequence of trigonometric polynomials such that for each positive integer n,
14. Let I E C(n. Since I is unifonnly continuous on [-n, n], if e > 0, there exists a 8 > 0 such that Vex) - l(y) I < e whenever Ix - y I < 8. By (2) of Exercise 12, it follows that Pn(y) - l(y)
=
1/2nfr-lt,lt)[f(Y - x) - 1(Y)]Pn(x)dx.
Show that limn-'>= IPn - II = = 0, and hence for each IE C(n and e > 0 there exists a trigonometric polynomial P such that I/(x) - P(x) I < e for each x E R. 15. Show that {un a basis for L l{n.
In E
Z} where Un is defined by un{x) = e inx for each x
E
R is
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11. Spaces of Functions
11.3 The Spaces U (11) and Banach Spaces Hilbert space norms are fairly specialized, being based on the concept of an inner product. This is what accounts for there being so few of them, essentially one for each cardinal. The LP -spaces are a special case of a more general class of spaces called the Banach spaces. These spaces derive their name from Stefan Banach who published a series of papers between 1922 and 1932 in several journals that culminated in the now famous Theorie des Operations lineaires (Monografje Matematyczne, Volume 1, Warsaw, 1932). In the special case p = 2, L 2 (11) is both a Hilbert space and a Banach space. As we shall see from the definition of Banach spaces, this is because all Hilbert spaces are Banach spaces. A normed linear space X is a vector space over the complex field C (i.e., an abelian group in which multiplication of group members by complex numbers, called scalar multiplication, is defined that satisfies the distributive laws) in which a non-negative real number Ixl (called the norm of x) is associated with each x E X that satisfies the following properties: Bl. Ixl = 0 if and only if x = O. B2. Ix + yl :0:; Ixl + Iyl for each pair x,y EX. B3.laxl = lallxl foreachxE XandaE C. Here, Ia I denotes the modulus of the complex number a. A metric can be defined in a normed linear space in the following way. Define d:X x X ~ [0,00) by d(x,y) = Ix - y I for each pair x, y E X. That d is actually a metric is left as an exercise. If the metric space (X, d) is complete, X is said to be a Banach space. The simplest Banach space is merely C itself, normed by absolute value, i.e., Ix I is simply the absolute value (modulus) of x for each x E C. One can also discuss real Banach spaces by restricting the field of scalars to R. The topology induced on X by d is called the norm topology and the set xl Ixl :0:; I} is the closed unit sphere in X. A mapping T of a normed linear space X into a normed linear space Y is said to be a linear transformation if T(x + y) = T(x) + T(y) and T(ax) = aT(x) for each pair x,y E X and a E C (linear transformations are sometimes called vector space homomorphisms - the definition does not depend on the norm, only on the vector spaces X and Y). Linear transformations are also commonly referred to as linear operators. In the special case the space Y is the Banach space C, Tis referred to as a linear functional. The kernel of a linear operator is the set of all elements in X that get mapped onto the zero element of Y, Le., Ker(1) = {x E X IT(x) = O}. The proof of the following proposition is left as an exercise. S(O,I) = {x E
PROPOSITION 11.11 The kernel of a linear operator T:X ~ Y from the linear space X into a linear space Y is a linear subspace ofX.
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11.3 The Spaces LP(fl.) and Banach Spaces
A linear transfonnation T is said to be bounded if there exists a real number a such that IT(x) I ~ a Ix I for each x E X. The smallest a with this property is called the norm of T and is denoted by ITI. It is easily seen that ITI = sup { IT(x)l/lxl I x E X and x ::f:. O}. Since IT(ax) I = I aT(x) I = Ia II T(x) I for each x E X and a E C, we could restrict ourselves to unit vectors (i.e., x E X such that Ix I = I) in the definition of the nonn of T. In this case, we would have ITI = sup{ IT(x)l/lxl I x E X and Ixl = I}. A bounded linear transfonnation T maps the closed unit sphere in X into the closed sphere 5(0,ITI)= lYE ylly-Ol ~ ITI}. To see this,letx E S(O,l)inX. Then Ixl ~ 1. Since IT(x) I ~ ITllxl, we have IT(x)l/lxl ~ ITI. Since Ixl ~ I this implies I T(x) I ~ ITI which in turn implies IT(x) - 0 I ~ I TI so T(x) E 5(0, ITI). THEOREM J 1.13 Let T:X ~ Y be a linear operator from a normed linear space X into a normed linear space Y. Then the following statements are equivalent: ( 1) T is bounded. (2) T is uniformly continuous. (3) T is continuous at some point of x. Proof: If T is bounded, IT(x) - T(y) I = IT(x - y) I ~ I Til x - y I for each pair x,y EX. Thenif£>Oand Ix-yl <£/ITI,wehave IT(x)-T(y)1 ~ ITI(£/ITI)=£ so Tis unifonnly continuous. Thus (I) ~ (2). That (2) ~ (3) is clear.
