Normed versus topological groups: Dichotomy and duality

86 N. H. Bingham and A. J. OstaszewskiProof. First suppose a n ∈ A ⊆ Ā with Ā compact. Without loss of generality a n →a 0 ∈ Ā. Hence z n := a n a −10 → e G . By Theorem 6.2 (the First Verification Theorem),ψ n (x) := z n x converges to the identity. Hence, for some a ∈ A and infinite M, we have{z m a : m ∈ M} ⊆ A. Taking t = a −10 a, we thus have a nt ∈ A and a n t → a ∈ A along M.Replace A by A −1 to obtain the other-handed result.The following theorem asserts that a ‘covering property modulo shift’ is satisfied bybounded shift-compact sets. It will be convenient to make the followingDefinitions. 1. Say that D:= {D 1 , ..., D h } shift-covers X, or is a shifted-cover of X if,for some d 1 , ..., d h in G,(D 1 − d 1 ) ∪ ... ∪ (D h − d h ) = X.Say that X is compactly shift-covered if every open cover U of X contains a finite subfamilyD which shift-covers X.2. For N a neighbourhood of e X say that D:= {D 1 , ..., D h } N-strongly shift-covers A, oris an N-strong shifted-cover of A, if there are d 1 , ..., d h in N such that(D 1 − d 1 ) ∪ ... ∪ (D h − d h ) ⊇ A.Say that A is compactly strongly shift-covered, or compactly shift-covered with arbitrarilysmall shifts if every open cover U of A contains for each neighbourhood N of e X a finitesubfamily D which N-strongly shift-covers A.Theorem 6.8 (Compactness Theorem – modulo shift, [BOst-StOstr]). Let A be a rightshiftcompact subset of a separable normed topological group G. Then A is compactlyshift-covered, i.e. for any norm-open cover U of A, there is a finite subset V of U, andfor each member of V a translator, such that the corresponding translates of V cover A.Proof. Let U be an open cover of A. Since G is second-countable we may assume that Uis a countable family. Write U = {U i : i ∈ ω}. Let Q = {q j : j ∈ ω} enumerate a densesubset of G. Suppose, contrary to the assertion, that there is no finite subset V of U suchthat elements of V, translated each by a corresponding member of Q, cover A. For eachn, choose a n ∈ A not covered by {U i − q j : i, j < n}. As noted earlier, A is precompact,so we may assume, by passing to a subsequence (if necessary), that a n converges to somepoint a 0 , and also that, for some t, the sequence a n t lies entirely in A. Let U i in U covera 0 t. Without loss of generality we may assume that a n t ∈ U i for all n. Thus a n ∈ U i t −1for all n. Thus we may select V := U i q j to be a translation of U i such that a n ∈ V = U i q jfor all n. But this is a contradiction, since a n is not covered by {U i ′q j ′ : i ′ , j ′ < n} forn > max{i, j}.The above proof of the compactness theorem for shift-covering may be improved tostrong shift-covering, with only a minor modification (replacing Q with a set Q ε = {q ε j :