Normed versus topological groups: Dichotomy and duality

78 N. H. Bingham and A. J. OstaszewskiRemark. For a similar approach to work in the normed group setting we would need toknow that the monotone correspondenceG(T ) := T ∩ ⋂ ⋃T · g m(T ), where g m (t) := t −1 zm −1 t,n∈ω m>ntakes Baire sets to Baire sets. Of course t ∈ G(T ) iff t ∈ T and t = t ′ mt −1m zm −1 t m ∈ T forsome t m , t ′ m ∈ T and so tt −1m z m t m = t ′ m ∈ T. To see the difficulty, write t m = w m t andcompute thatt ∈ G(T ) ⇐⇒ (∀n)(∃m > n)∃{w n }(∃u, s, t ′ )(∀k)[t ′ ∈ T & s ∈ T & d R (w k , e) ≤ 1/k & s = w m t & t = t ′ u & u = w m zm −1 wm−1 ].If the graph of the relation t = xy were analytic, we could deduce that G(T ) is analytic(see section 11 for definition) for T a G δ set (all that is needed for Th. 5.4); that in turnguarantees that G(T ) is Baire. However, if even the relation e = xy were analytic, thiswould imply that inversion is continuous and so the normed group would be topological(see Th. 3.41). We can nevertheless say a little more about G(T ).Theorem 5.7A (Non-meagreness of sequence embedding – normed groups). In a normedgroup X, for T ⊆ X almost complete in category, U open with T ∩ U non-meagre, andz n → e X , the set S U of t ∈ T ∩ U for which there exist points t m ∈ T with t m → R t andan infinite M t with{tt −1m z m t m : m ∈ M t } ⊆ Tis non-meagre.Proof. Suppose not; then there is an open set U such that S U is meagre. Letting H be ameagre F σ cover of S U , the set T ′ := (T \H) ∩ U is Baire and non-meagre. But then byTh. 5.6 there exists points t, t m ∈ T ′ and infinite set M t such thata contradiction.{tt −1m z m t m : m ∈ M t } ⊆ T ′ ⊆ T ∩ U,Theorem 5.8 (Squared Pettis Theorem). Let X be a topologically complete normed groupand A Baire non-meagre under the right norm topology. Then e X is an interior point of(AA −1 ) 2 .Proof. Suppose not. Then we may select z n ∈ B 1/n (e)\(AA −1 ) 2 . As z n → e, we applyTh. 5.6 to A, to find t ∈ A, M t infinite and t m ∈ A for m ∈ M t such that tt −1m z m t m ∈ Afor all m ∈ M t . So for m ∈ M tz m ∈ AA −1 AA −1 = (AA −1 ) 2 ,a contradiction.