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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). –  Amitesh Datta Jun 23 '11 at 3:40
    
@Amitesh Datta: thanks, if you know other way of proving this please post it as an answer so I can accept it. –  user10 Jun 23 '11 at 3:45
    
@user10: Why is it that being locally compact implies that there is an open n-hood of $x$ whose closure is compact? –  Weltschmerz Jun 23 '11 at 3:52
    
@Weltschmerz: by definition, see here: math.uiowa.edu/~jsimon/COURSES/M132Fall07/… –  user10 Jun 23 '11 at 3:53
    
@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. –  t.b. Jun 23 '11 at 3:56
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3 Answers

up vote 5 down vote accepted

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) –  t.b. Jun 23 '11 at 4:03
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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 $

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