User blog comment:Wythagoras/My Turing machines/@comment-6768393-20131012123348/@comment-5529393-20131013131946

@Wythagoras: I'm not sure that I agree with your descriptions. If the first two first level diagonalizers are $$\Omega, I$$, then naturally the next diagonalizer should be $$I_2$$, as $$I$$ collapses functions of $$\Omega$$ and $$I_2 $$ collapses functions of $$I$$. Then the sequence should continue $$I_3, I_4, ... $$.

Then we have $$\chi$$ and $$M$$, which diagonalize over the I hierarchy. So $$M$$ would be a second level diagonalzer. We then diagonalize over $$M$$ using $$I_{M+1}$$, establishing a whole new hierarchy of level one diagonalizers. We can then diagonalize over this hierarchy using $$M_2$$, the second level two diagonalizer. Then level two diagonalizers then continue $$M_3, M_4, \ldots $$, remembering that there is a entire hierarchy of level one diagonalizers between any two level two diagonalizers.

So, we next go to the ordinal that diagonalizes over the level two diagonalizers, which would be $$\Xi(3,0)$$. This is followed by a hierarchy of level two diagonalizers $$\Xi(2,\Xi(3,,0)+1), \Xi(2, \Xi(3,0) + 2), \ldots$$. (Each of which of course is separated by a hierarchy of level one diagonalizers.)  This takes us to the second level three diagonalizer, $$\Xi(3,1)$$, which diagonalizers over our brand new set of level two diagonalizers. Similarly we continue $$ \Xi(3,2), \Xi(3,3), \Xi(3,4)$$ and so on.

The fourth level diagonalizers are therefore $$\Xi(4,\alpha)$$, and the $$\alpha$$th level diagonalizers are $$ \Xi(\alpha, \beta)$$. K is a sort of hyper-diagonalizer, which diagonalizes over the levels of diagonalizers.