Summation over Interval equals Indexed Summation

Theorem

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

Let $a, b \in \Z$ be integers.

Let $\closedint a b$ be the integer interval between $a$ and $b$.

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

Then the summation over the finite set $\closedint a b$ equals the indexed summation from $a$ to $b$:

$\ds \sum_{k \mathop \in \closedint a b} \map f k = \sum_{k \mathop = a}^b \map f k$

Proof

By Cardinality of Integer Interval, $\closedint a b$ has cardinality $b - a + 1$.

By Translation of Integer Interval is Bijection, the mapping $T : \closedint 0 {b - a} \to \closedint a b$ defined as:

$\map T k = k + a$

is a bijection.

By definition of summation:

$\ds \sum_{k \mathop \in \closedint a b} \map f k = \sum_{k \mathop = 0}^{b - a} \map f {k + a}$

By Indexed Summation over Translated Interval:

$\ds \sum_{k \mathop = 0}^{b - a} \map f {k + a} = \sum_{k \mathop = a}^b \map f k$

$\blacksquare$