Poset where every monotonic function has a least fixed point

Question

Let $P$ be a poset such that every order-preserving map $f:P\to P$ has a least fixed point. Must $P$ be chain-complete?

Answer 1

The answer is yes. Fix any chain $C \subseteq P$. The general idea is to find an isotone map $f \colon P \to P$ having $\bigvee C$ as its least fixed point. If $C$ has maximal element then we are done. Let $U$ be the set of all upper bounds of $C$. Define $f(x) = x$ for all $x \in U$. Informally, for $x \in P \setminus U$ we define $f(x)$ to be the least $y\in C$ with $y \not\leq x$. However, since $C$ is not a well-ordering we can't take the least element. But using transfinite recursion we can define a well-ordered sequence of elements $x_\alpha$, where we choose some $x_{\beta} \in \{x \in C \mid \forall \alpha < \beta\ (x_\alpha < x)\}$ if this set is non-empty. This process terminates at some ordinal since $C$ is a set and ordinals form a proper class. This sequence forms a cofinal well-ordered subset $D \subseteq C$.

Formally, $f\colon P \to P$ is defined as follows: $$f(x) = \begin{cases} x,& \text{if } x \in U,\\ \text{the least $y \in D$ with $y\not\leq x$},& \text{otherwise.} \end{cases}$$

You can check the following:

1. $f$ is order-preserving. It is trivial for $x$ or $y$ in $U$. Take $x, y \in P \setminus U$ with $x \leq y$. We have $f(y) \not\leq x$, since otherwise $f(y) \leq x \leq y$, a contradiction with $f(y) \not \leq y$. Hence by definition of $f(x)$ we have $f(x) \leq f(y)$.

2. The set $P_f$ coincides with $U$ (since for $x \notin U$ we have $x \not\leq f(x)$).

By the assumption $P_f = U$ has the least element $\bigwedge U$ which by definition equals to $\bigvee C$.