Gram-Schmidt in Hilbert space? EDIT:
After some contemplation I decided to phrase the question better to avoid trivial answers.
Consider a Hilbert space with a basis $\{v_{i}\}$ where $i\in I$ an index set, which could be uncountably infinite.
We define a "Gram-Schmidt for infinite dimension" to consist of a total ordering $\leq$ on the index set $I$ and a set of orthonormal basis $\{b_{i}\}$ where $i\in I$, and the following condition is satisfied: for any $i\in I$, then $b_{i}$ can be written as a finite linear combination of $v_{i}$ and $b_{j}$ where $j<i$.
We define a "Gram-Schmidt for infinite dimension with series allowed" to be just like above, except that $b_{i}$ is allowed to be written as series instead of just finite linear combination.
So the question is, can "Gram-Schmidt for infinite dimension" be done in a general Hilbert space starting with an arbitrary basis $\{v_{i}\}$? What about "Gram-Schmidt for infinite dimension with series allowed"?
Clearly, if $I$ is countable, the answer is trivial. The interesting case is when $I$ is uncountable.
Thank you for your answer.
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Original question:
In Real Analysis for Graduate Student (Bass) chapter 19 (Hilbert space) page 188, wrote was "The Gram-Schmidt procedure from linear algebra also works in infinitely many dimensions".
So how exactly would such process get carried out?
Thank you for your answer.
 A: Formally, Gram-Schmidt is an algorithm when working in finite dimensions. In infinite dimensions it lacks an important property of algorithms: Termination after finitely many steps. However, do you see how we can handwave that restriction away with a limit process for the case of a Hilbert space and a countable Hilbert basis?

After your clarification of the question:
For a family $\{\,w_j\mid j\in J\,\}\subseteq V$ we let $\overline{Sp}(\{\,w_j\mid j\in J\,\})$ be the space of all $\sum_{j\in J}c_jw_j$ with at most countably many $c_j\ne 0$ and $\sum \lvert c_jw_j\rvert^2<\infty$.
Define a "partial Gram-Schmidt" as a subset $J\subseteq I$ together with a total order $\le$ on $J$ and orthonormal $\{\,b_j\mid j\in J\,\}$ such that $b_j\in\overline{Sp}(\{v_j\}\cup\{\,b_k\mid k\in J, k<j\,\})$ for all $j\in J$.
Then the set of partial Gram-Schmidts is inductively ordered in an obvious manner, hence ny Zorn's lemma there is a maximal one among them.
I claim that $J=I$ holds for this.
Assume otherwise, i.e. there exists $i\in I\setminus J$.
For $j\in J$ let $c_j=\langle b_j,v_i\rangle$.
If $S$ is any finite subset of $J$ verify that $\sum_{j\in S} c_jb_j$ is the unique vector in $\operatorname{span}(\{\,b_j\mid j\in S\,\})$ that is closest to $v_i$ and that it is shorter than $v_i$. Conclude that at most countably many $c_j$ are nonzero (otherwise one could find finite $S$ that would produce an $\sum_{j\in S} c_jb_j$ longer than $\lVert v_i\rVert$, for example) and that $\sum c_j^2\le \lVert v_i\rVert^2$. Subtract $\sum c_jb_j$ from $v_i$ and normalize to obtain $b_i$ and by letting $i>j$ for all $j\in J$ thus a partial Gram-Schmidt extended to $J\cup \{i\}$, contradicting maximiality of $J$.
Thus for "Gram-Schmidt with series allowed" the answer to your question is "yes" for arbitrary cardinalities.
A: Since at every step in the Gram Schmidt process you are in a finite dimensional space, there is absolutely no difference.
