Triangle Inequality for Summation over Finite Set

Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.

Let $S$ be a finite set.

Let $f : S \to \mathbb A$ be a mapping.

Let $\left\vert{\, \cdot\,}\right\vert$ denote the standard absolute value.

Let $\left\vert{f}\right\vert$ be the absoute value of $f$.

Then we have the inequality of summations on finite sets:
 * $\displaystyle \left\vert \sum_{s \mathop \in S} f(s) \right\vert \leq \sum_{s \mathop \in S} \vert f(s) \vert$

Outline of Proof
Using the definition of summation, we reduce this to Triangle Inequality for Indexed Summations.

Proof
Let $n$ be the cardinality of $S$.

Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.

By definition of summation, we have to prove the following inequality of indexed summations:
 * $\displaystyle \left\vert{\sum_{i \mathop = 0}^{n - 1} f \left({\sigma \left({i}\right)}\right)}\right\vert \le \sum_{i \mathop = 0}^{n-1} \left({\left\vert{f}\right\vert \circ \sigma}\right) \left({i}\right) $

By Absolute Value of Mapping Composed with Mapping:
 * $\left\vert{f}\right\vert \circ \sigma = \left\vert{f \circ \sigma}\right\vert$

The above equality now follows from Triangle Inequality for Indexed Summations.

Also see

 * Triangle Inequality for Series