Is $\bigcup_{n=1}^{\infty}\left ( -1+\frac{1}{n},1-\frac{1}{n} \right )$ open? This .pdf on Example 2 (page 4 on paper), it says that $$\bigcup_{n=1}^{\infty}\left ( -1+\frac{1}{n},1-\frac{1}{n} \right )=\{0\}\cup (-1/2,1/2)\cup\dots=(-1,1)$$
is open. Please check Theorem 1 on page 3. How can this Example be open, if $\{0\}$ is closed?
 A: There are two points here:


*

*$[0,1]\cup(-2,2)=(-2,2)$ is an open interval, which is open. The union of a closed set and an open set can be open, or closed, or neither (or both!).

*For $n=1$, you have $(0,0)=\varnothing$, and not $\{0\}$.
In any case, $(-1,1)$ is an open interval, the fact you could write it as the union of some sets which may or may not be open is meaningless to the fact that the interval is open.
A: Since $\left ( -1+\frac{1}{n},1-\frac{1}{n} \right)$ is open for any $n$,
$\bigcup\limits_{n=1}^{\infty}\left ( -1+\frac{1}{n},1-\frac{1}{n} \right )$ is open because union of open sets is always open (Theorem 1).
Clearly
$$
\bigcup\limits_{n=1}^{\infty}\left( -1+\frac{1}{n},1-\frac{1}{n} \right)\subset (-1,1)
$$
On the other hand, for any $x\in(-1,1)$, there is a $N$ such that $x<1/N$. So 
$$
(-1,1)\subset\bigcup\limits_{n=1}^{\infty}\left( -1+\frac{1}{n},1-\frac{1}{n} \right)
$$
This means
$$
\bigcup\limits_{n=1}^{\infty}\left( -1+\frac{1}{n},1-\frac{1}{n} \right)= (-1,1)
$$
Also since $0\in \left ( -1+\frac{1}{n},1-\frac{1}{n} \right)$ for any $n$ 
$$  0\in \bigcup_{n=1}^{\infty}\left ( -1+\frac{1}{n},1-\frac{1}{n} \right)=(-1,1)$$
 And thus 
$$\{0\}\cup (-1,1)=(-1,1)$$
