Disjoint closed sets in a metric space

Question

Give an example of a metric space $(X,d)$ and two closed sets $A,B$ with $A\cap B=\varnothing$ and $d(A,B)=0$.

Answer 1

A simple situation in $\Bbb R^2$ :

$A= \{ (x,y) \in\Bbb R^2 \mid y\geq e^x \}$

$B= \{ (x,y) \in\Bbb R^2 \mid y\leq 0 \}$

Answer 2

Try $\Bbb R^2$, $A=\{\,(x,y)\in\Bbb R^2\mid xy=0\,\}$, $B=\{\,(x,y)\in\Bbb R^2\mid xy=1\,\}$.

Answer 3

In $(0,1]$ with the usual Euclidean metric, let

$\qquad A = \{1/(2n)\mid n\in \Bbb N_+\}$,

$\qquad B = \{1/(2n+1)\mid n\in \Bbb N\}$.

Similarly, in $\Bbb R$, let

$\qquad A = \Bbb N_+$,

$\qquad B = \{n + \frac 1 {n+1} \mid n\in \Bbb N_+\}$.

Answer 4

It can be shown by some pigeonholing that if $\theta \in \mathbb{R} \setminus \mathbb{Q}$, then the sequence $\{ \theta n \mod{1} \}_{n = 1}^{\infty}$ is dense in $[0, 1]$. That is to say, for every $t \in [0, 1], \epsilon > 0$, there exists $n$ s.t. $| (\theta n \mod{1}) - t | < \epsilon$. This means, equivalently, that for some $N \in \mathbb{N}$, we have that $| (\theta n ) - (t + N) | < \epsilon$, and so we have some $n$ for which $d( \theta n, \mathbb{N} + t ) < \epsilon \Rightarrow d( \theta \mathbb{N} , \mathbb{N} + t ) < \epsilon$. So we can say that if $\theta$ is irrational, then $d ( \theta \mathbb{N}, \mathbb{N} ) < \epsilon$ for every positive $\epsilon > 0$, so $$d ( \theta \mathbb{N}, \mathbb{N} ) = 0$.

Alternatively, for an example in an incomplete space, let $ X = \mathbb{Q} \cap [0, + \infty) $. Let $A = \{ x \in X : x^{2} \leq 2 \}, B = \{ x \in X : x^{2} \geq 2 \}$. They're disjoint, as $\sqrt{2}$ is irrational. Moreover, they're closed (you can check that their complements are open). But then you can find arbitrarily good rational approximations from above and below, which will do it. What follows is an explicit construction of it. You NEED NOT read it if you already believe me.

Define a sequence $d(k)$ as follows: Let $d(1) = \max \{ j \in \{ 0, 1, \ldots, 9 \} : (1 + (j / 10))^{2} < 2 \}$. This is $4$, which is suspiciously enough the first digit after the dot in $\sqrt{2}$. Let $s(k) = 1 + \sum_{i = 1}^{k} d(k) 10^{-i}$, and set $d(k + 1) = \max \{ j \in \{0, 1, \ldots, 9 \} : ( s(k) + j / 10^{k + 1})^{2} < 2 \}$. Each $s(k)$ is $\sqrt{2}$ approximated to $k$ digits from below, but defined without reference to $\sqrt{2}$ (because we're in the rationals, so $\sqrt{2}$ is absent). Moreover, $s(k)$ is strictly in $A$. Let $t(k) = s(k) + 10^{-k}$. Then $t(k) \in B$. But $d(A, B) \leq |s(k) - t(k)| = 10^{-k}$ for every $k$, so since $10^{-k} \to 0$, we know that $d(A, B) = 0$.