Need help understanding this proof please
Theorem
If is a pointwise bounded sequence of complex functions on a countable set , then has a subsequence such that converges
Proof
Let , be the points of , arranged in a sequence. Since is bounded, a subsequence, which we shall denote by such that converges as
Let us now consider sequences , which we represent by the array
and which have the following properties:
(a) is a subsequence of for
(b) converges, as (the boundedness of makes it possible to choose 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 , they are in the same relation in every , 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
By (c), the sequence (except possibly its first terms) is a subsequence of , for Hece (b) implies that converges,, as ,
QED
Questions
1) For : ....
Are the terms of this sequence, S_m, the terms of the sequence , the convergent subsequence of the sequence ?
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 , 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 for all and some ?