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Why is: $$\bigcap_{n=0}^\infty\,\mathopen{]}0,e^{-n}\mathclose{[}\,=\emptyset\quad?$$ Indeed, $$\forall n\in\mathbb N, 0\in\mathopen{]}0,e^{-n}\mathclose{[},$$ and thus $0\in\bigcap_{n=0}^\infty\,\mathopen{]}0,e^{-n}\mathclose{[}$. So, what's wrong here ?

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    $\begingroup$ You are misreading the notation $]a,b[$. The reversed brackets indicate the endpoints are excluded, synonymous with $(a,b)$. $\endgroup$ – Erick Wong Aug 24 '15 at 8:49
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    $\begingroup$ I think $(a,b)$ is also preferable, because $]a,b[$ looks a bit messy somehow. $\endgroup$ – Matias Heikkilä Aug 24 '15 at 8:52
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    $\begingroup$ I've always used this notation, how is it messy? $\endgroup$ – Augustin Aug 24 '15 at 9:05
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    $\begingroup$ $]a,b[$ is a standard notation for French people. $\endgroup$ – Gabriel Romon Aug 24 '15 at 9:12
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    $\begingroup$ This notation was introduced by Bourbaki and became standard in France. See math.stackexchange.com/questions/430851/notation-for-intervals $\endgroup$ – Augustin Aug 24 '15 at 9:26
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$$\forall n\in\mathbb N, 0\in ]0,e^{-n}[$$

This is wrong. $0\notin ]0,e^{-n}[$.

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Hint: Try to show that if $\bigcap_{n=0}^\infty\neq \emptyset$ which means $\exists x\in\mathbb{R},\,x\in\bigcap_{n=0}^\infty]0,\,e^{-n}[$ then $\exists m\in\mathbb{N},\,e^{-m}<x$ and so $x\notin]0,\,e^{-m}[$ which leads to a contradiction. Note that since $x\in\bigcap_{n=0}^\infty]0,\,e^{-n}[$ then in particular $x\in ]0,1[$ and so $x>0$. Use the fact that $\lim\limits_{n\to +\infty}e^{-n}=0$.

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  • $\begingroup$ This answer actually does not adress the actual problem OP had. I.e., OP is confused why $0\in]0,e^{-n}[$ $\endgroup$ – 5xum Aug 24 '15 at 9:07
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    $\begingroup$ @5xum You are indeed right, however one can see it as an interesting complement to Augustin's answer. $\endgroup$ – Surb Aug 24 '15 at 9:10
  • $\begingroup$ @5xum I wrote this after Augustin's answer who already explained why $0$ isn't in that set. But here I explain why this set is $\emptyset$. $\endgroup$ – Scientifica Aug 24 '15 at 9:11

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