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.

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:

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:

Since $J_0$ is a bijection:

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

It follows that:

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

Thus by Cardinal Inequality implies Ordinal Inequality:
 * $\forall y, z \in x: J_0 \left({y, z}\right) < \left|{x}\right|$

It follows by Supremum Inequality for Ordinals that:
 * $J_0 \left({x \times x}\right) \subseteq \left|{x}\right|$

Hence:

But also by Set Less than Cardinal Product:
 * $\left|{x}\right| \le \left|{x \times x}\right|$

Thus:
 * $\left|{x}\right| = \left|{x \times x}\right|$