Sum over Complement of 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 $T \subseteq S$ be a subset.

Let $S \setminus T$ be its relative complement.

Then we have the equality of summations over finite sets:


 * $\displaystyle \sum_{s \mathop \in S \setminus T} f \left({s}\right) = \sum_{s \mathop \in S} f \left({s}\right) - \sum_{t \mathop \in T} f \left({t}\right)$

Proof
Note that by Subset of Finite Set is Finite, $T$ is indeed finite.

By Set is Disjoint Union of Subset and Relative Complement, $S$ is the disjoint union of $S \setminus T$ and $T$.

The result now follows from Sum over Disjoint Union of Finite Sets.

Also see

 * Sum over Union of Finite Sets