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

  • $\begingroup$ 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 $C$. $\endgroup$ – Henning Makholm Sep 1 '13 at 13:54
  • $\begingroup$ @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 $C$ $\endgroup$ – le duc quang Sep 1 '13 at 13:58
  • $\begingroup$ 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 $A$. $\endgroup$ – Henning Makholm Sep 1 '13 at 14:02
  • $\begingroup$ Oh sorry, I've got your point. Let me change the notion in the problem right now... $\endgroup$ – 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 $\overline A \subseteq C\setminus B \subseteq T$.

  • $\begingroup$ So great. It still seems not obvious at all, right...Thanks so much for your response. $\endgroup$ – le duc quang Sep 1 '13 at 13:51
  • $\begingroup$ @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 $C\setminus V$. $\endgroup$ – Henning Makholm Sep 1 '13 at 14:00
  • $\begingroup$ 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 ^^ $\endgroup$ – 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 $\overline{A} \subset V \subset T$.

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

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