Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 16H / b

Expand $L$ by a unary function symbol $f$. Let $T^{\ast }$ contain:

Every finite subset $S\subseteq T^{\ast }$ contains only finitely many size requirements, say no requirement beyond ${\theta }_N$. By hypothesis there is a finite nonrigid model $A\models T$ with at least $N$ elements. Interpret $f$ as a nonidentity automorphism of $A$. This expanded structure satisfies $S$; indeed it satisfies every automorphism axiom, including any finitely many that occur in $S$.

Thus every finite subset of $T^{\ast }$ is satisfiable. By the compactness theorem, $T^{\ast }$ has a model $M^{\ast }$. The sentences ${\theta }_n$ make its domain infinite, while $\exists x[f(x)\ne x]$ and the automorphism axioms make $f^{M^{\ast }}$ a nonidentity automorphism. The $L$-reduct $M$ is therefore an infinite nonrigid model of $T$. This is the compactness transfer of finite nonrigidity.

No. Take $L=\{<\}$ and let $T$ be the theory of linear orders. It has finite models of every positive size. Every finite linear order is a rigid first-order structure: an order automorphism fixes the least element, then the next element, and inductively every element.

However, $(\mathbb{Z},<)$ is an infinite model of $T$ and is not rigid. For example,is a nonidentity order automorphism. Hence arbitrarily large rigid finite models do not force all infinite models to be rigid.