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6 Compact Sets
6 Compact Sets A topological space X (not necessarily the subset of a TVS) is said to be compact if X is HausdorfT and if every open covering {Q,} of X contains a finite subcovering. The fact that {Q,} is an open covering of X means that each Q, is an open subset of X and the union of the sets Q, is equal to X.By a finite subcovering of the covering {QJ we mean a finite collection Q,, ,..., of sets Q, whose union is still equal to X. By going to the complements of the open sets Q, we obtain an equivalent definition of compactness: a HausdorfT space X is compact if every family of closed sets {Fi} whose intersection is empty contains a finite subfamily whose intersection is empty. In the sequel, we shall almost always be concerned with compact spaces which are subsets of a TVS and which carry the topology induced by the TVS in question; we shall then refer to them as compact sets. Let Y be a subset of a HausdorfT topological space X ; a subset of Y, B, is open in the sense of the topology induced by X if and only if there is an open subset A of X such that B = A n Y. In view of this, open coverings of Y are “induced” by families of open subsets of X whose union contains Y. Thus a subset K of X is compact if every family {Q,} of open subsets of X, whose union contains K, contains a finite subfamily whose union contains K. It should be pointed out that compactness is such that many properties of compact sets are independent, to a large extent, of the surrounding space. This will become apparent soon. We begin by stating without proof a few well-known properties of compact spaces (no linear structure is considered). If the student is not familiar with these properties, we strongly suggest that he proceed no further without having proved them by himself. QiV
PROPOSITION 6.1. A closed subset of a compact space is compact. PROPOSITION 6.2. Let f be a continuous mapping of a compact space X into a H a u s d o r - topological space Y . Then f ( X ) is a compact subset of Y. 50
51
COMPACT SETS
PROPOSITION 6.3. Let f be a one-to-one-continuous mapping of a compact space X onto a compact space Y . Then f is a homeomorphism (i.e., f -l is also continuous).
PROPOSITION 6.4. Let F, 7 be two Hausdorff topologies on a set X . Suppose that F isjiner than F’and that X , equipped with 9, is compact.
Then F = F’. Finite unions and arbitrary intersections of compact sets are compact. In a HausdorfT space X, every point is compact; every converging sequence is compact-provided that we include in it its limit point! The student should keep in mind that compact sets can be very complicated sets. Take for instance the real line R1.T h e BorelLebesgue-Heine theorem says that the compact subsets of R1are exactly the sets which are both closed and bounded. Note also that the Lebesgue measure of a sequence is equal to zero, and that if a set A is measurable, given any E > 0, there is a compact set K C A such that the measure of A n CK is <&. Take then the points x, with 0 x 1 , which are nonrational; they form a set of measure 1, since the rationals, which form a sequence, form a set of measure zero. This means that there are compact sets, contained in the interval [0, 11, which do not contain any rational number and whose Lebesgue measure is arbitrarily close to 1. Try to draw one of them! As the Weierstrass-Bolzano theorem shows, compact sets have interesting properties in relation with sequences of points. This extends to filters, as we are now going to see. In the immediate sequel, E is a HausdorfT topological space; when it is expressly mentioned, E is a TVS. T h e following terminology is useful:
< <
Definition 6.1. A point x of E is called an accumulation point of a filter 9 if x belongs to the closure of every set which belongs to 9. Let S = {xo , x1 ,...} be a sequence; a point x of E is often called an accumulation point of S if every neighborhood of x contains a point of S different from x. This terminology coincides with the one introduced by Definition 6.1 if we apply the latter to the filter Fsassociated with S (see p. 7). Let M e s s be arbitrary; M contains a subsequence of the form
s, = {% ,
,...I.
