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It is a theorem that every $\omega$-stable first-order theory is totally transcendental. The only proof I have seen of this theorem relies on the following fact:

Let $T$ be a first-order theory, $\mathbb{M}\models T$ a monster model, and $\text{RM}$ the Morley-rank function for $\mathbb{M}$. If $\phi$ is an ${\mathcal{L}}_{\mathbb{M}}$-formula with $\text{RM}(\phi)=\infty$, then there is an ${\mathcal{L}}_{\mathbb{M}}$-formula $\psi$ such that $\text{RM}(\phi\wedge\psi)=\text{RM}(\phi\wedge\neg\psi)=\infty$.

In other words, we can split any definable set of infinite Morley rank into two definable sets of infinite Morley rank. Intuitively this is plausible, but I don't see how to prove it. (The closest I can come is that for any ordinal $\alpha$, we can split a definable set of infinite Morley rank into a definable set of infinite Morley rank and a definable set of rank $\alpha$.) Marker's book and page 10 of these notes both argue as follows:

Let $\beta=\text{sup}\big\{\text{RM}(\psi):\psi\text{ an }{\mathcal{L}}_{\mathbb{M}}\text{-formula and RM}(\psi)<\infty\big\}$. Because $\text{RM}(\phi)=\infty\geq\beta+2$, we can find an ${\mathcal{L}}_{M}$-formula $\psi$ such that $\text{RM}(\phi\wedge\psi)\geq\beta+1$ and $\text{RM}(\phi\wedge\neg\psi)\geq\beta+1$. Then $\text{RM}(\phi\wedge\psi)=\text{RM}(\phi\wedge\neg\psi)=\infty$.

But unless we assume that $\beta<\infty$, the second sentence of this argument appears to beg the question. For if $\beta=\infty$, then second sentence simply reasserts that there is a $\psi$ with the desired property.

My question is, just how do we know that if $\text{RM}(\phi)=\infty$, then there is a $\psi$ with $\text{RM}(\phi\wedge\psi)=\text{RM}(\phi\wedge\neg\psi)=\infty$? Am I not understanding the quoted paragraph?

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    $\begingroup$ Correct me if I'm wrong, but isn't $\beta$ an ordinal? i.e $\beta < \infty$ $\endgroup$ – Jihoon Kang Jul 19 '19 at 0:12
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    $\begingroup$ Yes, as Andreas Blass's answer points out. That is what I was overlooking. $\endgroup$ – Anonymous Jul 19 '19 at 3:38
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$\beta$ is, as Jihoon Kang commented, intended to be a genuine ordinal, not $\infty$. It is defined, at the beginning of the quoted proof, as the supremum of the genuine ordinals that occur as Morley ranks of formulas. (So this depends on knowing that this supremum exists, which is true as long as your monster model is a set and not a proper class.) The next steps in the proof produce $\psi$ such that the Morley ranks of both $\phi\land\psi$ and $\phi\land\neg\psi$ are strictly bigger than $\beta$. If these two ranks were genuine ordinals, they'd be $\leq\beta$, by the definition of $\beta$ as a supremum. Since they're not $\leq\beta$, they can't be genuine ordinals, so they're $\infty$.

Here's another way to look at the situation. Instead of working with a particular $\beta$, consider completely arbitrary ordinals $\beta$ (but still genuine ordinals, not $\infty$). The argument gives you, for any $\beta$, a formula $\psi$ such that the Morley ranks of both $\phi\land\psi$ and $\phi\land\neg\psi$ are strictly bigger than $\beta$. But there's a proper class of ordinals $\beta$ and only a set of formulas $\psi$. So the same $\psi$ must occur for a proper class of $\beta$'s. For that $\psi$, the Morley ranks of both $\phi\land\psi$ and $\phi\land\neg\psi$ are $\infty$.

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  • $\begingroup$ Thanks, all is clear now. My mistake was to think that there could be L-M-formulae of arbitrarily large ordinal rank, which, as you point out, is not possible, because the L-M-formulae form a set. $\endgroup$ – Anonymous Jul 19 '19 at 3:27

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