Positive-Term Generalized Sum Converges iff Supremum

Theorem
Let $\left({G, \circ, \le}\right)$ be an abelian totally ordered group, considered under the order topology.

Let $\left\{ {x_i: i \in I}\right\}$ be an indexed set of positive elements of $G$.

Then:
 * the generalized sum $\displaystyle \sum \left\{ {x_i: i \in I}\right\}$ converges to a point $x \in G$


 * $x$ is the supremum of:
 * $P := \displaystyle \left\{{\sum_{i \mathop \in F} x_i: \text{$F \subseteq I$ and $F$ is finite}}\right\}$

Sufficient Condition
Let $\displaystyle \sum \left\{ {x_i: i \in I}\right\}$ converge to $x$.

We first show that $x$ is an upper bound of $P$.

Suppose for the sake of contradiction that for some finite subset $F$ of $I$:


 * $\displaystyle \sum_{i \mathop \in F} x_i > x$

Then the net of finite sums is eventually less than $\displaystyle \sum_{i \mathop \in F} x_i$.

Let $F \subseteq F' \subseteq I$.

Let $F'$ be finite.

Since finite sums are monotone:
 * $\displaystyle \sum_{i \mathop \in F'} x_i \ge \sum_{i \mathop \in F} x_i$

which is a contradiction.

Thus we conclude that $x$ is an upper bound of $P$.

Let $m \in G$ such that $m < x$.

Then the net of finite sums is eventually greater than $m$.

Thus $m$ is certainly not an upper bound of $P$.

So we have shown that $x$ is the supremum of $P$.

Necessary Condition
Let $x$ be the supremum of $P$.

Then:
 * $\forall b > x: \forall y \in P: b > y$

so the net of finite sums is always (hence eventually) less than $b$.

For all $a < x$, $a$ is not an upper bound of $P$.

Therefore there exists a finite subset $F$ of $I$ such that:
 * $\displaystyle \sum_{i \mathop \in F} x_i > a$

Since finite sums are monotone increasing, the net of finite sums is eventually greater than $a$.

Thus $\displaystyle \sum \left\{ {x_i: i \in I}\right\}$ converges to $x$.