Normed versus topological groups: Dichotomy and duality

Normed groups 121Proof. In the measure case the same approach may be used based now on the continuousfunction g(x 1 , ..., x d ) := x ε11 · ... · xε dd , ensuring that K is of positive measure (measuregreater than η). In the category case, if T ′ = T ε1 · ... · T ε dis non-meagre then, by theSteinhaus Theorem ([St], or [BGT, Cor. 1.1.3]), T ′ · T ′ has non-empty interior. Themeasure case may now be applied to T ′ in lieu of T. (Alternatively one may apply thePettis-Piccard Theorem, Th. 6.5, as in the Analytic Dichotomy Lemma, Th. 11.15.)Theorem 11.18 (Compact Spanning Approximation). In a connected, normed topologicalgroup X, for T analytic in X, if the span of T is non-null or is non-meagre, thenthere exists a compact subset of T which spans X.Proof. If T is non-null or non-meagre, then T spans X (by the Analytic DichotomyLemma, Th. 11.15); then for some ε i ∈ {±1}, T ε1 · ... · T ε dhas positive measure/ is nonmeagre.Hence for some K compact K ε1 · ... · K ε dhas positive measure/ is non-meagre.Hence K spans some and hence all of X.Theorem 11.19 (Analytic Covering Lemma – [Kucz, p. 227], cf. [Jones2, Th. 11]). Givennormed groups G and H, and T analytic in G, let f : G → H have continuous restrictionf|T. Then T is covered by a countable family of bounded analytic sets on each of whichf is bounded.Proof. For k ∈ ω define T k := {x ∈ T : ‖f(x)‖ < k} ∩ B k (e G ). Now {x ∈ T : ‖f(x)‖ < k}is relatively open and so takes the form T ∩ U k for some open subset U k of G. TheIntersection Theorem (Th. 11.3) shows this to be analytic since U k is an F σ set andhence Souslin-F.Theorem 11.20 (Expansion Lemma – [Jones2, Th. 4], [Kom2, Th. 2], and [Kucz, p.215]). Suppose that S is Souslin-H, i.e. of the formS = ⋃ α∈ω ω ∩∞ n=1H(α|n),with each H(α|n) ∈ H, for some family of analytic sets H on which f is bounded. If Sspans the normed group G, then, for each n, there are sets H 1 , ..., H k each of the formH(α|n), such that for some integers r 1 , ..., r kT = H 1 · ... · H khas positive measure/ is non-meagre, and so T · T has non-empty interior.Proof. For any n ∈ ω we haveS ⊆ ⋃ α∈ω ω H(α|n).