Absolute Net Convergence Equivalent to Absolute Convergence/Absolute Convergence implies Absolute Net Convergence

Theorem
Let $V$ be a Banach space.

Let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of elements in $V$.

Let $r \in \R_{\mathop \ge 0}$

Let the series $\ds \sum_{n \mathop = 1}^\infty v_n$ be absolutely convergent to $r$.

Then:
 * the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$.

Proof
Let $\epsilon \in \R_{\mathop \ge 0}$.

By definition of absolutely convergent:
 * $(2) \quad \exists N \in \N : \forall m \ge N : \size{\ds \sum_{n \mathop = 0}^m \norm{v_m} - r} < \dfrac \epsilon 3$

Let:
 * $F = \closedint 0 N$

Let:
 * $E \subseteq \N : E \supseteq F : E$ is finite.

Let:
 * $m = \max \set{n : n \in E}$

Let:
 * $G = \closedint 0 m$

We have:
 * $F = \closedint 0 N \subseteq E \subseteq \closedint 0 m = G$

From Set Difference and Intersection form Partition:
 * $E = F \cup E \setminus F$

and
 * $G = F \cup G \setminus F$

From Set Difference Intersection with Second Set is Empty Set:
 * $F \cap E \setminus F = \O$

and
 * $F \cap G \setminus F = \O$

From Set Difference over Subset:
 * $E \setminus F \subseteq G \setminus F$

We have:

Since $E$ was arbitrary, it follows:
 * $\exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $: \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$

Sine $\epsilon$ was arbitrary, it follows:
 * $\forall \epsilon \in \R_{\mathop > 0} : \exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $: \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$

From Characterization of Convergent Net in Metric Space:
 * the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$.