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!

It is relatively easy to show that if you have an infinite set $A$ and you write it in an increasing order, i.e.,
$$A=\{a_1<a_2<\dots<a_k<\dots\}$$
then
$$d(A)=\lim\limits_{k\to\infty} \frac{k}{a_k}.$$
If you apply this to your case, you get the required density. (This fact is also mentioned in the Wikipedia article.)
