Countable Union of Countable Sets is Countable/Proof 2

Theorem
The union of a countable number of countably infinite sets is countable.

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

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

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

All of these are countable by the fact that they are subsets of countable sets, and they have the same union $\displaystyle S = \bigcup_{i \in \N} {S_i^\prime}$.

Now we write the elements of $S_0', S_1', S_2', \ldots$ in the form of an 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 $\displaystyle S = \bigcup_{i \in \N} {S_i}$.

Now we can count the elements of $S$ by processing the table diagonally. First we pick $a_{00}$. Then we pick $a_{01}, a_{10}$. Then we pick $a_{02}, a_{11}, a_{20}$.

We can see that all the elements of $S$ will (eventually) be listed, and there is a specific number (element of $\N$) to index each of its elements with.

Thus we have the required one-to-one correspondence between $S$ and $\N$, and our assertion is proved.