Sum over Disjoint Union of Finite Sets

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

Let $S$ and $T$ be finite disjoint sets.

Let $S\cup T$ be their union.

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

Then we have the equality of summations over finite sets:


 * $\displaystyle \sum_{u \mathop \in S \cup T} f(u) = \sum_{s \mathop \in S} f(s) + \sum_{t \mathop \in T} f(t)$

Outline of Proof
We reduce to the case of indexed summations by choosing appropriate bijections in the definition of summation over finite set.

We thereby reduce the problem to Indexed Summation over Adjacent Intervals.

Proof
Note that by Union of Finite Sets is Finite, the union $S\cup T$ is finite.

Let $m$ be the cardinality of $S$ and $n$ be the cardinality of $T$.

Let $\N_{<m}$ denote an initial segment of the natural numbers.

Let $\sigma : \N_{<m} \to S$ and $\tau : \N_{<n} \to T$ be bijections.

Let $\alpha : \N_{<n} \to \left[{m \,.\,.\, m+n-1}\right]$ be the mapping defined as:
 * $\alpha (k) = k+m$

By Translation of Integer Interval is Bijection, $\alpha$ is a bijection.

By Composite of Bijections is Bijection and Disjoint Union of Bijections is Bijection, the union:
 * $\sigma \cup (\tau\circ \alpha) : \N_{<m} \cup \left[{m \,.\,.\, m+n-1}\right] \to S \cup T$ is a bijection.

By Union of Integer Intervals, $\N_{<m} \cup \left[{m \,.\,.\, m+n-1}\right] = \N_{<m+n}$.

We have:

Also see

 * Sum over Union of Finite Sets
 * Sum over Complement of Finite Set