If $S_n=\left[-\frac{n}{n+1},\frac{n}{n+1}\right]$ then $\bigcup\limits_{n\geq1}S_n=(-1,1)$ I was self reading Mathmatics for Economists by Simon and Blume.

Consider the closed sets $S_n=\left[-\frac{n}{n+1},\frac{n}{n+1}\right]$ for $n\geq1,n\in\mathbb N$. Then $$\bigcup_{n\geq1}S_n=(-1,1),$$ is an open interval. 

How to show that $\bigcup\limits_{n\geq1}S_n=(-1,1)$?

If $-1 < x < 1$, then $\exists N_1 \in \mathbb{N}$ such that
$$
x < 1-1/N_1
$$
and $N_2 \in \mathbb{N}$ such that
$$
x > -1 + 1/N_2
$$
Now take $n = \max\{N_1, N_2\}$. Then,
$$
-1 + 1/n < x < 1-1/n \Rightarrow x \in S_{n-1}
$$
Hence,
$$
(-1,1) \subset \bigcup_{n\geq 1} S_n
$$
Now $S_n \subset (-1,1)$ for all $n$, and hence $\supset$ also holds.
