Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $N\models ZFC$?

Thanks!

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
Thank you, I have been looking for this for a while. –  Kyle Gannon Jul 13 at 3:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.