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A topology question Modified:(which I think it is true in general but cannot prove.) Suppose S is a compact Metric space. 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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Seriously. Titles. Make them informative. – Asaf Karagila Oct 19 '12 at 13:16

The result is correct, but you can get it from much weaker hypotheses: you need only suppose that $S$ is a $T_3$-space with no isolated points.

You’re given a non-empty open set $P$ and a point $x\in S$. Since $P\ne\varnothing$, we can pick a point $y\in P$. $S$ has no isolated points, so $P\ne\{y\}$, and there is therefore another point $z\in P\setminus\{y\}$. At least one of the points $y$ and $z$ must be distinct from $x$, so without loss of generality assume that $y\ne x$. Let $F=\{x\}\cup(S\setminus P)$; $S\setminus P$ is closed, as is $\{x\}$, so $F$ is closed, and clearly $y\notin F$. Since $S$ is regular, there are disjoint open sets $U$ and $V$ such that $F\subseteq U$ and $y\in V$. Then $U$ is an open nbhd of $x$ disjoint from $V$, so $x\notin\operatorname{cl}V$, and by construction $V\subseteq P$, so we’re done.

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what do you mean by $S$ is regular? – Mathematics Oct 19 '12 at 13:46
@Mathematics: A space $S$ is regular if for any $p\in S$ and any closed $F\subseteq S$ with $p\notin F$ there are disjoint open sets $U$ and $V$ such that $p\in U$ and $F\subseteq V$. A $T_3$-space is regular and Hausdorff. See the Wikipedia article on separation axioms for more information. – Brian M. Scott Oct 19 '12 at 13:56
So, a compact set has the same property of a $T_3$ space which is regular? – Mathematics Oct 19 '12 at 14:10
@Mathematics: A compact Hausdorff space is not just $T_3$, but even $T_4$. A compact space that isn’t Hausdorff need not be regular. – Brian M. Scott Oct 19 '12 at 14:14
i mean compact metric space. Is it the same as Hausdorff space? – Mathematics Oct 19 '12 at 14:18

First let us have a look at the possibility that $x\in P$.

There exists an open ball $B(x,r)\subseteq P$

Since $x$ is not isolated, there exists at least one point $y\in B(x,r)$ such that $y\ne x$.

You want to choose a ball $V=B(y,r')$ around $y$ such that:

  • $x\notin\overline V$
  • $V\subseteq B(x,r)$

(It is useful to recall that $\overline V$ is a subset of the closed ball with the same center and radius.)

If you choose $r'$ is such way that $r'<d(x,y)$, the first condition will be fulfilled. If, in addition, you choose $r'$ is such way that $d(x,y)+r'<r$, then the second condition will hold too.

Now if $x\notin P$, take arbitrary $y\in P$ and $B(y,r)\subseteq P$. Then $V=B(y,r/2)$ works.

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Are $\bar V$ the closure of $V$ instead of closed ball with the same center ? – Mathematics Oct 19 '12 at 14:19
You are correct that the closure of open ball is not necessarily the same thing as the closed ball. I've corrected my post. Luckily, the inclusion which is needed there, is true. – Martin Sleziak Oct 19 '12 at 14:26

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