s P = f(ξ n )(x i x i 1 ). i=1

Transcription

1 Compactness and total boundedness via nets The aim of this chapter is to define the notion of a net (generalized sequence) and to characterize compactness and total boundedness by this important topological notion. Nets: definitions and main properties. Definition 0.1. A pre-ordering on a nonempty set I is a binary relation on I which is: (a) reflexive (i.e., α α for each α I); (b) transitive (i.e., if α β γ then α β). The couple (I, ) will then be called a pre-ordered set. Notice that the notion of a pre-ordered set is more general than the more common notion of a partially ordered set: a partially ordered set is a pre-ordered set whose pre-ordering is also antisymmetric (i.e., if α β and β α then α = β). Definition 0.2. A directed set (or upper-directed set ) is a pre-ordered set (I, ) such that α, β I γ I : γ α, γ β. For brevity, we shall write γ α, β to express γ α, γ β. Moreover, we shall use the following quite useful terminology. Let P be a property of elements of a directed set I. We shall say that: P holds eventually if there exists α 0 I such that P holds for each α α 0 ; P holds frequently if for each α I there exists β α satisfying P. Thus eventually means for all successors of some element, and frequently means for arbitrarily large elements. Definition 0.3. Let E be a set. A net (or generalized sequence ) in E is any mapping ϕ: I E where I is a directed set. Putting x α := ϕ(α) (α I), the net can be also denoted by (x α ) α I, or by (x α ). Definition 0.4. Let (x α ) α I be a net in a topological space X. We say that the net converges to a point x X if each neighborhood U of x contains x α eventually in I. In this case we write indifferently x α x or lim α x α = x. Examples 0.5. Let X be a topological space. (a) Each sequence in X is a net in X (with I = N), and its convergence as a net coincides with its convergence as a sequence. (b) Each function ϕ: R X is a net in X (since R is a directed set). What is the meaning of convergence for such a net? 1

2 2 (c) The family U(x) of all neighborhoods of x is directed by the inverse inclusion (U V U V ). Let ϕ: U(x) X be such that ϕ(u) U for every U U(x). Then the net ϕ converges to x. (d) Let f : [a, b] R be a Riemann integrable function. The set P of all partitions of [a, b] is directed by the inclusion. For every partition P = {a = x 0 < x 1 < < x n = b}, fix arbitrarily {ξ 1,..., ξ n } [a, b] such that ξ i [x i 1, x i ] (1 i n), and consider the corresponding Riemannian sum s P = n f(ξ n )(x i x i 1 ). i=1 Then {s P } P P is a net in R that converges to b a f(t) dt. (Verify the details!) (e) Consider a three-point set I = {a, b, c} with a binary relation defined by the following list: a a, b b, c c, a b, b c. Then I is a directed set and, if we put x a = 2, x b = 1, x c = 1, the net (x γ ) γ I converges (in R) to 1. Now, we are ready for the definition of a subnet of a net. Definition 0.6. Let (x α ) α I be a net in a set E. A subnet of (x α ) is any net of the form (x ψ(β) ) β J where J is a directed set and ψ : J I is a mapping that tends to infinity in the sense that for each α I, ψ(β) α eventually in J. In other words, a subnet of a net ϕ: I E is a composition ϕ ψ where ψ : J I is a mapping (from a directed set J) tending to infinity. Example 0.7. Any subsequence of a sequence is also its subnet. However, a sequence can have subnets that are not subsequences. For instance, consider the sequence (n 2 ) n N and its subnet (m 2 + 2mn + n 2 ) (m,n) N N (where N N is equipped with the coordinate-wise partial order; what is the function ψ in this case?). Moreover, a subnet of a sequence need not be indexed by a countable set; for example, the net ([t] 2 ) t R (where [t] denotes the lower integer part of t) is a subnet of the sequence (n 2 ) n N. Observation 0.8. Every subnet of a converging net (in a topological space) converges to the same limit. Now, let us state a useful lemma about existence of certain subnets. L:subnets Lemma 0.9. Let E be a set, and (x α ) α I a net in E. Let B be a family of subsets of E, satisfying: x α is contained frequently in each element of B, and the intersection of any two elements of B contains an element of B. Then (x α ) admits a subnet which is eventually contained in each element of B.

