# Does locally compact, not nowhere-dense topological subspace always contain open subset?

We have a topological space $$A$$. I want to prove that it's locally compact subspace $$B$$ always contains non-empty open subset if $$B$$ is not nowhere-dense.

I am not even sure is it true or not, may be it is possible to construct a counterexample.. But the statement in the title seems to be right to me.

There are no conditions on $$A$$, so it is not even Hausdorff.

I've tried to achieve a contradiction assuming that $$B$$ does not contain any open subset (so all points of B are boundary points) and it's closure contains at least one.

But I have no idea how to use locally compactness.

• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax. – dantopa Feb 25 at 18:11
• What is your definition of local compactness in general spaces? No Hausdorff etc. ? – Henno Brandsma Feb 25 at 18:35
• $B$ is locally compact if any point $x \in B$ has neighbourhood $B_x$ which closure is compact.. So $B$ has to be Hausdorff then? Ok, I just didn't see how it was needed in this definition. – Ahmad_Guner Feb 25 at 18:43
• @Ahmad_Guner no, I meant that for non Hausdorff spaces there are several nonequivalent definitions of local compactness. – Henno Brandsma Feb 25 at 20:34
• Oh I didn't know there were several of them... – Ahmad_Guner Feb 25 at 20:49

## 1 Answer

Let the integers Z have the cofinite topology.
Each open set, including Z, is compact and the
closure of any open set is either empty or Z, thus compact.
Therefore Z is locally compact.

The subset 2Z of even integers with the inherited topology is cofinite. In fact it is homeomorphic to Z, hence locally compact.
2Z is not a nowhere dense subset of Z.

There is no not empty, open subset of Z that is a subset of 2Z.
In general the answer to your question is no.
If the space A were Hausdorff, then your proposition might hold.