I am currently reading the chapter on Hilbert spaces in the book Functional Analysis by Peter D. Lax and I am confused about whether the orthonormal sets being alluded to are uncountable or not.
Specifically he states the following theorem
Theorem 9. Every Hilbert space contains an orthonormal basis.
The proof of this theorem never refers to whether the set obtained via an application of Zorn's lemma is countable or not. He then goes on to state that:
Suppose that $H$ is a separable Hilbert space; that is, it contains a denumerable set of points that is dense. In this case every orthogonal basis is denumerable, and the basis elements can be constructed without appealing to transcendental arguments such as Zorn's lemma...
so i assume that the theorem is valid for any Hilbert space. But then if $\{x_j\}$ is some orthonormal set in a Hilbert space then is it true, as is claimed in the book, that $\overline{\mathrm{span }\{x_j\}} = S$ where $S$ is the set of points \begin{align*} x = \sum (x,x_j)x_j,\quad \sum |(x,x_j)|^2<\infty. \end{align*} Are we to consider this as a transfinite sum?
Could we go about proving this in the following way:
Suppose that $\{x_j\}$ is uncountable and let $y\in \overline{\mathrm{span}\{x_j\}}$ then we could find a sequence $\{y_n\}$ together with an increasing sequence of finite index sets $\{F_n\}$ and scalars $\{c_{n,j}\}$ such that
$$y_n = \sum_{j\in F_n}c_{n,j}x_j$$
and $\|y_n-y\|\rightarrow 0$. By Pythagoras
$$\|y-y_n\|\geq \|y-\sum_{j\in F_n}(y,x_j)x_j\|$$
and therefore $\sum_{j\in F_n}(y,x_j)x_j\rightarrow y$ as $n\rightarrow \infty$. By Bessel's inequality we have for finite index sets $F$
$$\sum_{j\in F}|(y,x_j)|^2\leq \|y\|^2$$
and therefore
$$\sum |(y,x_j)|^2 = \sup_{F\text{ is finite}}\sum_{j\in F}|(y,(x_j))|^2\leq \|y\|^2$$
which implies that $y\in S$. But how is the sum $\sum(y,x_j)x_j$ defined? It is transfinite but not positive?
As Asaf Karagila stated in the comments the fact that $\sum |(y,x_j)|^2<\infty$ implies that at most countably many $(y,x_j)$ are nonzero.
Say that $G$ is a countable set which contains all indices $j$ such that $(y,x_j)\neq 0$ and let $G_n$ be some sequence of finite indices such that $\cup G_n = G$. We want to show that
$$\lim_n \sum_{G_n}(y,x_j)x_j = y.$$
Note that we can choose $G_n$ such that $F_n\setminus G_n$ only contains indices $j$ with the property that $(y,x_j) = 0$ then
$$\|y-y_n\|\geq \|y-\sum_{G_n}(y,x_j)x_j\|$$
which proves the result.
Let $S$ be a set of any cardinality, and consider $L^2(S,\delta)$, where $\delta$ is counting measure on $S$. For $s\in S$, let $e_s$ be the function that is $1$ at $s$ and $0$ otherwise. Each $\delta_s$ is a unit vector in $L^2(S,\delta)$, and $\langle e_s,e_s'\rangle=\delta_{s,s'}$. So $\{ e_s \}_{s\in S}$ is an orthonormal subset of $L^2(S,\delta)$. In fact, it is an orthonormal basis. Any other orthonormal basis of $L^2(S,\delta)$ must have the same cardinality as $\{ e_s \}_{s\in S}$. $L^2(S,\delta)$ is a Hilbert space, and all orthonormal bases of a Hilbert space must have the same cardinality, which means there is no countable orthonormal basis for $L^2(S,\delta)$ unless $S$ is countable.