A modified question of some properties of compact set with no isolated point 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 $V$
 A: 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.
A: 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.
