# Closed and bounded disjoint sets have positive distance

It is well known that disjoint closed sets in $\mathbb{R}^n$ can have distance equal to 0 (take, for example, the curves $y=\pm\frac{1}{x}$). If we add the requirement that the sets be bounded, this is no longer possible, since such sets are compact by the Heine-Borel theorem.

My question is whether there exists an elementary proof of this fact. I realize that I could surreptitiously inline the relevant parts of the proof of Heine-Borel, but that's not very interesting.

While thinking about this, I came to believe that it should be slightly easier to prove this for convex sets. The key here seems to be the proposition that two closed bounded convex sets can be separated by open balls. I'm not sure how much of a simplification this is as I don't have much background in geometry; the only relevant result that comes to mind is the separation of convex sets in Banach spaces, and this looks to be way overkill. I'd appreciate any input on this as well.

-
In what sense is the compactness argument not elementary? You need some analysis in the proof, since the result fails in $\mathbb{Q}$: $[0,\sqrt{2})\cap \mathbb{Q}$ and $(\sqrt{2},2]\cap \mathbb{Q}$ are closed, bounded and disjoint but distance $0$ apart. –  Chris Eagle Dec 27 '11 at 17:39
I was hoping for a proof with a somewhat geometric flavor. This is perhaps a lot to ask in general, so simplified variants, like the one in the third paragraph, are also welcome. –  Miha Habič Dec 27 '11 at 17:44
I'm not sure what you are aiming at. You need to do the Heine Borel thing probably at least once. If you have some elementary results about compact sets (e.g. continuos functions attain their extrema on compact sets) you can then build on them. In your case, e.g., you could use the fact that the continous distance function $d(.,.) :\mathbb{R}^n\times \mathbb{R}^n \rightarrow \mathbb{R}_{\ge 0}$ (with the first argument restricted to the first compact set, the second argument restricted to the other one) attains it's minimum on compact sets and you are done. –  user20266 Dec 27 '11 at 17:49
You can find points $x_n$ and $y_n$ in the two respective sets with $d(x_n, y_n) \lt \frac{1}{n}$ and take a common convergent subsequence of both, which would contradict their disjointness; compactness (here sequential compactness) is quite unavoidable, as others have pointed out as well. –  Henno Brandsma Dec 27 '11 at 17:52
@ChrisEagle They are not disjoint if $\mathbb{Q}$ is in both intersections. –  Don Larynx Oct 25 '13 at 20:16

Let ${\rm dist}\,(A,B) = \inf \{d(x,y): x \in A, y \in B\}$ denote the distance between sets.
Suppose $A_0$ and $B_0$ are closed bounded nonempty sets in ${\mathbb R}^n$ with $d(A_0,B_0) = 0$. $A_0$ is the union of finitely many closed bounded nonempty sets of diameter at most $1$ (e.g. take the intersections with rectangles of diameter at most $1$ that tile a large rectangle enclosing $A_0$), and at least one of these (call it $A_1$) must have ${\rm dist}\,(A_1, B_0) = 0$. Similarly find $B_1 \subset B_0$ of diameter at most 1 such that ${\rm dist}\,(A_1, B_1) = 0$.
Repeating this idea, inductively define sequences of closed, bounded nonempty sets $A_n$ and $B_n$ with $A_{n+1} \subset A_n$, $B_{n+1} \subset B_n$, such that ${\rm dist}\,(A_n, B_n) = 0$ and $A_n$ and $B_n$ have diameter at most $1/n$. If $x_n \in A_n$ and $y_n \in B_n$, then $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences and converge respectively to $x \in A_0$ and $y \in B_0$, and it's easy to see that $d(x,y) = 0$, so $x = y$. Thus $A_0$ and $B_0$ are not disjoint.