Normed versus topological groups: Dichotomy and duality

Normed groups 77cf. [Eng, 4.3.23].) Since this is an equivalent metric, for each a ∈ A and ε > 0, there isδ = δ(ε) > 0 such that B δ (a) ⊆ Bε ρ (a), where B δ (a) refers to the metric d X R .) Thus, bytaking ε = 2 −n−1 the δ-ball B δ (a) has ρ-diameter less than 2 −n .Working inductively in a normed-group setting, we define non-empty open subsets of A(of possible translators) B n of ρ-diameter less than 2 −n as follows; they are of courseBaire subsets of X. With n = 0, we take B 0 = A. Given n and B n open in A, chooseb n ∈ B n and N such that ‖z k ‖ < min{ 1 2 ‖x n‖, ε(B n )}, for all k > N. Let x n := z N ∈ Z;then by the Displacements Lemma B n ∩(B n b −1n x −1n b n ) is non-empty (and open). We maynow choose a non-empty subset B n+1 of A which is open in A with ρ-diameter less than2 −n−1 such that cl A B n+1 ⊂ B n ∩ (B n b −1n x −1n b n ) ⊆ B n . By completeness, the intersection⋂n∈N B n is non-empty. Lett ∈ ⋂ B n ⊂ A.n∈NNow tb −1n x n b n ∈ B n ⊂ A, as t ∈ B n+1 ⊂ B n b −1n x −1n b n , for each n. Hence M := {m :z m = x n for some n ∈ N} is infinite. Now b n ∈ B n so b n → R t, so w n := b n t −1 → e. Thustb −1n x n b n = wn−1we may write eitherorx n w n t, as b n = w n t. Moreover, if z m = x n , then adjusting the notation{tt −1m z m t m : m ∈ M t } ⊆ A,{w −1m z m w m t : m ∈ M t } ⊆ A.The latter shows that the right-shift ρ t underlies the conclusion of the theorem and nota left-shift.As for the topological group setting, the Displacements Lemma shows that we may passto the final conclusion by substituting e for b n to obtain{tz m : m ∈ M t } ⊆ A.We now apply Theorem 5.3 (Generic Dichotomy) to extend Theorem 5.6 from anexistence to a genericity statement, thus completing the proof of Theorem 5.1.Theorem 5.7 (Genericity of sequence embedding). In a normed topological group (resp.locally compact metric toplogical group) X, for T ⊆ X almost complete in category (resp.measure) and z n → e X , for generically all t ∈ T there exists an infinite M t such that{tz m : m ∈ M t } ⊆ T.Proof. Working as usual in d X R , the correspondenceF (T ) := ⋂ ⋃(T z−1 m )n∈ω m>ntakes Baire sets T to Baire sets and is monotonic. Here t ∈ F (T ) iff there exists aninfinite M t such that{tz m : m ∈ M t } ⊆ T. By Theorem 5.6 F (T ) ∩ T ≠ ∅, for T Bairenon-meagre, so by Generic Dichotomy F (T )∩T is quasi all of T (cf. Example 1 above).