Cardinal Product Equal to Maximum

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Theorem

Let $S$ and $T$ be sets that are equinumerous to their cardinal number.

Let $\left|{S}\right|$ denote the cardinal number of $S$.

Suppose $S$ is infinite.

Suppose $T > 0$.


Then:

$\left|{S \times T}\right| = \max\left({\left|{S}\right|, \left|{T}\right| }\right)$


Proof

Let $x$ denote $\max \left({ \left|{ S }\right| , \left|{ T }\right| }\right)$.

Then by Cartesian Product Preserves Cardinality:

$S \times T \sim \left|{ S }\right| \times \left|{ T }\right|$

Let $f : S \times T \to \left|{ S }\right| \times \left|{ T }\right|$ be a bijection.

It follows that $f : S \times T \to x \times x$ is an injection.


Hence:

\(\ds \left\vert{ S \times T }\right\vert\) \(\le\) \(\ds \left\vert{ x \times x }\right\vert\) Injection iff Cardinal Inequality
\(\ds \) \(=\) \(\ds \left\vert{ x }\right\vert\) Non-Finite Cardinal is equal to Cardinal Product
\(\ds \) \(\le\) \(\ds x\) Cardinal Number Less than Ordinal: Corollary

Therefore:

$\left|{ S \times T }\right| \le x$

$\Box$


Conversely:

$x = \left|{ S }\right|$ if $\left|{ T }\right| \le \left|{ S }\right|$

and:

$x = \left|{ T }\right|$ if $\left|{ S }\right| \le \left|{ T }\right|$

By Relation between Two Ordinals:

$x = \left|{ S }\right|$ or $x = \left|{ T }\right|$

It follows by Set Less than Cardinal Product that:

$x \le \left|{ S \times T }\right|$

$\Box$


Combining the two lemmas, it follows that:

$x = \left|{ S \times T }\right|$

$\blacksquare$


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