3 Proof. Clearly, the family B is directed by the inverse inclusion. Consider the set J = { (α, B) I B : x α B } equipped with the coordinate-wise pre-ordering. It is easy to see that J is a directed set (verify this!). The function ψ : J I, defined by ψ(α, B) = α, is nondecreasing and onto, an hence tends to infinity. Consequently, (x ψ(α,b) ) (α,b) J is a subnet of (x α ). Moreover, given A B, fix α 0 I so that x α0 A, and observe that if (β, B) (α 0, A) then x ψ(β,b) = x β B A. This completes the proof. 3 Basic topological properties of nets. Let us list some basic results about nets in topological spaces. We omit their relatively easy proofs; the interested reader can find them in the book [J.L. Kelley, General Topology]. Theorem Let X, Y be a topological spaces, A X, x 0 X. (a) x 0 A if and only if there is a net in A converging to x 0. (b) A is closed if and only if there in no net in A converging to a point of X \A. (c) X is Hausdorff if and only if each net converges to at most one point. (d) A mapping f : X Y is continuous at x 0 if and only if it satisfies the implication: X x α x 0 f(x α ) f(x 0 ). cpt Compactness and nets. Now, we are going to prove a well-known characterization of compactness via nets. The reader probably already knows its analogue for metric spaces and sequences (instead of nets). The main difference between the two cases is in their proofs. Let us recall that a family A of sets is said to be centered if each finite subfamily of A has nonempty intersection. It is an easy and well-known fact that a topological space X is compact if and only if every centered family of closed subsets of X has nonempty intersection. Theorem A topological space X is compact if and only if each net in X admits a subnet converging to a point of X. Proof. ( ) Let (x α ) α I be a net in a compact space X. Then the sets A α = {x β : β α} (α I) form a centered family; hence the family of their closures is centered as well. Since X is compact, there exists x α I A α. For each U U(x) (the family of all neighborhoods of x) and each α I, one has U A α, that is, x β U for some β α. But this means that the family U(x) satisfies the assumptions of Lemma 0.9. Hence there exists a subnet of (x α ) that is eventually contained in each element of U(x). In other words, this subnet converges to x. ( ) Assume that each net in X has a converging (in X) subnet. Let A be a centered family of closed sets in X. Then also the family B, formed by all finite intersections of elements of A, is centered, too. For each B B, choose an arbitrary x B B. Since B is clearly directed by the inverse inclusion, (x B ) B B is a net. α Let (x ψ(α) ) α I be a subnet of (x B ) such that x ψ(α) x X. Let us show that

4 4 x belongs to each B B. Given B, we have ψ(α) B eventually in I, that is, x ψ(α) ψ(α) B eventually in I. It follows that x = lim α x ψ(α) B = B, and the proof is complete. Total boundedness and nets. For defining the notion of total boundedness, we need more specific class of spaces than merely topological spaces. The most general appropriate structure is that of a uniform space, that includes metric spaces as well as topological linear spaces as particular cases. However, uniform spaces are not very common and working in such spaces is, in a sense, formally more difficult. Moreover, the theorem, saying that a metric space E is totally bounded if and only if each sequence in E has a Cauchy subsequence, is well-known and relatively easy, due to fact that it is relatively easy to work with sequences. From these reasons, here we confine ourselves to the case of subsets of a topological vector space. An introduction to the theory of uniform spaces can be found in the book [J.L. Kelley, General Topology]. Recall that a set E in a topological vector space is totally bounded (or precompact) if for each U U(0) (the set of neighborhoods of 0) there is a finite set E 0 E such that E E 0 + U. It is easy to see that in normed spaces (or in topological metric spaces) this definition coincides with the usual metric one: for each ε > 0 there is a finite set E 0 E such that dist(x, E 0 ) < ε for each x E. Definition Let X be a topological vector space. (a) A net (x α ) α I in X is said to be a Cauchy net if for each U U(0) there exists α 0 I such that x α x β U whenever α, β α 0. (b) A set E X is complete if each Cauchy net in E converges to a point of E. tot.bdd Theorem Let E be a set in a topological vector space. Then E is totally bounded if and only if each net in E admits a Cauchy subnet. Proof. Let (x α ) α I be a net in a totally bounded set E. The family A = {A E : x α A frequently} clearly contains E. Let B be a maximal subfamily of A that contains E and is closed under making finite intersections (existence of such B follows by Zorn s lemma). Let us show several properties of B. (a) If F is a finite subfamily of A such that F B, then F B. Let F = {A 1,..., A n }. We claim that, for some index k, A k B A for each B B. Indeed, if this is not the case, for each i {1,..., n} there exists B i B such that A i B i / A; but then we get B (A 1... A n ) n i=1 B i n i=1 (A i B i ) / A, a contradiction. Our claim implies that the family of all finite intersections of elements of B {A k } is closed under finite intersections and is contained in A. By maximality of B, we must have A k B. (b) For each set A E, the family B contains either A or E \ A. If A / A, then eventually x α E \A. Since the intersection of E \A with any element of B belongs to A, the family of all finite intersections of B {E \ A} is contained in A. Thus

