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I understand that if given a compact set K and a closed set C,that are disjoint, of a metric space then it follows that there is a minimum distance between them(You can prove this via a continuous function $\varphi_C:K\to R$ that contains the distance information and show that it obtains a minimum).

This may not be the case if K isn't compact. What I want to do is find an example of two disjoint closed sets $C_1$,and $C_2$ where the for every $\epsilon>0$ there exist elements $p\in C_1$ and $q\in C_2$ such that $d(p,q)< \epsilon.$

What I have tried

  • I have tried looking at closed sets $C$ of $\mathbb{R}$. But every noncompact closed set I can find are sets of the form $C=[a,\infty)$ or $C=(-\infty,b]$ where $a,b\in \mathbb{R}$. Which comes out looking something like:

    $\hskip1.1in$ enter image description here

$\hskip.26in$ But I really couldn't find a pair of closed set of $\mathbb{R}$ that has the desired property(i.e being                       disjoint but not having a positive min distance.) So I kinda gave up on that front.

  • I have come up with a picture very similar to the first one for $\mathbb{R^2}$, and reached a problem analogous of it's one dimensional counterpart. (Stuck)

Questions:

  1. Where should I be searching (Metric space wise)?
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  • $\begingroup$ "are disjoint" $\: \mapsto \:$ "are non-empty and disjoint" $\;\;\;\;$ $\endgroup$ – user57159 Jul 3 '15 at 7:08
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Let one set be $\mathbb{Z}$. Let the other one be $\{i+1/(i+1);\,i\in\mathbb{Z},i>0\}$. The main point here is that the two closed sets must both be unbounded to get what you want. Otherwise they will both be compact. (I'm talking about within $\mathbb{R}$ here.)

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    $\begingroup$ Thank you this capture the essense of the problem quite nicely. $\endgroup$ – user160110 Jul 3 '15 at 6:32
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The subsets \begin{align} A &= \{ (x, \frac{1}{x}): x > 0\}\,, \\ B &= \{ (x, \frac{1}{-x}): x < 0\} \end{align} of $\Bbb{R}^2$.

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