Are Schauder-like uncountable bases possible and of unique cardinality?

Question

Let me clarify what I mean by "uncountable" here, which differs from some other Q's with similar titles here. I mean the analogue of a Hamel basis, but allowing countable sums instead of finite linear combos. A Schauder basis is itself defined to be countable. But we also could consider an uncountable set $B$, from which we construct countable series. I.e., if $B$ is a basis of this sort for $V$, then every $v\in V$ uniquely can be written as some countable sum (i.e a series) $v=\sum_1^\infty c_i b_i$ (with the $b_i\in B$). We'd require a linear order on $B$ from which each such countable subset inherits its order (and thus becomes a sequence).

It's easy to show that if such a basis exists and $|V|>\aleph_1$, then $|B|=|V|$. However, if $|V|=\aleph_1$ we could have $|B|=\aleph_0$ or $|B|=\aleph_1$. In the former case, $B$ would be an ordinary Schauder basis. [As a technical point, I'm assuming $V$ is over a field of cardinality $\aleph_1$, such as $R$ or $C$).]

If everything ($V$, $B$, etc) is viewed as sitting in some big space, as $B$ increases in size the number of convergent expansions increases but the potential violations of uniqueness increase too.

My Q's:

1. Is it possible for a vector space $V$ (with $|V|\ge \aleph_1$) to have an uncountable basis (in this sense) at all? It seems like the answer should be yes, but it's also conceivable that uniqueness is an obstruction.

2. If the answer to (1) is yes, can $V$ (with $|V|=\aleph_1$) have both a Schauder basis AND one of these $\aleph_1$ bases? My intuition says no, but I'd like to prove it somehow.

3. If the answer to (1) is yes, is there a good reason we don't study such bases? I believe that Singer studied uncountable Schauder bases in the sense of uncountable sums. But I'm not aware of any work on countable sums drawn from an uncountable set? I assume there's a good reason for this?

Any help (or pointing out obvious misconceptions on my part or pointers to texts or papers) would be greatly appreciated.

Cheers, Ken

Answer 1

It seems this is nothing profound and just the result of competing definitions for "Schauder basis". Many people (including the wikipedia entry at this time) define it to be countable, while others allow the index set $B$ to be uncountable (but we still take countable sums). There is some discussion of this here.

Uncountable Schauder bases do in fact arise and are studied. Ex. non-separable Hilbert spaces can have them (though any Schauder basis for a separable Hilbert space must be countable).

So, for my Q's:

1. Yes.
2. This I remain unsure of. It seems likely that the answer is no, but I haven't seen a proof. Cardinality alone doesn't prevent it, because if $|V|=\aleph_1$, then $|B|$ can be either $\aleph_0$ or $\aleph_1$ (if $|V|>\aleph_1$ then $|V|=|B|$, so $|V|=\aleph_1$ is the only case which potentially admits both an uncountable and a countable Schauder basis).
3. They are indeed studied, but that fact can be obscured by the stricter definition some people adopt (and which I had encountered).