Normed versus topological groups: Dichotomy and duality

Normed groups 65Remark. In the penultimate step above with W = X, one can take ε > 0; then for0 < η ≤ ε we have d(t, y η (t)) ≤ η, and ω(y η ) < 3η ≤ 3ε, so the points {y η (t) : t ∈ X, 0 0, S(ε) = X, i.e. every point is an ε-shifting point for any ε > 0.In particular, X is a topological group.Proof. Suppose not. Then there is x and ε > 0 such that x is not ε-shifting, i.e. for eachn there is z n ∈ B 1/n (x) such thatd(x, xz n ) ≥ ε.Let η < ε/4. Since z n is null and B η (x) is open (so Baire) and non-meagre, by theassumed KBD there are t ∈ B η (x) and an infinite M t such that tz m ∈ B η (x) for m ∈ M t .So, since d(t, tz m ) < 2η, for any m ∈ M td R (x, xz m ) ≤ d R (x, t) + d R (t, tz m ) + d R (tz m , xz m )= 2d R (x, t) + d R (t, tz m ) < 4η < ε,a contradiction.Thus X = S(ε), for each ε > 0. By Th. 3. 49, X is a topological group.Theorem. 3.51 is a corollary of the Dense Oscillation Theorem (Th. 3.48) and indicatesa ‘Darboux-like’ pathology when the normed group is not topological.Theorem 3.51 (Pathology Theorem). If a normed group X is not a topological group,then there is an open set on which the oscillation function is uniformly bounded awayfrom 0.Proof. This follows from the continuity of ω at any point t where ω(t) > 0. This alsofollows from Th. 3.48, since for some ε > 0, the open set U := X\cl[Ω(ε)] is non-empty,and ω(t) ≥ ε for t ∈ U, as t /∈ Ω(ε).By way of a final clarification of our interest in ε-shifting points, we return to theliterature of ‘separate implies joint continuity’ and in particular to the key notion ofquasi-continuity, which we adapt here to a metric context (for further information seee.g. [Bou2]).