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Let $\;\;\; \big\langle V,+,\cdot, \;\; ||\cdot|| \;\; \big\rangle \;\;\;$ be a Banach space over the field $\mathbb{K}$, which is either $\mathbb{R}$ or $\mathbb{C}$.

Suppose there exists a subset $B\:$ of $V\:$ such that:


For all functions $\: f : B\to \mathbb{K} \:$, $\:$ if $\;\; \displaystyle\sum_{b\in B} \; (f(b) \cdot b) \;\;$ exists and is the zero vector, then $f\:$ is identically zero.

and

For all members $v$ of $V$, $\:$ there exists a function $\; f : B\to \mathbb{K} \;$ such that $\;\;\;\; \displaystyle\sum_{b\in B} \; (f(b) \cdot b) \;\; = \;\; v \;\;\;\;$.



Does it follow that all such subsets $B\:$ have the same cardinality?

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The sums are as defined at en.wikipedia.org/wiki/…;, using the topology induced by the norm. $\:$ If Countable Choice, then such a sum existing and being equal to $v$ is equivalent to [[$f$ has (at-most-)countable support] and [the (ordinary) series formed by the terms for which $f(b)$ is non-zero converge unconditionally] and [those series sum to $v$]]. $\;\;$ –  Ricky Demer Feb 23 '12 at 22:22
    
Your axioms mean nothing else that $B$ is a Schauder basis and all Schauder bases have the same cardinality, if I am not mistaken. –  Kofi Feb 23 '12 at 22:48
    
No, I'm allowing $B\:$ to be uncountable. $\;\;$ –  Ricky Demer Feb 23 '12 at 22:59
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up vote 2 down vote accepted

Well, Kofi is right. Your set $B$ is an unconditional basis in the extended sense. I think Singer treats this kind of basis in one of his volumes on basis theory. Anyway, standard techniques show that the (obviously well defined) biorthogonal functionals are continuous so that $B$ is a Markuschevich basis. It is essentially obvious that any two Markuschevich bases have the same cardinality (i.e., if you can prove that any two ON basis have the same cardinality, then you can prove that any two Markuschevich bases have the same cardinality).

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So, I've convinced myself that that proof uses AC and would work for any first-countable torsion-free $\hspace{.3 in}$ $(\text{T}_0)$ topological module over a non-trivial ring. $\:$ Do you know if those are right, and if so, whether a weaker version of choice could be used or a more general conclusion could be reached? $\;\;$ –  Ricky Demer Feb 24 '12 at 4:41
3  
I don't know, Ricky. For me, "a day without choice is like a day without wine". –  Bill Johnson Feb 24 '12 at 14:08
    
I can get by without uncountable wine, but my work would be held up without uncountable choice. All those weak-star cluster points, no longer guaranteed... –  Yemon Choi Feb 26 '12 at 22:50
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