TITLE: Rudin's proof theorem 7.23
QUESTION [8 upvotes]: I'm totally lost on that proof. Recall, the theorem is the following:

If $(f_n)_{n \in \mathbb{N}}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $(f_n)_{n \in \mathbb{N}}$ has a subsequence $(f_{n_k})_{k \in \mathbb{N}}$ such that $(f_{n_k}(x))_{k \in \mathbb{N}}$ converges for each $x \in E$.

I understand the beginning of the proof, i.e. defining $f_{1, k}$ a subsequence of functions s.t. $\lim_{k \rightarrow + \infty}f_{1, k}(x_1)$ exists. But once he defines his matrix of functions, I can't understand the reasons of the properties he mentioned.
Recall:
\begin{matrix}
   S_1: & f_{1, 1} & f_{1, 2} & \ldots \\
   S_2: & f_{2, 1} & f_{2, 2} & \ldots \\
   \vdots \\
   S_n: & f_{n, 1} & f_{n, 2} & \ldots \\
\end{matrix}
Why $S_n$ should be a subsequence of $S_{n-1}$? I can't get the point...
Thanks,

REPLY [2 votes]: Since $E$ is countable, we can rearrange $E = \{x_i\}_{i=1}^{\infty}$. Fix $x_1$. Then $\{f_n(x_1)\}$ is a bounded sequence, since $\{f_n\}$ is pointwise bounded. Thus, it has a subsequence $\{f_{n_k}(x_1)\}$ that is convergent. In addition, consider the sequence of functions $\{f_{n_k}\}$. Itself is also a pointwise bounded sequence. Thus, if we evaluate $\{f_{n_k}\}$ at another point say $x_2\in E$. Then $\{f_{n_k}(x_2)\}$ is also a bounded sequence, so that it has another convergent subsequence $\{f_{n_{k_\ell}}(x_2) \}$ of the subsequence $\{f_{n_k}(x_2)\}$. Therefore, $\{f_{n_{k_\ell}}\}$ is not only convergent sequence when evaluating at $x_2$ but also a subsequence of $\{f_{n_k}\}$. Note that since $\{f_{n_k}(x_1)\}$ is convergent, then every subsequence of $\{f_{n_k}(x_1)\}$ is convergent. Therefore, $\{f_{n_{k_\ell}}(x_1)\}$ is also convergent. Let $S_1 = \{f_{n_k}(x_1)\}$ and $S_2 = \{f_{n_{k_\ell}}(x_2)\}$ and so on.