Having such a dismantling process available it is reasonable to ask when it might stop (compare with the removal step for Xia's reduction algorithm and Proposition 2.21). The problems encountered at the limit ordinal are quite subtle (for example in their worst form encountered when trying to infinite-dismantle the one-way infinite fence) and are dealt with in -,  and , where conditions on the ordered set are given that ensure a smooth transition past the limit ordinal. We will consider the other possible stopping point here, namely the situation in which at a certain stage the set does not have a suitable retraction any more. (In finite ordered sets this is the only way the dismantling process stops.)
Easy examples show that a set can have many cores (every singleton subset of a finite fence is a -core of that fence), but one can show the -core of a finite set is unique up to isomorphism (cf. , ).
Proof: Let be such that . Then since the set of points that are not fixed by f has elements, there are distinct such that . Choose i<j both minimal such that . Then and . Moreover for all we have
Since for all fixed points of f, we are done \
Proof: For let be the smallest ordinal such that for all . (Such exists since the family is infinitely composable.) Now since , we have \
Part 1a is proved in , the finite
case being proved in  and in .
The proof is similar as the proof for part 2 that we are about
to present. Parts 1b and 1c
are easy as there is a largest up-retraction, resp. a
smallest down-retraction on every chain-complete ordered set.
For part 2 let be an -core such that there is an infinite -dismantling S of P onto C. Let be an infinite -dismantling, which is the infinite composition of the -retractions . We will prove inductively that for all we have . For this is trivial and for a limit ordinal this follows from Lemma 3.13. If is a successor ordinal, note that . Since is an -retraction and since is injective, we have that . Thus if , then is a retraction by Lemma 3.12 and . Since C is an -core, this means that , i.e., that .
Thus . Now suppose are -cores of P with being an infinite -dismantling to (i=1,2). Then by the above and , and hence and are isomorphic.
For part 3 let P be finite and let be an -core such that there is an -dismantling S of P onto C. Let be an -dismantling which is the composition of the -retractions , where is finite. We will prove inductively that for all there is an order-preserving mapping such that . For this is trivial (choose ) and since is finite we do not need to consider limit ordinals. If is a successor ordinal, note that . Since is an -retraction and since is injective, we have that . Thus if , then is a retraction by Lemma 3.12 and . Since C is an -core, this means that . Thus is the map with .
Since the dismantling of P stops after finitely many steps, there is an order-preserving map R such that . Now suppose are -cores of P with being a finite dismantling to (i=1,2). Then by the above there are order-preserving maps such that and . Hence and have the same number of elements. But then is an isomorphism from onto . \
Proof: Existence follows from the main theorem in , as for ordered sets with no one-way infinite fence and no infinite chains infinite -dismantlability is equivalent to the notion of dismantlability used in . \