15.80 (2002)

*Not necessarily. One example is $G = \prod_{\mathbb{N}} \mathbb{R}$ under the box topology. It is well known that $G$ does not have a countable dense subset (hence has no expansive sequence). However, the topology on $G$ has a basis having $2^{\aleph_0}$ elements, each element of the basis having $2^{\aleph_0}$ points, and one can partition $G$ into countably many dense subsets (via a transfinite induction of length $2^{\aleph_0}$). (Letter of S. Corson of 19 May 2025.) Another example: any infinite abelian group $G$ with finite set of elements of order 2 can be partitioned into countably many subsets dense in every non-discrete group topology on $G$ according to Corollary 12.21 in (Y. Zelenyuk, Ultrafilters and topologies on groups, De Gruyter, 2011) (Letter of I. V. Protasov of 19 May 2025).