# Local Compactness [proof verification]

Here is the question: Prove that a space $$X$$ is locally compact if and only if for each $$x$$ $$\in$$ $$X$$ there is a subspace $$A$$ of $$X$$ such that $$cl(A)$$ is compact and $$x \in int A$$.

My attempt:
Suppose $$X$$ is locally compact. Then for each $$x\in X$$, there exists some open $$U$$ containing $$x$$ such that $$cl(U)$$ is compact. Let $$A$$ be a subspace such that $$A=U$$ and $$x\in A$$. Then $$cl(A)=cl(U)$$ is compact by construction. Then since $$U$$ is open in $$X$$ and $$x\in A\cap U$$, then $$x$$ is in an open set in the subspace topology of $$A$$, so $$x\in int A$$.

Suppose for each $$x\in X$$, there is a subspace $$A$$ of $$X$$ such that $$cl(A)$$ is compact and $$x\in int A$$. Since $$x\in int A$$, then there exists some open $$U\subset A$$ such that $$x \in U$$. We can let $$U$$ be such that $$cl(U)=cl(A)$$. Then for any $$x\in X$$, there exists some $$U$$ containing $$x$$ such that $$cl(U)$$ is compact. $$X$$ is thus a locally compact space.

I am a bit concerned with the second half of my proof, specifically, with the statement "we can let $$U$$ be such that $$cl(U)=cl(A)$$". Is this statement always true? If not, then how should I proceed proving that direction?

Thanks!

1. "Let $$A$$ be a subspace such that $$A=U$$ and $$x\in A$$."
You have found $$U$$. Now simply define $$A = U$$. Clearly $$cl(A)=cl(U)$$ is compact by construction. Moreover, since $$A= U$$ is open, we have $$x \in A = int A$$.
2. "We can let $$U$$ be such that $$cl(U)=cl(A)$$."
You have found $$A$$. Now simply define $$U = int A$$. Then clearly $$x \in U$$. Moreover, $$cl(U) = cl(int A) \subset cl (A)$$, thus $$cl(U)$$ is a closed subset of the compact set $$cl (A)$$ and is therefore compact.
• Thank you, Paul, for the feedback! It's interesting that I had not considered that I was choosing $U = int A$... Really appreciate your help! May 8, 2021 at 5:17