 # Two disjoint closed sets with distance equal to zero

We take a metric space $$(E,d)$$ and consider two closed subsets $$A,B$$ having a distance $$d(A,B)$$ equal to zero. We raise the following question: can $$A$$ and $$B$$ be disjoint – $$A \cap B=\emptyset$$?

If $$A$$ or $$B$$ is compact, let’s say $$A$$, $$A \cap B$$ cannot be empty. The proof is quite simple. As $$d(A,B) = 0$$, for all $$n \in \mathbb{N}$$ we can pick up $$(a_n,b_n) \in A \times B$$ with $$d(a_n,b_n) \le \frac{1}{n+1}$$. As $$A$$ is compact, we can find a subsequence of $$(a_n)_{n \in \mathbb{N}}$$ converging to a point $$a \in A$$. Even if it means renumbering the subsequence, we can suppose that $$\lim\limits_{n \to +\infty} a_n = a$$. As $$d(a,b_n) \le d(a,a_n)+d(a_n,b_n)$$ and both terms of the right hand side of the inequality converge to zero, $$(b_n)_{n \in \mathbb{N}}$$ also converges to $$a$$. As $$B$$ is supposed to be closed, we have $$a \in B$$ and finally $$a \in A \cap B$$.

So we have to turn ourselves to the case where $$A$$ and $$B$$ are not compact to find a counterexample. If $$E = \mathbb{R}^n$$ with $$n \ge 1$$ that means $$A$$ and $$B$$ unbounded.

## An example in the real line

Take $$A=\{a_n = n ; n \in \mathbb{N}\}$$ and $$B=\{b_m = m + \frac{1}{m+2}; m \in \mathbb{N}\}$$. $$A$$ and $$B$$ are disjoint as if $$n=m + \frac{1}{m+2}$$ and $$n=m$$ we get the contradiction $$\frac{1}{m+2}=0$$ while if $$n \neq m$$ we get the contradiction $$1 \le \vert n-m \vert = \frac{1}{m+2} <1$$. However $$d(A,B) = 0$$ because $$\lim\limits_{n \to +\infty} d(a_n-b_n) =0$$.

## An example in the real plane

Take $$A=\{(x,y) ; y \ge 1/x > 0\}$$ and $$B=\{(x,0) ; x \ge 0\}$$.
$$A$$ and $$B$$ are closed subsets of the real plane and their intersection is empty. The proof is left to the reader.