Why why the natural density of the set $\{ n \mid n \equiv n_{0}\mod{m}\}$ is $\frac{1}{m}$. The natural density of a set $S$ is defined by  $\displaystyle\lim_{x \to{+}\infty}{\frac{\left | \{  n\le x \mid n\in S \} \right |}{x}}$.
This is maybe a silly question, but I got a confusion with this definition. And I really need understand why the natural density of the set $\{  n \mid n \equiv n_{0}\mod{m}\}$ is $\dfrac{1}{m}$.
Thanks!

I think you should add the condition $n\geq 0$ here. 
Let $n_{0}=pm+k$, where $0\leq k< m$. We know that $x\to\infty$. Let $x=rm+t$, where $0\leq t< m$. 
Then $\frac {n|n\equiv n_{0}\text { and } n\in S}{x}=\frac{r+1}{rm+t}$ if $t\geq k$, and $\frac {r}{rm+t}$ if $t<k$. 
In either case, as $x\to \infty$, the fractions simplify to $\frac {1}{m}$. 
