Let $A$ and $B$ be two sets of real numbers. Define the distance from $A$ to $B$ by $$\rho (A,B) = \inf \{ |a-b| : a \in A, b \in B\} \;.$$ Give an example to show that the distance between two closed sets can be $0$ even if the two sets are disjoint.
Asked
Active
Viewed 2.3k times
27
-
2Exercise for you: show that if this is true, $A$ or $B$ must be non-compact. – Neal May 16 '12 at 16:42
3 Answers
41
Let $A = \mathbb N$ and let $B = \left\{n+\frac{1}{2n} :n\in \mathbb N\right\}$. Then A and B are closed and disjoint, but $$\inf \{|a−b|:a \in A,b \in B\} = \inf \left | \frac{1}{2n}\right| = 0$$
Nishrito
- 526
-
1Hi. You may want to see some of the TeX edits I have made by clicking on the time stamp above my name. – Mar 29 '12 at 19:16
-
1And you prove that it is closed because it's the complement of an infinite amount of open intervals? – Thomas Ahle Aug 13 '14 at 14:46
-
@ThomasAhle the union of any number of open sets (which intervals can be regarded as) is an open set. The easiest way to see that $A$ and $B$ (and hence $A\cup B$) are each closed is to note that you can put an open interval around any point in the complement that resides entirely in the complement. – mathematrucker Jul 13 '17 at 16:17
12
Consider the sets $\mathbb N$ and $\mathbb N\pi = \{n\pi : n\in\mathbb N\}$. Then $\mathbb N\cap \mathbb N\pi=\emptyset$ as $\pi$ is irrational, but we have points in $\mathbb N\pi$ which lie arbitrarily close to the integers.
Alex Becker
- 60,569
-
8This is a nice example, but showing that $\mathbb{N}\pi$ is arbitrarily close to $\mathbb{N}$ is probably harder than the simplest example to original question :). – Theo May 16 '12 at 16:06
-
1I actually like this one a lot more than the $n+1/2n$ thing - seems more topologic to me, seems less DIY. – Anthony Jun 17 '19 at 22:01
-
@Ales Becker; as you said points in $\Bbb{N}\pi$ are arbitrary close to integers means $\Bbb{N}\pi$ is not closed but OP asked for closed set,I think your example is not correct! – Ibrahim Islam Mar 29 '21 at 07:28
-
@Ibs points in $\mathbb N \pi$ are arbitrarily close to $\mathbb N$, not to a specific integer n. The latter would mean it isn't closed, but this is not the case – Francisco José Letterio Jul 19 '21 at 23:34
8
Let $A$ be the set of positive integers, and let $B$ be the set of all numbers of the form $n+1/n$, where $n$ ranges over the integers $\gt 1$.
André Nicolas
- 507,029
-
2For $n=1$ you get $n+1/n=2$ - that's why in another answer $n+\frac1{2n}$ was taken. (Or you cut simply add the condition $n\ge 2$.) – Martin Sleziak May 16 '12 at 14:24
-
16@MartinSleziak: Thanks for spotting this. I had forgotten that $1+1=2$. – André Nicolas May 16 '12 at 14:28