4
$\begingroup$

One way to understand the Morley rank of a unary formula $\varphi(x)$ with $\text{RM}(\varphi(x)) < \omega$ is to imagine it as the height of an $\omega$-branching tree, where each node is mapped to a subset of an $\omega$-saturated model $M$. These sets are under the conditions that

  1. the set associated to the root is $\{m \in M : M \models \varphi(m)\}$;
  2. the set of a child is a subset of the parent;
  3. siblings have disjoint sets;
  4. no set is empty.

I have some difficulty extending this intuition to the case where $\text{RM}(\varphi) \geq \omega$, in particular when $\text{RM}(\varphi)$ gets "really" big, like $> \omega_1$. For $\text{RM}(\varphi) = \omega$ I imagine you have arbitrary finite height trees; for $\text{RM}(\varphi) = \omega + n$ I think of $n$-trees where the leaves can be made into arbitrary height trees; $\text{RM}(\varphi) = \omega + \omega$ is when you can find a sequence of arbitrary height trees for which every leaf can be made into an arbitrary height tree.

The fundamental difference with an infinite tree ($\text{RM}(\varphi) = \infty$) I understand as that in constructing arbitrary large finite trees we are allowed to change the formulas (sets) of the tree at each step, whereas in an infinite tree we fix it.

Are these intuitions correct? Can you improve them? More concretely I wonder:

  1. Can every ordinal occur as a Morley rank?
  2. Can you give concrete examples of a countable $M$ and $\varphi$ such that $\text{RM}(\varphi) = \omega+1, \omega+\omega, \omega_1$ (or more general)?
  3. In practice, what Morley ranks can you expect to find "in the wild"?
$\endgroup$
3
$\begingroup$

If $T$ is a totally transcendental theory (so every formula has Morley rank) in a language of cardinality $\kappa$, then the ordinals which appear as Morley ranks relative to $T$ are all less than $\kappa^+$ (i.e. they all have cardinality $\leq \kappa$).

Since we usually work with countable languages, this means that most of the Morley ranks you find "in the wild" are countable ordinals. In fact, most of the standard examples of totally transcendental theories have Morley ranks of formulas bounded above by $\omega^2$.

But it's possible to build a theory $T_\alpha$ in a language $L_\alpha$ for any ordinal $\alpha$ in which the formula $x = x$ (in one free variable $x$) has Morley rank $\alpha$. Here's a sketch of the inductive construction:

  • Let $L_0 = \emptyset$, and let $T_0$ be the theory of a singleton set: $\{\exists x\, \forall y\, (x = y)\}$.

  • Given an ordinal $\alpha$, the language $L_\alpha$, and the theory $T_\alpha$, let $L_{\alpha+1}$ consist of countably many new unary relation symbols denoted $P_n$ for $n\in \omega$, together with countably many disjoint copies of $L_\alpha$ denoted $L_{\alpha,n}$ for $n\in \omega$, so $$L_{\alpha+1} = \{P_n\mid n\in \omega\}\cup \bigcup_{n\in \omega} L_{\alpha,n}.$$ Let $T_{\alpha+1}$ be the theory which says that the relation symbols $P_n$ are pairwise disjoint, and that the elements of $P_n$ form a model of $T_{\alpha}$ in the language $L_{\alpha,n}$. So explicitly, we exchange all the $L_\alpha$-symbols in $T_\alpha$ for their corresponding $L_{\alpha,n}$-symbols, and then we relativize all quantifiers to $P_n$.

  • For a limit ordinal $\lambda$, given languages $L_\alpha$ and theories $T_\alpha$ for all $\alpha<\lambda$, let $L_\lambda$ consist of new unary relation symbols $P_\alpha$ for all $\alpha<\lambda$, together with a copy $L_{\alpha}^*$ of each language $L_\alpha$, so $$L_\lambda = \{P_\alpha\mid \alpha<\lambda\}\cup \bigcup_{\alpha<\lambda} L_\lambda^*.$$ Let $T_\lambda$ be the theory which says that the relation symbols $P_\alpha$ are pairwise disjoint, and that the elements of $P_\alpha$ form a model of $T_{\alpha}$ in the language $L_{\alpha}^*$. So explicitly, we exchange all the $L_\alpha$-symbols in $T_\alpha$ for their corresponding $L_{\alpha}^*$-symbols, and then we relativize all quantifiers to $P_\alpha$.

Note that this last step is the only one that possibly increases the cardinality of the language - and this only happens when $\lambda$ is a cardinal. In particular, $L_\alpha$ is countable for all countable ordinals $\alpha$.

Explicitly, $T_1$ is the theory of countably many disjoint singleton sets, $T_2$ is the theory of countably many disjoint sets, each of which is split into countably many disjoint singleton sets, and $T_n$ is the same, but $n$ levels deep. $T_\omega$ the theory of countably many disjoint sets, where the $n$th set is split as above $n$ levels deep. $T_{\omega+1}$ is the theory of countably many disjoint sets, each of which is split into countably many sets, where the $n$th set is split as above $n$ levels deep. Etc.

$\endgroup$
5
  • $\begingroup$ To show that $T_\alpha$ has Morley rank $\alpha$ prove QE by induction on $\alpha$. Then the result follows immediately. Correct? $\endgroup$ – Ibrahim Jun 25 '19 at 16:17
  • $\begingroup$ You've addressed all my concerns; only not the correctness of the intuition I stated in the question, do you agree with it, or is it flawed? (I know we've talked about this before, but I ask this for completeness as it is a separate question) $\endgroup$ – Ibrahim Jun 25 '19 at 16:20
  • $\begingroup$ Yes, I agree with your intuition, as long as "where the leaves can be made into arbitrary height trees" means what I think it does! $\endgroup$ – Alex Kruckman Jun 25 '19 at 16:27
  • $\begingroup$ And yes, the analysis is pretty easy given QE, but I wouldn't say "immediate". You need to argue that all of the unary relations have the right Morley rank, even when they're "embedded" in the bigger theory. $\endgroup$ – Alex Kruckman Jun 25 '19 at 16:29
  • $\begingroup$ The way to do this carefully would be to prove a lemma that whenever you create a theory by introducing a family of disjoint predicates and then "painting" each predicate $P_n$ by an $L_n$-theory $T_n$, then the translation of an $L_n$-formula $\varphi(x)$ to the new language (by relativizing to $P_n$) has the same Morley rank as $\varphi(x)$. The point is that the other symbols in the language (that didn't come from $L_n$) can't say anything non-trivial about elements of $P_n$. $\endgroup$ – Alex Kruckman Jun 25 '19 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.