Topologies that turn the real numbers into a compact Hausdorff topological group

Question

If I'm not mistaken, the question I put on the title used to be on this site, but I'm not being able to find it at all. I'm therefore reposting it so that someone can either give me the old link or help me answer this question anyway.

If $\tau$ is a topology on the real numbers, then saying that $(\mathbb{R}, \tau, +)$ is a compact Hausdorff topological group is equivalent to saying that $Hom_{\tau}(\mathbb{R}, S^1)$ is discrete, by Pontryagin Duality. I initially thought that we could just look at subgroups of $Hom(\mathbb{R}, S^1)$ and that their duals would give us what we want, but this probably doesn't work since the subspace topology of $Hom(\mathbb{R}, S^1)$ doesn't coincide with the topology of uniform convergence on compact sets, in the general case. So I'm lost.

I am convinced that one of the answers in the old link used this Pontryagin Duality approach though.

Answer 1

A compact abelian group $A$ is the same as the Pontryagin dual of a discrete abelian group $B$. The group $A$ is divisible ($nA=A$ for all $n\ge 1$) if and if $B$ is torsion-free, and $A$ is torsion-free iff $B$ is divisible. So $A$ is divisible torsion-free iff $B$ is divisible torsion-free, i.e., $B\simeq\mathbf{Q}^{(I)}$ for some set $I$. In this case, denoting by $P$ the Pontryagin dual of $\mathbf{Q}$, we have $A\simeq P^I$.

The cardinal of $P$ is continuum, so the cardinal of $P^I$ is continuum if $I$ is nonempty and countable (in particular, for $I$ singleton, i.e. for $A=P$). In this case, it is abstractly isomorphic to the group of real numbers. Of course, it is better not to identify $P$ with real numbers since the identification is very non-canonical (and not explicitable).