# Math Help - distance of two disjoint compact and closed set

1. ## distance of two disjoint compact and closed set

I am having troubles with one problem:
Let (M,d) be a complete metric space. For a compact set $K\subset{M}$, and for a closed set $F\subset{M}$ such that $K\cap{F}=\emptyset$

$\inf_{x\epsilon K,y\epsilon F}d(x,y)>0$ [

would that be true if K was closed but not compact?

OK, so far my idea is to show by contradiction

then we assume that the inf of the distance set is equal to zero, that implies x=y

now, $x\in{K}$ given K is compact and a compact set is bounded, however x is not necessary element of F (since a closed set might not be bounded)

but then, am I proving with that the statement is false?

2. I am sure as to exactly what you are asking.
Do you have the concept of distance from a point to a set?
$D(F;x) = \inf \left\{ {d(t,x):t \in F} \right\}$
Then show that a closed set is set of all points a distance of zero from the set.

So in this problem $\left( {\forall x \in K} \right)\left[ {D(F;x) > 0} \right]$
Because of compactness you can cover $K$ with finite collection of balls all disjoint from $F$. That should answer the first question.

In $R^2$ let $K=\{(x,y):x\le 0\}~\&~F=\{(1/n~,~n):n\in \mathbb{Z}^+\}$.
Both sets are closed and neither is compact.

3. Thanks, great feedback

I did not have the concept distance from a point to a set neither distance between two sets.

So, now I understand better. I want to check if the distance between a closed set and a compact set is greater than zero.

4. Originally Posted by morito14
Thanks, great feedback

I did not have the concept distance from a point to a set neither distance between two sets.

So, now I understand better. I want to check if the distance between a closed set and a compact set is greater than zero.
This shows a flaw in the effort of sites such as this tries to be.
We have no idea of the sequence of theorems you have been given.
Any time I have given a course on metric spaces the following is a theorm:
If H & K are disjoint closed sets then there are two disjoint open sets $O_H~\&~O_K$ such that $H \subseteq O_H ~\&~K \subseteq O_K$.