Prove that ⋃ n ∈ N ( − n , 1 n ] = (...

Step 1
Show mutual inclusion.
First suppose that $x\in \bigcup _{n\in \mathbb{N}}(-n,\frac{1}{n}]$. Choose $n\in \mathbb{N}$ such that $x\in (-n,\frac{1}{n}]$. Since $x\le \frac{1}{n}\le 1$, $x\in (-\mathrm{\infty },1]$. This shows that $\bigcup _{n\in \mathbb{N}}(-n,\frac{1}{n}]\subseteq (-\mathrm{\infty },1]$.
Step 2
Conversely, suppose that $x\in (-\mathrm{\infty },1]$. If $0\le x\le 1$, we may choose $n\in \mathbb{N}$ such that $x\le \frac{1}{n}$. Then $x\in (-n,\frac{1}{n}]$, so $x\in \bigcup _{n\in \mathbb{N}}(-n,\frac{1}{n}]$. On the other hand, if $x<0$, there exists $n\in \mathbb{N}$ such that $x>-n$. $x<0<\frac{1}{n}$, so $x\in (-n,\frac{1}{n}]$, so $x\bigcup _{n\in \mathbb{N}}(-n,\frac{1}{n}]$. This shows that $\bigcup _{n\in \mathbb{N}}(-n,\frac{1}{n}]\subseteq (-\mathrm{\infty },1]$.

Step 1
$x\in \bigcup (-n,\frac{1}{n}]\to \mathrm{\exists }n\in \mathbb{N}\text{s.t.}x\in (-n,\frac{1}{n}]$
as $n\in \mathbb{N}$ and as $-n>-\mathrm{\infty }$ then $x\in (-\mathrm{\infty },1]$
Step 2
For the second, we know that for every number a there exists number $b>a$ such that $b\in \mathbb{N}$, in the same way it works backwards, so that for every a there exists $b<a$ such that $b\in \mathbb{Z}$.
Step 3
If $0\le x\le 1$ we take $n=1$, else we take $n=y$ where y is a smaller number than x that is in $\mathbb{N}$ and satisfies that $y<x$, so we get that again $x\in (-y,\frac{1}{y}]$.