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In proving that a locally compact Hausdorff space $X$ is regular, we can consider the one-point compactification $X_\infty$ (this is not necessary, see the answer here, but bear with me). Since $X$ is locally compact Hausdorff, $X_\infty$ is compact Hausdorff. As a result, $X_\infty$ is normal.

Imitating the idea in my proof in the link above and taking into consideration the correction pointed out in the answer, let $A,B\subseteq X$ be two disjoint, closed sets in $X$, and consider $X_\infty$. Since $X_\infty$ is normal, there are disjoint open sets $U,V \subseteq X_\infty$ such that $A\subseteq U$ and $B \subseteq V$. Then considering $X$ as a subset of $X_\infty$, we can take the open sets $U \cap X$ and $V \cap X$ as disjoint, open subsets of $X$ that contain $A$ and $B$, respectively...or so I thought. I see from the answers to this question that this does not succeed in proving that a locally compact Hausdorff space is normal, since this is not true.

So my question is simply: why does the above proof fail?

Thanks.

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1 Answer 1

up vote 6 down vote accepted

$A$ and $B$ are closed in $X$. They need not be closed in the compactification $X_\infty$. You could try to fix this by replacing them with their closures in $X_\infty$, but then these need not be disjoint.

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A specific example: The Tikhonov plank is the one-point compactification of the deleted Tikhonov plank $X$, which is not normal: the sets $H$ and $K$ of this proof of non-normality are closed and disjoint in $X$, but the extra point of $X^*$ is in the closure in $X^*$ of each of them. –  Brian M. Scott Jan 29 '13 at 21:59
    
I have made that kind of mistake before. Maybe after making it this time I'll remember not to make it the next time! Hopefully. Thanks for the answer. –  Alex Petzke Jan 30 '13 at 2:00

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