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If $E$ and $F$ are disjoint closed subsets of a metric space $(X,d)$, then is $dist~(E,F) >0?$

My attempt:

Suppose $dist~(E,F)=0.$

Then $\exists~e \in E,f \in F$ such that $~\forall ~r>0,~d(e,f)<r.$

$\implies f$ is an accumulation for the set $E$ and $e$ is an accumulation point for $F$.

But, since, closed sets retain their accumulation points $\implies f \in E \bigcap F$ . But given that $E \bigcap F = \{\emptyset\}$ leading to a contradiction.

Hence, $dist~(E,F)>0.$

Could someone please have a look at my proof and point out flaws , if any.

Thank you very much for your help!

MathMan
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  • The points $e$ and $f$ with distance $< r$ depend on $r$. The distance between two disjoint closed sets can be $0$. – Daniel Fischer Aug 03 '15 at 13:46
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    See: http://math.stackexchange.com/questions/125709/example-to-show-the-distance-between-two-closed-sets-can-be-0-even-if-the-two-se – Hetebrij Aug 03 '15 at 13:48
  • @DanielFischer Got it. Thank you very much. – MathMan Aug 03 '15 at 13:57
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    If one of them is compact, yes. But this compactness can't be omitted. For instance, in $\Bbb R^2$ the graphs of $y=e^x$ and $y=-e^x$ tend to "meet" at $-\infty$ but they never do. – Vim Aug 03 '15 at 23:55

3 Answers3

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Consider $\mathbb R^2$ with standard metric and let $E=\{\,(x,y)\mid xy=1\,\}$, $F=\{\,(x,y)\mid xy=0\,\}$.

Your argument about $e,f$ being accumulation points is false. Instead of $$ \exists e\in E\exists f\in F\forall r>0\colon d(e,f)<r$$ we only have $$\forall r>0\exists e\in E\exists f\in F\colon d(e,f)<r $$

Hagen von Eitzen
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No. In fact, the conclusion

$\exists~e \in E,f \in F$ such that $~\forall ~r>0,~d(e,f)<r$

implies that $d(e, f) = 0$, and since $d$ is a metric, this is forces $e = f$ (and so that $X$ is a singleton).

The definition of distance between sets $E, F$ in a metric space $(X, d)$ is $$d(E, F) := \inf \{d(e, f) : e \in F, f \in F \},$$ so $d(E, F) = 0$ is equivalent to the existence of sequences $e_k$ in $E$ and $f_k$ in $F$ such that $\lim d(e_k, f_k) = 0$, that is, that $$\forall r > 0 \, \exists e \in E, f \in F : d(e, f) < r.$$

For an example, consider the metric space $$(\Bbb R_+, d),$$ where $$d(x, y) := \log \left\vert \frac{x}{y} \right\vert ,$$ take $E$ to be the set of positive, even numbers and $F$ the set of positive, odd numbers, which are closed and disjoint. For any positive integer $k$ we have $2 k \in E$ and $2 k - 1 \in F$, but $$\lim d(2k, 2k - 1) = \log \left\vert \frac{2 k}{2 k - 1} \right\vert = 0,$$ so $d(E, F) = 0.$

Travis Willse
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You can have $d(E,F)=0$.

Consider the real plane, $E=\{(x,y) \in \mathbb R^2 | x \ge 0, y \ge 0 \text{ and } xy \ge 1\}$ and $F$ the $x$-axis.

$P_n=(n,\frac{1}{n}) \in E$, $Q_n=(n,0) \in F$ for $n \ge 1$ and $\lim d(P_n,Q_n)=0$

mathcounterexamples.net
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