The dual group of a dense subgroup

summary:Throughout this abstract, $G$ is a topological Abelian group and $\widehat{G}$ is the space of continuous homomorphisms from $G$ into the circle group $\mathbb{T}$ in the compact-open topology. A dense subgroup $D$ of $G$ is said to determine $G$ if the (necessarily continuous) surjective isomorphism $\widehat{G}\twoheadrightarrow \widehat{D}$ given by $h\mapsto h\big |D$ 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 $\oplus _iD_i$ 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 $\mathop {\mathrm non}({\mathcal N})$ be the least cardinal $\kappa$ such that some $X \subseteq {\mathbb{T}}$ of cardinality $\kappa$ has positive outer measure. No compact $G$ with $w(G)\ge \mathop {\mathrm non}({\mathcal N})$ is determined; thus if $\mathop {\mathrm non}({\mathcal N})=\aleph _1$ (in particular if CH holds), an infinite compact group $G$ is determined if and only if $w(G)=\omega$. Question. Is there in ZFC a cardinal $\kappa$ such that a compact group $G$ is determined if and only if $w(G)<\kappa$? Is $\kappa =\mathop {\mathrm non}({\mathcal N})$? $\kappa =\aleph _1$