     Next: Clique Graphsk-null Ordered Up: Applications/New Directions Previous: The Cancellation Problem

## Results on the Structure of Fixed Point Sets

Aside from the determination whether an ordered set has the fixed point property it certainly is interesting to determine the structure of the fixed point sets. The earliest result in that direction is Theorem 1 in  (the fixed point sets of order-preserving maps of a complete lattice are complete lattices themselves) and another well-known result is Theorem 3 in  (in finite dismantlable ordered sets the fixed point sets of order-preserving mappings are dismantlable). Surprisingly, despite the apparent strength of the fixed point property, there are ordered sets with the fixed point property that have an order-preserving map whose set of fixed points is disconnected. Examples are the set P2 in  (here also depicted in Figure 3 as the set P) and the set in . The following is a sufficient criterion for the fixed point sets to be connected. Proof: Nonemptyness of follows for example from Theorem 4.27 or directly from Theorem 3.4 via an easy induction. The proof of connectedness is an induction on n:=|P|. For n=1 there is nothing to prove. For the step let P be an (n+1)-element connectedly collapsible ordered set and let be as in the definition of connected collapsibility. Let be a retraction and let b:=r(x). By definition and are connectedly collapsible and clearly both sets have elements. Thus is connected. Clearly , and thus must be one of the following four sets: H, , , . The case is trivial. In all the other cases we have to show that any two elements of are joined by a fence in .
In case we are trivially done if , so we will assume . Since r(f(b))=b and we infer f(b)=x. Moreover . Let and let be a fence in H. If , then and we are done. If , we can assume without loss of generality that there is exactly one index m such that and we can assume that . If f(x) is not related to x, then (since f(b)=x) f maps to itself and thus since is connectedly collapsible with elements, is connected. Hence there is a fence from to that lies entirely in and thus there is a fence from p to q in . If f(x) is related to x, there is a smallest fixed point of f that is above x or a largest fixed point of f that is below x. Call this fixed point c. Then and p and q are joined by a fence in .
In case again we will assume that (the case is treated when ). Again we infer f(b)=x. If , then we must have and we are done. Otherwise there is a fixed point of f that is related to b and not equal to b or x. Since f(b)=x, we have that d is related to x also. Let . Since d is related to x we can assume that . Let be a fence in H. Again we are done unless for exactly one m and (without loss of generality) . Since f(b)=x, we have and p and q are joined by a fence in .
Finally in case let us first assume . Then we are done if x is comparable to b. Otherwise f maps to itself, so x and b have a common upper bound in and thus is connected. In case , we have that . If f(b) is related to x (and thus to b), then there is a point in H that is above x or below x and thus is connected. If f(b) is not related to x, then f maps to itself. Hence again there is a point in H that is above x or below x and thus is connected. \ Proof: It is proved by Fofanova, Rival and Rutkowski in  that every ordered set of dimension 2 has a retractable point. The same result is proved for interval dimension 2 in . Since (interval) dimension is a hereditary concept this means that each ordered set of (interval) dimension 2 is collapsible. The result now follows from Theorem 4.27 and Proposition 5.6. \     Next: Clique Graphsk-null Ordered Up: Applications/New Directions Previous: The Cancellation Problem

Bernd.S.W.Schroder