Union of Countable Sets of Sets/Proof 3

Theorem
Let $\mathcal A$ and $\mathcal B$ be countable sets of sets.

Then:
 * $\left\{{A \cup B: A \in \mathcal A, B \in \mathcal B}\right\}$

is also countable.

Proof
Since $\mathcal A$ and $\mathcal B$ are countable, $\mathcal A \times \mathcal B$ is countable by Cartesian Product of Countable Sets is Countable.

Thus by Surjection from Natural Numbers iff Countable there is a surjection $f: \N \to \mathcal A \times \mathcal B$.

Let $\mathcal C = \left\{{A \cup B: A \in \mathcal A, B \in \mathcal B}\right\}$.

Let $g: \mathcal A \times \mathcal B \to \mathcal C$ be defined by letting $g \left({A, B}\right) = A \cup B$.

$g$ is a surjection:

Let $C \in \mathcal C$.

Then there exist $A \in \mathcal A$ and $B \in \mathcal B$ such that $C = A \cup B$.

By the definition of Cartesian product:
 * $\left({A, B}\right) \in \mathcal A \times \mathcal B$

Then:
 * $g \left({A, B}\right) = A \cup B = C$

It follows that $g$ is a surjection.

By Composite of Surjections is Surjection, $g \circ f: \N \to \mathcal C$ is surjective.

Hence $\mathcal C$ is countable.