Normed versus topological groups: Dichotomy and duality

58 N. H. Bingham and A. J. Ostaszewskiare closed, respectively open, if ρ z (x) = xz is continuous under d, and soΛ δ (ε) := {t : d R (t, tz) ≤ ε for all ‖z‖ ≤ δ} = ⋂ ‖z‖≤δ F ε(z)is closed. Evidently e X ∈ G ε (z) for ‖z‖ < ε.Proposition 3.42 (Uniform continuity of oscillation). For X a normed groupω(t) − 2‖s‖ ≤ ω(st) ≤ ω(t) + 2‖s‖, for all s, t ∈ X.Hence0 ≤ ω(s) ≤ 2‖s‖, for all s ∈ X,and the oscillation function is uniformly continuous and norm-bounded.Proof. We prove the right-hand side of the first inequality. Fix s, t. By the triangleinequality, for all 0 < δ < 1 and ‖z‖ ≤ δ we have that‖stzt −1 s −1 ‖ ≤ 2‖s‖ + 2‖t‖ + δ ≤ 2‖s‖ + 2‖t‖ + 1,which shows finiteness of ω δ (st) and ω δ (t), and likewise thatHence for all δ > 0Passing to the limit, one hasFrom herei.e.‖stzt −1 s −1 ‖ ≤ 2‖s‖ + ‖tzt −1 ‖ ≤ ω δ (t) + 2‖s‖.ω(st) ≤ ω δ (st) ≤ ω δ (t) + 2‖s‖.ω(st) ≤ ω(t) + 2‖s‖.ω(t) = ω(s −1 st) ≤ ω(st) + 2‖s −1 ‖,ω(t) − 2‖s‖ ≤ ω(st).Also since ω(e X ) = 0, the substitution t = e X gives ω(s) ≤ ω(e X ) + 2‖s‖, the finalinequality.Now, working in the right norm topology, let ε > 0 and put δ = ε/2. Fix x and considery ∈ B δ (x) = B δ (e X )x. Write y = wx with ‖w‖ ≤ δ; then taking s = w and t = x we havei.e.ω(x) − 2δ ≤ ω(y) ≤ ω(x) + 2δ,|ω(y) − ω(x)| ≤ ε, for all y ∈ B ε/2 (x).Thus the oscillation as a function from X to the additive reals R is bounded in thesense of the application discussed after Prop. 2.15.Our final group of results and later comments rely on density ideas and on the followingdefinition.