Cyclical relations

Question

A 1-cyclical relation is a reflexive relation. A 2-cyclical relation is a symmetric relation, and a 3 cyclical relation is one such that the conjunction of xRy and yRz imply zRx. A 4-cyclical relation is one such that the conjunction of xRy and yRz and zRw imply wRx. The pattern can be continued in the obvious way. I define a general cyclical relation to be an n-cyclical relation for some n. Is the class of cyclic relations first-order axiomatizable? That is, is there a set of first-order sentences involving a relation symbol R such that the cyclic relations and only the cyclic relations satisfy those axioms?

Answer 1

We can easily write in first-order that a relation is not $n$-cyclic for a specific $n$: by $\psi_n: \exists x_1, \ldots, x_n (\bigwedge_{i\neq j\le n}x_i \neq x_j)\wedge(\bigwedge_{i<n} x_i R x_{i+1}) \wedge \neg x_1 R x_n $.

But also there are $n$-cyclic relations for arbitrarily large $n$ which are not $m$-cyclic for any smaller $m<n$. Just take a directed ray of order type $\omega$ and fill forward the cords of length $n$.

Now what does compactness tell you?

Spoiler:

Suppose that $T$ is a collection of first-order sentences consistent with all cyclic relations (if the class of cyclic relations can be axiomatized, it must be axiomatized by such a $T$). Then we claim $T'=T\bigcup \{\psi_n\}_{n\in\mathbb{N}}$ is finitely consistent. Let $T_0\subseteq T'$ be finite, then we can take $n$ to be largest such that $\psi_n$ appears in $T_0$ (or $0$ if no such $\psi_n$ appear in $T_0$). Let $A$ be a model consisting of a $n+1$ cyclic relation which is not $m$-cyclic for any $m<n+1$. Then $A\models T$ because $A$ is a cyclic relation, and $A$ models the $\psi_k$ in $T_0$ since $A$ is not $k$ cyclic for any $k\le n$. So $A$ models $T_0$. So $T'$ is finitely consistent. So by compactness, $T'$ is consistent. So $T'$ has a model. But a model of $T'$ models $T$ since $T\subseteq T'$, yet cannot be $n$-cyclic for any $n\in \mathbb{N}$ since $\psi_n\in T'$ for all $n\in \mathbb{N}$, and so cannot be cyclic. So $T$ does not axiomatize all cyclic relations.