1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.


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.

$\endgroup$
  • $\begingroup$ Is the proof that $T_1$ implies Hausdorff that if $U$ is an open neighborhood of $e$ then for any $g \ne e$ we have that $W = U - \{ g,g^{-1} \}$ is also an open neighborhood of $e$ and that $gW$ is an open neighborhood of $g$ not containing $e$ so the topology is Hausdorff? $\endgroup$ – Daron Sep 23 '13 at 23:55
  • $\begingroup$ I don't think that gives you disjoint neighbourhoods. $\endgroup$ – Niels J. Diepeveen Sep 24 '13 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.