View source for 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 <cite> friedman1975:onehundred </cite> 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[924] 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 <cite> gaifman1976:models </cite>, Schmerl <cite> schmerl1981:recursively </cite> 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 <cite> kaufmann1977:arather </cite>, 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 <cite> hodges1985:building </cite>: 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 <cite> gaifman1976:models </cite> and Knight <cite> knight1976:hanf </cite> 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 <cite> kossak1995:four </cite>.

== Rigid models ==

There are $\aleph_1$-like rigid models of $PA$. Two different constructions are given in <cite> kossakschmerl1991:minimal </cite> and  <cite> koskot1992:game </cite> 

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 <cite> kaufmannschmerl1984:saturation </cite> (and also in <cite> kaufmannschmerl1987:remarks </cite>), 
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 
<cite> kaufmannschmerl1987:remarks </cite>, but the question remains open.

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