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.

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  • $\begingroup$ What is your definition of local compactness in general spaces? No Hausdorff etc. ? $\endgroup$ – Henno Brandsma Feb 25 at 18:35
  • $\begingroup$ $B$ 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. $\endgroup$ – Ahmad_Guner Feb 25 at 18:43
  • $\begingroup$ @Ahmad_Guner no, I meant that for non Hausdorff spaces there are several nonequivalent definitions of local compactness. $\endgroup$ – Henno Brandsma Feb 25 at 20:34
  • $\begingroup$ Oh I didn't know there were several of them... $\endgroup$ – Ahmad_Guner Feb 25 at 20:49

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.


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