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Let $X$ be a Hausdorff locally compact in $x \in X$. Show that for each open nbd $U$ of $x$ there exists an open nbd $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subset U$.

My work:

Since $X$ is Hausdorff and locally compact then $X$ is regular. Let $U$ be an open nbd of $x$. By assumption $X$ is locally compact so there exists some open nbd $W$ of $x$ such that $\overline{W}$ is compact. Now consider the open set $W \cap U$ this is non-empty since $x$ lies in the intersection. By regularity find an open set $V$ such that:

$x\in V \subset \overline{V} \subset W \cap U$

Then in particular $\overline{V} \subset U$. But also $\overline{V} \subset W \subset \overline{W}$. Since $\overline{W}$ is compact then $\overline{V}$ is a closed subset of a compact set, hence compact.

Is the above OK? Thank you.

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    Yes, your proof is fine and it is really nice to see that you have shown your working on the problem. You should prove that a locally compact Hausdorff space is regular if you have not seen the proof already (but if you have, then there is no need).2011-06-23
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    @Amitesh Datta: thanks, if you know other way of proving this please post it as an answer so I can accept it.2011-06-23
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    @user10: Why is it that being locally compact implies that there is an open n-hood of $x$ whose closure is compact?2011-06-23
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    @Weltschmerz: by definition, see here: http://www.math.uiowa.edu/~jsimon/COURSES/M132Fall07/LocallyCompactSpaces_part1.pdf2011-06-23
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    @Weltschmerz: That's the definition of local compactness: Every point has a compact neighborhood $K$. Since $X$ is Hausdorff, this neighborhood is closed (as it is compact) and its interior $U$ is an open set containing $x$. The closure $\overline{U}$ of $U$ is contained in $K$ and thus compact.2011-06-23
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    @Theo Buehler: what is your favorite topology textbook?2011-06-23
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    Dear user10, I have posted some exercises as an answer. From the theoretical point of view, there are proofs that are equivalent to the one you have given, in the sense of logical equivalence. However, from the practical point of view, your proof is the simplest and most natural.2011-06-23
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    If I may say so, my favorite topology textbook is James Munkres' *Topology: A First Course*.2011-06-23
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    @Amitesh Datta: thanks, that one is great, indeed!2011-06-23
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    @user10: It's a German textbook *Mengentheoretische Topologie* by "Boto von Querenburg" (in its original typewriter edition - it's written by Zieschang and his collaborators, the BOchumer TOpologiegruppe, located in Querenburg near the German city Bochum). It's essentially a very readable digest of Bourbaki's Topologie Générale. In English, I'd probably go for the classics by Dugundji or Kelley. I never more than leafed through Munkres (@Amitesh's favorite), but judging from all I hear it must be excellent.2011-06-23

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Since I have already commented regarding the validity of your proof, let me mention a couple of interesting relevant exercises:

Exercise 1: If $X$ is a locally compact Hausdorff space, then prove that $X$ is completely regular, that is, prove that for every point $x\in X$ and every closed set $C$ with $x\not\in C$, there exists a continuous function $F:X\to [0,1]$ such that $F(x)=0$ and $F(C)=\{1\}$. (Hint: examine the proof of Urysohn's lemma carefully.)

Exercise 2: If $X$ is a locally compact Hausdorff space and $\{\infty\}$ is a singleton set disjoint from $X$, then define $Y=X\cup \{\infty\}$. Furthermore, define a topology on $Y$ by dictating the following subsets of $Y$ to be open:

(1) If $U\subseteq X$ is open in $X$, then $U$ is open in $Y$.

(2) If $V=U\cup \{\infty\}$ where $U$ is open in $X$ and $X\setminus U$ is a compact subset of $X$, then $V$ is open in $Y$.

(a) Prove that $Y$ is a topological space.

(b) Prove that $Y$ is compact Hausdorff. (Hint: to prove that $Y$ is Hausdorff, you will need local compactness and to prove that $Y$ is compact, you will need to think about the role played by $\infty$ in the topological space $Y$.)

(c) If $X=\mathbb{N}$ with the discrete topology, then prove that $Y$ is homeomorphic to $\{0\}\cup \{\frac{1}{n}:n\in\mathbb{N}\}$.

(d) If $X=\mathbb{R}^n$ with the standard topology, then prove that $Y$ is homeomorphic to the sphere $S^n\subseteq \mathbb{R}^{n+1}$.

I hope these exercises are useful!

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    +1 for the usual reasons. Additional exercise 2 (3): Since $Y$ is Hausdorff, the set $\{\infty\}$ is closed. When is it open? (find a necessary and sufficient condition)2011-06-23
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I'll just point out that using the fact that $X$ is regular is a bit of overkill. Here's a simpler fact, which is adequate here:

Lemma. If $X$ is Hausdorff, $x \in X$, and $K \subset X$ is compact, then there exist disjoint open neighborhoods of $x$ and $K$.

Proof. Exercise.

Now, for your problem, assume without loss of generality that $U$ is precompact. (Otherwise, replace $U$ by its intersection with a precompact neighborhood of $x$.) Then the boundary $\partial U$ is compact. Let $V,W$ be disjoint open neighborhoods of $x$, $\partial U$. It's easy to verify that $V \cap U$ has the desired property. (Drawing a picture may help.)

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Solution. Let $U$ be an arbitrary nbd of $x$. By definition, there exist compact set $C\subset X$ with a nbd $W$ of $x$ such that $W\subset C$. Now put $Y=C\setminus \left( U\cap W\right) $. Since $C$ is closed ($X$ is Hausdorff), then $Y=C\setminus U\cap W$ must be also closed subset of $C$. Hence $Y$ is compact. Furthermore, $x\notin Y$. Due to lemma cited above, there are two disjoint open sets $O_{1}$ and $O_{2}$ such that $x\in O_{1}$, $Y\subset O_{2}$. Now if we put $V=O_{1}\cap U\cap W$, then $V$ suffices the desired properties.$\Box $