Prove that union of countable sets is countable

Question

Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.

Here are the relevant definitions and theorem.

Answer 1

Following the arrows you get a surjective function $f:\mathbb N\to\bigcup_{n=1}^\infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function $f:T\subseteq\mathbb N\to\bigcup_{n=1}^\infty E_n$.

Answer 2

What you have in (17) is a function $g:\mathbb N\times\mathbb N\to S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.

Because $g$ is surjective, for each $s\in S$ there exists $(x_1,y_1)\in\mathbb N\times\mathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let $$ T=\{(x_s,y_s):\ s\in S\}. $$ Then $S\sim T$ and $T\subset \mathbb N\times\mathbb N$. So $T$ is countable and $S$ is countable.

Answer 3

The sequence is a function from $\mathbb{N} \rightarrow$ S, where $i \mapsto x_i$. This function is surjective by construction. Since there may be double counting, there is a subset T $\subset \mathbb{N}$ such that T $\sim$ S.