**Proof:**
First note that since
is a retraction the condition
``*P* has the fixed point property" implies that
has the fixed point property for all .

For the other direction let be order-preserving.
Then has a fixed point *p*.
We will prove inductively for all
that the condition ``
has a fixed point p" implies
that
*f* has a fixed point.
For this is trivial.
Now let and assume the assertion holds for all
. First suppose is a
successor ordinal. If
has a
fixed point then
(since
satisfies a reflection condition)
has a
fixed point.
By induction hypothesis the latter implies that *f* has a fixed point.

Finally if is a limit ordinal
suppose that
has a fixed point *p*.
Then there is a such that
.
Thus has *p* as a fixed point and
by induction hypothesis *f* has a fixed point. \

Theorem 3.19 obviously leads to an algorithm for a sufficient
condition for the fixed point property:
Dismantle *P* via -retractions as far as possible and
verify a reflection condition at every step.
Any set for which this is possible with a core that has the fixed point
property has the fixed point property itself.
The difficulty in finding reflection conditions however is considerable
as can be seen in [118].
Infinite-dismantlability also gives an algorithmic insight
into a classical result about the fixed point property
for ordered sets of height 1 (recall that the height of an
ordered set *P* is the number of elements in the largest chain
in *P* minus 1):

**Proof:**
Assume *P* does not contain an irreducible point.
Let be arbitrary and minimal.
Let be an upper cover of .
Assume that have already been chosen and
is a fence.
We will assume without loss of generality that is minimal.
Then has an upper cover . Choose
. Then since *P* does not contain any crowns
is a fence.
In this fashion we can construct an infinite fence in *P*,
contradiction. \

**Proof:**
``1 2":
Necessity of connectedness is trivial.
If *P* contains a crown
*P* cannot have the fixed point property as is proved in
[95], p. 312 ff. (the argument given never uses that *P* is finite).
Thus *P* contains no crowns.
Were *P* to contain an infinite fence *F* and no crowns, then *F* is
an isometric spanning fence and hence by Theorem 2 in [34] it is
a retract (this is also easy to see directly).
This would again imply that *P* does not have the
fixed point property. Hence 2 must hold.

``2 3":
Let .
Suppose is an ordinal number and as in
Definition 3.7
have already been defined for .
If has a predecessor
such that is not a singleton, then there is a
point that is irreducible. Let be the retraction that
removes *x* and let .
If is a limit ordinal notice that since *P* does not contain
any infinite fences for each
the sequence as in Definition
3.7
must be constant for for some .
We define .
If is a singleton we stop.
The sequence thus generated is an infinite dismantling of *P*.

``3 1"
follows from Theorem 3.19. \

Theorem 3 in [82], which is used to prove Theorem 4 in [82] (which is Theorem 3.21 here) can now be obtained via one of the standard translation processes between ordered sets and graphs (similar to what is done in [82]).

**Proof:**
(Compare with the idea of the proof of
Theorem 2.14.)
Let *G*=(*V*,*E*) be a graph.
We define an ordered set of height 1 with underlying set
and with the order being
containment of sets.
Let be a graph endomorphism of *G*. Then

is an order-preserving map of *P*.
Moreover if
*F* has a fixed point
that is minimal in *P*, then
*f* fixes *v*. If *F*
fixes no singleton, but
has a fixed point
that is maximal in
*P*, then *f* fixes .
Thus if
*P* has the fixed point property, then every endomorphism of *G*
fixes an edge or a vertex.

Let be an order-preserving map of *P* that maps the
minimal elements to minimal elements.
If for we let *H*(*x*) be the unique element of
the singleton set *h*(*x*), then
*H* is a graph endomorphism of *G*.
Moreover if *H* has a fixed vertex, then
clearly *h* has a fixed point, and if
*H* has a fixed edge , then
and hence .
Thus if every endomorphism of
*G* has a fixed vertex or a fixed edge, then
every order-preserving map of *P* that maps minimal elements
to minimal elements has a fixed point and
consequently *P* has the fixed point property.

Hence *P* has the fixed point property iff
every endomorphism of *G* has a fixed edge or a fixed vertex.
However by Theorem 3.21 *P* has the fixed point
property iff *P* is connected and has no crowns and no infinite
fences.
This is the case exactly when *G* is connected and has no
cycles with elements and no infinite paths.

Finally note that
*G* is
-infinite-dismantlable
to a singleton iff *P* is.
\