### Transitive order type

An ordered set $A$ is called **(singly) transitive** if, for any $a\in A$ and $a'\in A$, there exists an order-automorphism $f:A\to A$ such that $f(a)=a'$. It is called **k-tuply transitive** if for any $r\leq k$ and any $2r$ elements of $A$ satisfying and there exists an order-automorphism $f:A\to A$ such that $f(a_i)=a'_i$ for all $1\leq i\leq k$. In general, if $\alpha$ is an arbitrary order type , we say that $A$ is **$\alpha$-transitive** if for any pair of suborderings $A_1,A_2\subset A$ of order type equal to a sub-order type of $\alpha$, there exists an automorphism $f:A\to A$ sending $A_1$ to $A_2$. (Rosenstein only defines k-tuple transitivity, but I've decided to generalize it to $\alpha$-transitivity in my personal notes, so that k-tuple transitivity is a special case with $\alpha=[k]$).

**Proposition 1.** A linear ordering $A$ is k-tuply transitive for all $k\geq 2$ if and only if it is k-tuply transitive for some $k\geq 2$.

*Proof.* Clearly being k-tuply transitive for all $k\geq 2$ implies k-tuple transitivity for *some* $k\geq 2$. Hence, we need only prove the reverse direction. Thus, we just need to prove the reverse direction, for which it suffices to notice that being k-tuply transitive for some $k > 2$ implies double transitivity (which follows directly from the definition), and double transitivity implies k-tuple transitivity for all $k\geq 2$.

So suppose that $A$ is doubly transitive. Given $2k$ elements of $A$ satisfying and define a sequence of functions by letting $f_i:A\to A$ be an automorphism sending $a_i\mapsto a'*i$ and $a*{i+1}\mapsto a'_{i+1}$. Then define an automorphism $f:A\to A$ piecewise as follows: This function clearly satisfies $f(a_i)=a'_i$ for all $1\leq i\leq k$. Hence, $A$ is k-tuply transitive, proving the theorem. $\blacksquare$

# order-theory

back to home page