locally compact Hausdorff

Question

A space $X$ is called locally compact if every point of $X$ has a compact neighbourhood.

I want to show that If $X$ is Hausdorff then $X$ is locally compact iff for every $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$.

My effort: It is clear that if for every $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ then $X$ is locally compact.

Now, assume $X$ is locally compact. Let $x\in X$ and $U$ is a neighbourhood of $x$. Since $X$ is locally compact, there exists a compact neighbourhood $W$ of $x$. I think that I have to show that $U \cap W$ is compact. But I do not know how to show it.

Please help me to show it

Answer 1

You might need a $T_3$ space. To show that this is sufficient, let $x\in\Omega$, with $\Omega$ open. Then $K=X\setminus\Omega$ is closed and $x\notin K$. So by the $T_3$ property, there are open sets $\Omega_1$ and $\Omega_2$ with $x\in\Omega_1$ and $K\subseteq\Omega_2$ and $\Omega_1\cap\Omega_2=\emptyset$. Let $K_2=X\setminus\Omega_2$. Then $K_2$ is closed and $x\in\Omega_1\subseteq K_2\subseteq\Omega$. If $x$ has a compact neighborhood $Q$, then I think $Q_2=Q\cap K_2$ is compact. Then $Q_2$ is a compact neighborhood of $x$ which is included in $\Omega$.

Now I think probably you can show that if every nbhd of $x$ includes a compact nbhd, then $X$ must be $T_3$. Or something like that.

PS. This question is fully covered in detail by Michael C. Gemignani, "Elementary topology", Second edition, Dover 1990. See pages 169–170. On page 169, he shows exactly what you ask for. However note that he defines the concepts in reverse. What you want to prove is what he calls "local compactness", and then he shows that this is equivalent to what we would call local compactness. His proof is essentially the same as yours, using regularity of a compact set in a Hausdorff space.

Even more interesting is that on page 170, he proves a theorem that any locally compact Hausdorff space is $T_3$, which is what I said. In other words, local compactness promotes $T_2$ to $T_3$, and that makes your theorem valid. Hence the validity of your theorem implies that the space is $T_3$.

Answer 2

I think I know how to prove it.

Let $U$ be an open neighbourhood of $x$ and $E$ be a compact neighbourhood of $x$. Then $E$ is regular. Since $U ∩ E$ is an open set in $E$ containing $x$, there exists a set $V$ open in $E$ such that $x \in V \subseteq cl_E(V)\subseteq U ∩ E \subseteq U$. $V$ is open in $E$ implies that $V = E \cap V'$ for some $V'$ open in $X$. This shows that $V$ is a neighbourhood of $x$ and hence $ cl_E(V)$ is also a neighbourhood of $x$. Since $ cl_E(V)$ is a closed subset of the compact subset $E$, $ cl_E(V)$ is compact.