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 $x$.

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

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.

Take $R_0$ to be the canonical order of $\operatorname{On}^2$.

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

It follows that:


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

But moreover:

Since $J_0$ is a bijection,

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

It follows that:

Therefore, $\left\vert{ R_0^{-1} \left({ y,z }\right) }\right\vert \lt \left|{ x }\right|$.

Thus, $J_0 \left({ y,z }\right) < \left|{ x }\right|$ by Cardinal Inequality implies Ordinal Inequality for all $y,z \in x$.

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

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

Thus, $\left|{ x }\right| = \left|{ x \times x }\right|$.