## Bolzano–Weierstrass for bounded functions

Need help understanding this proof please

Theorem
If $\displaystyle [f_n]$ is a pointwise bounded sequence of complex functions on a countable set $\displaystyle E$, then $\displaystyle [f_n]$ has a subsequence $\displaystyle [f_{n_k}]$ such that $\displaystyle [f_{n_k}(x)]$ converges $\displaystyle \forall{x\in{E}}$

Proof
Let $\displaystyle [x_i], i=1,2,3...$, be the points of $\displaystyle E$, arranged in a sequence. Since $\displaystyle [f_n(x_1)]$ is bounded, $\displaystyle \exists$ a subsequence, which we shall denote by $\displaystyle [f_{1,k}]$ such that $\displaystyle [f_{1,k}(x_1)]$ converges as $\displaystyle k\rightarrow\infty$

Let us now consider sequences $\displaystyle S_1,S_2,S_3,...$, which we represent by the array

$\displaystyle S_1$: $\displaystyle f_{1,1}$ $\displaystyle f_{1,2}$ $\displaystyle f_{1,3}$ $\displaystyle f_{1,4}$ ....
$\displaystyle S_2$: $\displaystyle f_{2,1}$ $\displaystyle f_{2,2}$ $\displaystyle f_{2,3}$ $\displaystyle f_{2,4}$ ....
$\displaystyle S_3$: $\displaystyle f_{3,1}$ $\displaystyle f_{3,2}$ $\displaystyle f_{3,3}$ $\displaystyle f_{3,4}$ ....
.................................

and which have the following properties:
(a) $\displaystyle S_n$ is a subsequence of $\displaystyle S_{n-1}$ for $\displaystyle n=2,3,4,...$

(b) $\displaystyle [f_{n,k}(x_n)]$ converges, as $\displaystyle k\rightarrow\infty$ (the boundedness of $\displaystyle [f_n(x_n)]$ makes it possible to choose $\displaystyle S_n$ in this way)

(c) The order in which the functions appear is the same in each sequence; i.e, if one function precedes another in $\displaystyle S_1$, they are in the same relation in every $\displaystyle S_n$, until one or the other is deleted. Hence, when going from one row in the above array to the next below, functions may move to the left but never to the right.

We now go down the diagonal of the array; i.e, we consider the sequence

$\displaystyle S$: $\displaystyle f_{1,1}$ $\displaystyle f_{2,2}$ $\displaystyle f_{3,3}$ $\displaystyle f_{4,4}$

By (c), the sequence $\displaystyle S$ (except possibly its first $\displaystyle n-1$ terms) is a subsequence of $\displaystyle S_n$, for $\displaystyle n=1,2,3,..$ Hece (b) implies that $\displaystyle [f_{n,n}(x_i)]$ converges,, as $\displaystyle n\rightarrow\infty$, $\displaystyle \forall{x_i}\in{E}$

QED

Questions
1) For $\displaystyle S_m$: $\displaystyle f_{m,1}$ $\displaystyle f_{m,2}$ $\displaystyle f_{m,3}$ $\displaystyle f_{m,4}$ ....
Are the terms of this sequence, S_m, the terms of the sequence $\displaystyle [f_{m,k}]$, the convergent subsequence of the sequence $\displaystyle [f_n(x_m)]$?

If question 1 is true

2) Regards to property (a), is this a property intrinsic to an array of sequences constructed in this manner; i.e an array of sequences made of the terms of $\displaystyle [f_{n,k}]$, or must the array be specially constructed to have this property?
If so, how do I guarantee this can always be done?

3) Would't property (a) imply $\displaystyle f_{n,j}=f_{n-1,p}$ for all $\displaystyle j=1,2,3...$ and some $\displaystyle p\in{\mathbb{N}}$?