Can a locally compact group with closed singleton be countable but not discrete?

Question

Problem: Prove if a locally compact group $(G,*)$ contains a closed singleton then it must be either discrete or uncountable

Proof Given: Assuming $G$ is countable we can write $G = \displaystyle \bigcup_{g \in G} {g} $. If every $\{ g\}$ has empty interior then this is $G$ expressed as a countable union of nowhere dense sets, which is not allowed. So there must be some $\{ g_0\}$ with nonempty interior, which implies $\{ g_0\}$ is open and hence each $\{ g\} = g g_0 ^{-1}\{ g_0\}$ is open and $G$ is discrete.

My problem with this proof is I don't see why $G$ cannot be expressed in this way, as a countable union of nowhere dense sets. I know if $G$ is hausdorff we can invoke a version of the Baire Category Theorem to get the result. But can we somehow prove $G$ must be hausdorff with the given information, or is there a version of the theorem with weaker requirements? Ideally someone could give me an example of a countable, non-discrete locally compact non-hausdorff group.

Answer 1

To start with your last question: a typical example of a countable, non-discrete locally compact non-Hausdorff group would be the integers with the indiscrete topology.

No example is likely to be more interesting than that, since every T₀ topological group is a Tychonoff space.

That also explains the proof: a topological group with a closed singleton is T₁ by homogeneity and therefore also T₂.

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Addendum: That T₀ implies T₂ can be proved as follows. Take $g \ne e$, such that $U$ is a neighbourhood of $e$ that does not include $g$. There is a neighbourhood $V \ni e$ such that $VV \subset U$. Then $V$ and $gV^{-1}$ are disjoint neighbourhoods of $e$ and $g$, and $g^{-1}V$ and $V^{-1}$ are disjoint neighbourhoods of $g^{-1}$ and $e$ respectively.