another proof for the countability of integers? 
Prove $\mathbb{Z}$ is countable.

Let $n \in \mathbb{N}$. Suppose you have a function $n-4$ so you have $N_4 = \{\mathbb{N} \cup [-4, -3, -2, -1, 0]\}$. Obviously the countable union of countable sets is countable, so $N_4 $ is too. Take $n$ to range over the set it is in, and you have a countable set.
Can we also say that anything that is discrete is countable? The real numbers need not be discrete. 
 A: A set $A$ is countable (or denumerable) if it is equipotent to $\Bbb N$, meaning there is a one to one correspondence from $A$ to $\Bbb N$. Thus, to prove $\Bbb Z$ is countable you should set up a bijection between it and $\Bbb N$. For example, say I map the negative integers as follows $-n\mapsto 2n$. Where would you map the positive integers to finish up?
A: Hint

Using the definition: a set $A$ is countable if there's a bijection $f:A\rightarrow \mathbb N$.

Prove that the map
$$f:\mathbb Z\rightarrow \mathbb N,\quad n\mapsto\left\{\begin{array}{cl}\\
2n& \text{if} \ n\geq 0\\
-2n-1& \text{if} \ n< 0\end{array}\right.$$
is a bijection.
A: There are discretely ordered sets of arbitrarily high cardinality. Just take any ordered set $A$, and the integers $\mathbb{Z}$. Form the direct product $A\times \mathbb{Z}$. For any two ordered pairs $(\alpha,m)$ and $(\beta,n)$, define $(\alpha,m)\lt (\beta,n)$ if $\alpha\lt \beta$ or $\alpha=\beta$ and $m\lt n$.  The reulting order is discrete, but $A$ can be chosen arbitrarily large.
