$T \vdash \forall x (f(x)=x \vee \dots \vee f^{n}(x)=x)$ for $\omega$-categorical theories

Question

Let $T$ a $L$-theory $\omega$-categorical such that $f \in L$ is a symbol unary of function. I want to show

\begin{equation} T \vdash \forall x (f(x)=x \vee \dots \vee f^{n}(x)=x) \end{equation} for some $n\in \omega$.

I know that $T \vdash \varphi$ iff $\mathcal{M} \vdash \varphi$ for every finite or countable model of $T$ for Löwenheim-Skolem. For every finite model is trivial. I don't sure if for compacteness the following thoery is consistent $$\Sigma:= T \cup \{f(c)=c \vee \dots \vee f^{n}(c)=c: n \in \omega\}$$ for $c$ a new constant symbol.

Answer 1

The statement you're trying to prove is false. For a counterexample, consider the theory which says:

1. There exist infinitely many $x$ such that $f(x) = x$.
2. There exist infinitely many $x$ such that $f(x) \neq x$.
3. For all $x$, if $f(x) \neq x$, then $f(f(x)) = f(x)$.
4. For all $x$, if $f(x) = x$, then there is exactly one $y\neq x$ such that $f(y) = x$.

A countable model for this theory consists of two countably infinite sets $X$ and $Y$, such that on $X$, $f$ is the identity, and on $Y$, $f$ is a bijection $Y\to X$. Clearly any two such models are isomorphic, so the theory is $\omega$-categorical. But it fails to prove that every element is part of a cycle of bounded length (in particular, if $y\in Y$, then $f^n(y) \neq y$ for all $n>0$).

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What is true is that if $T$ is $\omega$-categorical, then there is some $n$ such that $T\models \forall x\, \bigvee_{0\leq i<j\leq n} f^i(x)=f^j(x)$.

This is a consequence of the Ryll-Nardzewski theorem: Assume not for contradiction and try to find an infinite family of non-equivalent formulas in two free variables.

More generally, the models of an $\omega$-categorical theory $T$ are uniformly locally finite: for every $m$ there exists an $n$ such that every substructure of a model of $T$ generated by at most $m$ elements has size at most $n$.