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?
Best Answer
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