To show (3) ~ (I), assume T is continuous at z E X and suppose £ > O. Then there exists a 0 > 0 such that Ix - z I < 0 implies IT(x) - T(z) I < Eo Then Ix I ~ 0 implies I z + x - xl ~ 0 which in tum implies I T(z + x) - T(z) I < Eo But IT(z + x) - T(z) I = IT(x) I so Ix I < 0 which implies IT(x) I < £. Hence IT(x) 1/0 > Ix I/£ which implies I T(x) II Ix I > 0/£ which in tum implies IT(x) II Ix I < £/0. Therefore, Ix I < 0 implies IT(x) II Ix I <... £/0. Now let y E X and let a E C such that layl < Then IT(ay)l/layl < £/0 implies laIIT(y)l/lallyl < £/0 so IT(y) II Iy I < £/0. Thus T is bounded by £/0. •
o.
Let L(X, Y) denote the set of all bounded linear operators of a nonned linear space X into a nonned linear space Y. Define addition and scalar multiplication on L(X, Y) as follows: if 5, T E L(X, Y) and a E C, then 5 + Tis defined by (5 + T)(x) = 5(x) + T(x) for each x E X and aT is defined by (aT)(x) = aT(x) for each x E X. The following lemma is left as an exercise. LEMMA 11.3 L(X, Y) is a normed linear space with respect to
I I.
THEOREM 11.14 ffY is a Banach space then so is L(X, y). Proof: Let {Tn} be a Cauchy sequence in L(X, Y). For each x EX, {Tn(x)} is a sequence in Y. For x 0 E X, let E > O. Since {Tn} is Cauchy, there exists a positive integer N such that m, n > N implies I Tm - Tn I < E/lxo I. Now ITm(xo) - Tn(xo) I ~ ITm - Tn II Xo I < E so {Tn(xO)} is Cauchy in Yand therefore conver-
342
11. Spaces of Functions
ges to some xo' E Y. Define T:X ~ Y by T(x) = x' for each x converges to T(x) for each x E X.
E X.
Then {Tn(x) I
To show T E L(X, Y), we must show T is linear and bounded. To show Tis linear, let x, y E X and a E C. Then T(x + y) = (x + y)' and T(ax) = (ax)'. Let E > O. Since {Tn (x) I converges to T(x) for each x E X, there is a positive integer N such that n > N implies ITn(x) - x'i < E/2 and ITn(y) - y'l < E/2. Now ITn(x+y)-(x'+y')l = ITn(x)-x'+Tn(y)-y'l::; ITn(x)-x'l + ITn(y)-y'1 <E. Therefore, n > N implies I Tn(x + y) - (x' + y') I < E so {Tn(x + y) I converges to (x' + y'). Hence (x + y)' = (x' + y') so T(x + y) = T(x) + T(y). Also, there exists a positive integer M such that n > M implies ITn (x) - x'i < E/lal which in tum implies ITn(ax) - ax' I < E. Therefore, {Tn(ax) I converges to ax' so (ax)' = ax' which implies T(ax) = aT(x). Consequently, T is linear. To show that T is bounded, let E > O. Since {Tn I is Cauchy, there exists a positive integer M such that m, n > M implies ITm - Tn I < E. Pick m > M. Then ITml <<Xl. Foreachn>M, ITnl = ITn-ol::; ITn-Tml + ITm-ol <£+ ITml. Consequently, there exists a bound ~ such that ITn I < ~ for each n. Let x E X such that Ix I = 1. Then there exists a positive integer N such that n > N implies ITn(x)-T(x)1 <E. Now ITn (x) I ::; ITnl1xl <~and IT(x) I ::; IT(x)-Tn(x)1 + ITn (x) - 01 < E +~. Therefore, ITI = sup{ IT(x) I I Ixl = II < £ + ~ so Tis bounded. Consequently, T E L(X, Y). It remains to show that {Tn I converges to T with respect to the norm I I. For this, let E > O. Then there exists a positive integer M such that m, n > M implies ITm - Tn I < E/4. Suppose there is no n > M with ITn - TI < E/2. Then for each n > M, ITn - TI :::>: £/2. Let m > M. Then sup{ ITm(x) - T(x) I I Ixl = 1 I :::>: £/2 which implies there exists an x E X with Ixl = 1 such that ITm(x) - T(x) I :::>: £/2. Now, for each n > M, ITm(x) - Tn(x) I ::; ITm - Tn II X I < £/4. Therefore, for each n > M, 1 Tn (x) - T(x) I :::>: £/4 for otherwise, 1 Tm(x) - T(x) 1 ::; ITm(x) - Tn(x) I + ITn (x) - T(x) I < £/2 which is a contradiction. Hence {Tn (x) I does not converge to T(x) which is a contradiction. Therefore, there exists a k > M with Tk - TI < £/2. But then for eachn>M, ITn-TI::; 1Tn-Tkl + ITk-TI <Eso{TnlconvergestoT.1
