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Let's say we have E and F, two compact sets in a metric space (X,d). We remember that the distance between two sets is :

dist(E,F) = inf { d(x,y) | x belongs to E , y belongs to F }

Show that dist (E,F) > 0 if and only if the intersection between E and F is an empty set.

**I feel like I am able to prove this without using the fact that these are compact sets... Am I missing something ?

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  • $\begingroup$ If $X$ are the real numbers with the usual distance, and $E=(-\infty, 0), F= [0,\infty)$ they are disjoint but their distance is still 0. $\endgroup$ – Ingix Oct 15 '18 at 18:16
  • $\begingroup$ I see... how can compactness help us prove this $\endgroup$ – mimi Oct 15 '18 at 19:15
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    $\begingroup$ Are you familiar with the extreme value theorem: en.wikipedia.org/wiki/Extreme_value_theorem ? $\endgroup$ – Ingix Oct 15 '18 at 21:15
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Show that $\operatorname{dist} (E,F) > 0$ if and only if the intersection between $E$ and $F$ is an empty set.

If $\operatorname{dist} (E,F) > 0$ then $E$ and $F$ have no common points. Conversely, assume that $E$ and $F$ have no common points. Then $\operatorname{dist}|E\times F\to\Bbb R$, $(x,y)\mapsto \operatorname{dist}(x,y)$ is a continuous map of a compact set $E\times F$. So its image is a compact $K$ not containing the zero. Since $K$ is closed, the distance (in $\Bbb R$) between $K$ and the zero is bigger than $0$. And this distance is $\operatorname{dist} (E,F)$.

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inf { d(x,y) | x belongs to E , y belongs to F } >0 zero means there is now x in E and y in F such that d(x,y)=0. if x was in E and Y (meaning non empty intersection), then you would have d(x,x)=0, which was against the assumption, so the intersection is empty. (didnt make use of compact set)

Now, if they have an empty intersection, you could map see the distance as a continious function from E x F (which is compact) to the (positive) real line. And because of the intersection being empty, no pair of points is mapped to zero, furthermore the image is a compact set in the real line (continuous image of compact set). This set is closed, and it's infinum is contained in the set. So 0 is not in the set, meaning it could not be the infinum. Therefore the infinum is bigger than zero, meaning dist(E,F) >0.

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