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Let $A$ a closed subset of a metric space $E$ and let $x\in E-A$.

¿Is posible get two disjoint open sets U, V such that $A\subseteq U$ and $x\in V$?

If $A$ is a compact set I know if it is possible to demonstrate the exercise, but if $A$ If $ A $ is a compact set I know if it is possible to demonstrate the exercise, but if A is closed just not sure.

Roiner Segura Cubero
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1 Answers1

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$E - A$ is open, so $x$ is an interior point of $E-A$. So there is an open neighborhood $N_r(x)$ inside $E-A$ with radius $r > 0$. But instead of taking all of $N_r(x)$, just set $V = N_{r/2}(x)$.

Now, for every point $y \in A$, associate a neighborhood $N_{r/2}(y)$ of $y$. Take the union of these open sets for every point in $A$, so:

$$U = \bigcup_{y \in A} N_{r/2}(y)$$

Since this is a union of open sets, $U$ is open. Moreover, if $z \in U \cap V$, then $z$ is (less than) $r/2$ distance from some point of $A$ and $r/2$ distance from $x$. Now apply the triangle inequality and we have a contradiction.

A.S
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    Alternatively, one can let $U = E \setminus \overline{B}(x,\frac{r}{2})$, where $\overline{B}(x,\frac{r}{2})$ is the closed ball around $x$. – Caleb Stanford Jun 05 '13 at 22:44