Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,\rho)$ be a metric space. Two sets $A,B\subseteq X$ are separated if $\overline{A}\cap B=\varnothing$ and $\overline{B}\cap A=\varnothing$. Show that $A$ and $B$ are separated if and only if there exist open sets $U$ and $V$ with $A\subseteq U, B\subseteq V$ such that $U\cap V=\varnothing$.

I think I have a solution but I have a couple of questions about it.

Assume $A$ and $B$ are separated. If $A$ and $B$ are open, then we can take $U=A$ and $V=B$ and we're done. If either of $A$ or $B$ are closed, then we get the stronger condition that $\overline{A}\cap\overline{B}=\varnothing$. Let $\text{diam}(A)=\alpha$ and $\text{diam}(B)=\beta$. Let $x\in A, y\in B$ be such that $\rho(x,y)$ is minimal over all $x\in A,y\in B$. Then $x\in\overline{A}, y\in\overline{B}\Rightarrow x\neq y \Rightarrow \rho(x,y)=\epsilon>0$. The let $U,V$ be open sets such that $$ \text{diam}(U)=\alpha+\frac{\epsilon}{4} \qquad \text{diam}(V)=\beta+\frac{\epsilon}{4} $$ and $A\subseteq U, B\subseteq V$. Then $U\cap V=\varnothing$ and we're done.

Assume we have $U$ and $V$ as above. Because $\text{diam}(X)=\text{diam}(\overline{X})$ for any set $X$ in a metric space, $$ \text{diam}(\overline{A})\leq \text{diam}(U) \qquad \text{diam}(\overline{B})\leq \text{diam}(V) $$ Let $x\in\overline{A}$ and consider $\min\{\rho(x,y)\;|\;x\in\overline{A}, y\in B\}=:\epsilon$. We want to show that $\epsilon>0$. First notice that $\min\{\rho(u,v)\;|\;u\in U, v\in V\}>0$. Since $B\subseteq V,\min\{\rho(u,y)\;|\;u\in U, y\in B\}>0 $. But we also have that $u$ is at least as close to $B$ as $x\Rightarrow \epsilon>0$. Thus, $\overline{A}\cap B=\varnothing$. By the same argument, $\overline{B}\cap A=\varnothing$. Thus $A$ and $B$ are separated.

What happens if $A$ or $B$ is neither open nor closed though? I always confuse myself with these types of arguments because I'm assuming that a set is always either open or closed, but that's clearly not true. Is there a way to justify only considering these cases?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

$\newcommand{\cl}{\operatorname{cl}}$No, in general there isn’t. In particular, you can’t do it here.

Suppose that $A$ and $B$ are separated sets in $X$. For each $a\in A$ there is an $\epsilon_a>0$ such that $\rho(a,b)\ge 2\epsilon_a$ for each $b\in B$, and for each $b\in B$ there is an $\epsilon_b>0$ such that $\rho(b,a)\ge 2\epsilon_b$ for each $a\in A$. Let $$U=\bigcup_{a\in A}B(a,\epsilon_a)\quad\text{and}\quad V=\bigcup_{b\in B}B(b,\epsilon_b)\;.$$

If $x\in U\cap B$, then there are $a\in A$ and $b\in B$ such that $\rho(a,x)<\epsilon_a$ and $\rho(x,b)<\epsilon_b$. But then $$\rho(a,b)<\epsilon_a+\epsilon_b\le2\max\{\epsilon_a,\epsilon_b\}\;;$$ do you see the contradiction?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.