%l+l
If, then, x is an accumulation point of S , any neighborhood U of x contains some point xk with k arbitrarily large, in particular k 2 n. Thus U has a nonempty intersection with M, which means that x E i@. In
52
TOPOLOGICAL VECTOR SPACES
[Part I
other words, any accumulation point of the sequence S is an accumulation Conversely, if x is an accumulation point of the point of the filter YS. x belongs to the closure of all the sets belonging to Fs , in filter SS, particular to the closures of the sets S, , n = 0, 1,.... This means that given any neighborhood U of x and any integer n, there is k 3 n such that X ~ U. E
PROPOSITION 6.5. If afilter 9converges to apoint x , x is an accumulation point of 9. Indeed, suppose that x were not an accumulation point of 9. There would be a set M EF such that x 4 A?l. Hence the complement U of @, which is open, would be a neighborhood of x, and hence should belong to 9. But then we ought to have U n M # 0, vhence a contradiction. Of course, a filter might have more than one accumulation point. For instance, let 9 be the filter of all subsets of E containing a given subset A of E. Then every point of A is an accumulation point of 9. PROPOSITION 6.6. The following two conditions are equivalent: (a) x is an accumulation point of 9; which is finer than both 9 and the jilter of (b) there is a jilter F’ neighborhoods of x, 9 ( x ) , in other words: there is a filter 9‘ converging to x, which isfiner than 9. (a) => (b). Indeed, consider the family of subsets of E of the form U n M, where U varies over F ( x ) and M varies over F.These sets and they obviously are never empty if x is an accumulation point of 9, have Property (BF) of p. 7, hence they generate a filter 9’ which is obviously finer than 9 and 9 ( x ) . and if x’ is (b) 3 (a). If a filter 9 is less fine than another filter 9’ an accumulation point of F’,then x is also an accumulation point of 9. Thus it suffices to combine (b) with Proposition 6.5.
PROPOSITION 6.7. If a CauchyJilter 9on the TVS E has an accumulation point x , it converges to x. Let U be an arbitrary neighborhood of the origin and V another V C U. There is a set M E 9such neighborhood of zero such that V that M - M C V. On the other hand, V x intersects M , hence M - M n ( V x ) C V, or
+
+
MCV
+
+ M n (V + x) C V + V + x C U + x.
Q.E.D.
Chap. 6-41
53
COMPACT SETS
PROPOSITION 6.8. Let K be a H a u s d o e topological space. The following properties are equivalent: (a) K is compact; (b) every filter on K has at least one accumulation point. (a) s (b). Let F be any filter on K, and consider the family of closed sets fi when M varies over F. As no finite intersection of sets fi can be empty, neither can that be true of the intersection of all of them. (b) 3 (a). Let @ be a family of closed sets whose total intersection is empty. Suppose that @ does not contain any finite subfamily whose intersection is empty. Then take the family 0'of all the finite intersections of subsets belonging to @: it obviously forms a basis of a filter. This filter has an accumulation point, say x: thus x belongs to the closure of any subset belonging to the filter, in particular to any set belonging for these are closed. I n other words, x belongs to the intersection to of all the sets belonging to which is the same as the intersection of all the sets belonging to @. But the latter was supposed to be empty! @I,
@I,
Q.E.D.
COROLLARY 1. A compact subset K of a Hausdorf topological space E is closed. Proof. Let x E R; let S(x)IK be the filter on K generated by the sets U n K when U ranges over the filter of neighborhoods of x in E; that the sets U n K form the basis of a filter means precisely that x belongs to the closure of K. In view of Proposition 6.8, $(x)(K must have an accumulation point x1 E K. Necessarily x1 = x: otherwise we could find a neighborhood U of x whose complement in E is a neighborhood of x1 and we could certainly not have x1 E 0, even less x1 E U n K. Thus x belongs to K. COROLLARY 2. A compact subset of a Hausdog TVS is complete. It suffices to combine Propositions 6.7 and 6.8. COROLLARY 3. I n a compact topological space K , every sequence has an accumulation point. Definition 6.2. A subset A of a topological space X is said to be relatively compact if the closure A of A is compact. A converging sequence (without the limit point) is a relatively compact set. Dejinition 6.3. A subset A of a H a u s d o e T VS E is said to be precompact if A is relatively compact when viewed as a subset of the completion I? of E.
