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I'm reading a proof of a theorem, but an apparently trivial case has tripped me up. The theorem goes as

For any ordinal $\alpha$, there is a initial ordinal $\phi$ such that $i(\phi)=\alpha$. Here $i(\phi)$ is the order type of the set of all the initial ordinals in $\phi$.

Suppose otherwise, that there is an ordinal $\eta$ such that $\eta$ is not the index of any initial ordinal. Let $\eta$ be the least such ordinal. Case 1: $\eta=\beta+1$. This gives a contradiction. Case 2. $\eta$ is a limit ordinal...

I must be dense, but that's the contradiction exactly? If $\eta=\beta+1$, then $\beta<\eta$, and thus there exists some initial ordinal $\psi$ such that $\beta=i(\psi)$. Then $\eta=i(\psi)+1$. Is there some way to show that $\eta$ is actually the index of some initial ordinal to get a contradiction? I'm sure I'm overlooking something small. Thanks.

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up vote 7 down vote accepted

Some notation to clarify this answer (which I wrote somewhat in a hurry):

$\omega_\eta$ is the $\eta$-th initial ordinal, i.e. $i(\omega_\eta) = \eta$. For any ordinal $\alpha$, denote by $\alpha^+$ the least initial ordinal $\eta$ such that $\alpha<\eta$.

If $\eta=\beta+1$ then there is some $\omega_\beta$ and therefore there exists $\omega_\beta^+=\omega_\eta$.

If $\eta$ is a limit, you use the replacement axiom scheme to show that the initial ordinals $\omega_\beta$ for $\beta<\eta$ is a set, take its union and show it is an initial ordinal which is larger than all those that came before.

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So for the first case, is it the same as saying in my notation that $\eta=i(\psi+1)$? – Dani Hobbes May 4 '11 at 6:04
Yes. Some extra text here. – Asaf Karagila May 4 '11 at 6:12

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