Cardinality of Cartesian Product/Corollary

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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.

$\blacksquare$


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