Union of Countable Sets of Sets/Proof 1

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 $\mathcal A$ is countable, its contents can be arranged in a sequence:
 * $\mathcal A = \left\{{A_1, A_2, \ldots}\right\}$

Let $B \in \mathcal B$.

Consider the sequence of sets:
 * $\left \langle A_1 \cup B, A_2 \cup B, \ldots \right \rangle$

We may leave out any possible repetitions, and obtain a countable set:
 * $\left({A \cup B: A \in \mathcal A}\right\}$

for every $B \in \mathcal B$.

Thus as $B$ varies over all the elements of $\mathcal B$, we obtain the countable family:
 * $\left\{{A \cup B: A \in \mathcal A}\right\}_{\left({B \in \mathcal B}\right)}$

From Countable Union of Countable Sets is Countable, their union is countable.