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 [127]
(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 [30]
(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 [110]
(here also depicted in Figure 3
as the set P)
and the set
in [116]. 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 [44] that every ordered set of dimension 2 has a retractable point. The same result is proved for interval dimension 2 in [120]. 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. \
