A sentence that has infinite models, finite model, but no finite model above certain cardinality

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Let $T$ be a theory and $\sigma$ a sentence, such that

  • there exists infinite $\mathfrak{A} \models T + \sigma$.
  • there exists finite $\mathfrak{A} \models T + \sigma$.
  • there exists $n \in \mathbb{N}$, such that for all $\mathfrak{A}$ with $|\mathfrak{A}| > n$, $\mathfrak{A} \models T + \neg\sigma$.

Is this possible?

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Certainly.

Consider $\cal L$ to be the language containing one binary relation symbol $<$.

  1. $T$ is the theory stating that $<$ is a linear order (irreflexive, transitive and total).
  2. $\sigma$ is the statement that if there are $n$ different elements in the universe, then $<$ is unbounded. That is:$$\Big(\exists x_1\ldots\exists x_n(\bigvee_{i<n}x_i\neq x_{i+1})\Big)\rightarrow\forall x\exists y(x<y)$$

It's easy to see that $T$ has finite models of any cardinality, as well infinite models. But $\frak A\models\sigma$ then its universe infinite or has less than $n$ different objects.