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How can I to prove the equality between the following intersection of of open intervals with the following closed interval?

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

Thank you.

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    $\begingroup$ Prove that $[0,1]$ is included in the intersection, and no $-\epsilon$ nor $1+\epsilon$. $\endgroup$
    – user65203
    Oct 19 '18 at 10:16
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  1. If $x \in \bigcap_{n=1}^{\infty} \left(\frac{-1}n ; 1+\frac{1}n\right)$, then

$\frac{-1}n < x <1+\frac{1}n$ for all $n$. With $ n \to \infty$ we get $0 \le x \le 1$.

  1. If $0 \le x \le 1$, then $\frac{-1}n<0 \le x \le 1 <1+\frac{1}n$ for all $n$, hence $x \in \bigcap_{n=1}^{\infty} \left(\frac{-1}n ; 1+\frac{1}n\right)$ .
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Use double inclusion, clearly $[0,1]$ is contained in the intersetion since each point $P\in [0,1]$ lies in each open interval $(\frac{-1}n ; 1+\frac{1}n)$. For the other inclusion, pick a point $Q$ outside $[0,1]$ and you need find some $n$ such that $Q\notin (\frac{-1}n ; 1+\frac{1}n)$

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For all $n$,

$$\left(-\frac1n ; 1+\frac{1}n\right) \cap [0,1]=[0,1]$$

and for any positive $\epsilon$, if $n>\dfrac1\epsilon$,

$$0-\epsilon\notin\left(-\frac1n ; 1+\frac{1}n\right)$$

and

$$1+\epsilon\notin\left(-\frac1n ; 1+\frac{1}n\right).$$

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