properties of well-ordered sets
The purpose of this entry is to collect properties of well-orderedsets. We denote all orderings uniformly by . If you areinterested in history, have a look at [C].
The following properties are easy to see:
- •
If is a totally ordered set
such that for every subset the set of all elements of strictly greater thanthe elements of ,
is either empty or has a least element, then is well-ordered.
- •
Every subset of a well-ordered set is well-ordered.
- •
If is well-ordered and is a poset such that there is abijective
order morphism , then iswell-ordered.
Now we define an important ingredient for understanding the structureof well-ordered sets.
Definition 1 (section).
Let be well-ordered. Then for every we define thesection of :
A section is also known as an initial segment. We denote the set of all sections of by . This set isordered 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 acontradiction. Therefore is injective
. Now let, then there exists a such that , so is surjective.∎
Theorem 2.
Let and be well-ordered sets and abijective order morphism. Then there exists a bijective order morphism such that for all
Proof.
Setting is well-defined byTheorem 1. The rest of the theorem followssince and are bijective order morphisms.∎
Theorem 3.
Let be a well-ordered set and such that there is aninjective order morphism . Then .
Proof.
The image of a section of under has a maximal elementwhich in turn defines a smaller section of . We may thereforedefine the following two monotonically decreasing sequences of sets:
Now is well-ordered, so the set defined by the elements of thesequence has a minimal element, that is andhence 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 onebijective order morphism .
Proof.
Let be two bijective order morphisms. Let and set and . Then therestrictions and are bijective ordermorphisms, so the restriction of to is abijective 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 ordermorphism .
Proof.
Let and let be a section such that there is abijective order morphism . ByTheorem 3, is unique, and so is by Theorem 4. Defining by setting gives therefore awell-defined (by Theorem 1) and injective ordermorphism. But is also surjective, since any mapsuniquely to via , and back again by .∎
Theorem 6.
Let and be well-ordered sets. Then there is an injective ordermorphism or . If cannot be chosen bijective, then it can at least be chosen such thatits image is a section.
Proof.
Let be the set of sections of from which there is aninjective 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. Otherwisethe 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 isstrictly 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 forwhich there is no injective order morphism . Surely, is greater than all the elements from the images of thefunctions , but then there is a bijective ordermorphism from to byTheorem 5 which is acontradiction. Therefore, all sections of and itselfmap injectively and order-preserving to .∎
Theorem 7.
Let be a well-ordered set and a nonemptysubset. Then there is a bijective order morphism from to one ofthe 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 doesnot exist. Then, by virtue of Theorem 6, there isan injective but not surjective order morphism whoseimage is a section . The element defines a sectionin which is identical to byTheorem 3. Thus is surjective which isa contradiction.∎
References
- C G. Cantor, Beiträge zur Begründung dertransfiniten Mengenlehre (Zweiter Artikel), Math. Ann. 49,207–246 (1897).