Normed versus topological groups: Dichotomy and duality

Normed groups 117restricted to X\M for some meagre set M. As M may be included in a countable unionof closed nowhere dense sets N, f restricted to some non-meagre G δ -set is continuous (e.g.to X\N). Passing to a subset, there is a non-meagre G δ -set H in X with B X ε (H) ⊆ Gsuch that diam X (H) < ε/12 and diam Y (f(H)) < ε/4.Note that HH −1 HH −1 ⊆ B X ε (e X ) (as ‖h ′ h −1 ‖ ≤ ‖h ′ ‖ + ‖h‖) and likewisef(H)f(H) −1 f(H)f(H) −1 ⊆ B Y ε/3 (e Y ). By the Squared Pettis Theorem (Th. 6.5), thereis a non-empty open set U contained in HH −1 HH −1 . Fix h ∈ H and put V := Uh. ThenV = Uh ⊆ HH −1 HH −1 h ⊆ B ε (e X )h = B ε (h) ⊆ B X ε (H) ⊆ G,and so V ⊆ G; moreover, since f is a homomorphism,f(V ) = f(Uh) = f(U)f(h) ⊆ f(H)f(H) −1 f(H)f(H) −1 f(h) ⊆ B Y ε/3 (f(h))and so diam Y f(V ) < ε, as claimed.The final assertion now follows from the Banach-Neeb Theorem (Th. 11.7).The Souslin criterion and the next theorem together have as an immediate corollarythe classical Souslin-graph Theorem; in this connection recall (see the corollary of [HJ,Th. 2.3.6 p. 355] ) that a normed group which is Baire and analytic is Polish. Ourproof, which is for normed groups, is inspired by the topological vector space proof in[Jay-Rog, §2.10] of the Souslin-graph theorem; their proof may be construed as havingtwo steps: one establishing the Souslin criterion (Th. 11.9 as above), the other the Bairehomomorphism theorem. They state without proof the topological group analogue. (See[Ost-AB] for non-separable analogues.)Theorem 11.11 (Baire Homomorphism Theorem, cf. [Jay-Rog, §2.10]). Let X and Y benormed groups with X topologically complete. If f : X → Y is a Baire homomorphism,then f is continuous. In particular, if f is a homomorphism with a Souslin-F(X × Y )graph and Y is in addition a K-analytic space, then f is continuousProof. For f : X → Y the given homomorphism, it is enough to prove continuity at e X ,i.e. that for any ε > 0 there is δ > 0 such that B δ (e X ) ⊆ f −1 [B ε (e X )]. So let ε > 0. Wework with the right norm topology.Being K-analytic, Y is Lindelöf (cf. [Jay-Rog, Th. 2.7.1, p. 36]) and metric, so separable;so choose a countable dense set {y n } in f(X) and select a n ∈ f −1 (y n ). PutT := f −1 [B ε/4 (e Y )]. Since f is a homomorphism, f(T a n ) = f(T )f(a n ) = B ε/4 (e Y )y n .Note also that f(T −1 ) = f(T ) −1 , soT T −1 = f −1 [B ε/4 (e Y )]f −1 [B ε/4 (e Y ) −1 ] = f −1 [B ε/4 (e Y ) 2 ] ⊆ f −1 [B ε/2 (e Y )],by the triangle inequality.Nowsof(X) ⊆ ⋃ n B ε(e Y )y n ,X = f −1 (Y ) = ⋃ n T a n.