Local Compactness [proof verification]

Question

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!

Answer 1

Your approach to prove the theorem is correct, but I think you could improve the presentation.

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.