# Non-Finite Cardinal is equal to Cardinal Product

## Theorem

Let $\omega$ denote the minimally inductive set.

Let $x$ be an ordinal such that $x \ge \omega$.

Then:

$\card x = \card {x \times x}$

where $\times$ denotes the Cartesian product.

### Corollary

Let $S$ be a set that is equinumerous to its cardinal number.

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

Let:

$\card S \ge \omega$

where $\omega$ denotes the minimally inductive set.

Then:

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

## Proof

The proof shall proceed by Transfinite Induction on $x$.

Let:

$\forall y \in x: y < \omega \lor \card y = \card {y \times y}$

There are two cases:

### Case 1: $\card y = \card x$ for some $y \in x$

If this is so, then:

 $\ds \card x$ $=$ $\ds \card y$ $\ds$ $=$ $\ds \card {y \times y}$ Inductive hypothesis $\ds$ $=$ $\ds \card {x \times x}$

$\Box$

### Case 2: $\card y < \card x$ for all $y \in x$

We have that either:

$y < \omega$

or:

$\card y = \card {y + 1}$

In either case, we have that:

$\card {y + 1} < \card x$

and therefore:

$y + 1 \in x$

Therefore, $x$ is a limit ordinal.

Let $R_0$ denote the canonical order of $\On^2$.

Let $J_0$ be defined as the unique order isomorphism between $\On^2$ and $\On$ as defined in canonical order.

It follows that:

$\ds \map {J_0} {x \times x} = \bigcup_{y \mathop \in x} \bigcup_{z \mathop \in x} \map {J_0} {y, z}$

But moreover:

 $\ds \map {J_0} {y, z}$ $=$ $\ds \map {\in^{-1} } {\map {J_0} {y, z} }$ Definition of Preimage of Element under Mapping $\ds$ $=$ $\ds \map {J_0} {\map {R_0^{-1} } {y, z} }$ as $J_0$ forms an Order Isomorphism.

Since $J_0$ is a bijection:

 $\ds \card {\map {J_0} {y, z} }$ $=$ $\ds \card {\map {J_0} {\map {R_0^{-1} } {y, z} } }$ Equality shown above $\ds$ $=$ $\ds \card {\map {R_0^{-1} } {y, z} }$ Image of Bijection Cardinal Equality

Take $\max \set {y, z}$.

 $\ds \tuple {a, b}$ $\in$ $\ds \map {R_0^{-1} } {y, z}$ by hypothesis $\ds \leadstoandfrom \ \$ $\ds \tuple {a, b}$ $R_0$ $\ds \tuple {y, z}$ Definition of Image of Element under Mapping $\ds \leadstoandfrom \ \$ $\ds a, b$ $\le$ $\ds \max \set {y, z}$ Definition of Canonical Order $\ds \leadstoandfrom \ \$ $\ds a, b$ $\in$ $\ds \max \set {y, z} + 1$ Definition of Successor Set

It follows that:

 $\ds \card {\map {R_0^{-1} } {y, z} }$ $\le$ $\ds \card {\max \set {y, z} + 1 \times \max \set {y, z} + 1}$ $\ds$ $=$ $\ds \card {\max \set {y, z} + 1}$ Inductive hypothesis $\ds$ $<$ $\ds \card x$ by hypothesis for the case

Therefore:

$\card {\map {R_0^{-1} } {y, z} } < \card x$
$\forall y, z \in x: \map {J_0} {y, z} < \card x$

It follows by Supremum Inequality for Ordinals that:

$\map {J_0} {x \times x} \subseteq \card x$

Hence:

 $\ds \card {x \times x}$ $=$ $\ds \card {\map {J_0} {x \times x} }$ Image of Bijection Cardinal Equality $\ds$ $\le$ $\ds \card x$ Subset of Ordinal implies Cardinal Inequality

But also by Set Less than Cardinal Product:

$\card x \le \card {x \times x}$

Thus:

$\card x = \card {x \times x}$

$\blacksquare$