Gram-Schmidt for uncountable sets?

Question

I know that Gram Schmidt can be applied to countable linear independent sets on Hilbert spaces, but what happens if we apply it on uncountable sets? Obviously this process has to fail then (at least sometimes) cause we cannot have uncountable orthonormal systems in separable Hilbert spaces. But where exactly does it crack? And is there any structure in the set we construct via Gram Schmidt for uncountable systems?

Answer 1

One problem is that if you have a linear set, you don't really know where to start. Even if you decide on a starting point, it's not really that the algorithm cracks, it just does not give a result. It will construct an orthonormal system, of course, but it will not be complete. All it will be is a set of countably many vectors, all with norm $1$ and all orthogonal to each other.

Answer 2

There's also this problem. If an orthogonal set of vectors is summable, then all but at most countably many of them are zero.

Answer 3

Gram-Schmidt algorithm is constructed by induction (on natural numbers), so you can obtain at most countable ortonormal sets. In fact, existence of ortonormal bases in non-separable Hilbert spaces is shown by "transfinite induction", i.e. via Zorn' lemma.