0

Possible Duplicate:
If $A$ is compact and $B$ is closed, show $d(A,B)$ is achieved

Define the distance between two nonempty subsets $A$ and $B$ of $R^n$ by

$dist(A,B):=inf${$||x-y||:x∈A$ and $y∈B$}. Prove that if $A$ and $B$ are compact sets which satisfy $A∩B=∅$, then $dist(A,B)>0$.

(Is it obvious since $A∩B=∅$, then $inf${$||x-y||:x∈A$ and $y∈B$} must$>0$ since $x≠y$?)

Community
  • 1
i_a_n
  • 775
  • It isn't enough that $A\cap B=\emptyset$ - for instance $A=(0,1)$ and $B=(1,2)$ are disjoint and would have $\text{dist}(A,B)=0$. So you'll need to incorporate compactness in some way. – icurays1 Jan 28 '13 at 04:15

2 Answers2

0

It is not obvious in general, only as they are compact. Look at the easy example of $\left[-1, 0\right), \left(0, 1\right]$: the intervals don't interesect, but are at $0$ distance. Note they are not compact.

As to showing this, it's pretty much covered in this older MSE post: Disjoint compact sets in a Hausdorff space can be separated (which is assuming you have a Hausdorff space, as is likely).

Edit: as pointed out below by GM, we surely are in a Hausdorff Space, and the proof goes through verbatim defining open balls in the usual way.

Community
  • 1
gnometorule
  • 4,670
  • 1
    OP says we're in ${\bf R}^n$, presumaby with the usual metric topology, so, Hausdorff, for sure. But maybe that's overkill. – Gerry Myerson Jan 28 '13 at 04:54
  • @Gerry Myerson: Good point. As you know, proof is (or can be made to look) the same in the general setting, and that of $\mathbb{R^n}. However, I will add an edit reflecting your true observation. – gnometorule Jan 28 '13 at 04:59
0

Define $f:A\times B \to \mathbb{R}$ by setting $(x,y) \to \lVert x-y \rVert$. As $f$ is continuous and $A\times B$ is compact, then there must be $(x,y) \in A\times B$ such that $\operatorname{inf}f(A\times B)=dist(A,B)=f(x,y)$. So if $dist(A,B)=0$ then $f(x,y)=0$. But $f(x,y)=\lVert x-y\rVert=0$ means that $x \in cl(B)=B$, a contradiction.

Sayantan
  • 3,468