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

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
3  
Shoenfield absoluteness doesn't apply, because $\omega_1 \not\subset \mathfrak{M}$. –  Trevor Wilson May 8 at 16:45
    
You are right, I’ll fix it. –  Emil Jeřábek May 8 at 17:13
2  
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 at 18:11
    
Thank you, nice answer. –  user38200 May 9 at 10:01

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.