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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 ?

MickG
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idm
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    You are misreading the notation $]a,b[$. The reversed brackets indicate the endpoints are excluded, synonymous with $(a,b)$. – Erick Wong Aug 24 '15 at 08:49
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    I think $(a,b)$ is also preferable, because $]a,b[$ looks a bit messy somehow. – Matias Heikkilä Aug 24 '15 at 08:52
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    I've always used this notation, how is it messy? – Augustin Aug 24 '15 at 09:05
  • I also dislike the notation $]a,b[$. I think it probably looks "messy" because I'm not used to seeing anything bracketed with the brackets "pointing outwards". Similarly, despite it not having a mathematical meaning (at least not to me) $)a,b($ looks weird. – lokodiz Aug 24 '15 at 09:12
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    $]a,b[$ is a standard notation for French people. – Gabriel Romon Aug 24 '15 at 09:12
  • Unluckily, here in Italy many people use $]a,b[$ too. I prefer the notation $(a,b)$, but I do not believe that the former is "messy". – Siminore Aug 24 '15 at 09:20
  • As someone who had never encountered reversed-bracket notation before, my immediate thought was that $]a, b[$ must mean the complement of $(a, b)$: that is, it represented $(-\infty, a] \cup [b, \infty)$. (I am now aware that the notation is standard in places.) – Patrick Stevens Aug 24 '15 at 09:23
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    This notation was introduced by Bourbaki and became standard in France. See http://math.stackexchange.com/questions/430851/notation-for-intervals – Augustin Aug 24 '15 at 09:26

2 Answers2

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$$\forall n\in\mathbb N, 0\in ]0,e^{-n}[$$

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

Augustin
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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 $\lim\limits_{n\to +\infty}e^{-n}=0$.

Scientifica
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  • This answer actually does not adress the actual problem OP had. I.e., OP is confused why $0\in]0,e^{-n}[$ – 5xum Aug 24 '15 at 09:07
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    @5xum You are indeed right, however one can see it as an interesting complement to Augustin's answer. – Surb Aug 24 '15 at 09:10
  • @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 $\emptyset$. – Scientifica Aug 24 '15 at 09:11