# Prob. 10, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: If a Hausdorff space is locally compact at a point, then every neighborhood of the point ...

Here is Prob. 10, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:

Show that if $$X$$ is a Hausdorff space that is locally compact at the point $$x$$, then for each neighborhood $$U$$ of $$x$$, there is a neighborhood $$V$$ of $$x$$ such that $$\overline{V}$$ is compact and $$\overline{V} \subset U$$.

My Attempt:

Suppose that $$X$$ is a Hausdorff space, that $$x \in X$$, and that $$X$$ is locally compact at the point $$x$$. Let $$U$$ be any neighborhood of $$x$$ in $$X$$ (i.e. an open set containing $$x$$).

Since $$X$$ is locally compact at $$x$$, there exists a compact subspace $$C$$ of $$X$$ and a neighborhood $$U^\prime$$ of $$x$$ such that $$U^\prime \subset C. \tag{0}$$

Now let us put $$V \colon= U \cap U^\prime. \tag{Definition 0}$$ Then $$V$$ is a neighborhood of $$x$$, and (by (Definition 0) and (0) above), we obtain $$V \subset U^\prime \subset C,$$ and hence $$V \subset C. \tag{1}$$

Now as $$C$$ is a compact subspace of the Hausdorff space $$X$$, so $$C$$ is closed in $$X$$, by Theorem 26.3 in Munkres.

Moreover, as $$C$$ is a closed set in $$X$$ and as $$V \subset C$$ by (1) above, so we can also conclude that $$\overline{V} \subset C. \tag{2}$$

And, as $$\overline{V}$$ is a closed set in $$X$$ and as $$\overline{V} \subset C$$, so $$\overline{V} = \overline{V} \cap C,$$ and thus by Theorem 17.2 in Munkres we can also conclude that $$\overline{V}$$ is also closed in the subspace $$C$$ of $$X$$

Finally, as $$\overline{V}$$ is a closed set in the compact space $$C$$, so $$\overline{V}$$ is also compact, by Theorem 26.2 in Munkres; as $$\overline{V}$$ is compact as a subspace of $$C$$ and as $$C$$ is a subspace of $$X$$, so $$\overline{V}$$ is also compact as a subspace of $$X$$.

What next? How to proceed from here to show that $$\overline{V} \subset U$$ also?

You need to use that $$C$$, being compact in a Hausdorff space, is (a normal and hence) regular space, and in a regular space we have, whenever $$x \in O$$ open, an open $$O'$$ such that $$x \in O' \subseteq \overline{O'} \subseteq O$$.

Apply this to your $$U \cap U'$$ and inside we get the required $$V$$ almost for free from this regularity (open in open is open again, and $$\overline{V}$$ sits inside $$C$$ so is compact).

Without using regularity directly, in Lemma 26.4 Munkres shows that whenever $$Y$$ is a compact subset of a Hausdorff space $$X$$ and $$x_0 \notin Y$$, there are open and disjoint $$U$$ and $$V$$ containing $$x_0$$ resp. $$Y$$, which is all the regularity we need:

We have $$C$$ compact and $$U'$$ the open neighbourhood of $$x$$ inside $$C$$. Then $$C':=C \cap \left( X \setminus \left( U \cap U^\prime \right) \right)$$ is compact (being closed in $$C$$) and $$x \notin C'$$. So by 26.4 we have open $$x \in O_1$$ and $$O_2 \ \supseteq C'$$ that are disjoint.

It's then not too hard to see that $$V= U \cap U' \cap O_1$$ is as required, it's certainly an open neighbourhood of $$x$$ and $$\overline{V} \subseteq U$$.

(For the last inclusion, which is clear (for me at least) from a picture, a more formal proof:

Suppose $$y \in \overline{V}$$, then in particular $$y \in \overline{U'} \subseteq \overline{C}=C$$ (using Hausdorff again!; also when we applied 26.4 of course). Also $$y \in \overline{O_1}$$, so $$y \notin O_2$$ (any point in $$O_2$$ has a neighbourhood, namely $$O_2$$, that is disjoint from $$O_1$$ to guarantee that it's not in the closure of $$O_1$$), and so $$y \notin C'$$ either. Knowing the definition of $$C'$$ and $$y \in C$$ we see that $$y \notin X\setminus (U \cap U')$$ which implies $$y \in U' \cap U \subseteq U$$ as required.

The regularity route is easier, so that's why I chose it first, but Munkres treats regularity after compactness, hence this addition.

• regularity and normality are covered in Chap. 4 in Munkres. So can you please modify your proof to just use the material on compactness and local compactness as has been discussed by Munkres until Sec. 29? Apr 8, 2020 at 10:50
• @SaaqibMahmood Done. See last part. Apr 8, 2020 at 11:27