Positive-Term Generalized Sum Converges iff Supremum

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Theorem

Let $\struct {G, \circ, \le}$ be an abelian totally ordered group, considered under the order topology.

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

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Then:

the generalized sum $\ds \sum \set {x_i: i \in I}$ converges to a point $x \in G$

if and only if

$x$ is the supremum of:

$P := \ds \set {\sum_{i \mathop \in F} x_i: \text{$F \subseteq I$ and $F$ is finite} }$

Proof

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Sufficient Condition

Let $\ds \sum \set {x_i: i \in I}$ converge to $x$.

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

Aiming for a contradiction, suppose that for some finite subset $F$ of $I$:

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

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

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

Let $F'$ be finite.

Since finite sums are monotone:

$\ds \sum_{i \mathop \in F'} x_i \ge \sum_{i \mathop \in F} x_i$

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 $P$.

$\Box$

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:

$\ds \sum_{i \mathop \in F} x_i > a$

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

Thus $\ds \sum \set {x_i: i \in I}$ converges to $x$.

$\blacksquare$