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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?

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    $\begingroup$ How about total order on two elements? $\endgroup$
    – Wojowu
    Feb 1, 2019 at 9:35
  • $\begingroup$ Right :) you can post this as an answer, and I'll accept it - or I delete the question. Your choice! $\endgroup$
    – Dominic van der Zypen
    Feb 1, 2019 at 10:30

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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.

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