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arXiv · math/0204147

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 $Π_i G_i$. In particular, if each $G_i$ is compact then $\oplus_i G_i$ determines $Π_i G_i$. (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 $κ$ such that some $X \subseteq T}$ of cardinality $κ$ has positive outer measure. No compact $G$ with $w(G)\geq non(N)$ 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. Is there in ZFC a cardinal $κ$ such that a compact group $G$ is determined if and only if $w(G)<κ$? Is $κ=non(N)$? $κ=\aleph_1$?

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BibTeXRIS

W. W. Comfort, S. U. Raczkowski, F. Javier Trigos-Arrieta. 2002-06-03. Concerning the dual group of a dense subgroup. https://arxiv.org/abs/math/0204147

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