# Closure of a finite cover of sets

Suppose I had a compact set $X$ which can be disconnected into two closed in $X$ , nonempty, disjoint sets $A$ and $B$. Then $A$ would have an open cover of neighborhoods centered at points $a \in A$. That is, $A \subseteq \bigcup\limits_{i=1}^{N}N_X(a_i,\epsilon_i):=U$. We extract a similar finite cover for $B\subseteq \bigcup\limits_{i=1}^{M}N_X(b_i,\epsilon_i):=V$.

Is there a way I can guarantee that $U \cap V=\emptyset$?

• You can take the neighborhoods of the elements in $A$ as subsets of $A$ since $A$ is open as complement of closed $B$. The same is true for the neighborhoods in $B$. Then $U=A$ and $V=B$ so the sets are disjoint. This works in space $X$, but you are talking about a compact set $X$. Is $X$ a subset of some larger space $Y$ here? – drhab Dec 13 '13 at 10:25
• $X$ is just a compact subset of $R^2$. There was just a remark on the board as follows: $A \subseteq \bigcup\limits_{i=1}^{N}N_X(a_i,\epsilon_i)\subseteq\bigcup\limits_{i=1}^{N}Cl(N_X(a_i,\epsilon_i))\subseteq X\setminus B$. I was confused how that came about. But he remarked that this would allow us to know that $U \cap V=\emptyset$. – emka Dec 13 '13 at 10:34
• you can start with neighborhoods of elements of $A$ such that the closure of each does not meet $B$, Then there is a finite union $U$ that covers compact $A$ and the union of the closures is a closed set that does not meet $B$. Now the open complement (call it $V$) of this union is a neighborhood for all elements in $B$ and does not meet $U$. – drhab Dec 13 '13 at 10:57

$X\subset\mathbb{R}^{2}$ (see comment of OP) is compact, hence closed ($\mathbb{R}^{2}$ is Hausdorff).

Then the sets $A,B$ - closed in $X$ - are compact and closed.

Open $B^{c}$ contains $A$ so for every $a\in A$ there is some $\epsilon_{a}>0$ with $N\left(a,\epsilon_{a}\right)\cap B=\emptyset$.

Here $x\in N\left(a,\epsilon_{a}\right)\iff\left\Vert x-a\right\Vert <\epsilon_{a}$ and $x\in\overline{N\left(a,\epsilon_{a}\right)}\iff\left\Vert x-a\right\Vert \leq\epsilon_{a}$ so choosing $\epsilon$ small enough we have $\overline{N\left(a,\epsilon_{a}\right)}\cap B=\emptyset$.

Since $A$ is compact we can find $a_{1},\ldots,a_{n}\in A$ such that $A$ is covered by $U:=\cup_{i=1}^{n}N\left(a_{i},\epsilon_{a_{i}}\right)$. Note that $\overline{U}=\cup_{i=1}^{n}\overline{N\left(a_{i},\epsilon_{a_{i}}\right)}$ so that $\overline{U}\cap B=\emptyset$.

For every $b\in B$ some $\epsilon_{b}>0$ exists with $\overline{U}\cap N\left(b,\epsilon_{b}\right)=\emptyset$.

$B$ is compact so we can find $b_{1},\ldots,b_{m}\in B$ such that $B$ is covered by $V:=\cup_{i=1}^{m}N\left(b_{i},\epsilon_{b_{i}}\right)$.

Here $U\cap V\subset \overline{U}\cap V=\emptyset$ .