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I would like to show that a completely regular topological space is locally compact iff it is (weak-star) open in its Stone-Čech compactification.

Does this hold in general?

I.e given a compact subset $X$ of a normed space, then is it true that if $\overline{U}= X$ then, U is open iff U is locally compact?

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A Tikhonov space is locally compact if and only if it’s open in any of its compactifications.

Let $Y$ be dense in a compact Hausdorff space $X$. Suppose that $Y$ is open in $X$, and let $R=X\setminus Y$. Then $R$ is closed, so for each $y\in Y$ there is an open $V_y$ such that $y\in V_y\subseteq\operatorname{cl}V_y\subseteq X\setminus R=Y$; the closed nbhd $\operatorname{cl}V_y$ of $y$ is compact and Hausdorff, so $Y$ is locally compact.

Conversely, if $Y$ is locally compact, let $y\in Y$, and let $V$ be an open nbhd of $y$ such that $\operatorname{cl}_YV$ is compact. Let $W=\operatorname{int}_Y\operatorname{cl}_YV$; clearly $W$ is an open nbhd of $y$ in $Y$, and $\operatorname{cl}_YW=\operatorname{cl}_YV$. Now $\operatorname{cl}_YW$ is closed in $X$, so it’s equal to $\operatorname{cl}_XW$. Let $U=\operatorname{int}_X\operatorname{cl}_XW$; $Y$ is dense in $X$, so $Y\cap U=W$. But $U\subseteq\operatorname{cl}_XW=\operatorname{cl}_YW\subseteq Y$, so $Y\cap U=U$, and therefore $U=W$, and $W$ is open in $X$ as well as in $Y$. Since $y\in W\subseteq Y$, and $y\in Y$ was arbitrary, $Y$ is open in $X$.

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  • $\begingroup$ Thanks @Brian. How do we know the closure of $V_y$ is contained in $X\setminus R$ though? $\endgroup$ – user58514 Apr 6 '13 at 23:10
  • $\begingroup$ @rustyracketman: The closure in $Y$ of $V_y$ is compact, therefore it is already a closed set in $X$. $\endgroup$ – Brian M. Scott Apr 6 '13 at 23:15
  • $\begingroup$ Sorry. I am a little confused. For instance taking Y itself, then the closure of Y is X, which is certainly not disjoint from R. $\endgroup$ – user58514 Apr 6 '13 at 23:26
  • $\begingroup$ @rustyracketman: You have to pay attention to the space in which the closure is being taken. The closure of $V_y$ in Y is compact, therefore already closed in X as well. The closure of $Y$ in $Y$ is just $Y$, which isn’t compact and therefore is not closed in $X$. $\endgroup$ – Brian M. Scott Apr 6 '13 at 23:28
  • $\begingroup$ Yes, I am failing to see why $V_y$ is compact in Y. I mean if we know that, then that direction of the proof is done. $\endgroup$ – user58514 Apr 6 '13 at 23:37

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