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Given that $S$ is compact and it has no isolated point. Show that given any nonempty open set $P$ of $S$ and any point $x\in S$, there exists a nonempty open set $V\subset P $ such that $x\notin \bar V $ . I am not very sure what i have to do to finish it. My thought is that it for $x\in S$, it is not an isolated point so there exist an $y\in S$ but then not sure how to proceed, and not sure how to know the existence of $V$

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Is the space Hausdorff? Your question as it stands cannot be proved. Let $S = [0, 1]$ and $\tau = \{S, \emptyset\}$ the trivial topology. $S$ is compact and has no isolated points. Yet, you cannot find an open set $V \subset S$ such that $x \in S$ but $x \notin \overline{V}$. –  Ayman Hourieh Oct 19 '12 at 9:36
    
@AymanHourieh: You beat me to it by 19 seconds … –  Harald Hanche-Olsen Oct 19 '12 at 9:38
    
Are you using the term "compact" to mean "quasi-compact and Hausdorff"? (See en.wikipedia.org/wiki/Compact_set#Definition.) This is false unless you assume that the space is Hausdorff; e.g. it doesn't hold in an indiscrete space with more than one element. –  joriki Oct 19 '12 at 9:38
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There is more clustering of events in a Poisson process than you would naïvely expect. –  Harald Hanche-Olsen Oct 19 '12 at 9:40
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@Mathematics It's easy to prove in Hausdorff spaces, and all metric spaces are Hausdorff. Just Pick another point $y \in P$ and an open set $V$ that separates it from $x$. $V \cap P$ satisfies the requirements. –  Ayman Hourieh Oct 19 '12 at 13:18
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