Normed versus topological groups: Dichotomy and duality

76 N. H. Bingham and A. J. Ostaszewskihomeomorphism). Now for some r > 0, V ⊇ B r (e X ).Thus for x ∈ B r (e X )\N we have ‖x‖ < r and x /∈ N and, as e ∈ V \N and B r (x) =B r (e X )x, we have(B r (e X )\N) ∩ (B r (x)\Nx) ⊆ (V \N) ∩ (V x\Nx) = (Uy\My) ∩ (Uyx\Myx)⊆ Ay ∩ Ayx.Moreover if the intersection L := B r (e X )\N ∩ B r (x)\Nx is meagre, then, for s 0, if ‖y‖ ≤ δ, putting x = aya −1 we have ‖x‖ ≤ ε(A, a)and soA ∩ Ay is non-meagre for any y with ‖y‖ < δ.Theorem 5.5M (Displacements Lemma – measure case; [Kem] Th. 2.1 in R d withB i = E, a i = t, [WKh]). In a locally compact metric group with right-invariant Haarmeasure µ, if E is non-null Borel, then f(x) := µ[E ∩ (E + x)] is continuous at x = e X ,and so for some ε = ε(E) > 0E ∩ (Ex) is non-null, for ‖x‖ < ε.Proof. Apply Theorem 61.A of [Hal-M, Ch. XII, p. 266], which asserts that f(x) iscontinuous.Theorem 5.6 (Generalized BHW Lemma – Existence of sequence embedding; cf. [BHW,Lemma 2.2]). In a normed group (resp. locally compact metrizable topological group) X,for A almost complete Baire non-meagre (resp. non-null measurable) and a null sequencez n → e X , there exist t ∈ A, an infinite M t and points t m ∈ A such that t m → t and{tt −1m z m t m : m ∈ M t } ⊆ A.If X is a topological group, then there exist t ∈ A and an infinite M t such that{tz m : m ∈ M t } ⊆ A.Proof. The result is upward hereditary, so without loss of generality we may assume thatA is topologically complete Baire non-meagre (resp. measurable non-null) and completelymetrizable, say under a metric ρ = ρ A . (For A measurable non-null we may pass down toa compact non-null subset, and for A Baire non-meagre we simply take away a meagreset to leave a Baire non-meagre G δ subset; then A as a metrizable space is complete –