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.