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.

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:

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:

Since $J_0$ is a bijection:

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

It follows that:

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:

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

Thus:
 * $\card x = \card {x \times x}$