L(X, Y), where Y is a Banach space, is our first example of a Banach space whose elements are functions. There are many more function spaces that are Banach spaces. It is easily shown that for each 1 ::; p ::; <Xl that U (I!) is a Banach space with respect to the LP -norm. If Z is a dense linear subspace of X and Y is a Banach space, then by Theorem 1.16, if T E L(Z, y), T has a unique extension to a member T of L(X, Y). THEOREM 11.15
ITI = ITI.
Proof: Clearly, 1 TI ::; 1 T I from the definition of the I I-norm. Suppose ITI < 1T I. Then there exists an x E X - Z with 1x 1 = 1 such that 1T(x) I > ITI. Since Z is dense in X, there exists a sequence {zn I c Z that converges to x. For
11.3 The Spaces U()l) and Banach Spaces
343
each positive integer n, put Wn = anzn where an = 111 Zn I. Then {w n } is a sequence of unit vectors in Z that also converges to x. This follows from the fact that Ix - Wn I ~ Ix - Zn I + IZn - Wn I and both terms on the right hand side of this inequality can be made arbitrarily small if n is large enough. IZn - Wn I = 1(1 - an)zn I = 11 - an II Zn I = I(IZn I - 1) I and IZn I ~ IZn - x I + Ix - 0 I = Izn-xl + 1. This last term converges to 1 as n ---7 so {I Zn I} converges to 1 which implies { IZn - Wn I} converges to O. 00
Therefore, IT'(w n ) I = IT(w n ) I ~ ITI for each positive integer n. Since 1 is continuous, {T'(w n )) converges to T'(x). But this is impossible since IT'(x) I > ITI and IT'(w n ) I ~ ITI for each n. We conclude that 111 = ITI. The complex numbers C, normed by their absolute value form a Banach space. L(X, C) is called the dual space of X and is denoted by X*. L(X, C) is a Banach space by Theorem 11.14. The interplay between X and its dual space X* is the basis of much of that field of mathematics known as functional analysis. To explore this interplay, one needs the Hahn-Banach Theorem. The Hahn-Banach Theorem essentially says that if Y = C, then we can drop the assumption that Z be dense in X. Linear transformations from Z into C can then be extended to X in such a way that Theorem 11.15 still holds. By a real linear functional, we mean a linear operator from a real vector space (vector space over the real field). Letfbe a complex linear functional on a linear space X. Then for each x E X, f(x) = u(x) + iv(x) for some real valued functions u and v on X. Since X is a vector space over the complex field, it is clearly a vector space over the real field as well. It is easily seen that the linearity of f implies the linearity of u and v, i.e., u and v are real linear functionals. PROPOSITION 11.12 If X is a linear space and f is a linear functional on X then: (1) Ifu is the real part offthenj(x) = u(x) - iu(ix)for each x EX. (2) Ifu is a real linear functional on X andf(x) = u(x) - iu(x)for each x E X, then f is a (complex) linear functional on X. (3)1fX is a normed linear space andf(x) = u(x) - iu(ix)for each x E X, then If I = Iu I. Proof: If a, ~ E R and 'Y = a + i~, then the real part of i'Y is -~. Therefore, y Re(y) - iRe(iy) for each y E C. Then (1) follows with Y=f(x). To show (2), it is clear that f(x + Y) = f(x) + f(y) and that f(a.x) = af(x) for each a E R. But we must also show this second equation for a E C. It will suffice to show it for a = i. For this note thatf(ix) = u(ix) - iu( -x) = u(ix) + iu(x) = if(x).
