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

Question

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.

i think there's a few questions on the same topic, e.g. http://math.stackexchange.com/questions/250254/can-distance-between-two-closed-sets-be-zero — citedcorpse, Jun 14 '13 at 09:00
$\phi$ (\phi) is not the same symbol as $\emptyset$ (\emptyset) or $\varnothing$ (\varnothing). — Zev Chonoles, Jun 14 '13 at 09:03
@ZevChonoles: Yes I checked the edit and know what is the code. — Basil R, Jun 14 '13 at 09:04

Daniel Robert-Nicoud · Answer 1 · 2013-06-14T20:46:42.617

17

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.

edited Jun 14 '13 at 20:46

answered Jun 14 '13 at 09:07

Daniel Robert-Nicoud

30,663

2

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
3

@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

score 12 · Accepted Answer · answered Jun 14 '13 at 09:07

12

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 $\lim\frac n{a_n}=1$.

answered Jun 14 '13 at 09:07

Asaf Karagila

405,794

Thank you Asaf. It was a nightmare – Basil R Jun 14 '13 at 09:14
2

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

score 6 · Answer 3 · answered Jun 14 '13 at 11:04

6

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

answered Jun 14 '13 at 11:04

Seirios

34,083

score 3 · Answer 4 · answered Jun 14 '13 at 09:59

3

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

answered Jun 14 '13 at 09:59

Mikasa

67,942

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

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