TVS IV c Gabriel Nagy Topological Vector Spaces IV: Completeness and Metrizability Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we isolate two important features of topological vector spaces, which, when present, are very useful. With few exceptions, the material from sub-section A is is optional. Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. A. Completeness for Topological Vector Spaces This sub-section is aimed at treating completeness for most general types of topological vector spaces, beyond the traditional metric framework. Our approach here is essentially and adaptation of the the theory of uniform spaces1. As it turns out, a topological vector space is always a uniform space. Definitions. Suppose (X , T) is a topological vector space. A. A net (xλ)λ∈Λ ⊂ X is said to be T-Cauchy, if: (c) for every T- neighborhood V of 0, there exists λV ∈ Λ, such that: xλ − xµ ∈ V, ∀ λ, µ λV . This condition is equivalent to the fact that the double net (xλ −xµ)(λ,µ)∈Λ×Λ converges to 0. B. The linear topology T is said to be complete, if every T-Cauchy net in X is convergent. Example 1. Let I be an arbitrary non-empty set. If (Xi)i∈I is a collection of topological Q vector spaces, then the product space X = i∈I Xi is complete, when equipped with the product topology, if and only if all the Xi, i ∈ I, are complete. This follows from the fact i that a net (xλ)λ∈Λ in X , written in coordinates as xλ = (xλ)i∈I , ∀ λ, is Cauchy in X , if and i only if, for every i ∈ I, the net (xλ)λ∈Λ is Cauchy in Xi. In particular, all finite dimensional topological vector spaces are complete. 1 A uniform space is a pair (X, U) consisting of a non-empty set X and a filter U of subsets U ⊂ X × X, T subject to the following conditions: (i) U∈U U = {(x, x): x ∈ X} – the diagonal of X; (ii) if U belongs to U, then its inverse (as a relation) U −1 = {(y, x):(x, y) ∈ U} also belongs to U; (iii) for every U ∈ U there exists V ∈ U, such that the composition (as relations) V ◦ V is contained in U. The filter U is then called a uniformity on X. 1 Exercise 1. Prove that a topological vector space X is complete, if and only if every Cauchy net has a convergent sub-net. Comment. If one attempts to construct the completion of a topological vector space in a manner similar to the one used in the metric framework, then Definition B above is a bit useless, since there is no such thing as “the set of all nets.” Therefore, one would like to replace (for now) nets with filters2, thus the following definition is natural. Definition. Suppose (X , T) is a topological vector space. A filter F in X is called Cauchy, if for every T- neighborhood V of 0, there exists F ∈ F, such that: x − y ∈ V, ∀ x, y ∈ F. With all the preparations in place, the exercises below will guide the reader towards build- ing the completion of a topological vector space (X , T). Except for Notations, Definitions, and Lemma 1, the remainder of this sub-section is optional. Exercises 2-8. These three exercises clarify the connections between Cauchy filters and Cauchy nets. Fix a topological vector space X . 2. Prove that, if a filter F is convergent, that is, there exists x ∈ X , such that, for every neighborhood V of x, there exists F ∈ F with F ⊂ V, then F is Cauchy. 3. Prove that, for a net (xλ)λ∈Λ in X , the following are equivalent: (i) (xλ)λ∈Λ is Cauchy; (ii) the tail filter F = {Fλ}λ∈Λ, defined by Fλ = {xµ : µ λ}, is Cauchy. 4*. Prove that, for a filter F on X , the following are equivalent: (i) F is Cauchy; (ii) all F-section nets3 are Cauchy. (Hint: For (ii) ⇒ (i), equip each F ∈ F with a well ordering F , and consider the set Λ = {(F, z): F ∈ F, z ∈ F}, equipped with the following order: (F1, z1) (F2, z2), if either F1 ( F2, or F1 = F2 and z1 F1 z2. Show that Λ is directed. Consider the obvious map Λ 3 (F, z) 7−→ z ∈ X as a net (xλ)λ∈Λ. Show that this net is an F-section, and prove that by condition (ii) the filter F is Cauchy.) 5. Prove that if F is a Cauchy filter, then so is any filter F0 ⊃ F. In particular, all ultrafilters that contain F are Cauchy. 