# Non-Finite Cardinal is equal to Cardinal Product

## Theorem

Let $\omega$ denote the minimal infinite successor 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 $\left|{ S }\right|$ denote the cardinal number of $S$.

Let:

- $\left|{S}\right| \ge \omega$

where $\omega$ denotes the minimal infinite successor set.

Then:

- $\left|{S \times S}\right| = \left|{S}\right|$

## 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.

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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:

- $\card {\map {R_0^{-1} } {y, z} } < \card x$

Thus by Cardinal Inequality implies Ordinal Inequality:

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

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:

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

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 10.33$