12
$\begingroup$

When I learn the below theorem:

If $I_n$ is closed interval, and $I_{n+1} \subset I_n$, then $$\bigcap I_n \ne \varnothing$$

and someone says if we replace closed interval with open interval, can construct counter-example.

So I have tried to construct the one: Does $$\bigcap_{n=1}^{+\infty}\left(-\frac{1}{n},\frac{1}{n}\right) = \varnothing\quad?$$

Thanks very much.

$\endgroup$
  • 8
    $\begingroup$ I am 100% certain that this question was asked at least twice before. $\endgroup$ – Asaf Karagila Jan 5 '13 at 14:21
  • 1
    $\begingroup$ Yes and I am searching for that link, Asaf. $\endgroup$ – mrs Jan 5 '13 at 14:24
  • 19
    $\begingroup$ 10 answers? really? $\endgroup$ – Julien Jan 5 '13 at 20:17
  • $\begingroup$ its obviously non-empty? $\endgroup$ – user85461 Jul 19 '13 at 18:41
  • $\begingroup$ Related post: math.stackexchange.com/questions/1304402/… $\endgroup$ – Martin Sleziak Jan 8 '16 at 12:40
20
$\begingroup$

No, it is not empty, but in fact, if $I_n$ is closed and $I_{n+1}\subseteq I_{n}$, that is not enough to ensure that $$\bigcap_{n=1}^\infty I_n\neq\varnothing.$$ As an example, take the closed intervals $[n,\infty)$. In order for what you have written to hold true, one needs that at least one of the $I_n$ is bounded.

Finally, as others have pointed out, the intersection of open sets you have is nonempty since $0\in\left(-\frac{1}{n},\frac{1}{n}\right)$ for all $n\in\mathbb{N}$. As a counterexample, one can consider $$\bigcap_{n=1}^\infty\left(0,\frac{1}{n}\right).$$

$\endgroup$
  • $\begingroup$ thank you, I have made a mistake. yes, $I_n$ need to be bounded. $\endgroup$ – Laura Jan 5 '13 at 14:50
  • $\begingroup$ @Tai: You're welcome, I just wanted to be sure you didn't have the wrong 'theorem' in mind. $\endgroup$ – Clayton Jan 5 '13 at 14:51
  • 2
    $\begingroup$ +1 for explaining how the problem should be solved as well. $\endgroup$ – Joe Z. Jan 5 '13 at 16:16
  • $\begingroup$ +1 nice approach. $\endgroup$ – mrs Feb 3 '13 at 3:26
  • $\begingroup$ @BabakSorouh: Thanks! (And for all the +1's :) ) $\endgroup$ – Clayton Feb 3 '13 at 3:27
31
$\begingroup$

No. It is not empty. Since $0\in (-1/n, 1/n)$ for all $n$, so $0\in\bigcap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$.

$\endgroup$
  • $\begingroup$ thank you ,but if $x \in (-\frac{1}{n},\frac{1}{n})$, then $-\frac{1}{n}<x<\frac{1}{n}$ let $n \rightarrow \infty$, then $0<x<0$, so $x$ dose not exist. but apparently,your statement is right. so confuse me. $\endgroup$ – Laura Jan 5 '13 at 14:24
  • 17
    $\begingroup$ @Tai: When we take $n\to\infty$, we can no longer ensure that the inequalities remain strict, so we get $0\leq x\leq 0$. $\endgroup$ – Cameron Buie Jan 5 '13 at 14:29
  • 1
    $\begingroup$ oh,I make a mistake, when let $n \rightarrow +\infty$, $0\leq x \leq 0$ $\endgroup$ – Laura Jan 5 '13 at 14:30
12
$\begingroup$

As you already have a bunch of nice answers to your question, I just wanted to comment on the relation with the theorem you have referred to. Note that $J_n\subset I_n$ where $J_n = [-\frac{1}{2}n,\frac{1}{2}n]$ and thus we know that $\bigcap\limits_n J_n$ is not empty. However, $\bigcap\limits_n J_n\subset \bigcap\limits_n I_n$ and the latter is hence non-empty as well.

