# Uncountable models with interesting second-order properties

## Friedman's 14th problem

Let $T$ be a completion of PA. Let ${\rm Ot}(T)$ be the spectrum of order types of nonstandard models of $T$. In  Harvey Friedman asked: Does ${\rm Ot(T)}$ depend on $T$?

Shelah has recently announced that the answer is negative, but there is no paper available yet. See Sh for an interesting effort in the opposite direction.

## Rather classless models

$X\subseteq M\models PA$ is a class if for all $a\in M$, $\{x\in X: x<a\}$ is definable (coded) in $M$. $M$ is rather classless if each class of $M$ is definable.

Using the fact that every model of $PA$ has a conservative elementary end extension , Schmerl  proved that for each cardinal $\kappa$ such that ${\rm cf}(\kappa)>\aleph_0$, there are $\kappa$-like rather classless models of $PA$. A model is $\kappa$-like if it is of cardinality $\kappa$ and each of its proper initial segments is of smaller cardinality.

Kaufmann , assuming $\lozenge$, proves that there are recursively saturated $\aleph_1$-like rather classless models. Later Shelah showed that $\lozenge$ can be eliminated from the proof. Nevertheless one can still ask, as Hodges did in : Prove the existence of rather classless recursively saturated models of $PA$ in cardinality $\aleph_1$ without assuming diamond at any stage of the proof.

## Jonsson models

A model $\mathfrak B$ is Jonsson if $|{\mathfrak B}|>\aleph_0$ and for every ${\mathfrak A}\prec {\mathfrak B}$, if $|{\mathfrak A}|=|{\mathfrak B}|$, then ${\mathfrak A}={\mathfrak B}$.

Gaifman  and Knight  independently showed that there are Jonsson models of $PA$. . Jonsson models $M$ of $PA$ of cardinality $\aleph_1$ are either $\aleph_1$-like or are short. A model $M$ of $PA$ is short there is an $a\in M$ such that the Skolem closure of $a$ is cofinal in $M$. Each known Jonsson model realizes uncountably many complete types.

Kossak has asked: Is there an $\aleph_1$-like Jonsson model $M\models PA$ such that $|\{{\rm tp}(a): a\in M\}|=\aleph_0$?

If $M\models PA$ is $\aleph_1$-like and recursively saturated, then $|\{{\rm tp}(a): a\in M\}|=\aleph_0$, but $M$ is not Jonsson. Therefore, another related question is: Is there a weakly Jonsson model $M\models PA$, i.e. a recursively saturated model $M\models PA$ such that for every recursively saturated $K\prec M$, if $|K|=|M|$, then $K=M$? The problem was posed in .

## Rigid models

There are $\aleph_1$-like rigid models of $PA$. Two different constructions are given in  and 

Problem: Are there rigid recursively saturated $M\models PA$ such that ${\rm cf}(M)=\aleph_0$?

## Lofty models

A model $M$ of PA is lofty if there is $e \in M$ such that whenever $a \in M$ and $\Phi = \{\varphi_i(a,x) : i < \omega\}$ is a recursive set of formulas that is finitely satisfiable in $M$, then there is a definable $D \subseteq M$ such that $M \models |D| \leq e$ and $\Phi \cup \{x \in D\}$ is finitely satisfiable in $M$. This notion was defined and studied by Kaufmann and Schmerl in  (and also in ), where the following was proved: If $M$ is a countable model of PA, then $M$ is lofty iff $M$ has a simple elementary extension (i.e. generated by a single element) that is recursively saturated. The question was asked whether the modifier "countable" could be eliminated. Further progress was made in , but the question remains open.