Closed subsets of a metric space Let $F_1$ and $F_2$ be two closed subsets of a metric space $X$.
I need to show that if $F$ is their union, it is possible to find two closed subsets of $X$: $A$ and $B$ such that the union of $A$ and $B$ is $X$ and $F_1$ is the intersection of $F$ and $A$, $F_2$ is the intersection of $F$ and $B$.
Corrected it now. Sorry.
Thanks in advance.
 A: For each $x\in F\setminus F_1$ there is an $\epsilon_x>0$ such that $B(x,2\epsilon_x)\cap F_1=\varnothing$; let $$U_1=\bigcup_{x\in F\setminus F_1}B(x,\epsilon_x)\;.$$ Similarly, for each $x\in F\setminus F_2$ there is an $\epsilon_x>0$ such that $B(x,2\epsilon_x)\cap F_2=\varnothing$; let $$U_2=\bigcup_{x\in F\setminus F_1}B(x,\epsilon_x)\;.$$
Clearly $U_1\cap F=F\setminus F_1$ and $U_2\cap F=F\setminus F_2$.


*

*Show that $U_1\cap U_2=\varnothing$. HINT: Assume that there is a $y\in U_1\cap U_2$ and get a contradiction. Then let $A=X\setminus U_1$ and $B=X\setminus U_2$, and show that $A$ and $B$ have the desired properties.


If you get completely stuck, I’ve considerably extended the hint in the spoiler-protected field below; mouse-over to see it.

 If $y\in U_1\cap U_2$, there are $x_1\in F\setminus F_1$ and $x_2\in F\setminus F_2$ such that $y\in B(x_1,\epsilon_{x_1})\cap B(x_2,\epsilon_{x_2})$. Let $\epsilon=\min\{\epsilon_{x_1},\epsilon_{x_2}\}$; say $\epsilon=\epsilon_{x_1}$. Then $d(x_1,x_2)\le d(x_1,y)+d(y,x_2)<\epsilon+\epsilon\le 2\epsilon_1$; why is this impossible?

