User blog comment:Wythagoras/All my stuff/@comment-7484840-20130708082913/@comment-5529393-20130708140221

You guys aren't using I correctly. I is not $$\Omega_{\Omega_{Omega_\ldots}}$$, it is inconcievably larger than anything you can construct using the Omega function or any recursive extension of it, just as Omega is larger than any recursive notation of countable ordinals. The psi function "collapses" Omega down phi(2,0), but one must be careful not to equate Omega with phi(2,0) - for example, $$\psi(\Omega*2)$$ is not phi(2,0)*2, but rather phi(2,1). Similarly, I does not match up with a specific smaller ordinal like $$\Omega_{\Omega_{Omega_\ldots}}$$, but rather works as a diagonlization operator over the $$\alpha \mapsto \Omega_\alpha$$ function.

The first fixed point of $$\alpha \mapsto \Omega_\alpha$$ is $$\psi_I(0)$$. The second fixed point is $$\psi_I(1)$$, and the 1+alphath fixed point is $$\psi_I(\alpha)$$. So what you are calling $$I_{1 + \alpha}$$ is actually $$\psi_I(\alpha)$$. $$\psi_I(I)$$ diagonalizes over $$\psi_I(\alpha)$$, i.e. it is the first ordinal alpha such that $$\psi_I(\alpha) = \alpha$$. So your kappa is actually $$\psi_I(I)$$.

$$\psi_I(I \alpha + \beta) = \lambda (1 + \alpha, \beta) $$.

I hope that clarifies things.