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.
$\begingroup$
$\endgroup$
1
asked Mar 29, 2012 at 1:58
-
2$\begingroup$ Exercise for you: show that if this is true, $A$ or $B$ must be non-compact. $\endgroup$– NealMay 16, 2012 at 16:42
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
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 $$\inf \{|a−b|:a \in A,b \in B\} = \inf \left | \frac{1}{2n}\right| = 0$$
-
1$\begingroup$ Hi. You may want to see some of the TeX edits I have made by clicking on the time stamp above my name. $\endgroup$– user21436Mar 29, 2012 at 19:16
-
1$\begingroup$ And you prove that it is closed because it's the complement of an infinite amount of open intervals? $\endgroup$ Aug 13, 2014 at 14:46
-
$\begingroup$ @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. $\endgroup$ Jul 13, 2017 at 16:17
$\begingroup$
$\endgroup$
4
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.
answered Mar 29, 2012 at 2:03
-
8$\begingroup$ This 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 :). $\endgroup$– TheoMay 16, 2012 at 16:06
-
1$\begingroup$ I actually like this one a lot more than the $n+1/2n$ thing - seems more topologic to me, seems less DIY. $\endgroup$– AnthonyJun 17, 2019 at 22:01
-
$\begingroup$ @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! $\endgroup$ Mar 29, 2021 at 7:28
-
$\begingroup$ @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 $\endgroup$ Jul 19, 2021 at 23:34
$\begingroup$
$\endgroup$
2
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 $\gt 1$.
answered Mar 29, 2012 at 2:01
-
2$\begingroup$ For $n=1$ you get $n+1/n=2$ - that's why in another answer $n+\frac1{2n}$ was taken. (Or you cut simply add the condition $n\ge 2$.) $\endgroup$ May 16, 2012 at 14:24
-
16$\begingroup$ @MartinSleziak: Thanks for spotting this. I had forgotten that $1+1=2$. $\endgroup$ May 16, 2012 at 14:28