54
TOPOLOGICAL VECTOR SPACES
[Part I
A Cauchy sequence in E is precompact; it is not necessarily relatively compact! For this would mean that it converges (Proposition 6.7). Another example illustrating the difference between relatively compact sets and precompact sets is the following one: let SZ be an open subset of R”, different from R”. In virtue of the Heine-Borel-Lebesgue theorem, every bounded open subset of Q is precompact; but an open subset Q‘ of SZ is relatively compact, in Q, if and only if at the same time 52’ is bounded and its closure is contained in SZ. A subset K of the Hausdorff TVS E is compact if and only if K is both complete and precompact. Indeed, if K is compact when viewed as subset of E, K is still compact when viewed as subset of 8:therefore, by Corollaries I and 2 of Proposition 6.8, we know that K is closed in E, hence complete; of course its closure in I?, identical to K itself, is compact. Conversely, if K is complete, we have I? = K and therefore, if I? is compact, K is also compact. Our purpose is to prove a criterion of precompactness which is to be used later. T h e proof of it is made very easy if we use the notion of ultrafilter: Definition 6.4. A filter U on a set A is called an ultrafilter ;f every filter on A which isfiner than U is identical to U.
LEMMA 6.1. Let 9 be a filter on a set A ; there is at least one ultrafilter on A which isfiner than 9. Proof. Let @ be the family of all filters on A finer than 9, ordered by the relation “to be finer than,” and 0’a subfamily of @ totally ordered for this relation. The elements of @’ are filters, that is to say subsets of the set ofsubsets V ( A )of A ;we may therefore consider their union 9’. I t is immediately seen that 9‘ is a filter on A, obviously finer than 9. We may therefore apply Zorn’s lemma to the family @, whence Lemma 6.1.
LEMMA 6.2. Let A be a topological space; ;f an ultrafilter U on A has an accumulation point in A, U converges to x. Proof. We apply Proposition 6.6: if x is an accumulation point of LI, there is a filter 9 which is finer than U and converges to x . As U is an ultrafilter, we must have S = U. LEMMA6.3. A Hausdor- topological space K is compact ;f and only ;f every ultrajilter on K converges.
Proof. If K is compact, every filter on K has an accumulation point
Chap. 6-61
55
COMPACT SETS
(Proposition 6.8), therefore every ultrafilter converges, by Lemma 6.2. Conversely, suppose that every ultrafilter converges in K and let 9 be some filter on K. By Lemma 6.1, there is an ultrafilter U on K which is finer than 9; u converges to some point x. By Proposition 6.6, x is an Q.E.D. accumulation point of 9. We may now state and prove the announced criterion of precompactness:
6.9. The following properties of a subset K of a H a u s d o g PROPOSITION T V S E are equivalent: (a)
K is precompact; given any neighborhood of the origin V in E , there is aJinite family (b) of points of K , x1 ,..., x, , such that the sets xi V form a covering of K , i.e., such that
+
KC(x,+ V)u.-u(x,+
V).
Proof. (a) implies (b). Let U be an open neighborhood of zero in E, contained in V. There exists an open neighborhood of zero in 8,0, such that U = tf n E. Consider the family of sets x 0 when x varies over K. They form an open covering of the closure R of K in E. Indeed, let 9 be an arbitrary point of R and let W be a neighborhood of of zero in E such that @ = - W C tf. Then there is x E K such that xE9 p,i.e., 9 E x W C x 0. This open covering of the compact set I? contains a finite subcovering, x1 tf,..., x, 0.We have:
+
+
+
K
+
+
+
+ 0)n E ] n .-.n [(x, c (XI + V ) u ... u (x, + V ) .
=
R n E C [(xl
10)n E ]
(b) implies (a). If K possesses Property (b), its closure Iz in I? possesses the same property in 8.Indeed, let P be an arbitrary neighborhood of zero in 8;let W be a closed neighborhood of zero in 8,contained in P. There is a finite number of points of K , x1 ,..., x,, such that KC(x,
+ m)u
**.
u (x,
+ @).
