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If $G$ is a compact (Hausdorff) topological group with a dense cyclic subgroup, is it necessarily true that $G$ is first countable? This claim seems to be implicit in a paper that I am reading at the moment (the paper asserts the existence of a translation-invariant metric which induces the topology of $G$, which is equivalent by the Birkhoff-Kakutani theorem).

Previously posted at https://math.stackexchange.com/questions/852364/is-every-compact-monothetic-group-metrizable

Martin Sleziak
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Saul
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  • Did you try using Pontryagin duality? – Benjamin Steinberg Jul 30 '14 at 16:10
  • @BenjaminSteinberg There is a natural map $f:\mathbb Z \to G$ sending 1 to a generator of the dense cyclic subgroup. The monomorphism $\widehat f: \widehat{G} \to \widehat{\mathbb{Z}}$ induced by duality identifies $\widehat G$ with a subgroup of $\widehat{\mathbb{Z}} \cong S^1$ (I use the word "identify" loosely, as the inverse of this monomorphism need not be continuous). Moreover, this subgroup is countable by separability of G. In particular, by Pontryagin duality, $G\cong \widehat{\widehat{G}}$ is isomorphic to the dual of a countable group. Does this imply the desired metrizability? – Saul Jul 30 '14 at 16:44
  • $G$ was not assumed separable. Indeed, if $S^1_\delta$ is $S^1$ endowed with the discrete topology, then its Pontryagin dual $G$ is a compact group with a dense cyclic subgroup, that is not first countable; since $S^1_\delta$ is isomorphic to $\mathbf{Q}^{(c)}\times\mathbf{Q}/\mathbf{Z}$, the group $G$ is isomorphic to $\hat{\mathbf{Q}}^{c}\times\hat{\mathbf{Z}}$, where $c$ is the continuum cardinal. In particular $(S^1)^c$ has a dense cyclic subgroup but is not first countable. – YCor Jul 30 '14 at 18:19
  • On the other hand if $G$ is both locally compact with a dense cyclic subgroup and totally disconnected then $G$ is either $\mathbf{Z}$ or a quotient of $\hat{Z}$, and hence is second countable. – YCor Jul 30 '14 at 18:20
  • The dual of a countable group is metrizable because it embeds in a countable direct product of $S^1$ and a countable product of metric spaces is metric. – Benjamin Steinberg Jul 30 '14 at 18:20
  • Thank you, I see my mistake. When I say "$X$ is separable," I mean "$X$ enjoys a countable dense subset," but the claim I made in my post used the existence of a countable base for the topology, which need not follow from the first notion of separability in a non-metrizable space. – Saul Jul 30 '14 at 18:29
  • Your group G is metrizable iff $\widehat{G}$ is countable. – Benjamin Steinberg Jul 30 '14 at 21:34
  • Basically monothetic compact groups correspond to subgroups of $S^1_\delta$ and the metrizable ones correspond to countable subgroups. – Benjamin Steinberg Jul 30 '14 at 21:36
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    I am probably being stupid in my sleep-deprivation, but regarding the original question, isn't the Bohr compactification of the integers non-metrizable? Cf. http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups Maybe this has already been pointed out in the comments, in which case my apologies to Yves and Benjamin – Yemon Choi Jul 31 '14 at 01:33
  • @Jack yes, this was originally given in the MathSE post https://math.stackexchange.com/a/1141513/35400 – YCor May 01 '19 at 15:23

1 Answers1

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No, a compact separable group need not be second countable and may contain ${\bf Z}$ as a dense subgroup. A simple example is a torus of infinite uncountable dimension.

So let $(1, \theta_i)_I$ be an algebraic basis of ${\bf R}$ over ${\bf Q}$. We consider the compact group $G = ({\bf R / Z})^I$ endowed with the product topology. $G$ is separable but not second countable. This follows from the following classical results (see e.g. Dugundji, "Topology").

Theorem
A product of Hausdorff spaces (each with more than a point) is second countable if and only if the product is at most countable and all spaces are second countable.

Theorem
A product of Hausdorff spaces (each with more than a point) is separable if and only if the product is at most the cardinality of the continuum and all spaces are separable.

Now let us consider the element $\theta = (\theta_i)_{i\in I} \in G$. The group ${\bf Z}\theta$ generated by $\theta$ is dense in $G$. Indeed this comes from the following facts.

  • For all finite $F \subset I$, the family $(1, \theta_i)_{i\in F}$ is free over ${\bf Q}$ hence the group generated by $(\theta_i)_{i\in F}$ is dense in $({\bf R / Z})^F$.

  • An open set in $G$ contains a pullback (via the canonical projection) of an open set in $({\bf R / Z})^F$ for some finite $F\subset I$.

Beware that this does not mean that any $y\in G$ can be approximated by a sequence of multiples of $\theta$. The group $G$ is neither second countable, first countable (By the Birkhoff-Kakutani theorem, as pointed out by YCor) nor sequentially compact. Being a cluster point does not allow to extract some converging subsequence.

coudy
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  • Additional keyword: "Bohr compactification of $\mathbf{Z}$". – YCor May 23 '17 at 22:53
  • Note that you're using that a Hausdorff topological group is first-countable iff it's metrizable, by Birkhoff-Kakutani's theorem (and for a compact group this is equivalent to being second-countable). – YCor May 23 '17 at 22:58
  • @YCor Indeed, I edited the answer. – coudy May 24 '17 at 06:29