Concerning the dual group of a dense subgroup

Details Concerning the dual group of a dense subgroup Concerning the dual group of a dense subgroup

2002-04-10

arxiv.org/abs/math/0204147v2

Proceedings of the Ninth Prague Topological Symposium, (Prague, 2001), pp. 23--35, Topology Atlas, Toronto, 2002

Mathematics

General Topology

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

Scientific paper

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 $w(G)\geq non(N)$ 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)=\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 $\kappa=non(N)$? $\kappa=\aleph_1$?

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