5
$\begingroup$

Consider a complete first order theory $T$ whose language contains a binary predicate $\leq$. Assume that $T$ has an uncountable model that is well-ordered by $\leq$ so that this question isn't stupid and assume for simplicity that $T$ is countable.

If we want to restrict our attention to models $\mathcal{M}$ of $T$ for which $\leq$ actually well-orders $M$ we immediately get a kind of downward Löwenheim–Skolem theorem since any sub-order of a well-order is automatically well-ordered. The naive analog of the upward Löwenheim–Skolem theorem is false because the only well-ordered model of $\text{Th}(\mathbb{N})$ is the standard one, but there is still the possibility of analogous statements requiring the existence of uncountable well-ordered models.

Ultraproducts of well-ordered sets with regards to countably complete ultrafilters are well-ordered, so assuming that they exist (i.e. assuming the existence of a measurable cardinal) and assuming $T$ has a well-ordered model whose cardinality can be increased by a countably complete ultrapower, then we get a strictly larger well-ordered model. (I assume that this works for all uncountable cardinals or maybe for all cardinals with uncountable cofinality?) It doesn't necessarily follow that we have arbitrarily large well-ordered models because unions of elementary chains of well-ordered sets aren't necessarily well-ordered, unless countably complete ultrapowers of ordinals are always end-extensions, which I don't know one way or the other.

There are a lot of questions: Is there a sufficient condition for the existence of arbitrarily large well-ordered models of $T$ (assuming the existence of a measurable cardinal if necessary)? Is there a way of doing it without measurable cardinals? Is it possible to characterize which order types well-ordered models of $T$ can be? When do models of $T$ have proper elementary end extensions? (And incidentally when do well-ordered models of $T$ have initial segments that are elementary sub-structures?)

$\endgroup$
8
$\begingroup$

A very nice collection of questions.

Here are a few things one can say to get started.

  • If $\kappa$ is a measurable cardinal and $T$ has a well-ordered model of size at least $\kappa$, then it has arbitrarily large well-ordered models. To see this, suppose that $M$ is a model of $T$ in which $\leq$ is a well-order and $M$ has size at least $\kappa$. Since $\kappa$ is measurable, we may by iterating the ultrapower maps find an elementary embedding $j:V\to N$ into a transitive class $N$ with critical point $\kappa$ and $j(\kappa)$ as large as desired. So $j(M)$ is a model of $T$, with the order $j(\leq)$ of size at least $j(\kappa)$ and a well-order in $N$ and hence also still a well-order in $V$.

  • One can do it with less than a measurable. Specifically, if $\kappa$ is merely an unfoldable cardinal (this is consistent with $V=L$, so much smaller than measurable, but above indescribable), and $T$ has a well-ordered model of size at least $\kappa$, then it has arbitrarily large well-ordered models. To see this, note first that a downward Löwenheim-Skolem argument shows that $M$ has well-ordered models of size exactly $\kappa$. Now, we can place this model into a transitive model $N$ of size $\kappa$, and then apply unfoldability to get elementary embeddings $j:N\to \bar N$ with $j(\kappa)$ as large as desired. The same reasoning as before shows that $j(M)$ is a well-ordered model of large size.

You can get a kind of converse from the same idea, using the extension property characterization of unfoldability, if one allows the language to become larger.

Theorem. A cardinal $\kappa$ is unfoldable if and only if every theory $T$ of size at most $\kappa$ with a model of size at least $\kappa$ in which $\leq$ is a well-order, has well-ordered models of arbitrarily large size.

The idea is to write down the diagram of $\langle V_\kappa,\in,A,\leq\rangle$, where $\leq$ is a well-order of $V_\kappa$ and $A\subset V_\kappa$, and then get arbitrarily large extensions $\langle N,\in,A^*,\leq^*\rangle$, which will give you the extension version of the unfoldability property.

It seems to me that your questions are very closely related to how one often thinks about unfoldable cardinals.

If you are only seeking end-extensions, rather than arbitrarily large extensions, then it is weak compactness, a weaker large cardinal, that you will want. An inaccessible cardinal $\kappa$ is weakly compact if and only if for every set $A\subset V_\kappa$, there is an elementary extension of $\langle V_\kappa,\in,A\rangle$ to a taller well-founded model $\langle M,\in,A^*\rangle$.

$\endgroup$

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.