# assumed distance in metric spaces

Determine whether or not the distance between nonempty $A,B\subset X$ for metric space $X$ is assumed if A and B are closed.

The definition of distance between sets A and B is $d(A,B)=\inf\{d(a,b)|a\in A, b\in B\}$ and the distance is "assumed" if $\exists a_o\in A, b_o\in B$ : $d(A,B)=d(a_o,b_o)$.

I am trying to see how it could be the case that $A,B$ closed $\implies$ $d(A,B)$ assumed. Is it because closed sets contain all their accumulation points?

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It is not the case. Consider the two curves in $\mathbf{R}^2$, $A=\{(x,-1/x)\mid x<0\}$ and $B=\{(x,1/x)\mid x>0\}$. $d(A,B)=0$ but $d(u,v)>0$ for all $u\in A$ and $v \in B$.

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Why is $d(A,B) = 0$? It seems to be finite, in fact $2\sqrt 2$.. – VSJ Feb 12 '12 at 23:17
It is the case if $A$ and $B$ are compact. Maybe even if $A$ is compact and $B$ is clased, I don't know. – Stefan Smith Feb 12 '12 at 23:18
It makes sense now, have an upvote :) – VSJ Feb 12 '12 at 23:26
@user20520: both $A$ and $B$ have to be compact. Let $X$ be $\mathbb{R}^2$ with the origin removed, let $A$ be the set $\{(x, |x|+1) : x \in \mathbb{R}\}$, which is compact, and let $B$ be the punctured $x$ axis, which is closed in $X$. Note that $X$ is also locally compact, so it is not going to help to assume $B$ is locally compact. – Carl Mummert Feb 12 '12 at 23:26
The set should be $\{(x, |x|+1) : x \in [-1,1]\}$. Or we could just let the set be the one point $(0,1)$. – Carl Mummert Feb 12 '12 at 23:58