Definition 1 (section)   Let be well-ordered. Then for every we define the section of :
A section is also known as an initial segment. We denote the set of all sections of by . This set is ordered by inclusion.

Theorem 1   Let be a well-ordered set. Then the mapping defined by is a bijective order morphism. In particular, is well-ordered.

Proof. Let with . Then , so is an order morphism. Now assume that . If didn't equal , we would have , leading to a contradiction. Therefore is injective. Now let , then there exists a such that , so is surjective.

Theorem 2   Let and be well-ordered sets and a bijective order morphism. Then there exists a bijective order morphism such that for all

Proof. Setting is well-defined by Theorem 1. The rest of the theorem follows since and are bijective order morphisms.

Theorem 3   Let be a well-ordered set and such that there is an injective order morphism . Then .

Proof. The image of a section of under has a maximal element which in turn defines a smaller section of . We may therefore define the following two monotonically decreasing sequences of sets:
Now is well-ordered, so the set defined by the elements of the sequence has a minimal element, that is and hence for some sufficiently large . Applying times to the latter equation yields , that is is the maximal element of , and thus of .

Theorem 4   Let and be well-ordered sets. Then there exists at most one bijective order morphism .

Proof. Let be two bijective order morphisms. Let and set and . Then the restrictions and are bijective order morphisms, so the restriction of to is a bijective order morphism to . Now either or , so by Theorem 3 , hence and thus .

Theorem 5   Let and be well-ordered sets such that for every section there is a bijective order morphism to a section and vice-versa, then there is a bijective order morphism .

Proof. Let and let be a section such that there is a bijective order morphism . By Theorem 3, is unique, and so is by Theorem 4. Defining by setting gives therefore a well-defined (by Theorem ) and injective order morphism. But is also surjective, since any maps uniquely to via , and back again by .

Theorem 6   Let and be well-ordered sets. Then there is an injective order morphism or . If cannot be chosen bijective, then it can at least be chosen such that its image is a section.

Proof. Let be the set of sections of from which there is an injective order morphism to . If is the empty set, then must be empty, since otherwise we could map the least element of to . If is not empty, we may consider the set . If , nothing remains to be shown. Otherwise the set is nonempty an hence has a least element . By construction, there is no injective order morphism from to , but there is an injective order morphism from for every element which is strictly smaller than . Now assume there is an element such that there is no injective order morphism from . Then we can similarly construct a least element for which there is no injective order morphism . Surely, is greater than all the elements from the images of the functions , but then there is a bijective order morphism from to by Theorem 5 which is a contradiction. Therefore, all sections of and itself map injectively and order-preserving to .

Theorem 7   Let be a well-ordered set and a nonempty subset. Then there is a bijective order morphism from to one of the sets in .

Proof. The set is well-ordered with respect to the order induced by . Assume a bijective order morphism as stated by the theorem does not exist. Then, by virtue of Theorem 6, there is an injective but not surjective order morphism whose image is a section . The element defines a section in which is identical to by Theorem 3. Thus is surjective which is a contradiction.