Cardinal of Cardinal Equal to Cardinal/Corollary
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Theorem
Let $\NN$ denote the class of all cardinal numbers.
Let $x$ be an ordinal.
Then:
- $x \in \NN \iff x = \card x$
Proof
Necessary Condition
Suppose $x = \card x$.
Then $x = \card y$ for some $y$ by Existential Generalisation.
By definition of class of all cardinals:
- $\NN = \set {x \in \On: \exists y: x = \card y}$
It follows that $x \in \NN$.
$\Box$
Sufficient Condition
\(\ds x\) | \(\in\) | \(\ds \NN\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \card y\) | Definition of Class of All Cardinals | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \card x\) | \(=\) | \(\ds \card y\) | Cardinal of Cardinal Equal to Cardinal | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \card x\) | Equality is Transitive |
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.38$