Please note that this might be some confusion on my part about the work surrounding Vaught's conjecture.

First of all, Vaught Conjecture states that if a first-order complete theory $T$ in a countable language has infinitely many models of size $\aleph_0$, then either $I(T,\aleph_0)=\aleph_0$ or $I(T,\aleph_0)=2^{\aleph_0}$. We will denote this conjecture as $VC$.

It has been shown that if $I(T,\aleph_0)> \aleph_1$, then $I(T,\aleph_0)=2^{\aleph_0}$. Therefore, we know that $ZFC+CH \vdash VC$.

If one wants to prove that $VC$ is false, then one must construct a model $M$ such that $M\models ZFC$ and a theory $T_1$, such that $M\models ZFC+ \neg CH$ and $M \models I(T_1,\aleph_0)=\aleph_1$.

But here lies my problem: Let $M_1,M_2 \models ZFC + \neg CH$. Suppose that we can construct first order theories $T_1$ and $T_2$ such that $M_1 \models I(T_1,\aleph_0)=\aleph_1$ and $M_2 \models I(T_2,\aleph_0)=\aleph_1$. Is it possible that $M_1\models I(T _2,\aleph_0)=2^{\aleph_0}$ and $M_2\models I(T_2,\aleph_0)=2^{\aleph_0}$?

Better yet, we can simply ask: if $T$ is a counterexample to Vaught's Conjecture in some model $M$ of $ZFC$ must it be the case that $N\models I(T,\aleph_0)=\aleph_1$ for any $N$ such that $N\models ZFC$?



The answer is yes. See section 5 of the paper "Bounds on weak scattering" by Sacks (http://www.math.harvard.edu/~sacks/bws.pdf); he cites Morley 1970 ("The number of countable models," http://www.jstor.org/stable/2271150) as the original proof.


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.