Let $\;\;\; \big\langle V,+,\cdot, \;\; ||\cdot|| \;\; \big\rangle \;\;\;$ be a [Banach space](http://en.wikipedia.org/wiki/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](http://en.wiktionary.org/wiki/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](http://en.wikipedia.org/wiki/Cardinality)?
  
  




Uniqueness of dimension in Banach spaces
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).


