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Definition: the cardinality of a set $A, |A|$ is the least ordinal s.t. $A \sim \alpha$

Definition: We define a cardinal to be an ordinal $\alpha$ s.t. $\alpha = |\alpha|.$ i.e, an ordinal s.th. $\alpha \nsim \beta$ for all $\beta < \alpha.$

My question: If $\alpha$ is an ordinal and not a cardinal, then why $|\alpha| < \alpha?$

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  • $\begingroup$ Well, I know either $|\alpha| < \alpha$ or $|\alpha| > \alpha$ $\endgroup$ – Nafeekh Oct 25 '13 at 0:07
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One can show that if $\alpha$ is a cardinal then it does not inject into any of its members, or in other words into any smaller ordinal.

Since given two ordinals $\alpha,\beta$ one is a member of the other, if $\alpha$ is not a cardinal it injects into a smaller ordinal, and the least such ordinal must be $|\alpha|$.

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For any ordinal $\alpha$, we have $|\alpha| \leq \alpha$, because there's trivially an enumeration of a set of size $|\alpha|$ in order-type $\alpha$ - namely, the ordinal $\alpha$ itself with the usual well-ordering of its members by inclusion. $|\alpha| \lt \alpha$ here is then just the consequence of $|\alpha| \leq\alpha$ and $|\alpha|\neq\alpha$ (i.e., that $\alpha$ isn't a cardinal).

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Because the ordinals are wellordered, every class of ordinals has a least member. Let $X$ be $\{\beta \in On: \beta \sim \alpha\}$; by definition, $|\alpha|$ is the least member of $X$. $\alpha$ is obviously in $X$, and if it is not a cardinal it can only be greater than $|\alpha|$.

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