Normed versus topological groups: Dichotomy and duality

120 N. H. Bingham and A. J. Ostaszewskiconsidering S, the subgroup generated by T ; since T is analytic, S is analytic and henceBaire, and, being non-meagre, is clopen and hence all of G, as the latter is a connectedgroup.In the measure case, by the Steinhaus Theorem, Th. 6.10 ([St], [BGT, Th. 1.1.1],[BOst-StOstr]), T 2 has non-empty interior, hence is non-meagre. The result now followsfrom the category case.Our next result follows directly from Choquet’s Capacitability Theorem [Choq] (seeespecially [Del2, p. 186], and [Kech, Ch. III 30.C]). For completeness, we include the briefproof. Incidentally, the argument we employ goes back to Choquet’s theorem, and indeedfurther, to [RODav] (see e.g. [Del1, p. 43]).Theorem 11.16 (Compact Contraction Lemma). In a normed topological group carryinga Radon measure, for T analytic, if T · T has positive Radon measure, then for somecompact subset S of T , S · S has positive measure.Proof. We present a direct proof (see below for our original inspiration in Choquet’sTheorem). As T 2 is analytic, we may write ([Jay-Rog]) T 2 = h(H), for some continuoush and some K σδ subset of the reals, e.g. the set H of the irrationals, so that H =⋂i⋃jd(i, j), where d(i, j) are compact and, without loss of generality, the unions areeach increasing: d(i, j) ⊆ d(i, j + 1). The map g(x, y) := xy is continuous and hence sois the composition f = g ◦ h. Thus T · T = f(H) is analytic. Suppose that T · T is ofpositive measure. Hence, by the capacitability argument for analytic sets ([Choq], or [Si,Th. 4.2 p. 774], or [Rog1, p. 90], there referred to as an ‘Increasing sets lemma’), for somecompact set A, the set f(A) has positive measure. Indeed if |f(H)| > η > 0, then theset A may be taken in the form ⋂ i d(i, j i), where the indices j i are chosen inductively,by reference to the increasing union, so that |f[H ∩ ⋂ i η, for each k. (ThusA ⊆ H and f(A) = ⋂ i f[H ∩ ⋂ i