To show (3) note that Iu(x) I ~ If(x) I for each x E X which implies Iu I ~ E x. Put ~ = f(x)/lf(x) I. Then I ~ I = 1 and ~ If(x) I = f(x). Put a = Ial = 1 and af(x) = If(x) I. Hence If(x) I =f( ax) which implies
If I. Let x ~-l. Then
11. Spaces of Functions
344
f(ax) isreal sof(ax) = u(ax)-S: lullaxl. Then If I =sup{lf(x)lllxl =1}-S: Iu I. Therefore, If I = Iu I. One of the most important theorems in the theory of Banach spaces is the
Hahn-Banach Theorem. It allows us to extend bounded linear functionals on subspaces of a normed linear space in such a way that the norm is preserved. THEOREM 1I.16 (S. Banach, H Hahn, 1932) ffY is a subspace ofa normed linear space X and iff is a bounded linear functional on Y, thenf can be extended to a bounded linear functional F on X such that IF I = If I. Proof: We first prove the theorem assuming the field of scalars to be real, i.e.,
we assume X is a real normed linear space and f is a real bounded linear functional on Y eX. If If I = 0 then clearly, the desired extension is F = O. If If I > 0, we may assume, without loss of generality, that If I = 1 since if If I of. 1, there exists an a E R with Iallfl = 1 which implies Iafl = 1. We can then prove the theorem for af and simply divide the extension F by Ia I. Assuming If I = 1, pick z E X - Y and let N be the subspace of X spanned by z and Y (i.e., N = {x + azlx E Yand a E R}). DefinefN:N ~ R by fN(X + az) =f(x) + aA for any fixed A E R. It is left as an exercise to show that fN is a linear functional on N such that fN restricted to Y is f. We next show that it is possible to pick A such that IfN I = 1. For this, first note that by the definition of I I, that If I -s: IfN I. Also,
so that if If(x) + al..I -s: Ix + azl for each x E Y and a E R, then IfNI -s: 1. Hence If(x) + al..I + Ix + azl for each x E Yand a E R which implies IfN I = 1. Therefore, it suffices to show that A can be chosen such that If(x) + aA I -s: Ix+az I for each x E Yand a E R. This can be shown if we choose A such that If (x)+aA
Ia
I
I
-s:
I Haz I Ia I
for each x E Y and a E R.
But this is equivalent to showing that we can find a A such that If(x/a) + AI -s: Ix/a + zl which in tum is equivalent to finding aWE Y such that If(w) - AI -s: IW - z I in view of the substitution w = -x/a. For each W E Y put )'(w) = few) Iw-zl and P(w) =f(w) + Iw-zl. If),(w)-S:A-s:P(w),then If(w)-AI-s: Iw-zl. Hence IfN I = 1 if )'(w) -s: A -s: pew) for each W E Y To show this, let I(w) be the interval [)'(w), pew)] for each w E Y. Then IfN I = 1 if n{ I(w) IWE y} of. 0, or equivalently if )'(w) ~ P(v) for each pair W,v E Y. Now)'(w)-P(w)=f(w-v)-Iw-zl-Iz-vl. Sincef(w-v)~ If(w-v)l:5 If I Iw - v I = I w - v I ~ Iw - z I + Iz - v I, )'(w) - P(v) ~ 0 so )'(w) ~ ~(v) for each
11.3 The Spaces LP(Il) and Banach Spaces
345
pair w,v E Y. Hence n{I(w) IWE Y} # 0. Pick A E n{I(w) IWE Y}. Then y(w) :<=;; A:<=;; ~(w) for each WE Y so IfN I = 1. We have shown that f can be extended to N such that the extension fN has the same norm as f. Let P be the collection of all subspaces N such that YeN and there exists a real linear extension fN of f on N with IfN I = 1. Define a partial order:<=;; on P by N :<=;; S if N c Sand fN(X) = fs(x) for each x E N. By the Hausdorff Maximal Principle (an equivalent form of the Axiom of Choice), there exists a maximal totally ordered subcollection Q of P. Put A = u {N IN E Q}. It is easily shown that A is a subspace of X. For each x E A, define F(x) = fN(x) such that x E N for some N E Q. Clearly, F is well defined since fN(X) = fs(x) if N c S and Q is totally ordered. It is left as an exercise to show that F is a linear functional on A. To show that IFI = 1, note first that If I :<=;; IFI so that 1 :<=;; IFI. Suppose Then there exists an x E A such that IF(x) I > 1. But x E A implies x E N for some N E Q so IfN(X) I > 1 which is a contradiction since IfN I = 1. It is easily seen that A = X, for otherwise, there would be Y E X - A and as we have already seen, the space Y spanned by y and A would be a subspace of X larger than A and F could be extended to a linear functional G on Y such that IG I = 1. But then Y E Q which implies YeA which is a contradiction. Therefore, F is the desired extension off to X.
