Notation of a sum over an arbitrary index The usual definition of a series $\sum_{n=0}^{\infty }a_n$, is the limit of the partial sums $S_k$ when $k\to \infty $
For the summation of families of non-negative numbers over arbitrary index sets there's the next definition:
When summing a family $\left\{a_i \right\}$, $i\in I$, of non-negative numbers, one may define
$$
\sum_{i \in I}^{} a_i = \sup \left\{\sum_{i \in A}^{} a_i \mid A \text{ finite}, A \subset I \right\} \in \left [ 0, \infty  \right ]$$
$A \subset I$ is a subset of $I$ and it is finite. What I don't know is what elements are in the subset. As it is written I can take any elements from the index set $I$ to form the subset. For instance if $I= \mathbb{N}$, the subset $A$ could be $\left\{9,11,14 \right\}$.  How many subsets are there and how do you deduce it from the notation? If there is more than a subset, are they disjoint? I don't understand this notation.
 A: When the definition says "$A$ finite, $A \subset I$", it means exactly what it says:
it includes every finite subset of $I$. Really.
You may be troubled because you are trying to reconcile this definition
with the prior understanding of $\sum_{n=0}^\infty a_n$ involving the convergence of a sequence of partial sums.
But the definition using $\mathrm{sup}$ is not based on the convergence of a sequence partial sums.
When we define the set
$$ T = \left\{\sum_{i \in A}^{} a_i \mid A \text{ finite}, A \subset I \right\}, $$
we have something involving sums of only some of the $a_i$s
(just as partial sums involve only some of the $a_i$s),
and each sum is a number,
but the numbers are not in sequence.
Since the numbers are not in sequence, there is no way that they can converge to anything.
That's why we take the ${\mathrm{sup}}$ -- it's not about convergence, it's about putting an upper bound on the numbers (if there is one) and choosing the least possible upper bound out of all possible upper bounds.
You can't "break" the ${\mathrm{sup}}$ of a set by adding numbers to the set that are equal to or less than numbers already in the set. The ${\mathrm{sup}}$ already had to be at least as great as the greater numbers.
For example, if $I = \mathbb N,$
then the fact that you are already planning to include
$\sum_{i\in\{1,2,3,4,5,6,7\}} a_i$ in your calculations means it doesn't matter
whether you consider $\sum_{i\in\{1,7\}} a_i$;
since the $a_i$s are all non-negative,
$\sum_{i\in\{1,7\}} a_i$ cannot be greater than $\sum_{i\in\{1,2,3,4,5,6,7\}} a_i.$
In short, the fact that the new idea allows us to count sums over subsets of indices with "holes" in the set, and not just sums over subsets without "holes"
(subsets of the form "the first $k$ integers"), has no impact on the final answer.
A: Altering the notation slightly to put the collection of sets beneath,
$$
\sum_{i \in I} a_i 
= \sup_{\substack{A \subseteq I \\ A \text{ finite}}} 
\sum_{i \in A} a_i
$$
emphasizes the fact that the supremum is over all finite sets of indices. Let’s call this quantity $S_{\text{fin}}$.
In order to compare this to the usual definition, we must assume that $a_i \geq 0$ for all $i \in I$. Now, the usual series definition involves partial sums, which are sums over particular sets of indices:
$$
\begin{align}
s_1 &= a_1 & A &= \{1\} \\
s_2 &= a_1 + a_2 & A &= \{1, 2\} \\
&\;\,\vdots &&\;\,\vdots \\
s_k &= a_1 + a_2 + \cdots + a_k & A &= \{1, 2, \dots, k\}
\end{align}
$$
Let’s call this usual sum $S = \smash{\displaystyle\lim_{k \to \infty}} s_k$.
We have to convince ourselves that $S_{\text{fin}} = S$ when $I = \mathbb{N}$. For this, we rely on the fact that for non-negative numbers $\{a_i\}$,
$$
\sum_{i \in I} a_i \leq \sum_{i \in I’} a_i 
\quad \text{when } I \subseteq I’ 
\label{subsum}\tag{1}
$$
For any finite set $A \subset \mathbb{N}$, we have
$$
A \subseteq \{1, 2, \dots, k\}, 
\quad \text{where } k = \max A, 
$$
so $S_{\text{fin}} \leq S$ by (\ref{subsum}).
In the other direction, observe that $\{1, 2, \dots, k\}$ is a finite set $A$ for each $k \geq 1$, so we just need to convince ourselves that for collections of sets of indices,
$$
\sup_{A \in \mathcal{A}} \sum_{a \in A} \leq 
\sup_{A \in \mathcal{A}’} \sum_{a \in A} 
\quad \text{when } \mathcal{A} \subseteq \mathcal{A}’,
$$
but this is clear, so $S \leq S_{\text{fin}}$.
