Countable Union of Countable Sets is Countable/Proof 1

Theorem
Let the axiom of countable choice be accepted.

Then it follows that a countable union of countable sets is countable.

Proof
Let $\left\langle{S_n}\right\rangle_{n \in \N}$ be a sequence of countable sets.

Define:
 * $\displaystyle S = \bigcup_{n \mathop \in \N} S_n$

For all $n \in \N$, let $\mathcal F_n$ denote the set of all injections from $S_n$ to $\N$.

Since $S_n$ is countable, $\mathcal F_n$ is non-empty.

Using the axiom of countable choice, there exists a sequence $\left\langle{f_n}\right\rangle_{n \in \N}$ such that $f_n \in \mathcal F_n$ for all $n \in \N$.

Let $\phi: S \to \N \times \N$, where $\times$ denotes the cartesian product, be the mapping defined by:
 * $\phi \left({x}\right) = \left({n, f_n \left({x}\right)}\right)$

where $n$ is the (unique) smallest natural number such that $x \in S_n$.

From the Well-Ordering Principle, such an $n$ exists; hence, the mapping $\phi$ exists.

Since each $f_n$ is an injection, it (trivially) follows that $\phi$ is an injection.

Since $\N \times \N$ is countable, there exists an injection $\alpha: \N \times \N \to \N$.

From Composite of Injections is Injection, the mapping $\alpha \circ \phi: S \to \N$ is an injection.

Hence, $S$ is countable.