$\endgroup$
  • 1
    $\begingroup$ (+1) I think this connection is likely to be helpful to the OP. $\endgroup$ – cardinal Jan 5 '13 at 20:58
12
$\begingroup$

NO!

$$\bigcap_{n=1}^{+\infty}\,\left(-\frac{1}{n},\frac{1}{n}\right)\,\neq\, \varnothing$$


WHY NOT?:
Note that as $\,n \to \infty,\,$ the endpoints of the intervals get increasingly close to $\,0$, but never reach $0$, hence every interval contains $0\,$: $$\;0 \in \left(-\dfrac1n, \cfrac1n\right)\;\; \forall n \in \mathbb{N}.$$ Indeed,

$$\bigcap_{n=1}^{+\infty}\,\left(-\frac{1}{n},\frac{1}{n}\right) = \{0\}.$$


You might be interested to know that, e.g., $$\bigcap_{n=1}^{+\infty}\,\left(0, \frac1n\right)\,=\, \varnothing$$ It' simply a matter of choosing the correct endpoints of the open intervals.

$\endgroup$
  • $\begingroup$ And, of course, $$\bigcap_{n=1}^{+\infty}\left(-\frac{1}{n},\frac{1}{n}\right)\subseteq \{0\}.$$ If $x\in\bigcap_{n=1}^{+\infty}\left(-\frac{1}{n},\frac{1}{n}\right)$ it means $$|x|\lt \frac1{n}\quad\forall n\in\Bbb N$$ so $x=0$. $\endgroup$ – leo Jan 5 '13 at 16:25
  • $\begingroup$ Oh well... the thing was prove that the intersection is not empty. It suffices with what you've done. $\endgroup$ – leo Jan 5 '13 at 16:30
7
$\begingroup$

No. As others have mentioned, it contains $0$.

But consider $I_n = (1-\frac{1}{n},1)$. Or, more closely to your example, $I_n = (0,\frac{1}{n})$.

$\endgroup$
3
$\begingroup$

$\bigcap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)=\{0\}\neq\emptyset$.

$\endgroup$
  • 6
    $\begingroup$ What does it add to the answers that are already given? $\endgroup$ – Ilya Jan 5 '13 at 15:00
  • 3
    $\begingroup$ @Ilya: As far as I can tell, at the time this answer was posted, it was the only one that actually gave a complete, explicit description of the set in question. $\endgroup$ – cardinal Jan 5 '13 at 20:54
1
$\begingroup$

The theorem also requires that the diameter of the sets $A_n$ go to zero $$\mathrm{Diam}A_n= \sup_ { x,y \in A_n} |x-y| $$ for example take $B_n = [n,\infty)$ it is easy to verify $$\bigcap _ {i=1}^{\infty}B_n= \not 0 $$

$\endgroup$
  • $\begingroup$ Just bounded is diameter is enough. $\endgroup$ – MichalisN Jan 5 '13 at 15:40
  • $\begingroup$ @Michalis I was not talking about $\mathbb{R}$ with the usual metric but generally for a complete metric space, do you claim it suffices that the diameter is bounded for the general case? $\endgroup$ – clark Jan 5 '13 at 15:54
  • $\begingroup$ This post seems better suited as a comment than as an answer. $\endgroup$ – Jesse Madnick Jan 6 '13 at 3:03
0
$\begingroup$

By definition of intersection, a $x\in\displaystyle\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right)$ if, only if, $x\in \left(-\frac{1}{n},+\frac{1}{n}\right)$ for all $n\in\mathbb{N}\backslash\{0\}$. And $\displaystyle 0\in\left(-\frac{1}{n},+\frac{1}{n}\right)$ for all $n\in\mathbb{N}\backslash\{0\}$. Then

$$ 0\in\displaystyle\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right). $$ If $x\in \left(-\frac{1}{n},+\frac{1}{n}\right)$ for all $n\in\mathbb{N}\backslash\{0\}$ then $-\frac{1}{n}<x<+\frac{1}{n}$ for all $n\in\mathbb{N}\backslash\{0\}$. By Archimedean property or Supremum Axiom we have that $x=0$. Then $$ \{0\}=\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.