Bessel's inequality in Rudin's RCA

Question

I am reading Theorem 4.14 on Rudin's RCA. The assertion is here:
Let $\{u_\alpha : \alpha \in A\}$ be an orthonormal set in a Hilbert space $H$, and let $P$ be the space of all finite linear combinations of the vectors $u_\alpha$.
Then the inequality $$ \sum_{\alpha \in A} |<x, u_\alpha>|^2\leq ||x||^2$$ holds for every $x\in H$.
Rudin said that since we have checked that inequality holds if $A$ is finite. I have understood that inequality holds if $A$ is finite but I don't know why this inequality holds if $A$ is countable and if $A$ is uncountable.

Answer 1

By definition if $I$ is a non-empty set and $(a_i)_{i \in I}$ is a collection of non-negative numbers indexed by $I$ then $\sum_{i\in I} a_i $ is the supremum of all finite sums of $a_i$'s. So if the inequality holds when $A$ is replaced by any finite subset of $A$ then it holds as stated.

Answer 2

This is almost a comment given Kavi Rama Murthy's answer - however!

First of all, there used to be an argument - almost as a side-comment - in the book that 'uncountable' is not an issue. I haven't checked whether it's still there in the latest edition - but here it is, using Kavi Rama Murthy's notation:

Say that the supremum $M = \sum_{i \in I} a_i$ is finite ($a_i$ positive). Then consider the set $I_n=\{i\mid a_i > 1/n\}$, and write $I_\infty=\{i \mid a_i \not= 0\}= \cup I_n$. Then $I_\infty$ is countable, as $I_n$ must be finite, since $$(\# I_n/ n) \le \sum_{i \in I_n} a_i \le M.$$

That said, don't forget that, algebraically speaking, addition is a finite operation - one cannot 'add' infinitely many terms. In all cases, when one writes an infinite sum, one means a limit of finite sums. In particular, here, with $a_i$ positive, the expression $$M=\sum_{i \in I} a_i $$ is, by definition, the supremum of all sums with a finite number of summands - taken over $i\in I$. (Think Lebesgue integration...) Therefore if Rudin's inequality holds for all finite sums, it holds for $M$.