[FOM] FOM query
Gregory Taylor
Gregory.Taylor at baruch.cuny.edu
Thu Jun 21 07:53:27 EDT 2018
Dear Professor Davis,
Please consider posting the following query under heading "Flawed inductive technique". (I am a FOM subscriber.) Thank you for your consideration.
****************BEGINNING OF QUERY
I wish to claim the following, omitting details.
CLAIM: where $\theta$ is a strong inaccessible and $\gamma < \theta$,
there exists a formula
and a sequence of domain elements
such that the expansion of said formula modulo saturation by said sequence is of rank no less than $\gamma$. In other words, the ranks of formulae exceed any bound below $\theta$.
The simple proof proceeds by transfinite induction on $\gamma < \theta$. The cases whereby $\gamma$ is $0$ or a successor are trivial. So suppose that $\gamma$ is a limit ordinal. Let $\delta$ be an arbitrary ordinal less than $\gamma$. By induction hypothesis, there exists some formula
and some sequence such that
the expansion of said formula with respect to said sequence is of rank no less than $\delta$.
Moreover, without loss of generality, we may assume that, for $\delta < \gamma$, the assumed formulae
may be chosen so that
their expansions modulo saturation are pairwise distinct.
(For $\delta = 0$ choose some atomic formula, for $\delta = 1$ its negation, for $\delta = 2$ the negation of that negation, and so forth; for $\delta = \omega + 1$ choose the disjunction of all the foregoing, for $\delta = \omega + 2$ the negation of that disjunction, and so forth.)
In that case, the expansion of the infinite disjunction of all the assumed formulae
is of rank not less than $\gamma$ itself.
In connection with something like this proof, an anonymous referee of an earlier paper insisted, long ago, that this technique will not work beyond $\gamma =\omega^2$ but did not explain why. If that is correct, can anyone explain why this technique will not work? Or was the referee wrong?
Gregory Taylor
****************************END OF QUERY?
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