In one of the proofs of Rudin's real complex analysis, the following implication seems to be assumed:

Let $X$ be a locally compact Hausdorff space. Let $V$ be an open subset of $X$. Let $x\in V$. Then there exists an open set $W\subseteq X$ such that:

1) $x\in W$

2) The closure of $W$ is compact

3) $\overline{W}\subseteq V$

I would like to see a proof for the above implication.

I can show that there exists an open set $W$ that satisfies conditions 1,2.The problem is with the third condition. It seems I am missing something obvious. I am not sure.

Thank you

  • $\begingroup$ And locally compact is defined as "each point has a compact neighborhood" ? $\endgroup$ Sep 22, 2013 at 22:17
  • $\begingroup$ @StefanH. Yes ${}{}$ $\endgroup$
    – Amr
    Sep 22, 2013 at 22:17
  • $\begingroup$ Let $K$ be a compact neighbourhood of $x$. Let $U \subset K$ be an open neighbourhood of $x$. Then $K\setminus U$ is compact, and can be separated from $x$ by disjoint open neighbourhoods $V$ of $K\setminus U$ and $W$ of $x$ in the compact space $K$. Then $\overline{W}$ does not intersect $K\setminus U$, hence $\overline{W}\subset U$. $\endgroup$ Sep 22, 2013 at 22:20
  • $\begingroup$ @DanielFischer OK. Thank you $\endgroup$
    – Amr
    Sep 22, 2013 at 22:25

1 Answer 1


There is a compact nbh $K$ of $x$ and $K\cap V$ is a neighborhood of $x$ in $K$. Since $K$ is a compact Hausdorff space with the subspace topology, it is regular, so there is a nbh $C\subseteq K\cap V$ of $x$ which is closed relative to $K$, thus compact and closed. Now, let $W=\text{int}(C)$. Then $\overline W\subseteq C\subseteq K\cap V\subseteq V$ and $\overline W$ is compact.

This actually shows that if each point has a compact Hausdorff neighborhood, then $X$ has a local base of open sets with compact closures for each point.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .