Series of complex numbers over an uncountable set 
Question: Let I be an arbitary index set and $(a_i)_{i\in I} \subset \Bbb C$ such that $\sum_{i \in I}a_i$ converges. Show that there exists a countable set $I_0 \subset I$ such that $a_i=0$ for all $i \in I\setminus I_0$.  

I think I am able to prove this if all the real and imaginary parts of $a_i$ are positive but for any sequence $a_i$ I'm struggling. I feel like I want to use the Cauchy criterion, namely that since we have convergence, we also have that for all $\epsilon>0$ there exists a set $J_{\epsilon}$ such that $$\bigg|\sum_{i\in J\setminus J_{\epsilon}}a_i\bigg|<\epsilon$$ for all $J\supset J_{\epsilon}$ where $J\subset I$ is finite. We could then choose $\epsilon=\frac{1}{n}$ to generate a family of $J_n$ and possibly take the union over $n$ to create the countable set I'm after. But beyond that I'm stuck. 
Any help is much appreciated.
 A: Note that positivity isn't really the assumption you need for the easier argument you mention - rather, you just want all the real parts to have the same sign and all the imaginary parts to have the same sign. As long as this happens, you're still happy since all the nonzero terms still "point away from zero in the same way."
Now think about how the four quadrants partition the complex plane. If I have an uncountable set of nonzero points $\{a_i: i\in I\}$, then there must be an uncountable $J\subseteq I$ such that one of the following happens:


*

*If $a+bi, c+di\in J$ then both $a$ and $b$ are $\ge 0$ and both $c$ and $d$ are $\ge 0$.

*If $a+bi, c+di\in J$ then both $a$ and $b$ are $\ge 0$ and both $c$ and $d$ are $\le 0$.

*If $a+bi, c+di\in J$ then both $a$ and $b$ are $\le 0$ and both $c$ and $d$ are $\ge 0$.

*If $a+bi, c+di\in J$ then both $a$ and $b$ are $\le 0$ and both $c$ and $d$ are $\le 0$.
(There could be more than one such $J$, and there could be $J$s exhibiting different behaviors; that's fine.) You should be able to show that $\sum_{i\in J}a_i$ diverges. Now, can you show that $\sum_{i\in I}a_i$ diverges since $J\subseteq I$ (and get from there to the theorem you want)?
A: You do not say what is your definition of the convergence of $\sum_{i \in I} a_i$, but I guess it is this:
$\sum_{i \in I} a_i = a$ if for each $\epsilon >0$ there exists a finite $I_\epsilon \subset I$ such that for all finite $J \subset I$ which contain $I_\epsilon$ one has $\lvert \sum_{i \in J} a_i - a \rvert < \epsilon$.
This implies that for all $j \notin I_\epsilon$
$$\lvert a_j \rvert = \lvert \sum_{i \in I_\epsilon \cup \{j \}} a_i - \sum_{i \in I_\epsilon } a_i \rvert = \lvert \sum_{i \in I_\epsilon \cup \{j \}} a_i - a + a - \sum_{i \in I_\epsilon } a_i \rvert \\ \le \lvert \sum_{i \in I_\epsilon \cup \{j \}} a_i - a \rvert + \lvert a - \sum_{i \in I_\epsilon } a_i \rvert  < 2\epsilon .$$
Let $J_n = \{ i \in I \mid \lvert a_i \rvert \ge 1/n \}$ and $I_0 = \{ i \in I \mid a_i \ne 0 \}$. We have $I_0 = \bigcup_{n=1}^\infty J_n$.
Assume that some $J_n$ is infinite. For $\epsilon = 1/2n$ choose $I_\epsilon$ as above. Pick any $j \in J_n \setminus I_\epsilon$. Then $\lvert a_j \rvert < 2 \cdot 1/2n = 1/n$ which is a contradiction.
Hence all $J_n$ are finite and $I_0$ is countable.
Note that this proof can be generalized to arbitrary normed linear spaces.
