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Let $\mathfrak{M}$ be a countable transitive model of set theory.

Let $L$ be some countable language and $A$ be a countable (in $\mathfrak{M}$) $L$-structure. My question is:

  1. In $\mathfrak{M}$ can we carry the construction of Scott sentence of $A$ $\sigma(A)^\mathfrak{M}$?

  2. Is $\sigma(A)^\mathfrak{M}$ identical with the Scott sentence of $A$ in the real world?

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up vote 7 down vote accepted
  1. Of course.

  2. The Scott sentence is not syntactically unique, it is only defined up to equivalence. Its defining property ($A\models\sigma$, and every countable model of $\sigma$ is isomorphic to $A$) is $\Pi^1_2$, hence any sentence satisfying it in $\mathfrak M$ will also satisfy it in the real world by Shoenfield’s absoluteness theorem if $\omega_1\subseteq\mathfrak M$. The same is in fact true for every transitive model $\mathfrak M$ of ZFC, but this seems to require a more complicated argument.

    First, notice that equivalence of $L_{\omega_1,\omega}$-sentences (which is the same as equivalence on countable models, by Löwenheim–Skolem) is $\Pi^1_1$, hence it is absolute irrespective of $\omega_1\subseteq\mathfrak M$. It thus suffices to show the result for a particular Scott sentence. One construction of Scott sentences yields a sentence of the form $$\sigma=\sigma_\varnothing\land\bigwedge_{\vec a}\forall\vec x\,\Bigl(\sigma_{\vec a}(\vec x)\to\bigwedge_c\exists y\,\sigma_{\vec a,c}(\vec x,y)\land\forall y\,\bigvee_c\sigma_{\vec a,c}(\vec x,y)\Bigr),$$ where $\vec a$ runs over finite sequences of elements of $A$, $\vec x$ has matching length, $c\in A$, and $\sigma_{\vec a}(\vec x)$ is an $L_{\omega_1,\omega}$-formula such that $A\models\sigma_{\vec a}(\vec a)$, and $\sigma_{\vec a}(\vec x)$ implies the diagram of $\vec a$. Assume that we have carried out this construction in $\mathfrak M$, and let $B$ be a model of $\sigma$ in the real world. Then $\{\langle\vec a,\vec b\rangle:B\models\sigma_{\vec a}(\vec b)\}$ is a back-and-forth system between $A$ and $B$, hence $A\equiv_{L_{\infty,\omega}}B$; in particular, if $B$ is countable, then $A$ and $B$ are isomorphic.

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Shoenfield absoluteness doesn't apply, because $\omega_1 \not\subset \mathfrak{M}$. – Trevor Wilson May 8 '14 at 16:45
You are right, I’ll fix it. – Emil Jeřábek May 8 '14 at 17:13
In case it’s relevant, the argument does not need full choice: everything works fine if $V$ and $\mathfrak M$ satisfy just ZF + DC. – Emil Jeřábek May 8 '14 at 18:11
Thank you, nice answer. – user38200 May 9 '14 at 10:01

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