Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 6 down vote accepted

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 $v \in B$.

share|cite|improve this answer
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

Your Answer


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.