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 $V$
Separating points from open sets in a compact space without isolated points
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general-topology
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4Is 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 $x \notin \overline{V}$. – 2012-10-19
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0@AymanHourieh: You beat me to it by 19 seconds … – 2012-10-19
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0Are you using the term "compact" to mean "quasi-compact and Hausdorff"? (See http://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. – 2012-10-19
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0Weird, the question was sitting there for an hour, and then three identical comments within a minute :-) – 2012-10-19
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2There is more clustering of events in a Poisson process than you would naïvely expect. – 2012-10-19
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0I think it is true for $S$ is a metric space but i am not sure how to prove it – 2012-10-19
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1@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. – 2012-10-19