Normed versus topological groups: Dichotomy and duality

Normed groups 79Remarks. 1. See [Fol] for an early use of a similar, doubled ‘difference set’ and [Hen]for the consequences of higher order versions in connection with uniform boundedness.2. One might have assumed less and required that A be almost complete; but we havefairly general applications in mind. In fact one may assume almost completeness of X inplace of topological completeness. The proof above merely needs the Baire non-meagreset A to contain an almost complete subset, but that turns out to be equivalent to Xbeing almost complete. (See [Ost-LB3] Th. 2 for the separable case and [Ost-AB] for thenon-separable case).3. This one-sided result will be used in Section 11 (Th. 11.11) to show that Borel homomorphismsof topologically complete normed separable groups are continuous. When Xis a topological group, there is no need to square (and the order AA −1 may be commutedto A −1 A, since A is then Baire non-meagre iff A is); this follows from Th. 5.6, but wedelay this derivation to an alternative bi-topological space setting.We close this section with a KBD-like result for normed groups. Thereafter we shallbe concerned mostly (though not exclusively) with topological normed groups. The resultis striking, since under a weak assumption it permits some non-trivial ‘left-right transfer’.We do not know whether this assumption implies that the normed group is topological.We need a definition.Definition. Say that a group-norm is density-preserving if under one (or other) of thenorm topologies, for each dense set D, the set γ g (D) is dense for each conjugacy γ g . See[Ost-AB] for an application.Note that D is dense in X under d R iff D −1 is dense in X under d L , since d R (x, d) =d L (x −1 , d −1 ). Thus density preservation under d R is equivalent to density preservationunder d L .Proposition 5.9. If the group-norm on X is density-preserving, then under the rightnorm topology the left-shift gD of any dense set D is dense. Likewise for the left normdensity and right-shifts.Proof. Fix a dense set D, a point g, and ε > 0. For any x ∈ X, put y = xg −1 . Since γ g (D)is dense we may find d ∈ D such that d R (y, gdg −1 ) < ε; then d R (x, gd) = d R (yg, gd) =d R (y, gdg −1 ) < ε. Thus gD is dense.Remarks. 1. The result shows that density preservation can be defined equivalently byreference to appropriate shifts.2. If D −1 is dense under d L , then so is aD −1 (since λ a (t) is a homeomorphism). However,this does not mean that aD is dense under d R , so the definition of density preservationasks for more.