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Normed versus topological groups: Dichotomy and duality

Normed versus topological groups: Dichotomy and duality

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62 N. H. Bingham <strong>and</strong> A. J. OstaszewskiRemarks. 1. A sequential version may be formulated: call t an ε-shifting point for thenull sequence z n if there exists N(ε) such that for m > N(ε)d R (t, tz m ) < ε.Then t is an ε-shifting point iff t is ε-shifting for every null sequence. Indeed, if t isnot an ε-shifting point, then for each δ = 1/n there is z n with ‖z n ‖ < 1/n such thatd R (t, tz n ) ≥ ε, so t is not ε-shifting for this null sequence.2. In the notation associated with oscillation, t is ε-shifting for {z n } ift ∈ H ε ({z n }) := ⋃ n G ε(z n ).3. Evidently, if t is an ε-shifting point for each ε > 0, then γ t is continuous (being continuousat e) <strong>and</strong> so a member of the <strong>topological</strong> centre Z Γ (X) of the normed group.4. The sequential version is motivated by the Kestelman-Borwein-Ditor Theorem of Section1 (Th. 1.2) which, roughly speaking, says that tz n → t generically. (See Cor. 3.50.)5. In referring to this property, the theorem which follows assumes something less thanthat the centre Z Γ is dense, only that the open set H ε ({z n }) is dense for each ε > 0 <strong>and</strong>each {z n }.Theorem 3.48 (Dense Oscillation Theorem). In a normed group X⋂cl [Ω(1/n)] = ⋂ Ω(1/n) = Z Γ.n∈N n∈NHence, if for each ε > 0 the ε-shifting points are dense, equivalently Ω(ε) = {t : ω(t) < ε}is dense for each ε > 0, then the normed group is <strong>topological</strong>.More generally, if for some open W <strong>and</strong> all ε > 0 the set Ω(ε) ∩ W is dense in W, thenω = 0 on W ; in particular,(i) if e X ∈ W <strong>and</strong> X is connected <strong>and</strong> Baire under its norm topology, then X is a<strong>topological</strong> group,(ii) if X is separable, connected <strong>and</strong> <strong>topological</strong>ly complete in its norm topology, then Xis a <strong>topological</strong> group.Proof. The opening assertion follows from the continuity of ω. For ε > 0, if Ω(ε) is denseon W, then clW ⊆clΩ(ε). Hence, if Ω(ε) is dense on W for all ε > 0, clW ⊆ Z Γ . So ifW = X = Z Γ , i.e. γ s is continuous for all s ∈ X, then the conclusion follows from theEquivalence Theorem (Th. 3.4). For a more general W, the conclusion follows from Th.3.45.Remark. It is instructive to see how the density property of the last theorem bestowsthe ε-shifting property to nearby points. Fix s <strong>and</strong> ε > 0. For n > 1/ε, let t ∈ Ω(1/n) ∩B ε (s).Then for some δ = δ(n) we have ω δ (t) ≤ 1/n, equivalently d(tz, t) ≤ 1/n for‖z‖ ≤ δ, <strong>and</strong> so for such zd R (sz, s) ≤ d R (sz, tz) + d R (tz, t) + d R (t, s)≤ 2d R (s, t) + 1/n ≤ 3ε.

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