# 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:

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

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:

$\forall y \in x: y < \omega \lor \left|{ y }\right| = \left|{ y \times y }\right|$

There are two cases:

### Case 1: $\left|{ y }\right| = \left|{ x }\right|$ for some $y \in x$

If this is so, then:

 $\displaystyle \left\vert{ x }\right\vert$ $=$ $\displaystyle \left\vert{ y }\right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert{ y \times y }\right\vert$ Inductive hypothesis $\displaystyle$ $=$ $\displaystyle \left\vert{ x \times x }\right\vert$

$\Box$

### Case 2: $\left|{ y }\right| < \left|{ x }\right|$ for all $y \in x$

We have that either:

$y < \omega$

or:

$\left|{ y }\right| = \left|{ y + 1 }\right|$

In either case, we have that:

$\left|{y + 1}\right < \left|{x}\right|$

and therefore:

$y + 1 \in x$

Therefore, $x$ is a limit ordinal.

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

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

It follows that:

$J_0 \left({x \times x}\right) = \bigcup_{y \mathop \in x} \bigcup_{z \mathop \in x} J_0 \left({y, z}\right)$

But moreover:

 $\displaystyle J_0 \left({ y,z }\right)$ $=$ $\displaystyle \in^{-1} \left({ J_0 \left({ y,z }\right)}\right)$ Definition of Preimage $\displaystyle$ $=$ $\displaystyle J_0 \left({ R_0^{-1} \left({ y,z }\right) }\right)$ as $J_0$ forms an Order Isomorphism.

Since $J_0$ is a bijection:

 $\displaystyle \left\vert{J_0 \left({y, z}\right) }\right\vert$ $=$ $\displaystyle \left\vert{J_0 \left({R_0^{-1} \left({y, z}\right) }\right) }\right\vert$ Equality shown above $\displaystyle$ $=$ $\displaystyle \left\vert{ R_0^{-1} \left({y, z}\right) }\right\vert$ Image of Bijection Cardinal Equality

Take $\max \left({y, z}\right)$.

 $\displaystyle \left({a, b}\right)$ $\in$ $\displaystyle R_0^{-1} \left({y, z}\right)$ hypothesis $\displaystyle \iff \ \$ $\displaystyle \left({a, b}\right)$ $R_0$ $\displaystyle \left({y, z}\right)$ Definition of Image $\displaystyle \iff \ \$ $\displaystyle a, b$ $\le$ $\displaystyle \max \left({y, z}\right)$ Definition of Canonical Order $\displaystyle \iff \ \$ $\displaystyle a, b$ $\in$ $\displaystyle \max \left({y, z}\right) + 1$ Definition of Successor

It follows that:

 $\displaystyle \left\vert{R_0^{-1} \left({y, z}\right) }\right\vert$ $\le$ $\displaystyle \left\vert{\max \left({y, z}\right) + 1 \times \max \left({y, z}\right) + 1}\right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert{\max \left({y, z}\right) + 1}\right\vert$ Inductive hypothesis $\displaystyle$ $<$ $\displaystyle \left\vert{x}\right\vert$ Hypothesis for the case

Therefore:

$\left\vert{ R_0^{-1} \left({y, z}\right) }\right\vert < \left|{ x }\right|$
$\forall y, z \in x: J_0 \left({y, z}\right) < \left|{x}\right|$

It follows by Supremum Inequality for Ordinals that:

$J_0 \left({x \times x}\right) \subseteq \left|{x}\right|$

Hence:

 $\displaystyle \left\vert{x \times x}\right\vert$ $=$ $\displaystyle \left\vert{J_0 \left({x \times x}\right) }\right\vert$ Image of Bijection Cardinal Equality $\displaystyle$ $\le$ $\displaystyle \left\vert{x}\right\vert$ Subset of Ordinal implies Cardinal Inequality

But also by Set Less than Cardinal Product:

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

Thus:

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

$\blacksquare$