Normed versus topological groups: Dichotomy and duality

34 N. H. Bingham and A. J. OstaszewskiTheorem 3.13. The family H u (T ) of bi-uniformly continuous bounded homeomorphismsof a complete metric space T is a complete topological group under the symmetrizedsupremum metric. Consequently, under the supremum metric it is a topological groupand is topologically complete.Proof. Suppose that T is metrized by a complete metric d. The bounded homeomorphismsof T , i.e. those homeomorphisms h for which sup d(h(t), t) < ∞, form a group H = H(T )under composition. The subgroupH u = {h ∈ H : h and h −1 is uniformly continuous}is complete under the supremum metric ˆd(h, h ′ ) = sup d(h(t), h ′ (t)), by the standard 3εargument. It is a topological semigroup since the composition map (h, h ′ ) → h ◦ h ′ iscontinuous. Indeed, as in the proof of Proposition 2.13, in view of the inequalityd(h ◦ h ′ (t), H ◦ H ′ (t)) ≤ d(h ◦ h ′ (t), H ◦ h ′ (t)) + d(H ◦ h ′ (t), H ◦ H ′ (t))≤ ˆd(h, H) + d(H ◦ h ′ (t), H ◦ H ′ (t)),for each ε > 0 there is δ = δ(H, ε) < ε such that for ˆd(h ′ , H ′ ) < δ and ˆd(h, H) < ε,ˆd(h ◦ h ′ , H ◦ H ′ ) ≤ 2ε.Likewise, mutatis mutandis, for their inverses; to be explicit, writing g = h ′−1 , G = H ′−1etc, for each ε > 0 there is δ ′ = δ(G, ε) = δ(H ′−1 , ε) such that for ˆd(g ′ , G ′ ) < δ ′ andˆd(g, G) < ε,ˆd(g ◦ g ′ , G ◦ G ′ ) ≤ 2ε.Set η = min{δ, δ ′ } < ε. So for max{ ˆd(h ′ , H ′ ), ˆd(g, G)} < η and max{ ˆd(h, H), ˆd(g ′ , G ′ )}