Normed versus topological groups: Dichotomy and duality

122 N. H. Bingham and A. J. OstaszewskiEnumerate the countable family {H(α|n) : α ∈ ω n } as {T h : h ∈ ω}. Since S spans G,we haveG = ⋃ ⋃(Th∈ω k∈N h k 1· ... · T kh ) .As each T k is analytic, so too is the continuous imageT k1 · ... · T kh ,which is thus measurable. Hence, for some h ∈ N and k ∈ N h the setT k1 · ... · T khhas positive measure/ is non-meagre.Definition. We say that S is a pre-compact set if its closure is compact. We will saythat f is a pre-compact function if f(S) is pre-compact for each pre-compact set S.Theorem 11.21 (Jones-Kominek Analytic Automaticity Theorem for Metric Groups).Let G be either a non-meagre normed topological group, or a topological group supportinga Radon measure, and let H be K-analytic (hence Lindelöf, and so second countable inour metric setting). Let h : G → H be a homomorphism between metric groups and let Tbe an analytic set in G which finitely generates G.(i) (Jones condition) If h is continuous on T, then h is continuous.(ii) (Kominek condition) If h is pre-compact on T, then h is precompact.Proof. As in the Analytic Covering Lemma (Th. 11.19), writeT = ⋃ k∈N T k.(i) If h is not continuous, suppose that x n → x 0 but h(x n ) does not converge to h(x 0 ).SinceG = ⋃ ⋃T (m)m∈N k∈Nk,G is a union of analytic sets and hence analytic ([Jay-Rog, Th. 2.5.4 p. 23]). Now, forsome m, k the m-span T (m)kis non-meagre, as is the m-span S (m)kof some compact subsetS k ⊆ T k . So for some shifted subsequence tx n → tx 0 , where t and x 0 lie in S (m)k. Thusthere is an infinite set M such that, for n ∈ M,tx n = t 1 n...t m n with t i n ∈ S k .Without loss of generality, as S k is compact,and sot (i)n→ t (i)0 ∈ S k ⊂ T,tx n = t 1 n...t m n → t 1 0...t m 0 = tx 0 with t i 0 ∈ S k ⊂ T.