But as the right-hand side is a closed subset of 8,it also contains the closure of K in 8,whence our assertion. I n view of this, it will suffice to prove that if a closed subset K of a complete Hausdorif T V S E has Property (b), it is compact. We shall apply Lemma 6.3 and show that an arbitrary ultrafilter U on K converges to a point of K.
56
TOPOLOGICAL VECTOR SPACES
Let V be an arbitrary neighborhood of zero in E, and x1 ,..., xr a finite family of points of K such that the sets (xj V) form a covering of K. We contend that at least one of these sets xi V belongs to the filter U. First of all, we show that at least one of these sets intersects every set belonging to U. If this were not true, for each j = 1,...,r, we would be able to find a set MiE U which does not intersect xj V; then the intersection M I n n M , would be empty, contrary to the fact that U is a filter, since it would not intersect any one of the sets xi V and V, say xi V, since these form a covering of K. Thus one of the sets xi intersects all the sets which belong to U. This means that, if we consider the family of subsets of K of the form M n (xi V ) , where M runs over U, it is the basis of a filter on K. This filter is obviously finer than, therefore equal to, the ultrafilter U. In other words, the set (xi V) belongs to U. If y , z are two elements of this set, we have y - z E V. In other words, we have proved that, given any neighborhood of zero V in E, there is a set MEU such that M - M C V : U is a Cauchy filter. But as E is complete, U must converge to some point x E E; as K is Q.E.D. closed, x E K.
+ +
+
+
+
+
+
+
Exercises 6.1. By using the compactness criterion provided by Lemma 6.3, prove the following version of Tychonoff’s theorem:
ni,,
THEOREM 6.1. Let {Ei} ( i E I ) be a famiry of Hausdorff TVS, and E = Ei their product (equipped with the product TVS structure). Let A ibe a subset of Eifor each index i, and A = A i the product of the Ails, regarded as a subset of E. Then A is compact in E ;f and only $,for every i E I , A‘ is compact in Ei .
niE~
6.2. Prove that the balanced hull of a compact subset (i.e., the smallest balanced set containing K)is compact.
K of a Hausdorff TVS E
6.3. Prove that a TVS E is compact if and only if it consists of a single element, 0. 6.4. Consider the TVS Fc(N;C) of complex functions on the set N of nonnegative integers with the topology defined in Exercise 3.5. Prove that every converging sequence in Sc(N;C) must be contained in some space Sc(n)(see Exercise 3.6). Derive from this the fact that every compact subset of Sc(N, C) is contained in a finite dimensional linear subspace. 6.5. Let E be any one of the TVS, (a)-(d), of Exercise 3.4. Prove that any infinite dimensional linear subspace M of E contains a sequence which converges in E and which is not contained in any finite dimensional linear subspace of M .
PROPOSITION 6.1. A closed subset of a compact space is compact. PROPOSITION 6.2. Let f be a continuous mapping of a compact space X into a H a u s d o r - topological space Y . Then f ( X ) is a compact subset of Y. 50
51
COMPACT SETS
PROPOSITION 6.3. Let f be a one-to-one-continuous mapping of a compact space X onto a compact space Y . Then f is a homeomorphism (i.e., f -l is also continuous).
PROPOSITION 6.4. Let F, 7 be two Hausdorff topologies on a set X . Suppose that F isjiner than F’and that X , equipped with 9, is compact.
