Cardinal Product Equinumerous to Ordinal Product

From ProofWiki
Jump to navigation Jump to search


Let $S$ and $T$ be sets that are equivalent to their cardinal numbers.

Let $\card S$ denote the cardinal number of $S$.

Let $\cdot$ denote ordinal multiplication and let $\times$ denote the Cartesian product.


$S \times T \sim \card S \cdot \card T$


Let $f: S \to \card S$ and $g: T \to \card T$ be bijections.

Let $\cdot$ denote ordinal multiplication, while $\times$ shall denote the Cartesian product.

Define the function $F$ to be:

$\forall x \in S, y \in T: \map F {x, y} = \card S \cdot \map g y + \map f x$

Suppose $\map F {x_1, y_1} = \map F {x_2, y_2}$.

\(\ds \card S \cdot \map g {y_1} + \map f {x_1}\) \(=\) \(\ds \card S \cdot \map g {y_2} + \map f {x_2}\)
\(\ds \leadsto \ \ \) \(\ds \map f {x_1}, \map f {x_2}\) \(\in\) \(\ds \card S\) Definition of Mapping $f$
\(\ds \leadsto \ \ \) \(\ds \map f {x_1}\) \(=\) \(\ds \map f {x_2}\) Division Theorem for Ordinals
\(\, \ds \land \, \) \(\ds \map g {y_1}\) \(=\) \(\ds \map g {y_2}\)
\(\ds \leadsto \ \ \) \(\ds x_1\) \(=\) \(\ds x_2\) Definition of Bijection
\(\, \ds \land \, \) \(\ds y_1\) \(=\) \(\ds y_2\)

It follows that $F: S \times T \to \card S \cdot \card T$ is a injection.


Furthermore, take any $z \in \card S \cdot \card T$.

By Division Theorem for Ordinals, it follows that:

$z = \card S \cdot a + b$ for some unique $a$ and $b \in \card S$.

Moreover, $a \in \card T$ since otherwise, this would contradict the fact that $z \in \card S \cdot \card T$.

Therefore, $a = \map f x$ for some $x \in S$ and $b = \map g y$ for some $y \in T$.

Thus, $F : S \times T \to \card S \cdot \card T$ is a surjection.


Therefore, $F : S \times T \to \card S \cdot \card T$ is a bijection.

By Condition for Set Equivalent to Cardinal Number, it follows that:

$S \times T \sim \card S \cdot \card T$