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I'm having trouble understanding initial ordinals. In particular, I can't prove a seemingly trivial theorem about them.

Def: An ordinal $\kappa$ is an initial ordinal iff $\forall \delta < \kappa \, \vert \delta \vert < \vert \kappa \vert$.

Proposition. For all ordinals $\alpha$ there is an initial ordinal $\kappa$ such that $\vert \alpha \vert = \vert \kappa \vert$. This proposition does not us AC.

I have tried so many different permutations for this proposition, but none seem to work. This means there is something that I don't understand concerning initial ordinals. My first approach was to let $\kappa$ be the least ordinal such that $\vert \alpha \vert = \vert \kappa \vert$, and then show that $\kappa$ is initial. The proof must be something along these lines. Any help would be greatly appreciated.

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Your initial approach is correct. Since the identity function is a bijection between $\alpha$ and itself, $|\alpha|=|\alpha|$. Therefore the set $\{\beta\leq\alpha\mid |\beta|=|\alpha|\}$ is non-empty. So it has a least element.

Call that least element $\kappa$. To show that it is an initial ordinal, if $\delta<\kappa$, and there is a bijection between $\delta$ and $\kappa$, show that it means that there is one between $\delta$ and $\alpha$. This contradicts the minimality of $\kappa$.


Note that choice was not used at all. We utilize the fact that $\alpha$, or rather $\alpha+1$ is already a naturally well-ordered set. We only use the definition of an ordinal.

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Let $\alpha$ be an ordinal such that for every $ \beta<\alpha$ there is an initial ordinal $\kappa$ such that $|\beta|=|\kappa|$.

If $|\alpha|=|\beta|$ for some $\beta<\alpha$ then initial ordinal $|k|$ exists with $|\alpha|=|\beta|=|\kappa|$.

If no such $\beta$ exists then it can be deduced that $|\beta|<|\alpha|$ for every $\beta<\alpha$ so that $\alpha$ is an initial ordinal itself.

Proved is now by induction that for every ordinal $\alpha$ there is an initial ordinal $\kappa$ such that $|\alpha|=|\kappa|$.

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