# Cardinality of Cartesian Product/Corollary

## Corollary to Cardinality of Cartesian Product

Let $S \times T$ be the cartesian product of two sets $S$ and $T$ which are both finite.

Then:

$\card {S \times T} = \card {T \times S}$

where $\card {S \times T}$ denotes the cardinality of $S \times T$.

## Proof 1

 $\ds \card {S \times T}$ $=$ $\ds \card S \times \card T$ Cardinality of Cartesian Product $\ds$ $=$ $\ds \card T \times \card S$ Integer Multiplication is Commutative $\ds$ $=$ $\ds \card {T \times S}$ Cardinality of Cartesian Product

$\blacksquare$

## Proof 2

Let $f: S \times T \to T \times S$ be the mapping defined as:

$\forall \tuple {s, t} \in S \times T: \map f {s, t} = \tuple {t, s}$

which is shown to be bijective as follows:

 $\ds \map f {s_1, t_1}$ $=$ $\ds \map f {s_2, t_2}$ $\ds \leadsto \ \$ $\ds \tuple {t_1, s_1}$ $=$ $\ds \tuple {t_2, s_2}$ Definition of $f$ $\ds \leadsto \ \$ $\ds \tuple {s_1, t_1}$ $=$ $\ds \tuple {s_2, t_2}$ Equality of Ordered Pairs

showing $f$ is an injection.

Let $\tuple {t, s} \in T \times S$.

Then:

$\exists \tuple {s, t} \in S \times T: \map f {s, t} = \tuple {t, s}$

showing that $f$ is a surjection.

So we have demonstrated that there exists a bijection from $S \times T$ to $T \times S$.

The result follows by definition of set equivalence.

$\blacksquare$