# If a space is regular and every point has a compact neighborhood, is it locally compact?

I attempted to find sufficient conditions for a space to be locally compact given that every point has a compact neighborhood, and I found that being regular is sufficient. I'm not sure my proof is correct so I will write it down. Thanks for your help.

Suppose that $(X,T)$ is a regular topological space in which every point has a compact neighborhood. Then $(X,T)$ is locally compact.

Let $x\in X$ and $U$ be a neighborhood of $x$. We know that there exists a compact neighborhood $K$ of $x$. We have to prove that there exists a compact neighborhood $V$ of $x$ such that $V \subset U$. Since $X$ is regular, there exists an open neighborhood $O$ of $x$ such that $\overline O \subset U$. Let $V = \overline O \cap K$. Then $V$ is a closed subset of $K$, so it is a compact subset of $K$, therefore $V$ is a compact subset of $X$. Since $K$ is a neighborhood of $x$, there exists an open neighborhood $O'$ of $x$ such that $O' \subset K$. Then $O\cap O'$ is an open neighborhood of $x$ and $O\cap O' \subset \overline O \cap K = V$, so $V$ is a neighborhood of $x$, it is compact, and $V \subset U$ since $\overline O \subset U$.

• What is your definition of local compactness? I was under the impression that "every point has a compact neighborhood" = local compactness, regular or not, but I also know terminology tends to be inconsistent in topology. – Kaj Hansen Nov 19 '16 at 9:49
• @KajHansen The definition of local compactness is "for every open set $U$, and every point $x \in U$, there exists a compact neighbourhood of $x$ contained in $U$". – Crostul Nov 19 '16 at 9:54
• @KajHansen I'm using Crostul's definition - or the equivalent one that every point has a local base consisting of compact sets (called a "compact local base") – Cauchy Nov 19 '16 at 9:56
• Your proof seems fine for me. I have checked it and it has no flaws. – Crostul Nov 19 '16 at 10:21
• @Crostul: There are several definitions of local compactness; yours is one, and Kaj’s is another. – Brian M. Scott Nov 19 '16 at 19:29