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 $\ds S = \bigcup_{i \mathop \in \N} {S_i}$.
Assume that none of these sets have any elements in common.
Otherwise, we can consider the sets $S_0' = S_0, S_1' = S_1 \setminus S_0, S_2' = S_2 \setminus \paren {S_0 \cup S_1}, \ldots$
All of these are countable by Subset of Countable Set is Countable, and they have the same union $\ds S = \bigcup_{i \mathop \in \N} {S_i'}$.
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 $\ds S = \bigcup_{i \mathop \in \N} {S_i}$.
Furthermore, we have that $\phi: S \to \N \times \N, a_{ij} \mapsto \tuple {i, j}$ 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.
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Sources
- 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 2.2$: Countable sets: Theorem $2$