Cardinality of the set of all pairs of integers The set $S$ of all pairs of integers can be represented as $\{i \ | \ i \in \mathbb{Z} \} \times \{j\ | \ j \in \mathbb{Z}\}$. In other words, all coordinates on the cartesian plane where $x, y$ are integers.
I also know that a set is countable when $|S|\leq |\mathbb{N}^+|$. I attempted to map out a bijective function, $f : \mathbb{N}^+ \rightarrow S$.
$1 \rightarrow (1,1) \\
2 \rightarrow (1,2)\\
3 \rightarrow (1,3) \\
\quad \vdots
$
I determined from this that the natural numbers can only keep up with $(1,*)$. But there is the ordered pairs where $x=2,3,4,\cdots$ not to mention the negative integers. In other words, $|S|\geq |\mathbb{N}^+|$ and therefore $S$ is not countably infinite.
Is this correct? (I don't think it is... Something to do with my understanding of infinite sets)

Define $\sigma: \Bbb Z \times \Bbb Z \to  \Bbb Z \times \Bbb Z$ by
$$   
    \sigma(m,n) = \left\{\begin{array}{lr}
        (1,-1), & \text{for } (m,n) = (0,0)\\
        (m, n+1), & \text{for } m \gt 0 \, \land \, -m \le n \lt m\\
        (m-1, n), & \text{for } n \gt 0 \, \land \, -n \lt m \le n\\
        (m, n-1), & \text{for } m \lt 0 \, \land \, m \lt n \le -m\\
        (m+1, n), & \text{for } n \lt 0 \, \land \, n \le m \lt -n-1\\
        (m+2,n-1), & \text{for } n \lt 0 \, \land \, m = -n-1
        \end{array}\right\} 
$$
Exercise: Show that $n \mapsto \sigma^n(0,0)$ is a bijective mapping between $\{0,1,2,3,...\}$ and $\Bbb Z \times\Bbb Z$.
