Discreteness of Pontryagin dual of compact abelian group

Question

We have the following theorem:

If $G$ is a compact (Hausdorff) group, then the Pontryagin dual $\widehat{G}$ is discrete.

Does this also imply that $\widehat{G}$ is countable? Or is it possible that, for example, $\widehat{G}$ is the unit circle $\mathbb{T}$ endowed with the discrete topology instead of the standard Euclidian topology? I could not think of any examples where $\widehat{G}$ is an uncountable discrete group.

Answer 1

No, there are uncountable discrete roups, just as there are non-metrizable compact groups.

Start with any non-metrizable compact group. Its dual is an uncountable discrete group.

Example. "The Tubby Torus"
Let $$ G = \mathbb T^\Gamma $$ where $\mathbb T$ is the unit circle, and $\Gamma$ is an uncountable set. Use the usual topology and group structure for $\mathbb T$, and use the product group structure and product topology for $G$. In fact, the dual $\widehat{G}$ is the uncountable direct sum $$ \bigoplus_{\gamma \in \Gamma} \mathbb Z $$ as a discrete group.

This works the other way, too. Take any uncountable discrete group $D$, then $\widehat{D}$ is a non-metrizable compact group.