Abstract
A subgroup $H$ of a topological group $G$ is (weakly) totally dense in $G$ if for each closed (normal) subgroup $N$ of $G$ the set $H \cap N$ is dense in $N$. We show that no compact (or more generally, $\omega$-bounded) group contains a proper, totally dense, countably compact subgroup. This yields that a countably compact Abelian group $G$ is compact if and only if each continuous homomorphism $\pi :G \to H$ of $G$ onto a topological group $H$ is open. Here "Abelian" cannot be dropped. A connected, compact group contains a proper, weakly totally dense, countably compact subgroup if and only if its center is not a ${G_\delta }$-subgroup. If a topological group contains a proper, totally dense, pseudocompact subgroup, then none of its closed, normal ${G_\delta }$-subgroups is torsion. Under Lusinâs hypothesis ${2^{{\omega _1}}} = {2^\omega }$ the converse is true for a compact Abelian group $G$. If $G$ is a compact Abelian group with nonmetrizable connected component of zero, then there are a dense, countably compact subgroup $K$ of $G$ and a proper, totally dense subgroup $H$ of $G$ with $K \subseteq H$ (in particular, $H$ is pseudocompact).
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