DLO is unstable

Question

I'm trying to prove that $DLO$ is unstable (without invoking general theorems on theories defining an infinite linear order). I'll describe my attempt, although I'm now convinced that it cannot work. The proof that $DLO$ is not $\omega$-stable relies on the observation that there are at least as many $1$-types over $\mathbb{Q}$ as cuts, and there are $2^{\aleph_0}$ cuts. I've tried to mimic this for $\kappa>\aleph_0$ as follows: let $M=\kappa\times\mathbb{Q}$ ordered lexicographically. It is a model of $DLO$ of cardinality $\kappa$. Let $N$ be the set of all initial segments of $M$ ordered by inclusion. It is also a model of $DLO$ and $M$ can be embedded into $N$ as a dense subset. If I could prove that $|N|>|M|$, then I would be done, since I can separate any two points of $N$ by a point of $M$ and hence get at least as many types over $M$ as elements of $N$. In $N$ every subset which is bounded from above has a sup (the union), so I thought I'd try to mimic Cantor's proof of the uncountability of $\mathbb{R}$. However, I'm not convinced that this can work for any $\kappa$, i.e. I'm not convinced that in general $|N|>\kappa$. Any thoughts on an alternative proof?

Answer 1

In my opinion, the best definition of stability is via the order property:

• Definition: A formula $\varphi(x;y)$ has the order property if there exist $(a_i)_{i\in \omega}$ and $(b_j)_{j\in \omega}$ in $M\models T$ such that $M\models \varphi(a_i;b_j)$ if and only if $i\leq j$.
• Definition: $T$ is stable if no formula $\varphi(x;y)$ has the order property.

Now it's a theorem that $T$ is unstable if and only if for every infinite cardinal $\kappa$, there exists a set $A\subseteq M\models T$ with $|A| \leq \kappa$ such that $|S_1(A)|>\kappa$, where $S_1(A)$ is the space of $1$-types over $A$.

Since $x<y$ obviously has the order property in DLO, DLO is unstable, so you can follow the proof of this theorem to find your set $A$ with lots of types over $A$.

Here's what the proof gives you in the case of DLO. The main idea of the proof is a cardinal arithmetic trick: Since $2^\kappa>\kappa$, we can let $\mu$ be the least cardinal such that $2^\mu>\kappa$, and $\mu\leq \kappa$. Then if we define $2^{<\mu} = \bigcup_{\beta<\mu} 2^\beta$ (the $\beta$ are ordinals here), we have $|2^{<\mu}| \leq \kappa$. Why? $\mu\leq \kappa$, and $|2^\beta| = 2^{|\beta|} \leq \kappa$, since $|\beta|<\mu$, and a $\kappa$-sized union of $\kappa$-sized sets has size $\kappa$.

Now we can view $2^{<\mu}$ as a binary tree of height $\mu$ and find elements $a_{\eta}$ in a model of DLO for all $\eta\in 2^{<\mu}$ such that $a_\eta<a_{\eta'}$ if and only if $\eta$ is left of $\eta'$ in the tree. Let $A = \{a_\eta\mid \eta\in 2^{<\mu}\}$. Then $|A|\leq \kappa$ by the computation above. But every element of $2^\mu$ defines a path through the tree, which defines a cut in $A$, and distinct paths give distinct cuts, and hence distinct $1$-types over $A$, and there are $|2^\mu| > \kappa$ of these.

This is basically the proof in the general unstable case - we didn't gain much by restricting to DLO.