# Countable Union of Countable Sets is Countable/Informal Proof

## Theorem

Let the axiom of countable choice be accepted.

Then it can be proved that a countable union of countable sets is countable.

## Proof

Consider the countable sets $S_0, S_1, S_2, \ldots$ where $\displaystyle S = \bigcup_{i \mathop \in \N} {S_i}$.

Assume that none of these sets have any elements in common.

Otherwise, we can consider the sets $S_0' = S_0, S_1' = S_1 \setminus S_0, S_2' = S_2 \setminus \left({S_0 \cup S_1}\right), \ldots$

All of these are countable by Subset of Countable Set is Countable, and they have the same union $\displaystyle S = \bigcup_{i \mathop \in \N} {S_i'}$.

Now we write the elements of $S_0', S_1', S_2', \ldots$ in the form of a (possibly infinite) table:

- $\begin{array} {*{4}c} {a_{00}} & {a_{01}} & {a_{02}} & \cdots \\ {a_{10}} & {a_{11}} & {a_{12}} & \cdots \\ {a_{20}} & {a_{21}} & {a_{22}} & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}$

where $a_{ij}$ is the $j$th element of set $S_i$.

This table clearly contains all the elements of $\displaystyle S = \bigcup_{i \mathop \in \N} {S_i}$.

Furthermore, we have that $\phi: S \to \N \times \N, a_{ij} \mapsto \left({i, j}\right)$ is an injection.

By Cartesian Product of Countable Sets is Countable, there exists an injection $\psi: \N \times \N \to \N$.

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

That is to say, $S$ is countable.

$\blacksquare$

## Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice, by way of [[**Choosing infinitely many enumerations (one for each $S_i$)**]].

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

## Sources

- 1968: A.N. Kolmogorov and S.V. Fomin:
*Introductory Real Analysis*... (previous) ... (next): $\S 2.2$: Countable sets: Theorem $2$