How can I to prove the equality of intervals of open intervals with the equality of the closed interval

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.

• Prove that $[0,1]$ is included in the intersection, and no $-\epsilon$ nor $1+\epsilon$.
– user65203
Oct 19 '18 at 10:16

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 $$x \in \bigcap_{n=1}^{\infty} \left(\frac{-1}n ; 1+\frac{1}n\right)$$ .

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 $$Q\notin (\frac{-1}n ; 1+\frac{1}n)$$

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

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