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I'd appreciate verification of this proof's validity (or lack thereof). The only slight concern I have is in the second to last sentence. This actually seems to be a nice example of the sort of thing I was looking for in this thread

"Let $X$ be a locally compact Hausdorff space, $A$ a closed subset of $X$, and $p$ a point not in $A$. Prove that there are disjoint open sets $U$ and $V$ in $X$ such that $p \in U$ and $A \subset V$."

Let $X_\infty$ denote the one-point compactification of $X$. Since $A$ is a closed subset of this compact space, it is compact itself in $X_\infty$. Since $X$ is Hausdorff and locally compact, $X_\infty$ is Hausdorff and so there are disjoint open sets in $X_\infty$ containing the disjoint compact subsets $\{p\}$ and $A$ by Theorem 6.5*. Since the open sets of $X$ are open considered as subsets of $X_\infty$, these disjoint open sets can come from $X$. Therefore, there are disjoint open sets $U$ and $V$ in $X$ with $p \in U$ and $A \subset V$.

*Theorem 6.5: "If $A$ and $B$ are disjoint compact subsets of a Hausdorff space $X$, then there exist disjoint open sets $U$ and $V$ in $X$ such that $A \subset U$ and $B \subset V$."

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"Since $A$ is a closed subset of this compact space". Why? For example: $A = [0,\infty)$ is not closed in $X_\infty$ for $X = \mathbb R$. I claim: a set $A \subseteq X$ is closed in $X_\infty$ iff $A$ is compact in $X$. This shows that you need to modify your argument a little bit (not much). –  commenter Oct 9 '12 at 2:51
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$\newcommand{\cl}{\operatorname{cl}}$The argument can be made to work, but not quite as you’ve stated it. The problem is that you have no reason to think that the open sets $U$ and $V$ given you by Theorem 6.5 do lie in $X$, so one further step is necessary: the definition of $X_\infty$ ensures that $U\cap X$ and $V\cap X$ are open in $X$, and of course $(U\cap X)\cap(V\cap X)=U\cap V\cap X=\varnothing$, $p\in U\cap X$, and $A\subseteq V\cap X$. Thus, $U\cap X$ and $V\cap X$ give you the desired separation in $X$.

It’s by no means necessary to invoke $X_\infty$, howver, to prove that $X$ is $T_3$. Just let $U=X\setminus A$, let $V$ be an open nbhd of $p$ with compact closure, and let $W=U\cap V$. Then $\cl W$ is compact and Hausdorff and therefore $T_3$, and $W$ is open in $\cl W$, so there is a set $G$ open in $\cl W$ such that $$p\in G\subseteq\cl G\subseteq W\;.$$ It’s easy to verify that $G$ is actually open in $X$ (why?), so $G$ and $X\setminus\cl G$ form the desired separation.

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Sorry for the late reply, I'm at a conference and have had limited internet time. I see why that's the case and have worked things out. As for your second part: my book has not introduced the separation axioms yet, and gave using $X_\infty$ as a hint with this problem, I imagine to show an application of using the compactification. I do see how your proof works though. Thanks for the help. –  Alex Petzke Oct 11 '12 at 19:58
    
@Alex: You’re welcome. (I’m surprised about the separation axioms, though: they’re usually introduced quite early.) –  Brian M. Scott Oct 12 '12 at 6:56
    
Interesting. In case you're curious, I'm using Principles of Topology by Croom. It was recommended to me by someone helping me with my studies early on. It basically goes 'the line and the plane,' metric spaces, topological spaces, connectedness, compactness, product and quotient spaces, separation axioms and metrization, and then the fundamental group. –  Alex Petzke Oct 12 '12 at 23:01
    
@Alex: Very idiosyncratic. (And an order that’s not at all to my taste, but that’s a separate issue!) –  Brian M. Scott Oct 13 '12 at 4:08
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Yes, that second-to-last sentence is troublesome: you have no reason to believe that the open subsets of $X_\infty$ that Theorem 6.5 hands you are contained in $X$.

But given neighborhoods of $p$ and $A$ in $X_\infty$, can you think of a way to get (possibly smaller) neighborhoods of $p$ and $A$ in $X$?

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