Concerning the dual group of a dense subgroup

Abstract

Throughout this Abstract, $G$ is a topological Abelian group and $\hat{G}$ is the space of continuous homomorphisms from $G$ into $T$ in the compact-open topology. A dense subgroup $D$ of $G$ determines $G$ if the (necessarily continuous) surjective isomorphism $\hat{G}\twoheadrightarrow \hat{D}$ given by $h\mapsto h|D$ is a homeomorphism, and $G$ is determined if each dense subgroup of $G$ determines $G$. The principal result in this area, obtained independently by L. Aussenhofer and M. J. Chasco}, is the following: Every metrizable group is determined. The authors offer several related results, including these. (1) There are (many) nonmetrizable, noncompact, determined groups. (2) If the dense subgroup $D_i$ determines $G_i$ with $G_i$ compact, then ${\oplus }_i{D}_i$ determines ${\Pi }_i{G}_i$. In particular, if each $G_i$ is compact then ${\oplus }_i{G}_i$ determines ${\Pi }_i{G}_i$. (3) Let $G$ be a locally bounded group and let $G^{+}$ denote $G$ with its Bohr topology. Then $G$ is determined if and only if $G^{+}$ is determined. (4) Let $non(N)$ be the least cardinal $\kappa$ such that some $X \subseteq T}$ of cardinality $\kappa$ has positive outer measure. No compact $G$ with $w(G)\ge non(N)$ is determined; thus if $non(N)={\aleph }_1$ (in particular if CH holds), an infinite compact group $G$ is determined if and only if w(G)=ω$.Question.IsthereinZFCacardinal$κ$suchthatacompactgroup$G$isdeterminedifandonlyif$w(G)<κ$?Is$κ=non(N)$?$κ=\aleph_1$?

13 pages. A full version of this article, with complete proofs, will be submitted for publication elsewhere. Typos corrected