Then F = F’. Finite unions and arbitrary intersections of compact sets are compact. In a HausdorfT space X, every point is compact; every converging sequence is compact-provided that we include in it its limit point! The student should keep in mind that compact sets can be very complicated sets. Take for instance the real line R1.T h e BorelLebesgue-Heine theorem says that the compact subsets of R1are exactly the sets which are both closed and bounded. Note also that the Lebesgue measure of a sequence is equal to zero, and that if a set A is measurable, given any E > 0, there is a compact set K C A such that the measure of A n CK is <&. Take then the points x, with 0 x 1 , which are nonrational; they form a set of measure 1, since the rationals, which form a sequence, form a set of measure zero. This means that there are compact sets, contained in the interval [0, 11, which do not contain any rational number and whose Lebesgue measure is arbitrarily close to 1. Try to draw one of them! As the Weierstrass-Bolzano theorem shows, compact sets have interesting properties in relation with sequences of points. This extends to filters, as we are now going to see. In the immediate sequel, E is a HausdorfT topological space; when it is expressly mentioned, E is a TVS. T h e following terminology is useful:
< <
Definition 6.1. A point x of E is called an accumulation point of a filter 9 if x belongs to the closure of every set which belongs to 9. Let S = {xo , x1 ,...} be a sequence; a point x of E is often called an accumulation point of S if every neighborhood of x contains a point of S different from x. This terminology coincides with the one introduced by Definition 6.1 if we apply the latter to the filter Fsassociated with S (see p. 7). Let M e s s be arbitrary; M contains a subsequence of the form
s, = {% ,
,...I.
%l+l
If, then, x is an accumulation point of S , any neighborhood U of x contains some point xk with k arbitrarily large, in particular k 2 n. Thus U has a nonempty intersection with M, which means that x E i@. In
52
TOPOLOGICAL VECTOR SPACES
[Part I
other words, any accumulation point of the sequence S is an accumulation Conversely, if x is an accumulation point of the point of the filter YS. x belongs to the closure of all the sets belonging to Fs , in filter SS, particular to the closures of the sets S, , n = 0, 1,.... This means that given any neighborhood U of x and any integer n, there is k 3 n such that X ~ U. E
PROPOSITION 6.5. If afilter 9converges to apoint x , x is an accumulation point of 9. Indeed, suppose that x were not an accumulation point of 9. There would be a set M EF such that x 4 A?l. Hence the complement U of @, which is open, would be a neighborhood of x, and hence should belong to 9. But then we ought to have U n M # 0, vhence a contradiction. Of course, a filter might have more than one accumulation point. For instance, let 9 be the filter of all subsets of E containing a given subset A of E. Then every point of A is an accumulation point of 9. PROPOSITION 6.6. The following two conditions are equivalent: (a) x is an accumulation point of 9; which is finer than both 9 and the jilter of (b) there is a jilter F’ neighborhoods of x, 9 ( x ) , in other words: there is a filter 9‘ converging to x, which isfiner than 9. (a) => (b). Indeed, consider the family of subsets of E of the form U n M, where U varies over F ( x ) and M varies over F.These sets and they obviously are never empty if x is an accumulation point of 9, have Property (BF) of p. 7, hence they generate a filter 9’ which is obviously finer than 9 and 9 ( x ) . and if x’ is (b) 3 (a). If a filter 9 is less fine than another filter 9’ an accumulation point of F’,then x is also an accumulation point of 9. Thus it suffices to combine (b) with Proposition 6.5.
PROPOSITION 6.7. If a CauchyJilter 9on the TVS E has an accumulation point x , it converges to x. Let U be an arbitrary neighborhood of the origin and V another V C U. There is a set M E 9such neighborhood of zero such that V that M - M C V. On the other hand, V x intersects M , hence M - M n ( V x ) C V, or
+
+
MCV
+
+ M n (V + x) C V + V + x C U + x.
Q.E.D.
Chap. 6-41
53
COMPACT SETS
PROPOSITION 6.8. Let K be a H a u s d o e topological space. The following properties are equivalent: (a) K is compact; (b) every filter on K has at least one accumulation point. (a) s (b). Let F be any filter on K, and consider the family of closed sets fi when M varies over F. As no finite intersection of sets fi can be empty, neither can that be true of the intersection of all of them. (b) 3 (a). Let @ be a family of closed sets whose total intersection is empty. Suppose that @ does not contain any finite subfamily whose intersection is empty. Then take the family 0'of all the finite intersections of subsets belonging to @: it obviously forms a basis of a filter. This filter has an accumulation point, say x: thus x belongs to the closure of any subset belonging to the filter, in particular to any set belonging for these are closed. I n other words, x belongs to the intersection to of all the sets belonging to which is the same as the intersection of all the sets belonging to @. But the latter was supposed to be empty! @I,
@I,
Q.E.D.
