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,
Best Answer
It is entirely possible to create such sequences. As to the reason, here it is.
Consider the sequence
$f_{1,1}(x_2),f_{1,2}(x_2),\ldots$
This is a bounded sequence since the functions are pointwise bounded.
Thus, it contains a convergent subsequence, $f_{1,n_i}(x_2)$
This subsequence of functions, we denote by $f_{2,i}$.
You can go through such a process repeatedly to get $S_n\subset S_{n-1}$ with the required properties.