Why is a summable family at most countable? 
Why is a summable family at most countable ?
Let $(a_i)_{n\in I}$($a_i\in [0,\infty],\forall i$) is summable then, $\{i\in I:a_i≠0\}$ is at most countable.
a family is said to be summable if $\sum_{i\in I}a_i<\infty$

Can we always choose a countable set s.t., every member of this set is bigger then some $\epsilon$, but how ?
 A: For any positive $\varepsilon$ denote by $I_\varepsilon$ the set of all indices for which $a_i$ is greater than $\varepsilon$: $I_\varepsilon=\{i \in I \mid a_i > \varepsilon\}$.
Now it boils down to two things:


*

*$I_\varepsilon$ is finite for every $\varepsilon > 0$. This is the part where it is necessary to use that the sum $\sum_{i \in I} a_i$ is finite.

*Realize that your set $\{i \in I \mid a_i > 0\}$ can be represented as a union:
$$
    \{i \in I \mid a_i > 0\} = \bigcup_{n \in \mathbb{N}} I_{1/n}
$$
Any union of a countable family of finite sets is at most countable, qed.


PS: There was no definition given for the sum $\sum_{i \in I}a_i$ in the question. But, since all $a_i$ are nonnegative, I just assumed that you mean $\sum_{i \in I}a_i = \sup_J \sum_{j \in J} a_j$, where $J$ iterates over all finite subsets of $I$.
A: Note that
$$
\{i :\ a_i\ne0\}=\bigcup_n\{i: \ |a_i|> 1/n\}.
$$
If  the set on the left is uncountable,  then at least one set on the right is infinite. 
Now, when considering a series indexed by a net, one has an issue with the definition.  The usual way to define the symbol $\sum_{i\in I}a_i$ is by taking the limit along the net of finite subsets of $I$, ordered by inclusion. This implies that absolute convergence is required. 
In the case above,  if $|a_i|>1/n$ for infinitely many $i$, then $\sum_i|a_i|=\infty$, and so the series cannot be convergent. 
