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Let $K\subset X$ a compact set of a metric space. Given that $x\in K^c$, show that exists open sets $U$ and $V$ such that $x\in V$, $K\subset U$ and $U\cap V=\emptyset$.

Well if $K$ is compact, then $K$ is closed and bounded, so $K^c$ is a open set. If $U\cap V=\emptyset$ then $U$ and $V$ are disjoint sets.

Then I need to find $U$ and $V$ such that $K^c\cap V\neq\emptyset$ (because $x\in K^c$ and $x\in V$) then $V\cap K=\emptyset$ (because $K$ is subset of $U$). How I can find such $U$ and $V$?

Sahiba Arora
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Roland
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  • Assuming we're dealing with metric spaces, you can find an open ball centered on $x$ inside $K^c$, with radius $r$. Take V to be the open ball centered on $x$ with radius $r/2$ and let U be the complement of its closure. – Ben G. Jul 01 '17 at 22:46
  • Why don't you tell us what $X$ is? – Friedrich Philipp Jul 01 '17 at 22:57
  • @FriedrichPhilipp The question doesn't say nothing, but maybe it is a metric space. – Roland Jul 01 '17 at 23:02

2 Answers2

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For every $y\in K$, there are an open neighborhood $U_y$ of $y$ and an open neighborhood $V_y$ of $x$ such that $U_y\cap V_y=\emptyset$ (take open balls, if you want).

The family $(U_y)_{y\in K}$ is an open cover of $K$, so…

egreg
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Let $r=dist(x,K)=\inf\{d(x,y):y \in K\}>0$. As $K$ is compact, it is totally bounded. Hence, there exists $\{y_1,y_2, \cdots,y_n\}\subseteq K$ such that $$K\subseteq \bigcup_{i=1}^n B\left(y_i,\frac r4\right)$$

Now, let $U= \bigcup_{i=1}^n B\left(y_i,\frac r4\right)$ and $V=B(x,\frac r4)$. Then $U$ and $V$ are open, $x \in V$ and $K \subseteq U$.

If possible, $U \cap V \neq \emptyset$, then there exists $z \in U \cap V$. Thus, there exists $i$ such that $$d(z,y_i)<\frac r4$$ Also, $$d(x,z)<\frac r4$$ This implies $$d(x,y_i)<\frac r2<r$$ This gives us a contradiction. Hence, $U \cap V=\emptyset$.

Sahiba Arora
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