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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.

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    $\begingroup$ I am 100% certain that this question was asked at least twice before. $\endgroup$
    – Asaf Karagila
    Jan 5 '13 at 14:21
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    $\begingroup$ Yes and I am searching for that link, Asaf. $\endgroup$
    – Mikasa
    Jan 5 '13 at 14:24
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    $\begingroup$ 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/… $\endgroup$ Jan 8 '16 at 12:40
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No. It is not empty. Since $0\in (-1/n, 1/n)$ for all $n$, so $0\in\bigcap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$.

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  • $\begingroup$ 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
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    $\begingroup$ @Tai: When we take $n\to\infty$, we can no longer ensure that the inequalities remain strict, so we get $0\leq x\leq 0$. $\endgroup$ Jan 5 '13 at 14:29
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    $\begingroup$ oh,I make a mistake, when let $n \rightarrow +\infty$, $0\leq x \leq 0$ $\endgroup$
    – Laura
    Jan 5 '13 at 14:30
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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 $$\bigcap_{n=1}^\infty\left(0,\frac{1}{n}\right).$$

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  • $\begingroup$ 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
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    $\begingroup$ +1 for explaining how the problem should be solved as well. $\endgroup$
    – Joe Z.
    Jan 5 '13 at 16:16
  • $\begingroup$ +1 nice approach. $\endgroup$
    – Mikasa
    Feb 3 '13 at 3:26
  • $\begingroup$ @BabakSorouh: Thanks! (And for all the +1's :) ) $\endgroup$
    – Clayton
    Feb 3 '13 at 3:27
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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.

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

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

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

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

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    $\begingroup$ What does it add to the answers that are already given? $\endgroup$
    – Ilya
    Jan 5 '13 at 15:00
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    $\begingroup$ @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
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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 $$\bigcap _ {i=1}^{\infty}B_n= \not 0 $$

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  • $\begingroup$ 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. $\endgroup$ Jan 6 '13 at 3:03
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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 $$ \{0\}=\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right) $$

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