# Find two closed subsets or real numbers such that $d(A,B)=0$ but $A\cap B=\varnothing$

Here is my problem:

Find two closed subsets or real numbers such that $d(A,B)=0$ but $A\cap B=\varnothing$.

I tried to use the definition of being close for subsets like intervals but I couldn't find any closed sets. Any hint? Thank you.

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i think there's a few questions on the same topic, e.g. math.stackexchange.com/questions/250254/… – citedcorpse Jun 14 '13 at 9:00
$\phi$ (\phi) is not the same symbol as $\emptyset$ (\emptyset) or $\varnothing$ (\varnothing). – Zev Chonoles Jun 14 '13 at 9:03
@ZevChonoles: Yes I checked the edit and know what is the code. – Basil R Jun 14 '13 at 9:04

Hint: You cannot do this for bounded closed sets. Think about $\Bbb N$ and a set $\{a_n\mid n\in\Bbb N\}$ such that $\lim\frac n{a_n}=1$.

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Thank you Asaf. It was a nightmare – Basil R Jun 14 '13 at 9:14
The condition that $\lim (n-a_n)=0$ seems more promising than $\lim\frac{n}{a_n}=1$. – Did Jun 14 '13 at 11:06

Take the following two sets:

$$A = \mathbb{N}$$ $$B = \{n + 2^{-n}:n\in\mathbb{N}\}$$

Then both are obviously closed subsets of the reals, and for any $\varepsilon>0$ you have some $n\in\mathbb{N}$ such that $\frac{1}{n}<\varepsilon$, and thus $d(A,B) = 0$.

Also notice that if you took one of the two sets to be bounded (say $A$), then it would be compact. In that case if $d(A,B) = 0$ you would have two sequences of points $S_A=\{a_n\}\subseteq A$ and $S_B=\{b_n\}\subseteq B$ such that $d(a_n,b_n)\rightarrow 0$. Then $S_B$ must be bounded by some $M$. $C = B\cap [-M,M]\supseteq S_B$ is compact. By compactness we get two convergent subsequences of $S_A$ and $S_B$. Their limits are contained in $A$ and $B$ respectively (since the sets are closed) and must be the same point since $d(a_n,b_n)\rightarrow 0$. This is a contradiction, and thus neither one of the sets can be bounded.

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(+1) nice answer. – Mhenni Benghorbal Jun 14 '13 at 9:10
One tiny detail: $2$ is in both your $A$ and $B$ ($2 = 1 + \frac11$). Try replacing $B$ with $C = \{n + 2^{-n} \in \mathbb R \mid n \in \mathbb N\}$. – kahen Jun 14 '13 at 12:21
@kahen Wouldn't 1 be in both your sets assuming 0 is included in the natural numbers? – MatsT Jun 14 '13 at 14:27
@MatsT Only dirty rotten set theorists (hi, Asaf :D) think that $0$ should be in $\mathbb N$. – kahen Jun 14 '13 at 14:34
@MatsT: I usually denote by $\mathbb{N}_0 = \mathbb{N}\cup\{0\}$. – Daniel Robert-Nicoud Jun 14 '13 at 20:45

In $\mathbb{R}^2$, a visual example is $A= \mathbb{R} \times \{0\}$ and $B= \{(x,e^x) \mid x \in \mathbb{R} \}$.

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Consider the following two sets: $$A=\{2,3,4,5,...\},~~~B=\{2\frac{1}{2},3\frac{1}{3},4\frac{1}{4},...\}$$

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Consider $A=\{\frac{1}{2k} : k\in \mathbb{N}\}$ and $B=\{\frac{1}{2k+1} : k\in \mathbb{N}\}$.

Clearly $\frac{1}{2k} - \frac{1}{2k+1} = \frac{1}{4k^2 +2k}$ tends to $0$.

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Neither of these are closed, since $0$ is a limit point of both but isn't contained in either. – huon Jun 14 '13 at 14:47
Ah, yes. I thought for a moment that none of them have limit points... – Xena Jun 14 '13 at 14:57