# Why are compact subsets of a non-locally-compact Hausdorff topological group nowhere dense?

In the EOM entry for nowhere dense set it is stated that

In an infinite-dimensional Hilbert space, every compact subset is nowhere dense. The same holds for infinite-dimensional Banach spaces, non-locally-compact Hausdorff topological groups, and products of infinitely many non-compact Hausdorff topological spaces.

First of all, some generalizations are possible:

1. For an infinite-dimensional Banach space, even relatively compact subsets are nowhere dense (see for example here, the proof of the accepted answer also works in a general infinite-dimensional Banach space)
2. The statement regarding infinite products also holds without the Hausdorff assumption (see for example Theorem 16 in Kelley).

My question is about the remaining statement about topological groups, which I'm not very familiar with:

Could anybody please provide a proof? Is the Hausdorff assumption necessary here and can compactness be weakened to relative compactness?

In other words, I'm looking for the weakest condition on a topological group such that every (relatively) compact subset is nowhere dense

Edit: The proof of the accepted answer below (with a slight and straightforward modification*) shows that every relatively compact subset of a non-locally-compact Hausdorff topological group is nowhere dense.

• A relatively compact subset which is not nowhere dense has an interior point, of which the closure of the subset is a compact neighbourhood.
• You should add your definition of local compactness for non-Hausdorff spaces. – Paul Frost Jul 19 '20 at 8:24
• @PaulFrost you are of course right, thanks for pointing it out! The conclusion holds without the Hausdorff assumption if relative compactness is defined in terms of compact neighbourhoods. This is enough for my purposes, but maybe it can be generalized to one of the weaker definitions. – r_faszanatas Jul 19 '20 at 9:59
• relative compactness of course means local compactness – r_faszanatas Jul 19 '20 at 10:44
• In non-Hausdorff spaces your definition of local compactness may have undesirable consequences. You cannot be sure that any neighborhood of a point $x$ contains a compact neigborhood of $x$. Also note that compact subsets are not necesarily relatively compact. – Paul Frost Jul 19 '20 at 10:57
• @PaulFrost Indeed, so the proof below works for Hausdorff topological groups. I have adjusted the edit to my answer. Thanks again! – r_faszanatas Jul 19 '20 at 12:26

## 1 Answer

Suppose a compact set $$K$$ is not nowhere dense. Then it has an interior point say $$g$$. Note that $$K$$ is a compact neighborhood of $$g$$. For any other point $$h$$ apply the homeomorphism $$x \to hg^{-1}x$$ to see that $$h$$ also has compact neighborhood. Hence the space is locally compact.

• Thanks a lot, in fact your proof works for relative compact sets as well (see the edit of my answer)! – r_faszanatas Jul 19 '20 at 10:00
• Your proof works only when $K$ is closed. This is true in Hausdorff $X$. If $K$ is not closed, you cannot be sure that $\overline K$ is compact. – Paul Frost Jul 19 '20 at 10:48
• @PaulFrost the title says 'non-locally compact Hausdorff space'. I think this means Hausdorff but not locally compact. – Kavi Rama Murthy Jul 19 '20 at 11:38
• @KaviRamaMurthy You are of course right. My comment concerned the OP's question "Is the Hausdorff assumption necessary here and can compactness be weakened to relative compactness? In other words, I'm looking for the weakest condition on a topological group such that every (relatively) compact subset is nowhere dense". – Paul Frost Jul 19 '20 at 13:40