Prove that if $A$, $B$ are countable, then $A \times B$ is countable? Is $A\times B$ referring to the axis here? So an $X$ and $Y$ coordinate plane?
$A$ is countable, therefore a bijection occurs from $A \rightarrow \mathbb{N}$.
$B$ is countable, therefore a bijection occurs from $B \rightarrow \mathbb{N}$.
If these statements hold true then any elements belonging to the sets can be linked and ordered one-by-one from any natural number, thus any coordinate $(a,b)$ would be countable as the elements it represents ($x$ and $y $ coordinates) are countable. Together they preserve their nature of cardinality.
Am I wrong?

Let $f:\mathbb{N} \to A, g:\mathbb{N} \to B$ be the two bijections, and define $h: \mathbb{N}\times \mathbb{N} \to A\times B$ as $h(m,n) = (f(m),g(n))$. You can show easily that $h$ is bijective. Now you need to show there is bijection from $\mathbb{N}$ to $\mathbb{N}\times \mathbb{N}$. We define $k: \mathbb{N}\times \mathbb{N} \to \mathbb{N}$ as $k(m,n) = (2m-1)\cdot 2^{n-1}$. Check that $k$ is bijective, thus $k^{-1}$ is also bijective.Thus the function $l = h\circ k^{-1}$ is a bijection from $\mathbb{N}$ to $A\times B$, hence $A\times B$ is countable.