5 E \ A B by the maximality of B. In the same way we get that if E \ A / A then A B. Finally, if both A and E \ A belong to A then one of them belongs to B by (a) (since E B). (c) B contains arbitrarily small elements, in the sense that for each U U(0) there exists B B such that B B U. Given U U(0), there exists V U(0) with V V U. By total boundedness, there exists a finite set F := {y 1,..., y n } E such that E F + V. Denoting E i = (y i + V ) E (i = 1,..., n), we have E = n i=1 E i. Consider the set P = {i {1,..., n} : E i A} and its complement P c in {1,..., n}. Since C := i P c E i / A, we must have P. Let A = i P E i. Then E \ A / A (since E \ A C). By (b), we must have A B. By (a), there exists k P with E k B. Notice that E k E k V V U. To conclude the proof of this implication, notice that the family B satisfies the assumptions of Lemma 0.9. Hence there exists a subnet of (x α ) that is eventually contained in each element of B. By (c), this subnet is Cauchy. Assume that E is not totally bounded. There exists U U(0) such that E \ (F + U) for each finite set F U. An easy inductive construction gives a sequence (x n ) in E such that x n+1 / {x 1,..., x n } + U for each n. Since for each two indexes m > n we have x m x n / U, our sequence has no Cauchy subnets. The proof is complete. The above theorem easily implies the following characterization of compactness. 5 cpt.tvs Theorem A set E in a topological vector space is compact if and only if E is totally bounded and complete. Proof. Let E be compact. Given an open V U(0), the open cover {y + V : y E} of E admits a finite subcover. This proves that E is totally bounded. Let (x α ) be a Cauchy net in E. By Theorem 0.11, (x α ) admits a subnet converging to a point of E. It easily follows that the net (x α ) converges to the same limit. Assume E is totally bounded and complete. Given a net (x α ) in E, it admits a Cauchy subnet by Theorem Since E is complete, this subnet converges to a point of E. Again, it follows that (x α ) converges to the same point. By Theorem 0.11, E is compact. Theorem 0.14 is an analogue of the well-known result about metric spaces: a metric space is compact if and only if it is totally bounded and (sequentially) complete. Let us show that, in topological vector spaces with a countable base of neighborhoods of 0, sequential completeness and completeness coincide. Recall that a family B U(0) is a base of neighborhoods of 0 if each element of U(0) contains some element of B. Each t.v.s. which is metrizable clearly admits a countable base of U(0). It is a well-known theorem that a topological vector space is metrizable if and only if it is Hausdorff and admits a countable base of neighborhoods of 0 (and, in this case, the metric can be chosen to be translation invariant) (see [W. Rudin, Functional Analysis]). However, we shall not need this fact.

6 6 Proposition Let X be a topological vector space with a countable base of neighborhoods of 0. A set E X is complete if and only if E is sequentially complete. Proof. Let B = {V n : n N} be a countable base of U(0). We can assume that V 1 V 2... ; indeed, otherwise we can substitute B with the base {V 1, V 1 V 2, V 1 V 2 V 3,...}. Let E be complete, and (x n ) a Cauchy sequence in E. There exists a subnet (x ϕ(α) ) α I converging to a point x E. Let us construct inductively a sequence (α k ) in I. Choose α 1 so that x ϕ(α) x + V 1 for each α α 1. If we already have α 1,..., α k, choose α k+1 α k so that ϕ(α k+1 ) ϕ(α k ) + 1 and x ϕ(α) x + V k+1 for each α α k+1. It is easy to verify that (x ϕ(αk )) k N is a subsequence of (x n ) that converges to x. Let E be sequentially complete, and (x α ) α I a Cauchy net in E. Let us construct inductively a sequence (α k ) in I. Choose α 1 so that x α x α1 V 1 for each α α 1. If we already have α 1,..., α k, choose α k+1 α k so that x α x αk+1 V k+1 for each α α k+1. Then (x αn ) is a Cauchy sequence since x αm x αn V n whenever m n. Consequently, (x αn ) converges to a point x E. Now, it is easy to show that (x α ) converges to x, too.