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If $X$ is a locally compact( every point has a compact nbd) $T_2$ space then $X^*$ is a compact $T_2$ space such that, if X is not compact, then $X$ is dense in $X^*$.

Earlier in the theorem is established that $X^*/ X$ is one point and $X^*$ is a compact $T_2$ space.

We will assume that X is not compact, thus there is a cover that it does not have a finite subcover, and will have to show that $X$ is dense in $X^*$. Let U be an open subset of $X^*$, I need to show that $U\cap X\neq \emptyset$. I need help starting the proof. Thank you for your patience. The text defines $X^*=X\cup \{ \infty \}$, my teacher said just another point out of $X$

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What is your question? – Chris Eagle Jan 29 '13 at 18:35
You should say what $X^*$ is. Presumably it is the one point compactification of $X$. – Miha Habič Jan 29 '13 at 18:38
@Miha Yes it is the one point compactification of X – Klara Jan 29 '13 at 18:39
up vote 3 down vote accepted

The only non-empty set that could miss $X$ would be $\{p\} = X^\ast \setminus X$. But this set is not open (the only open sets that contain $p$ must have a compact complement in $X$ and $X$ is not compact itself). So all non-empty open sets intersect $X$.

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HINT: Let $p$ be the one point of $X^*$ that is not in $X$. The only subset of $X^*$ that does not meet $X$ is $\{p\}$. If $\{p\}$ were open in $X^*$, $X^*\setminus\{p\}=X$ would be closed in $X^*$. What do you know about a closed subset of a compact space?

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X would be compact, as a closed subspace of a compact set, which is not true. – Klara Jan 29 '13 at 18:52
@Klara: Exactly. And you’ve excluded that, so $\{p\}$ can’t be open. – Brian M. Scott Jan 29 '13 at 18:53
@ Brian and then? – Klara Jan 29 '13 at 18:55
@Klara: That means that if $U$ is a non-empty open subset of $X^*$, $U\ne\{p\}$, and therefore $U\cap X\ne\varnothing$. Remember, $X^*\setminus X=\{p\}$. – Brian M. Scott Jan 29 '13 at 18:56
@ Brian, I now this should be easy to see, but where does this definition come from?.. – Klara Jan 29 '13 at 18:58

Let $p$ be such that $X^*=X \cup \{p\}$ (so $p \notin X$ since $X$ is not compact). By definition of $X^*$, any neighborhood $U$ of $p$ contains $X \backslash K$ for some compact $K$ in $X$; therefore, $U \cap X \neq \emptyset$ since $X$ is not compact. You deduce that $X$ is dense in $X^*$.

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