# Small question on proof that any ordinal is the index of some initial ordinal

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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$\omega_\eta$ 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.
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