COROLLARY 1. A compact subset K of a Hausdorf topological space E is closed. Proof. Let x E R; let S(x)IK be the filter on K generated by the sets U n K when U ranges over the filter of neighborhoods of x in E; that the sets U n K form the basis of a filter means precisely that x belongs to the closure of K. In view of Proposition 6.8, $(x)(K must have an accumulation point x1 E K. Necessarily x1 = x: otherwise we could find a neighborhood U of x whose complement in E is a neighborhood of x1 and we could certainly not have x1 E 0, even less x1 E U n K. Thus x belongs to K. COROLLARY 2. A compact subset of a Hausdog TVS is complete. It suffices to combine Propositions 6.7 and 6.8. COROLLARY 3. I n a compact topological space K , every sequence has an accumulation point. Definition 6.2. A subset A of a topological space X is said to be relatively compact if the closure A of A is compact. A converging sequence (without the limit point) is a relatively compact set. Dejinition 6.3. A subset A of a H a u s d o e T VS E is said to be precompact if A is relatively compact when viewed as a subset of the completion I? of E.
54
TOPOLOGICAL VECTOR SPACES
[Part I
A Cauchy sequence in E is precompact; it is not necessarily relatively compact! For this would mean that it converges (Proposition 6.7). Another example illustrating the difference between relatively compact sets and precompact sets is the following one: let SZ be an open subset of R”, different from R”. In virtue of the Heine-Borel-Lebesgue theorem, every bounded open subset of Q is precompact; but an open subset Q‘ of SZ is relatively compact, in Q, if and only if at the same time 52’ is bounded and its closure is contained in SZ. A subset K of the Hausdorff TVS E is compact if and only if K is both complete and precompact. Indeed, if K is compact when viewed as subset of E, K is still compact when viewed as subset of 8:therefore, by Corollaries I and 2 of Proposition 6.8, we know that K is closed in E, hence complete; of course its closure in I?, identical to K itself, is compact. Conversely, if K is complete, we have I? = K and therefore, if I? is compact, K is also compact. Our purpose is to prove a criterion of precompactness which is to be used later. T h e proof of it is made very easy if we use the notion of ultrafilter: Definition 6.4. A filter U on a set A is called an ultrafilter ;f every filter on A which isfiner than U is identical to U.
LEMMA 6.1. Let 9 be a filter on a set A ; there is at least one ultrafilter on A which isfiner than 9. Proof. Let @ be the family of all filters on A finer than 9, ordered by the relation “to be finer than,” and 0’a subfamily of @ totally ordered for this relation. The elements of @’ are filters, that is to say subsets of the set ofsubsets V ( A )of A ;we may therefore consider their union 9’. I t is immediately seen that 9‘ is a filter on A, obviously finer than 9. We may therefore apply Zorn’s lemma to the family @, whence Lemma 6.1.
LEMMA 6.2. Let A be a topological space; ;f an ultrafilter U on A has an accumulation point in A, U converges to x. Proof. We apply Proposition 6.6: if x is an accumulation point of LI, there is a filter 9 which is finer than U and converges to x . As U is an ultrafilter, we must have S = U. LEMMA6.3. A Hausdor- topological space K is compact ;f and only ;f every ultrajilter on K converges.
Proof. If K is compact, every filter on K has an accumulation point
Chap. 6-61
55
COMPACT SETS
(Proposition 6.8), therefore every ultrafilter converges, by Lemma 6.2. Conversely, suppose that every ultrafilter converges in K and let 9 be some filter on K. By Lemma 6.1, there is an ultrafilter U on K which is finer than 9; u converges to some point x. By Proposition 6.6, x is an Q.E.D. accumulation point of 9. We may now state and prove the announced criterion of precompactness:
6.9. The following properties of a subset K of a H a u s d o g PROPOSITION T V S E are equivalent: (a)
K is precompact; given any neighborhood of the origin V in E , there is aJinite family (b) of points of K , x1 ,..., x, , such that the sets xi V form a covering of K , i.e., such that
+
KC(x,+ V)u.-u(x,+
V).
