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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 $N\models ZFC$?


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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.

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Thank you, I have been looking for this for a while. –  Kyle Gannon Jul 13 '14 at 3:17

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