# How can I prove that if $\alpha$ is an ordinal, then there is an initial ordinal $\kappa$ such that $|\alpha|=|\kappa|$?

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.

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.

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 $$|\alpha|=|\kappa|$$.