Proof. (a) implies (b). Let U be an open neighborhood of zero in E, contained in V. There exists an open neighborhood of zero in 8,0, such that U = tf n E. Consider the family of sets x 0 when x varies over K. They form an open covering of the closure R of K in E. Indeed, let 9 be an arbitrary point of R and let W be a neighborhood of of zero in E such that @ = - W C tf. Then there is x E K such that xE9 p,i.e., 9 E x W C x 0. This open covering of the compact set I? contains a finite subcovering, x1 tf,..., x, 0.We have:
+
+
+
K
+
+
+
+ 0)n E ] n .-.n [(x, c (XI + V ) u ... u (x, + V ) .
=
R n E C [(xl
10)n E ]
(b) implies (a). If K possesses Property (b), its closure Iz in I? possesses the same property in 8.Indeed, let P be an arbitrary neighborhood of zero in 8;let W be a closed neighborhood of zero in 8,contained in P. There is a finite number of points of K , x1 ,..., x,, such that KC(x,
+ m)u
**.
u (x,
+ @).
But as the right-hand side is a closed subset of 8,it also contains the closure of K in 8,whence our assertion. I n view of this, it will suffice to prove that if a closed subset K of a complete Hausdorif T V S E has Property (b), it is compact. We shall apply Lemma 6.3 and show that an arbitrary ultrafilter U on K converges to a point of K.
56
TOPOLOGICAL VECTOR SPACES
Let V be an arbitrary neighborhood of zero in E, and x1 ,..., xr a finite family of points of K such that the sets (xj V) form a covering of K. We contend that at least one of these sets xi V belongs to the filter U. First of all, we show that at least one of these sets intersects every set belonging to U. If this were not true, for each j = 1,...,r, we would be able to find a set MiE U which does not intersect xj V; then the intersection M I n n M , would be empty, contrary to the fact that U is a filter, since it would not intersect any one of the sets xi V and V, say xi V, since these form a covering of K. Thus one of the sets xi intersects all the sets which belong to U. This means that, if we consider the family of subsets of K of the form M n (xi V ) , where M runs over U, it is the basis of a filter on K. This filter is obviously finer than, therefore equal to, the ultrafilter U. In other words, the set (xi V) belongs to U. If y , z are two elements of this set, we have y - z E V. In other words, we have proved that, given any neighborhood of zero V in E, there is a set MEU such that M - M C V : U is a Cauchy filter. But as E is complete, U must converge to some point x E E; as K is Q.E.D. closed, x E K.
+ +
+
+
+
+
+
+
Exercises 6.1. By using the compactness criterion provided by Lemma 6.3, prove the following version of Tychonoff’s theorem:
ni,,
THEOREM 6.1. Let {Ei} ( i E I ) be a famiry of Hausdorff TVS, and E = Ei their product (equipped with the product TVS structure). Let A ibe a subset of Eifor each index i, and A = A i the product of the Ails, regarded as a subset of E. Then A is compact in E ;f and only $,for every i E I , A‘ is compact in Ei .
niE~
6.2. Prove that the balanced hull of a compact subset (i.e., the smallest balanced set containing K)is compact.
K of a Hausdorff TVS E
6.3. Prove that a TVS E is compact if and only if it consists of a single element, 0. 6.4. Consider the TVS Fc(N;C) of complex functions on the set N of nonnegative integers with the topology defined in Exercise 3.5. Prove that every converging sequence in Sc(N;C) must be contained in some space Sc(n)(see Exercise 3.6). Derive from this the fact that every compact subset of Sc(N, C) is contained in a finite dimensional linear subspace. 6.5. Let E be any one of the TVS, (a)-(d), of Exercise 3.4. Prove that any infinite dimensional linear subspace M of E contains a sequence which converges in E and which is not contained in any finite dimensional linear subspace of M .