What is the result of $\bigcap_{n=1}^{\infty}{(-1/n; 1/n)}$ I would like to know the intersection of $(-1/n ; 1/n), \forall n \in N$. I am in trouble thinking it could be $\{0\}$ or $\emptyset$. Can anyone help me?
 A: Actually you are right with your first suspicion:
$$
\{0\} = \bigcap_{n=1}^\infty (-1/n, 1/n)
$$
We prove this the standard way by showing "$\subset$" and "$\supset$"
"$\subset$":
For every $n \in \mathbb{N}$ $0 \in (-1/n, 1/n)$. 
"$\supset$":
Let $x \in \bigcap_{n=1}^\infty (-1/n, 1/n)$. This means that $x \in (-1/n, 1/n)$ for all $n \in \mathbb{N}$. This are actually two inequalities:
$\tag{1} x > -1/n$
and
$\tag{2} x < 1/n.$
Both hold for all $n \in \mathbb{N}$. Taking the limit $\lim_{n \to \infty}$ shows
$x \geq 0$ and $x \leq 0$ which proves $x = 0$.
A: $$x\in \bigcap_{n\geq 1}\left(\frac{-1}{n},\frac{1}{n}\right)\iff \forall n, |x|<\frac{1}{n}\underset{(*)}{\iff} \forall \varepsilon>0, |x|<\varepsilon\iff x=0$$
The way $\Leftarrow$ of $(*)$ is maybe less obvious. Let's prove it. Let $\varepsilon>0$. Since $\frac{1}{n}\underset{n\to\infty }{\longrightarrow }0$, there is a $N$ s.t. $|x|<\frac{1}{N}\leq \varepsilon$. Therefore $|x|<\varepsilon$ for all $\varepsilon>0$ (since $x$ doesn't depend on $N$).
