Unconditional Schauder Basis Convergence? Let $X$ be a Banach space and $\{e_n\}$ be a unconditional Schauder basis for $X$, i.e., if for every $x \in X$, its expansion 
$$x = \sum_i a_i e_i$$ 
converges unconditionally. 
(Recall that a series $\sum_i x_i$ in a Banach space $X$ is unconditionally convergent if $\sum_i \epsilon_i x_i$ converges for all choices of signs $\epsilon = \pm 1$.)
Claim: 
For given $A \subseteq \mathbb{N}$, $\sum_{i \in A} a_i e_i$ converges whenever $\sum_{i=1}^{\infty} a_i e_i$ converges in $X$. 
It looks simple, but I couldn't check this. 
Any help will be appreciated. 
 A: Let $\epsilon_i = 1$ when $i\in A$ and $\epsilon_1=-1$ otherwise. By assumption, $\sum \epsilon_i a_i x_i$ converges. We also know $\sum a_i x_i$ converges. The sum of two convergent series converges, and this sum is $2\sum_{i\in A} a_i e_i$. 
Note on the order of summation: in the above argument I suppose it is fixed somehow, probably by having $i$ running through natural numbers. Otherwise, the word "unconditional" would not be needed at all. 
Indeed, consider a Schauder basis an with arbitrary index sets $J$. In this case, with no preferred order of summation on $J$, convergence $\sum_{j\in J} a_ix_i = x$ has to be understood as: for every $\epsilon>0$ there exists a finite set $E \subset J$ such that for every finite set $F$ containing $E$ we have 
$$\left\|\sum_{i\in F}a_i x_i - x\right\| < \epsilon$$
In this form, convergence is automatically unconditional. One does not need any assumption with $\epsilon$ to conclude that the sum $\sum_{i\in A} a_i x_i$ converges: just use the definition. Given $\epsilon$, there exists $E$ as above... then let $E'=E\cap A$ and this set works for summation over $A$. 
(Aside: to me, unconditional convergence really means what's written in the preceding paragraph, not the $\epsilon_i$ thing. They are equivalent, though.)
