Union of Countable Sets of Sets/Proof 2

Proof
Both $\mathcal A$ and $\mathcal B$ are countable.

Enumerate their elements by the sequences $\left({A_n}\right)_{n \mathop \in \N}$ and $\left({B_n}\right)_{n \mathop \in \N}$ respectively.

When one of $\mathcal A$ and $\mathcal B$ would be finite, achieve such sequences by allowing elements to occur multiple times.

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

For each $C \in \mathcal C$, define the set $N_C := \left\{{ \left({i, j}\right) \in \N \times \N: A_i \cup B_j = C}\right\}$.

The definition of $\mathcal C$ ensures that $N_C \ne \varnothing$ for all $C \in \mathcal C$.

Consider the lexicographic ordering $\preceq$ on $\N \times \N$.

Combining the Well-Ordering Principle with Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering, $\preceq$ is a well-ordering on $\N \times \N$.

As $N_C \subseteq \N \times \N$ is nonempty, it thus has a smallest element; call it $n_C$.

Define now the map $\phi: \mathcal C \to \N \times \N$ by $\phi \left({C}\right) = n_C$.

Well-Ordering Minimal Elements are Unique ensures that $\phi$ is well-defined.

Now suppose that $C_1, C_2 \in \mathcal C$ satisfy:


 * $\phi \left({C_1}\right) = \phi \left({C_2}\right)$

that is, $n_{C_1} = n_{C_2}$, which in turn equal some $\left({i, j}\right) \in \N \times \N$.

As $\left({i, j}\right) \in N_{C_1}$ and $\left({i, j}\right) \in N_{C_2}$, $C_1 = A_i \cup B_j = C_2$.

It follows that $\phi$ is an injection.

Finally, $\N \times \N$ is countable by Cartesian Product of Countable Sets is Countable.

Let $\psi: \N \times \N \to \N$ be an injection.

Then $\psi \circ \phi: \mathcal C \to \N$ is also an injection by Composite of Injections is Injection.

Hence $\mathcal C$ is countable.