Are generalized transitive relations an axiomatizable class? Let $R$ be a binary relation and let $n$ be a positive integer. $R$ is said to be $n$-transitive iff for all $x_1$,..,$x_{n+2}$, $(R(x_1,x_2) \land ... \land R(x_{n+1}, x_{n+2})) \to R(x_1, x_{n+2})$. A generalized transitive relation is an $n$-transitive relation for some $n$. Is the class of all generalized transitive relations a first-order axiomatizable class?
 A: The answer is no for the following reason:
For each $n \in \mathbb N$ fix a model $(M_n; R_n)$ such that $R_n \subseteq M_n \times M_n$ is $n$-transitive but not $m$-transitive for all $m < n$. (I leave it to you to prove that such models exist.) Now fix a non-principal ultrafilter $U$ on $\mathbb N$ and consider the ultrapower
$$
(M; R) = \prod_{n \in \mathbb N} (M_n; R_n) \mod U.
$$ 
If generalized transitive relations were first order definable, say by the theory $T_0$, then Łoś's theorem would imply that $(M; R)$ is $n$-transitive for some $n \in \mathbb N$. But, again by Łoś's theorem, it's easy to see that this is not the case:
Fix $n \in \mathbb N$. For all $m > n$ there is a sequence $(x^m_1, \ldots, x^m_{n+2})$ such that 
$$R_m(x_i^m,x_{i+1}^m)$$
for $i = 1, \ldots, n+1$ but 
$$
\neg R_m(x^m_1, x^m_{n+2}).
$$
For $m \le n$ fix any sequence $(x_1^m, \ldots, x_{n+2}^m)$ in $M_m$. Then, for $i = 1, \ldots, n+2$, let
$$
x_i \colon \mathbb N \to \bigcup_{m \in \mathbb N} M_m, \ m \mapsto x^m_i.
$$
By Łoś's theorem we get that
$$
R(x_i, x_{i+1})
$$
for all $i = 1 ,\ldots, n+1$ but
$$
\neg R(x_1, x_{n+2}).
$$
A: No. For every $n$, there is a sentence $\varphi_n$ asserting that a structure is not $n$-transitive (it's the negation of the sentence you wrote defining $n$-transitivity). Now suppose the first-order theory $T$ axiomatizes the generalized transitive relations. Consider $T\cup \{\varphi_n\mid n\geq 1\}$. Since for any $N$ there is a structure which is generalized transitive but not $m$-transitive for any $m\leq N$, this theory is consistent (by the compactness theorem). A model satisfies $T$ but is not $n$-transitive for any $n$, which is a contradiction.

Note that I barely used anything about the notion of generalized transitivity - just that notion is an infinite disjunction of first-order properties, and no finite number of these imply the rest. This is a completely standard kind of argument that can be applied in a wide variety of situations.
