Definability of sets of a certain form in an ultraproduct

Question

I want to look at countable ultraproducts in a concrete setting.

Specifically, for each $n\in \mathbb{N}$, let $S_n = (A_n, B_n)$ be a finite model where $B_n$ is a binary relation. Let $U$ be a nonprincipal ultrafilter on $\mathbb{N}$.

Suppose that we have a subset $T$ of the ultraproduct $$S:= \prod_{n\in \mathbb{N}} {S_n}/{\mathcal{U}}$$ such that $T= \prod_{n\in \mathbb{N}} {T_n}/{\mathcal{U}}$ for some $T_n \subset S_n$.

Is $T$ necessarily definable in first-order logic with equality? I can't think of a reason why it must be (although, conversely, any definable subset of $S$ can be written as the ultraproduct of a sequence of sub-models of the $S_n$).

But I'm also having trouble coming up with a counterexample. Any help would be appreciated.

Answer 1

No. For instance, if $B_n$ is always equality, the only subsets of $T$ that are definable (with parameters) are finite and cofinite sets (because we have no relations besides equality). But if, say, $|S_n|=2n$ for each $n$ and $|T_n|=n$ for each $n$, then your $T$ is neither finite nor cofinite in $S$.

Answer 2

Just to add to Eric's answer, a sufficient condition for the set $\prod T_n/\mathcal{U}$ to be definable in the ultraproduct $\prod S_n/\mathcal{U}$ is that the sets $T_n$ are uniformly definable in the structures $S_n$. That is, there is a single formula $\varphi(x,y)$ and parameters $b_n\in S_n$ for all $n$ such that $T_n = \{a\in S_n\mid S_n\models \varphi(a,b_n)\}$.

If the $T_n$ are a family of finite sets, then each $T_n$ will be definable by a formula of the form $\bigvee_{i=1}^k x = b_i$, where $T_n = \{b_1,\dots,b_k\}$. If there is a single $N$ such that $|T_n|\leq N$ for all $n$, then the $T_n$ are uniformly definable by instances of $\bigvee_{i=1}^N x = y_i$. But if the $T_n$ have arbitrarily large finite size, no single formula in the language of equality suffices to define the $T_n$.