Triangle Inequality for Generalized Sums

Theorem
Let $V$ be a Banach space.

Let $\norm {\,\cdot\,}$ denote the norm on $V$.

Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$.

Let the generalized sum $\ds \sum \set {v_i: i \in I}$ converge absolutely.

Then:
 * $\ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$

Proof
First of all, note that Absolutely Convergent Generalized Sum Converges assures us that the left expression in above equality is defined.

there exists an $\epsilon > 0$ such that:
 * $\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I} + \epsilon$

This supposition is seen to be equivalent to:
 * $\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I}$

Then, by definition of a generalized sum, there necessarily exists a finite subset $F$ of $I$ with:


 * $\ds \norm {\sum_{i \in F} v_i} > \norm {\sum \set {v_i: i \in I} } - \epsilon > \sum \set {\norm {v_i}: i \in I}$

However, using the standard triangle equality on this finite sum (that is,, repetitively), we also have:


 * $\ds \norm {\sum_{i \in F} v_i} \le \sum_{i \in F} \norm {v_i} \le \sum \set {\norm {v_i}: i \in I}$

Here the second inequality follows from Generalized Sum is Monotone.

These two estimates constitute a contradiction, and therefore such an $\epsilon$ cannot exist.

Hence:
 * $\ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$