Let $A$ be a compact set and $B$ a closed set ($\varnothing\ne A,B\subseteq \mathbb{R}^n$). Prove there's a minimum distance between $A$ and $B$.

In class we've seen that there's a minimum distance between a compact set $A$, and a point $x_0\notin A$. I thought about utilizing it as a generalization.

First we may assume the points (if exist) must be on the spheres of the sets. For each $x_0$ in the sphere of $B$ there's a point $y_0$ in the sphere of $A$ such that $\forall y\in A: \|y_0-x_0\| \le \|y-x_0\|$.

So we define $f:A\to \mathbb{R}$ such that $f(x) = \text{minimumDistance(x,B)}$.

Is that a good start? How should I proceed?

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    $\begingroup$ By "sphere" do you mean "boundary"? $\endgroup$ – BaronVT Jan 12 '15 at 17:40
  • $\begingroup$ math.stackexchange.com/questions/48714/… $\endgroup$ – Seth Jan 12 '15 at 17:42
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    $\begingroup$ @BaronVT, I guess so (I thought a "Sphere" is a well-known term) $\endgroup$ – AlonAlon Jan 12 '15 at 17:42
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    $\begingroup$ Shouldn't $B$ be nonempty? $\endgroup$ – Hagen von Eitzen Jan 12 '15 at 17:50
  • $\begingroup$ @HagenvonEitzen, that wan't mentioned but I guess we may assume this, otherwise this question has no meaning. $\endgroup$ – AlonAlon Jan 12 '15 at 17:51

The reason is that in $\mathbb R^n$, closed bounded sets are exactly the compact sets.

$A$ is bounded, so we can set $B' = B\cap [-K,K]^n$ for sufficiently large $K$ so that all points in $B\setminus B'$ are far enough from $A$ (i.e. so that $d(A, B\setminus B')>d(a,b)$ for some fixed $a\in A$ and $b\in B$).

$B'$ is closed and bounded, so it's compact and there's a minimal distance between $A$ and $B'$ which remains minimal between $A$ and $B$ from the choice of $K$.

Far enough means that

  • $\begingroup$ I see you narrowed the problem to: distance between two compact sets and concluded immediately that there's a minimum between them. Why? $\endgroup$ – AlonAlon Jan 12 '15 at 18:01
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    $\begingroup$ @AlonAlon Or maybe you use that continuous functions on a compact set attain a minimum? $\endgroup$ – user2345215 Jan 12 '15 at 18:10
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    $\begingroup$ @AlonAlon It's similar to the proof that a point has a minimum distance from a compact set. In this case, set $f(a)=d(a,B)$, then it has a minimum since $A$ is compact. You need that $d(a,B)$ is continuous in $a$, which follows from the generalized triangle inequality $d(a_1,B)\le d(a_1,a_2)+d(a_2,B)$. $\endgroup$ – user2345215 Jan 12 '15 at 18:16
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    $\begingroup$ @AlonAlon I added the condition you need. Imagine $A$ sitting in a ball of radius $R$, then increase this radius by $d(a,b)$ for arbitrary $a\in A, b\in B$. This gives you the radius you need to ensure that $B$ outside of that is too far from $A$. $\endgroup$ – user2345215 Jan 12 '15 at 18:25
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    $\begingroup$ @AlonAlon As I said, prove the triangle inequality I wrote about. It automatically implies the continuity. $\endgroup$ – user2345215 Jan 12 '15 at 19:18

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