The question is exactly that of the title. Suppose $\varphi\in V$ is an $\mathcal{L}_{\infty\omega}$-sentence, and $W$ is an inner model of $V$ such that $\varphi\in W$. Is the statement

$\varphi$ has a model

absolute between $V$ and $W$? It is clearly upwards absolute, by induction on rank, so I'm really asking about the downwards direction.

The main obstacle that I see right now is the lack of a Lowenheim-Skolem theorem, which prevents us from using Shoenfield absoluteness. As a specific example, let $A$ be $\omega_1^W$ as a linear order, and suppose $V$ is a forcing extension of $W$ in which $\omega_1$ is collapsed. In $V$, there is a sentence $\varphi\in\mathcal{L}_{\omega_1\omega}\subset\mathcal{L}_{\infty\omega}$ whose only countable model up to isomoprhism is $A$: the Scott sentence of $A$. Now there is a formula $\psi\in \mathcal{L}_{\infty\omega}\cap W$ such that, in $V$, $\psi$ and $\varphi$ are equivalent; this is a roundabout way of saying "$\varphi\in W$," which is not exactly true. But then $\psi$ has no countable models in $W$, since such a model would have to be isomorphic to $A$ in $V$.

I assume this is well-known, but I haven't been able to find an answer myself.


1 Answer 1


Here is a somewhat easier counterexample.

Let $I$ be a countable set, which is uncountable in $W$. Let $c_n$ and $d_\alpha$ be constant symbols, for $n\in\mathbb{N}$ and $\alpha\in I$. Consider the formula $\varphi$ that asserts that all the $d_\alpha$'s are different, but that every $d_\alpha$ is equal to some $c_n$. That is, $$\varphi=\bigl(\bigwedge_{\alpha\neq\beta\in I}d_\alpha\neq d_\beta\bigr)\wedge\bigl(\bigwedge_\alpha\bigvee_n d_\alpha=c_n\bigr).$$ A model of this sentence is essentially providing an injective function from $I$ to $\mathbb{N}$, mapping $\alpha\mapsto n$ when $n$ is least for which $d_\alpha=c_n$. Thus, the sentence is not satisfiable in $W$, since $I$ is uncountable there, but it is satisfiable in $V$, since $I$ is countable in $V$.

Using the same idea, one can make a counterexample in a finite signature language as follows. Suppose that $P(\mathbb{N})^W$ is countable in $V$. First, we can make a sentence that asserts that we have a copy of $\langle\mathbb{N},S\rangle$, by asserting that everything in that sort is a finite successor of $0$. Next, we can have another sort that will be a copy of $P(\mathbb{N})^W$, each subset represented by a vertex pointing exactly at its elements. In a single formula (of size continuum in $W$), we can say that every subset of $\mathbb{N}$ that it is represented by some vertex. Finally, with one more binary relation, we can say that every subset is associated with a distinct natural number. The point now is that the overall assertion is not satisfiable in $W$, since $P(\mathbb{N})^W$ is uncountable in $W$, but it will be satisfiable in $V$, since the set is countable there.

Update. Meanwhile, the satisfiability of sentences in $\cal{L}_{\omega_1,\omega}$ logic, that is, using only countable meets and joins, is a $\Sigma^1_1$ assertion and hence absolute between models with the same countable ordinals.

  • $\begingroup$ This is lovely, and very simple - thanks! $\endgroup$ Oct 23, 2013 at 2:27
  • $\begingroup$ With a bit of work, this can even be turned into a complete $\mathcal{L}_{\infty\omega}$ sentence, as well. $\endgroup$ Oct 23, 2013 at 3:25
  • $\begingroup$ This is a comment on JDH's "Update" rather than the original question: For $L_{\omega_1,\omega}$ sentences, model existence in $\aleph_0$ and $\aleph_1$ is absolute. $\endgroup$ Jul 28, 2015 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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