IF I > 1.
To prove the theorem for a complex bounded linear functional f on a subspace Y of a complex normed linear space X, let u be the real part of f. As shown above, u can be extended to a real linear functional V on X such that IV I = Iu I. Define F:X -t C by F(x) = Vex) - iU(ix) for each x E X. By Proposition 11.12, F is a complex linear functional on X such that IF I = IV I = Iu I = If I. Moreover, if x E Y then F(x) = Vex) - iU(ix) = u(x) - iu(ix) = f(x) so F is an extension off. COROLLARY 11.4 If Y is a linear subspace of a normed linear space X and u is a vector in X - Y with distance d(u, Y) > 0 from u to Y, then there exists a bounded (continuous) linear functional f:X -t C such that If I = l,f(Y) = 0 andf(u) = d(u, Y).
Proof: Let K = {x E X I Ix I = O}. Then Z = Y + K = {y + x lYE Y and x E K} is a linear subspace of X. Since Y c Z it follows that d(u, Z) :<=;; d(u, Y). Let e > O. From the definition of d(u, Z), there exists a vector y + x E Z with Y E Y and x E K such that Iu - (y + x) I < d(u, Z) + e. Hence
Iu - y I
:<=;;
Iu - (y + x) I + I(y + x) - y I =
Iu - (y + x) I + Ix I
=
Iu - (y + x) I + 0
d(u, y) :<=;;
< d(u, Z) + e.
Since this holds for each e > 0, we have d(u, Y) :<=;; d(u, Z)' But then d(u, Y) =
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11. Spaces of Functions
d(u,Z). Let V denote the linear subspace of X spanned by u and Z, i.e., V = {cu + zlc E C and z E Z}. Define g:V ~ C by g(cu + z) = cd(u, Y). Clearly g is a
linear mapping. To show that g is a bounded linear functional let cu + z be an arbitrary vector in V. If c = 0 then g(cu + z) = O. If c "- 0, put w = -zlc. Then w E Z and Icu + zl = cllu + zlcl = Icllu - wi ~ cd(u, Z) = Icld(u, Y) > O. Hence Ig(cu+z) I = Ic Id(u, Y) ~ Icu + z I so that g is bounded. This also shows that Igl ~ 1. To show Ig I = 1, let r E (0,1). Then there exists a z E Z such that Iu - zl < d(u, Z)lr which implies d(u, Z) > Iu - z Ir, Now Ig(u - z) I = d(u, Z) > r Iu - z I so Ig I > r. Since this is true for each r E (0, 1) we have Ig I ~ 1. But then Ig I = 1. By the Hahn-Banach Theorem, g can be extended to a bounded linear functional f:X ~ C such that If I = Ig I = 1. Since f is an extension of g and Y c V we have f(Y) = g(Y) = O. Finally,f(u) = g(u) = d(u, Y).COROLLARY J J.5 For each vector u in a normed linear space X with
Iu I > 0, there exists a continuous linear functional f:X ~ C with If I = J and f(u) = I u I.
Returning now to the dual space X* of a normed linear space X, we note that it is possible to form the second dual space of X or bidual space or the second conjugate space (X*)* = L(X*, C). (X*)* is often denoted by X**. Again by Theorem 11.14, X** is a Banach space. Define a function n:x ~ X** as follows. If x E X, put n(x) =