Cardinal Equal to Collection of All Dominated Ordinals

From ProofWiki
Jump to navigation Jump to search

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 $f : x \to S$


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

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 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 cardinals.

$\Box$


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

Therefore, no injection can exist.

$\blacksquare$


Sources