6. Prove that the following conditions are equivalent: (i) X is complete; (ii) every Cauchy filter in X is convergent; 2 Recall that, given some non-empty set X, a filter in X is a collection F of non-empty subsets, which is directed with respect to the reverse inclusion, that is, for every F1,F2 ∈ F, there exists F ∈ F, with F ⊂ F1 ∩ F2. 3 Recall that a net (xλ)λ∈Λ is called an F-section, if for every F ∈ F, there exists λF ∈ Λ, such that xλ ∈ F, ∀ λ λF . 2 (iii) every Cauchy ultrafilter in X is convergent. (Hints: Prove the implications (i) ⇒ (ii) ⇒ (iii) ⇒ (i). For (i) ⇒ (ii) use Exercise 4. For (iii) ⇒ (i), start with a Cauchy net (xλ) and choose an ultrafilter U that contains the tail filter. By Exercises 3 and 5, U is Cauchy, so assuming (iii), the ultrafilter U converges to some x. Prove that xλ → x.) 7. Prove that if F1 and F2 are Cauchy filters, then {F1 + F2 : F1 ∈ F1, F2 ∈ F2} is a Cauchy filter. By an abuse of notation, this filter will be denoted by F1 + F2. 8. Prove that if F is a Cauchy filter, then for every α ∈ K, {αF : F ∈ F} is a Cauchy filter. By an abuse of notation, this filter will be denoted by αF. Notation. We denote by cf(X ) the set of all Cauchy filters in (X , T). Equip cf(X ) with the following relation (use the notations from Exercises 5 and 6): F ≈ G ⇔ F + (−1)G converges to 0. Exercise 9. Prove the following (i) ≈ is an equivalence relation on cf(X ). (ii) If F is a Cauchy filter, and F0 is a filter that contains F (which is automatically Cauchy by Exercise 5), then F ≈ F0. (iii) If F1 ≈ G1 and F2 ≈ G2, then F1 + F2 ≈ G1 + G2. (iv) If F ≈ G, then αF ≈ αG, ∀ α ∈ K. Notations. For each F ∈ cf(X ), let [F] denote its equivalence class. Given a non-empty set S ⊂ X , define S˜ = {[F]: F ∈ cf(X ), F 3 S}. Using the preceding Exercise, we see that one has two operations X˜ × X˜ −−−−→addition X˜, multiplication K × X˜ −−−−−−−→ X˜, correctly defined on equivalence classes by [F1] + [F2] = [F1 + F2] and α[F] = [αF]. Exercise 10. Prove that the above two operations define a vector space structure on the quotient space X˜. Exercise 11. Define, for every x ∈ X , the trivial filter Fx = {x} . Prove that: (i) For every x ∈ X , the filter Fx is convergent, thus Cauchy. ˜ (ii) The map J : X 3 x 7−→ [Fx] ∈ X is an injective linear map. Exercise 12. Consider the collection B = {N˜ : N ⊂ X balanced T-neighborhood of 0}. Prove that: 3 (i) B is a filter; (ii) all sets in B are balanced and absorbing; (iii) for any V ∈ B, there exists W ∈ B, such that W + W ⊂ V; T (iv) V∈B V = {0}. Conclude, using Corollary 1 from TVS I, that there exists a unique Hausdorff linear topology T˜ on X˜, so that B is a basic system of T˜-neighborhoods of 0. Exercise 13. Consider the injective linear map J : X → X˜ defined in Exercise 10. (i) Prove that the linear subspace R = Range J ⊂ X˜ is dense in X˜. (ii) Prove that, if (xλ)λ∈Λ is a Cauchy net in X , and we consider its tail filter F defined in ˜ ˜ Exercise 1, then J(xλ) → [F] in (X , T). (iii) Prove that any point in X˜ can be realized using the construction from (ii), namely, for ˜ every v ∈ X , there exists a Cauchy net (xλ)λ∈Λ in X , such that J(xλ) → v. ˜ ∼ (iv) Prove that, when we equip R with the induced topology T R, the map J : X −→R becomes a linear homeomorphism. (v) Prove that, if (X , T) is already complete, then J :(X , T) → (X˜, T˜) is a linear homeo- morphism. (Hint: For (iv) show that if V is a neighborhood of 0 in X , then V ⊂ J −1(V˜) ⊂ V.) Exercise 14*. Prove that (X˜, T˜) is a complete topological vector space. (Hint: Con- sider the filter V = {V˜ : V neighborhood of 0 in X}. Show that if G is a Cauchy filter in X˜, then G + V is again a Cauchy filter in X˜, equivalent to G. Using the preceding Exercise, it follows that the collection F = {J −1(A): A ∈ G0} is a filter in X . Show that F is Cauchy, and then prove that G0 converges to v = [F] in X˜. Conclude that, since G is equivalent to G0, it also converges to v.) Definition. The topological vector space (X˜, T˜) is called the completion of (X , T). Comment. The construction of the completion is borrowed (word for word) from the theory of uniform spaces in Topology.