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I want to find the infimum of the set:

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

Would this infimum exist and be equal to $1$ or would it not exist, and why?

My intuition is going nuts because of the infinite intersection and I'm having doubts towards both answers.

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    $\begingroup$ What is your definition of infimum? $\endgroup$
    – Sean
    Oct 21, 2015 at 23:25
  • $\begingroup$ @Sean For a non-empty subset $S$ of the real numbers, the infimum is the greatest lower bound of $S$. Would this in any way tie into the fact that this infinite intersection produces the empty set? $\endgroup$
    – Areyouheaty
    Oct 21, 2015 at 23:28
  • $\begingroup$ Note: $n$ is not the bound variable of the conjunction series; that is $i$. Is this a typo? $\endgroup$
    – Graham Kemp
    Oct 21, 2015 at 23:30
  • $\begingroup$ @GrahamKemp Yes, thank you. $\endgroup$
    – Areyouheaty
    Oct 21, 2015 at 23:31

2 Answers 2

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Note $I:=\bigcap_{n=1}^\infty \left(1-\frac{1}{n},1+\frac{1}{n}\right)=\{1\}$

Clearly 1 belongs to all $\left(1-\frac{1}{n},1+\frac{1}{n}\right)$, so $1\in I$. On the other hand, for all $a<1$, there is some $n$ such that $a<1-\frac{1}{n}$. Then $a\notin\left(1-\frac{1}{n},1+\frac{1}{n}\right)$, so $a\notin I$. For the same reason, for all $a>1$, $a\notin I$

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    $\begingroup$ It seems I have misunderstood the idea of infinite unions/intersections. Thank you though! $\endgroup$
    – Areyouheaty
    Oct 21, 2015 at 23:47
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$\bigcap_{i=1}^\infty (1-\frac{1}{n},1+\frac{1}{n})=1$.It is clear that $1$ belong the intersection. If $x>1\Rightarrow \epsilon =x-1>o$ then since $\frac{1}{n}\to 0$ there is $n_0$ such that $\frac{1}{n_0}<x-1\Rightarrow 1+\frac{1}{n_0}<x$ so $x\notin (1-\frac{1}{n_0},1+\frac{1}{n_0}) $ similarly for $x<1$.

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