# Exchange of Order of Summations over Finite Sets/Cartesian Product

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## Theorem

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

Let $S, T$ be finite sets.

Let $S \times T$ be their cartesian product.

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

Then we have an equality of summations over finite sets:

- $\displaystyle \sum_{s \mathop \in S} \sum_{t \mathop \in T} \map f {s, t} = \sum_{t \mathop \in T} \sum_{s \mathop \in S} \map f {s, t}$

## Proof 1

## Proof 2

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.

We have:

\(\displaystyle \sum_{s \mathop \in S} \sum_{t \mathop \in T} \map f {s, t}\) | \(=\) | \(\displaystyle \sum_{s \mathop \in S} \sum_{j \mathop = 0}^{n - 1} \map f {s, \map \tau j}\) | Definition of Summation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 0}^{m - 1} \sum_{j \mathop = 0}^{n - 1} \map f {\map \sigma i, \map \tau j}\) | Definition of Summation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 0}^{n - 1} \sum_{i \mathop = 0}^{m - 1} \map f {\map \sigma i, \map \tau j}\) | Exchange of Order of Indexed Summations over Rectangular Domain | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{t \mathop \in T} \sum_{i \mathop = 0}^{m - 1} \map f {\map \sigma i, t}\) | Definition of Summation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{t \mathop \in T} \sum_{s \mathop \in S} \map f {s, t}\) | Definition of Summation |

$\blacksquare$

## Proof 3

Let $n$ be the cardinality of $T$.

The proof goes by induction on $n$.

### Basis for the Induction

Let $n = 0$.

### Induction Step

Let $n > 0$.

Let $t \in T$.

Use Cardinality of Set minus Singleton