Let $X$ be a finite partially ordered set, let $f\colon X\to X$ be an order-preserving map [edit: meaning $x\le y\implies f(x)\le f(y)$], and let $x\_0$ be an initial point. Define $x\_n = f(x\_{n-1})$ for all $n$; then the sequence $(x\_n)$ is ultimately periodic. What is its worst-case period? I.e. what are the minimal $k<\ell$ such that $x\_k=x\_\ell$?


With no order assumption, one could have $\ell=\#X$ since the map $f$ could cycle through all of $X$. I'm particularly interested in the case when $X$ is the family of subsets of a set $Y$, and $x\_0$ is a small subset. Does there then exist a non-trivial bound, say polynomial in $\#Y$ and exponential in $\#x\_0$?



Maximal order of an order-preserving map
This is a comment about a special case that I think gives a negative answer to the last question. Let $\varOmega$ be a set of size $n$ and let $B$ be the lattice of its subsets with the inclusion order.


Now let $\phi:\varOmega\to\varOmega$ be any function. For $X\subseteq\varOmega$, define $X^\phi=\lbrace \phi(x) \mid x\in X\rbrace$ (duplicates removed of course).


Now it is elementary that $f:B\to B$ by $X\mapsto X^\phi$ is order-preserving. The question is: what can be said about the eventual cycle length of the trajectory of an element of $B$?


Consider the case that $\phi$ is a permutation. Take $x\_0\in B$ to have one element from each of the cycles of $\phi$. Then the trajectory of $x\_0$ is a cycle of length equal to the order of $\phi$ (as an element of the symmetric group).


For example, if $n=2 + 3 + \cdots + p\_k$ (sum of first $k$ primes) and $\phi$ has one cycle of each prime size, then $\#x\_0=k$ and the length of the cycle is $N = 2\times3\times\cdots\times p\_k$.


I understand that
$N = \exp\bigl((1 + o(1)) k \log k\bigr)$.
So $N$ is larger than exponential in $\# x\_0$.


Also, $n\sim \frac12 k^2\log k$. If I got the inversion right, $\# x\_0=\Theta(\sqrt{n/\log n})$ and $N=\exp(\Theta(n^{1/2}\log n))$.


