Cardinal Equal to Collection of All Dominated Ordinals

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In particular: The use of the classes of ordinals and cardinals is needlessly obfuscating. We can avoid these constructs altogether and just talk about ordinals and cardinals
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Theorem

Let $S$ be a set.

Let $\preccurlyeq$ denote the dominance relation.

Let $\On$ denote the class of all ordinals.

Let $x = \set {y \in \On: y \preccurlyeq S}$


Then:

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


Proof

$x$ 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.

$\Box$


$x$ is a cardinal

Aiming for a contradiction, suppose $x$ is not a cardinal.

It follows that $\size x \in x$ and $x \sim \set x$ by corollary to Cardinal of Cardinal Equal to Cardinal and Ordinal Number Equivalent to Cardinal Number.


Let $g: x \to \size x$ be a bijection.

Let $f: \size x \to S$ 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 Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, which is a contradiction.

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

$\Box$


No injection $f: x \to S$

Aiming for a contradiction, 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 Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, which is a contradiction.

Therefore, no injection can exist.

$\blacksquare$


Sources

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