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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?

user2345215
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AlonAlon
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1 Answers1

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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

user2345215
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  • I see you narrowed the problem to: distance between two compact sets and concluded immediately that there's a minimum between them. Why? – AlonAlon Jan 12 '15 at 18:01
  • We've proved there's a minimum distance between a compact set and point. – AlonAlon Jan 12 '15 at 18:02
  • Maybe we could define $f:A\to B'$ which returns the point in $B$ where the distance is minimal. – AlonAlon Jan 12 '15 at 18:04
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    @AlonAlon Or maybe you use that continuous functions on a compact set attain a minimum? – user2345215 Jan 12 '15 at 18:10
  • Yes! Actually it was mentioned. – AlonAlon Jan 12 '15 at 18:11
  • But what would be this function in our case? – AlonAlon Jan 12 '15 at 18:12
  • By the way, how do you define $B'$ formally? (instead of "all points in $B- B'$ are far enough)? – AlonAlon Jan 12 '15 at 18:14
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    @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 $d(a_1,B)\le d(a_1,a_2)+d(a_2,B)$. – user2345215 Jan 12 '15 at 18:16
  • OK, thank you @user2345215 – AlonAlon Jan 12 '15 at 18:19
  • But one more thing, how do you define $B'$ rigorously? – AlonAlon Jan 12 '15 at 18:21
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    @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 $A$. – user2345215 Jan 12 '15 at 18:25
  • There's a technical problem; How do you define the function $f(a)=d(a,B)$? Is it the minimal distance or just distance? What is the domain? – AlonAlon Jan 12 '15 at 19:14
  • Anyhow, I'm trying to prove $d(x,B)$ is continuous, but having hard times formalize this :/ – AlonAlon Jan 12 '15 at 19:15
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    @AlonAlon As I said, prove the triangle inequality I wrote about. It automatically implies the continuity. – user2345215 Jan 12 '15 at 19:18