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 $|\cdot|$ denote the standard absolute value.

Let $\vert f \vert$ 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 on finite set, 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 on finite set, we have to prove the following inequality of indexed summations:
 * $\displaystyle \left\vert \sum_{i \mathop = 0}^{n-1} f(\sigma(i)) \right\vert \leq \sum_{i \mathop = 0}^{n-1} (\vert f \vert \circ \sigma) (i) $

By Absolute Value of Mapping Composed with Mapping, $|f|\circ\sigma = |f\circ\sigma|$.

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

Also see

 * Triangle Inequality for Series