A theorem of Shelah (which has been quoted several times on this website) states that if $T$ is a countable unstable theory and $\kappa$ is an uncountable cardinal, then $T$ has $2^\kappa$ non-isomorphic models of cardinality $\kappa$. I was wondering if there is an elementary way ($\mathit{i.e.}$ requiring very little model and set theory) to illustrate this theorem in the special case of dense linear orders. In particular, I would like to construct explicitly $2^{\mathfrak{c}}$ pairwise non-isomorphic dense linear orders of cardinality $\mathfrak{c}$. My starting point was to consider the real line (“horizontally”) and to attach to every real number on the line a “vertical” fiber, which would be either $\mathbb{R}$ or $\mathbb{Q}$, and then order the whole subset of $\mathbb{R}^2$ obtained this way by the lexicographic order. Given any $A\subseteq\mathbb{R}$, I would then consider the dense linear order $\mathcal{M}_A$ which has $\mathbb{R}$ as a vertical fiber over every $x\in A$ and $\mathbb{Q}$ as a vertical fiber over every $x\notin A$. I have produced $2^{\mathfrak{c}}$ models of cardinality $\mathfrak{c}$, which unfortunately are not pairwise non-isomorphic (for example, whenever $A$ and $B$ are singletons, $\mathcal{M}_A$ and $\mathcal{M}_B$ are isomorphic). So I'm not sure this is the way to go. Any thoughts?
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Start with $\kappa$ and replace each ordinal with either $(\omega_1^*+\omega_1)\times \mathbb{Q}$ or $(1+\omega_1^*+\omega_1)\times \mathbb{Q}$. This gives $2^{\kappa}$ many nonisomorphic dense orders.
answered Dec 26, 2017 at 16:11
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1$\begingroup$ @R.Harlow 1. The basic point is that $\omega_1+1+\omega_1^*$ is not isomorphic to $\omega_1 + \omega_1^*$. In the first, the cut of uncountable cofinality is filled, and in the second it isn't. On the other hand, $\omega_1 + 1+\omega_1$ is isomorphic to $\omega_1 + \omega_1$, so the argument wouldn't work. 2. The $\times\mathbb{Q}$ is just there to make the order dense. 3. The base is taken to be well-ordered so that it's rigid. If you start with $\mathbb{R}$, you have the problem that the base order already has lots of automorphisms, which makes things more complicated. $\endgroup$ Dec 27, 2017 at 16:41
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1$\begingroup$ 4. Let $O_1$ and $O_2$ be two orders constructed as above, and suppose they are isomorphic. The isomorphism establishes a bijection between the cuts of uncountable cofinality in $O_1$ and those in $O_2$. These cuts are ordered as $(\kappa,<)$, and $(\kappa,<)$ is rigid, so there is only one choice for the induced bijection on cuts. Hence a cut in $O_1$ is filled if and only if the corresponding cut in $O_2$ is filled, so $O_1$ and $O_2$ were constructed from the same subset of $\kappa$. $\endgroup$ Dec 27, 2017 at 16:47
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1$\begingroup$ @R.Harlow No, because for any two countable linear orders $L$ and $L'$, $L\times \mathbb{Q}$ and $L\times \mathbb{Q}$ are isomorphic (being countable dense linear orders). $\endgroup$ Dec 27, 2017 at 20:15
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2$\begingroup$ No, this is where the uncountability assumption comes in. Remember you are replacing the ordinals by copies of the rational, so you need gaps of uncountable cofinality. $\endgroup$ Dec 27, 2017 at 20:16
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1$\begingroup$ @ratalan You can find the definitions at the beginning of this MO question. I was being a bit sloppy with language - I should have referred to cuts of cofinality type $(\aleph_1,\aleph_1)$. $\endgroup$ Dec 29, 2017 at 18:16