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Let $k^+=\{\alpha : \alpha \hspace{2mm} \text{is an ordinal,and} \hspace{2mm} |\alpha|<\kappa\}$ where $k^+$ is an infinite cardinal.

I am trying to understand the proof the the following statement:

If $\kappa$ is a cardinal, then $\kappa^+$ is the smallest cardinal which is bigger than $\kappa$.

The proof begins with the following statement,

If $\alpha < \kappa^+$, then $|\alpha| \le \alpha < \kappa^+$.

I am not sure how to interpret $| \alpha| \le \alpha$. The cardinality of an ordinal is less than the ordinal? So, the cardinal $\kappa$ for which there exists a bijection $f: \kappa \to \alpha$ is such that $\kappa \subseteq \alpha$?

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Apparently, the cardinals are (not unusual) the smallest ordinal having a given cardinality. Thus $\lvert\alpha\rvert$ is the smallest ordinal having the same cardinality as $\alpha$, and thus $\lvert\alpha\rvert \leqslant \alpha$. – Daniel Fischer Apr 29 '14 at 14:08
Note that $|\alpha|$ is defined as the smallest of ordinals equipotent with $\alpha$. Clearly, $\alpha$ is a member of this set. – Marcin Łoś Apr 29 '14 at 14:09
But, in the definition of $k^+$, it ought to be $\lvert\alpha\rvert \leqslant \kappa$, not $< \kappa$. – Daniel Fischer Apr 29 '14 at 14:11
You must consider the infinite case : $\omega$, $\omega + 1$, $ω.2$, $ω^2$ are all different but they have all the same cardinality : $\aleph_0$. – Mauro ALLEGRANZA Apr 29 '14 at 15:14

First of all, recall that $|\alpha|$ is an ordinal itself which satisfy that for all $\beta<|\alpha|$, there is no bijection from $\beta$ onto $|\alpha|$.

And by definition $|\alpha|$ is also the least ordinal such that $|\alpha|$ and $\alpha$ have a bijection. In particular it cannot be the case that $\alpha<|\alpha|$. Therefore $|\alpha|\leq\alpha$.

(That been said, in the definition of $\kappa^+$ it should really be $|\alpha|\leq\kappa$, as Daniel Fischer remarks in the comments.)

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