Given a poset $(P,\leq)$ the *interval topology* $\tau\_i(P)$ on $P$ is generated by
$$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$
where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.


We say that a poset $(P,\leq)$ has the *fixed point property (FPP)* if for every order-preserving map $f:P\to P$ there is $x\in P$ such that $f(x) = x$.


**Question.** If $(P,\leq)$ has the (FPP), does every *continuous* map from the topological space $(P,\tau\_i(P))$ to itself have a fixed point?



Fixed point property and interval topology
Consider a totally ordered set $P$ on two elements $x<y$. Clearly it has the FPP. The interval topology on $P$ is discrete, so the map swapping $x$ with $y$ is continuous, but has no fixed point.


