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Given $(X,d)$ a metric space for subsets $A,B$ of $X$, define $$d(A,B)=\inf\{d(a,b):a\in A,b\in B\}$$

could any one confirm me with logic which of the following are/is true and false?

  1. if $\bar{A}\cap\bar{B}=\phi$, then $d(A,B)>0$

  2. if $d(A,B)>0$ then there exists open sets $U\supsetneq A,V\supsetneq B$, $U\cap V=\phi$

  3. $d(A,B)=0$ iff there exist a sequence of points $\{x_n\}$ in $A$ converging to a point in $B$.

well, I took several example from $\mathbb{R}$ and found that $1$ is true, as any metric space is normal $2$ is also true, $3$ is true as in that case $x\in \bar{A}\cap\bar{B}$ and hence by the definition it is true. Thank you.

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Think about asymptotes. – Andrés E. Caicedo Jan 20 '13 at 7:30
I don't think 3 is true. Let $A = (-\infty, 0)$ and $B = (0, \infty)$ in $\mathbb{R}$ with its standard metric. – Michael Albanese Jan 20 '13 at 7:36
To kind of make Andres' comment a little more specific, think about $\mathbb{R}^2$, the "line" $y=0$, and its asymptote, $y = 1/x$. – Anon Jan 20 '13 at 7:37
up vote 1 down vote accepted

(1) and (3) are not true in general. (1) would hold if one of $\overline A$ or $\overline B$ were compact. As Andres and Anon mentioned in comments, you can get counterexamples to (1) by taking a line and a curve that asymptotically approaches the line, getting arbitrarily close as the line and curve "go to infinity". Such an example would also answer (3) negatively, but Michael Albanese's comment gives a simpler example for this case, namely two disjoint open intervals in $\mathbb R$ with a common missing endpoint.

For (2), you could make it a little clearer how you want to apply normality, because $A$ and $B$ are not assumed to be closed. An extra step would be to show that the closures are disjoint, which is straightforward. (That is, the converse to (1) is true.) Or you could argue more directly: Set $D=d(A,B)$, let $U=\{x\in X:d(x,A)<D/2\}$ and $V=\{x\in X:d(x,B)<D/2\}$.

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