Theorem: Absolutely Convergent Generalized Sum Converges

Let $V$ be a Banach space.
Let $\norm {\, \cdot \,}$ denote the norm on $V$.
Let $d$ denote the corresponding induced metric.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$ such that the generalized sum $\ds \sum \set {v_i: i \in I }$ converges absolutely.
Then the generalized sum $\ds \sum \set {v_i: i \in I}$ converges.

Proof:
The proof proceeds in two stages:
:$(1): \quad$ Finding a candidate $v \in V$ where the sum might converge to
:$(2): \quad$ Showing that the candidate is indeed sought limit.
That $\ds \sum \set {v_i: i \mathop \in I}$ converges absolutely means that $\ds \sum \set {\norm {v_i}: i \mathop \in I}$ converges.
Now, for all $n \in \N$, let $F_n \subseteq I$ be finite such that:
:$\ds \sum_{i \mathop \in G} \norm {v_i} > \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n}$
for all finite $G$ with $F_n \subseteq G \subseteq I$
It may be arranged that $n \ge m \implies F_m \subseteq F_n$ by passing over to $\ds F'_n = \bigcup_{m \mathop = 1}^n F_m$ if necessary.
Define:
:$\ds v_n = \sum_{i \mathop \in F_n} v_i$
Next, we need to show that the sequence $\sequence {v_n}_{n \mathop \in \N}$ is Cauchy.
So let $\epsilon > 0$, and let $N \in \N$ be such that:
:$2^{-N} < \epsilon$.
Then for $m \ge n \ge N$, we have:
{{begin-eqn}}
{{eqn | l = \map d {v_m, v_n}
      | r = \norm {\paren {\sum_{i \mathop \in F_m} v_i} - \paren {\sum_{i \mathop \in F_n} v_i} }
      | c = {{Defof|Metric Induced by Norm}}
}}
{{eqn | r = \norm {\sum_{i \mathop \in F_m \setminus F_n} v_i}
      | c = $F_n \subseteq F_m$
}}
{{eqn | o = \le
      | r = \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
      | c = Triangle Inequality for $\norm {\, \cdot \,}$
}}
{{end-eqn}}
Now to estimate this last quantity, observe:
{{begin-eqn}}
{{eqn | l = \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n} + \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
      | o = <
      | r = \sum_{i \mathop \in F_n} \norm {v_i} + \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
      | c = Defining property of $F_n$
}}
{{eqn | r = \sum_{i \mathop \in F_m} \norm {v_i}
      | c = Union with Relative Complement, $F_n \subseteq F_m$
}}
{{eqn | o = \le
      | r = \sum \set {\norm {v_i}: i \mathop \in I}
      | c = Generalized Sum is Monotone
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
      | o = <
      | r = 2^{-n}
}}
{{end-eqn}}
Finally, by the defining property of $N$, as $n \ge N$:
:$2^{-n} < 2^{-N} < \epsilon$
Combining all of these estimates leads to:
:$\map d {v_m, v_n} < \epsilon$
It follows that $\sequence {v_n}_{n \mathop \in \N}$ is a Cauchy sequence.
As $V$ is a Banach space:
:$\ds \exists v \in V: \lim_{n \mathop \to \infty} v_n = v$
Having identified a candidate $v$ for the sum $\ds \sum \set {v_i: i \in I}$ to converge to, it remains to verify that this is indeed the case.
According to the definition of considered sum, the convergence is convergence of a net.
Next, Metric Induces Topology ensures that we can limit the choice of opens $U$ containing $v$ to open balls centered at $v$.
Now let $\epsilon > 0$ be arbitrary.
We want to find a finite $F \subseteq I$ such that:
:$\map d {\ds \sum_{i \mathop \in G} v_i, v} < \epsilon$
for all finite $G$ with $F \subseteq G \subseteq I$.
Now let $N \in \N$ such that:
:$\forall n \ge N: \map d {v_n, v} < \dfrac \epsilon 2$
with the $v_n$ as above.
By taking a larger $N$ if necessary, we need to ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well.
We need to verify that the set $F_N$ defined above has sought properties.
So let $G$ be finite, such that:
:$F_N \subseteq G \subseteq I$
Then:
{{begin-eqn}}
{{eqn | l = \map d {\sum_{i \mathop \in G} v_i, v}
      | r = \norm {\paren {\sum_{i \mathop \in G} v_i} - v}
      | c = {{Defof|Metric Induced by Norm}}
}}
{{eqn | o = \le
      | r = \norm {\paren {\sum_{i \mathop \in G} v_i} - \paren {\sum_{i \mathop \in F_N} v_i} } + \norm {\paren {\sum_{i \mathop \in F_N} v_i} - v}
      | c = Triangle inequality for $\norm {\, \cdot \,}$
}}
{{eqn | o = <
      | r = \norm {\sum_{i \mathop \in G \setminus F_N} v_i} + \frac \epsilon 2
      | c = $F_N \subseteq G$, defining property of $N$
}}
{{eqn | o = \le
      | r = \sum_{i \mathop \in G \setminus F_N} \norm {v_i} + \frac \epsilon 2
      | c = Triangle inequality for $\norm {\, \cdot \,}$
}}
{{end-eqn}}
For the first of these terms, observe:
{{begin-eqn}}
{{eqn | l = \sum \set {\norm {v_i}: i \in I} - 2^{-N} + \sum_{i \mathop \in G \setminus F_N} \norm {v_i}
      | o = <
      | r = \sum_{i \mathop \in F_N} \norm {v_i} + \sum_{i \mathop \in G \setminus F_N} \norm {v_i}
      | c = defining property of $F_N$
}}
{{eqn | r = \sum_{i \mathop \in G} \norm {v_i}
      | c = Union with Relative Complement, $F_N \subseteq G$
}}
{{eqn | o = \le
      | r = \sum \set {\norm {v_i}: i \in I}
      | c = Generalized Sum is Monotone
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop \in G \setminus F_N} \norm {v_i}
      | o = <
      | r = 2^{-N}
}}
{{end-eqn}}
Note that:
:$2^{-N} < \dfrac \epsilon 2$
Combine these inequalities leads to:
:$\ds \map d {\sum_{i \mathop \in G} v_i, v} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
By definition of convergence of a net, it follows that:
:$\ds \sum \set {v_i: i \in I} = v$
{{qed}}