an interesting statement about Hausdorff locally compact space

I'm doing this exercise in Munkres' book and have some point stuck in this problem. Hope someone can help me to clear it out.

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

My proving is that: Assume $x \in U \subset C$, here $U$ is open and $C$ is compact in $X$. Suppose $T$ is a neighborhood of $x$. Let $V = T \bigcap U$. Obviously, $V$ is open. So $C - V$ is closed in compact space $C$, so $C - V$ is compact. Then we can choose $A$ is neighborhood of $x$ and $B$ is open such that $C - V \subset B$ that $A$ and $B$ are disjoint. Therefore, $A$ is the neighborhood of $x$, which $\overline{A}$ is compact, and $A \subset T$.

Here is where I've got stuck. How can I assert that $\overline{A} \subset T$? In the proving of Theorem 29.2 (page 185), the author takes this point for granted, but I think it's not obvious. Can anyone help me. Thanks

• Not directly related to the question, but it looks like you ought to swap $T$ and $U$ in your proof relative to the exercise. Otherwise if the enemy chooses a large enough $U$, it may not have any compact superset $C$. – Henning Makholm Sep 1 '13 at 13:54
• @HenningMakholm:oh, $U$ is just the set I've got from the definition of locally compactness. Here $U$ must be specific, not for every open sets, but there must exist one such open $U$ which there exists compact superset $C$ – le duc quang Sep 1 '13 at 13:58
• x @leduc: In the problem statement $U$ is not something you can decide what is -- you must provide a proof that work for every neighborhood $U$ of $x$. On the other hand you're free to choose $T$ as the neighborhood that has a compact superset. Also, what the problem statement calls $V$ is not what your proof calls $V$, but is what the proof calls $A$. – Henning Makholm Sep 1 '13 at 14:02
• Oh sorry, I've got your point. Let me change the notion in the problem right now... – le duc quang Sep 1 '13 at 14:03

Since $A\subseteq C$ (right?) and $A$ is disjoint from $B$ we must have $A\subseteq C\setminus B$. Also $C\setminus V\subseteq B$ is the same as $C\subseteq V\cup B$ which is the same as $C\setminus B\subseteq V$. So we have

$$A \subseteq C\setminus B \subseteq V \subseteq T$$ Any limit point of $A$ is also a limit point of $C\setminus B$, and since $C\setminus B$ is closed, we have $\overline A \subseteq C\setminus B \subseteq T$.

• So great. It still seems not obvious at all, right...Thanks so much for your response. – le duc quang Sep 1 '13 at 13:51
• @leducquang: It feels reasonably intuitive to me, especially if you sketch a diagram. The intuition is that when we extend $C\setminus V$ to the open $B$, the intersection $B\cup V$ forms a "demilitarized zone" that separates $C\setminus B$ from $C\setminus V$ and prevents any limit point of $A$ from sticking into $C\setminus V$. – Henning Makholm Sep 1 '13 at 14:00
• Yes, you're quite right. With so much open and compact sets, I have to sketch a diagram to make it easier. Still topology is very interesting with me ^^ – le duc quang Sep 1 '13 at 14:02

You can in general choose $A$ so that $\overline{A} \not\subset T$. But, and that choice may be explicit or implicit in the book (if the latter, that is not a good thing), you can choose the neighbourhood $A$ of $x$ to be contained in $V$.

Then $\overline{A} \subset C$, and since $B$ is open, $\overline{A} \cap B = \varnothing$ hence a fortiori $\overline{A} \cap (C\setminus V) = \varnothing$, and the latter, together with $\overline{A}\subset C$ implies $\overline{A} \subset V \subset T$.

• Sorry, I don't get your point. Why $\overline{A} \subset C$ and $B$ is open, you can conclude that $\overline{A} \bigcap B = \varnothing$? – le duc quang Sep 1 '13 at 14:08
• You have $A\cap B = \varnothing$ by premise. Since $B$ is open, you have $\overline{A} \cap B = \varnothing$, since $X\setminus B$ is closed and contains $A$. – Daniel Fischer Sep 1 '13 at 14:11
• Oh, right. Thanks so much. A very easy step that I forgot to lead to complete solution... – le duc quang Sep 1 '13 at 14:13