Absolutely Convergent Generalized Sum Converges

Theorem
Let $V$ be a Banach space; let $\left\Vert{\cdot}\right\Vert$ denote its norm and $d$ the corresponding induced metric.

Let $\left({v_i}\right)_{i\in I}$ be an indexed subset of $V$ such that the generalized sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ converges absolutely.

Then the generalized sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ converges.

Proof
The proof proceeds in two stages:


 * 1) Finding a candidate $v \in V$ where the sum might converge to
 * 2) Showing that the candidate is indeed sought limit

That $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ converges absolutely means that $\displaystyle \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$ converges.

As for all $i \in I$, $\left\Vert{v_i}\right\Vert \in \R_{\ge 0}$, the set $I_{>0} := \left\{{i \in I: \left\Vert{v_i}\right\Vert > 0}\right\}$ is countable by Convergent Generalized Sum of Positive Reals has Countably Many Non-Zero Terms.

Now axiom N1 for a norm implies that $I_{>0} = \left\{{i \in I: v_i \ne \mathbf{0}_V}\right\}$.

It follows that $\displaystyle \sum \left\{{v_i: i \in I}\right\} = \sum \left\{{v_i: i \in I_{>0}}\right\}$.

Now as $I_{>0}$ is countable, there exists an injection $\phi: I_{>0} \to \N$.

Define the sequence $\left({v_n}\right)_{n \in \N}$ by $\displaystyle v_n = \begin{cases}v_i & \text{if } n = \phi \left({i}\right) \\ \mathbf{0}_V & \text{otherwise}\end{cases}$.

Next, it is to be shown that the series $\displaystyle \sum_{n=1}^\infty v_n$ is Cauchy.

Denote the partial sums by $\displaystyle s_l = \sum_{n=1}^l v_n$.

Now let $\epsilon > 0$.

As $\displaystyle \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I_{>0}}\right\} = \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$ converges, there exists a finite subset $F$ of $I_{>0}$ such that:


 * $\displaystyle \sum_{i \in F} \left\Vert{v_i}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I_{>0}}\right\} - \epsilon$

Let $N = \max \phi \left({F}\right)$; this maximum exists as $F$ is finite.

Let $l \ge m \ge N$. Then:

Let $J = \left\{{i \in I_{>0}: l+1 \le \phi \left({i}\right) \le m}\right\}$.

Then the preceding expression is seen to equal $\displaystyle \sum_{i \in J} \left\Vert{v_i}\right\Vert$.

Furthermore, as for $i \in F$, $\phi \left({i}\right) \le N \le l+1$, necessarily $F \cap J = \varnothing$.

This subsequently means that:

That is, the sequence $\left({s_n}\right)_{n \in \N}$ is Cauchy.

Now $V$ is a Banach space, so it is complete; thus there is a $v \in V$ such that $\displaystyle \lim_{n \to \infty} s_n = v$.

Having identified a potential candidate $v$ for $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ to converge to, we can now verify that this is indeed the case.

According to the definition of considered sum, the convergence is convergence of a net.

Next, Metric Induces a Topology ensures that we can limit the choice of opens $U$ containing $v$ to neighborhoods of $v$.

Now let $\epsilon > 0$.

We want to find a finite $G \subseteq I_{>0}$ such that for all finite $G \subseteq G' \subseteq I_{>0}$, $d \left({\displaystyle \sum_{i \in G'} v_i, v}\right) < \epsilon$.

Let $N \in \N$ be such that for $n \ge N$, we have $d \left({s_n, v}\right) < \dfrac \epsilon 2$, with the $s_n$ as above.

Let $N' \in \N$ be such that for $n \ge N'$, we have $\displaystyle \sum_{k=1}^n \left\Vert{v_k}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert : i \in I}\right\} - \frac \epsilon 2$.

Let $N'' = \max \left({N, N'}\right)$.

Define $G = \left\{{i \in I_{>0}: \phi \left({i}\right) \le N''}\right\}$.

Now let $G \subseteq G' \subseteq I_{>0}$ be finite.

Then we have, letting $G'' = G' \setminus G$:

Concluding that $G$ has the properties sought after, it follows that $\displaystyle \sum \left\{{v_i: i \in I}\right\} = \sum \left\{{v_i: i \in I_{>0}}\right\} = v$.