Recall Vaught conjecture: the number of countable models of a first-order complete theory in a countable language is finite or $\aleph_0$ or $2^{\aleph_0}.$

Now let $\lambda$ be an uncountable cardinal. Is the following version of Vaught conjecture true:

Vaught conjecture for uncountable languages. If $T$ is a complete theory in a language of size $\lambda,$ and if $T$ has more than $\lambda-$many non-isomorphic models of size $\lambda$, then $T$ has $2^\lambda$ many non-isomorphic models models of size $\lambda.$

Also, do we have the following version of Moreley theorem:

Morley theorem for uncountable languages. If $T$ is a complete theory in a language of size $\lambda,$ and if $T$ has more than $\lambda^+-$many non-isomorphic models of size $\lambda$, then $T$ has $2^\lambda$ many non-isomorphic models models of size $\lambda.$

  • $\begingroup$ I am wondering about your formulation of "Morley's theorem for uncountable languages". In fact, there is a theorem with this name and it has been proven by Shelah. It states: If a theory in a language with cardinality $\kappa \geq \omega$ is categorical in some cardinality greater than $\kappa$, then it is categorical in every cardinality greater than $\kappa$. However, I cannot immediately see how this fits to your variant of Morley's theorem. $\endgroup$ – user42061 Oct 30 '13 at 13:29
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    $\begingroup$ tobi, I think he's referring to the following theorem of Morley (which can be found in chapter 4 of Marker's book): Let $T$ be a complete theory in a countable language. If $I(T,\aleph_0)>\aleph_1$ then $I(T,\aleph_0)=2^{\aleph_0}$. $\endgroup$ – Haim Oct 30 '13 at 18:52
  • $\begingroup$ Thanks Haim, now it makes more sense to me! I did not knew this theorem till now. $\endgroup$ – user42061 Oct 30 '13 at 20:31
  • $\begingroup$ By the way, it's interesting to note that the proof of Morley's theorem relies on the fact that if $X$ is countable and $Y\subseteq 2^{X}$ is analytic, then either $|Y|\leq \aleph_0$ or $|Y|=2^{\aleph_0}$. So, towards a proof of a generalized Morley theorem, it would be natural to expect a generalization of this classical result from descriptive set theory. Perhaps something like that was already done by Sy Friedman and his collaborators? I don't know. $\endgroup$ – Haim Oct 31 '13 at 0:20

Suppose that $2^{\aleph_{1}} > 2^{\aleph_{0}} > \aleph_{2}$, and consider the language with unary predicates $P_{n}$ $(n \in \omega)$ and $Q_{\alpha}$ $(\alpha < \omega_{1})$. Consider the theory $T$ which says that (1) each $P_{n}$ is satisfied by infinitely many points (2) no point satisfies more than one $P_{n}$ (3) no point satisfies any $Q_{\alpha}$.

This appears to me to be complete, as for each $n \in \omega$ you can argue by complexity of formulas that for any $k$-ary formula $\phi$ not using any of $P_{m}$ ($m \geq n$), any two models $M$ and $N$ of $T$ and any two tuples $\langle a_{0},\ldots,a_{k-1} \rangle$ from $M$ and $\langle b_{0},\ldots,b_{k-1} \rangle$ from $N$, if $M \models P_{p}(a_{i})$ if and only if $N \models P_{p}(b_{i})$ for each $p < n$ and each $i < k$ then $M \models \phi(a_{0},\ldots,a_{k-1})$ if and only if $N \models \phi(b_{0},\ldots,b_{k-1})$. Using this fact, one can show that the second player wins the back-and-forth game using $M$ and $N$ of any fixed finite length, restricting to any finite set of the $P_{n}$'s.

Now $T$ appears to have exactly continuum many models of cardinality $\aleph_{1}$ up to isomorphism, determined by how many points satisfy each $P_{n}$ (two options in each case), and how many satisfy none of them (countably many options).

  • $\begingroup$ This is a counterexample to VC as stated, but this is why we usually use $|T|$ (the number of $T$-inequivalent formulas) instead of $|L|$ (the size of the language). Because of course this is a countable theory; the $Q_\alpha$ do nothing. Can this be modified to be a "truly" uncountable language and still work? $\endgroup$ – Richard Rast Sep 20 '14 at 13:12
  • $\begingroup$ It seems that there are of lots of things you could do to make the $Q_{\alpha}$'s act nontrivially but harmlessly, like making them all be satisfied by a unique point not satisfying any $P_{n}$ or any other $Q_{\beta}$. Or, more generally, letting them act on a separate domain in a manner producing at most $\lambda$ many models. I imagine that this is still not very interesting. Can one give a precise meaning for "truly uncountable"? $\endgroup$ – Paul Larson Sep 20 '14 at 18:02
  • $\begingroup$ Sure you can- the typical definition is to say there are uncountably many T-inequivalent formulas. I think your patch would work, so the example should satisfy as a counterexample to this conjecture. $\endgroup$ – Richard Rast Sep 21 '14 at 3:42

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