Cardinal Equal to Collection of All Dominated Ordinals

Theorem
Let $S$ be a set.

Let $\preccurlyeq$ denote the dominance relation.

Let $\operatorname{On}$ denote the class of all ordinals.

Set $x = \left\{ y \in \operatorname{On} : y \preccurlyeq S \right\}$

Then:


 * 1) $x$ is an element of the class of cardinals.
 * 2) There is no injection $f$ such that $f : x \to S$

$x$ is an ordinal
$x$ is clearly a subset of the class of all ordinals.

Moreover, suppose $y \in x$ and $z \in y$.

Then $z \subseteq y$ by the fact that $y$ is an ordinal.

$y \in x$ means that $f : y \to S$ for some injective mapping $f$ by the definition of dominance.

But then, $f \restriction_z : z \to S$ is also an injection by Restriction of Injection is Injection.

Thus, $z \in x$ by the definition of $x$, so $x$ is transitive.

Therefore, $x$ is an ordinal, since it is a transitive subset of the ordinal class.

$x$ is a cardinal
Assume $x$ is not a cardinal.

It follows that $\left|{ x }\right| \in x$ and $x \sim \left|{ x }\right|$ by Cardinal of Cardinal Equal to Cardinal/Corollary and Ordinal Number Equivalent to Cardinal Number.

Let $g : x \to \left|{ x }\right|$ be a bijection and let $f : \left|{ x }\right| \to S$ be an injection.

Then $f \circ g : x \to S$ is an injection by Composite of Injections is Injection.

It follows that $x \preccurlyeq S$, so $x \in x$.

This means that $x \subsetneq x$ by Ordinal Proper Subset Membership, which is a contradiction.

Therefore, $x$ must be an element of the class of cardinals.

No injection $f : x \to S$
Suppose that there is an injection $f : x \to S$.

By the definition of $x$, it follows that $x \in x$.

This means that $x \subsetneq x$ by Ordinal Proper Subset Membership, which is a contradiction.

Therefore, no injection can exist.