# Fixed point property and interval topology

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?

• How about total order on two elements? – Wojowu Feb 1 '19 at 9:35
• Right :) you can post this as an answer, and I'll accept it - or I delete the question. Your choice! – Dominic van der Zypen Feb 1 '19 at 10:30

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