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单词 PropertiesOfWellorderedSets
释义

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 A is a totally ordered setMathworldPlanetmath such that for every subsetBA the set of all elements of A strictly greater thanthe elements of B,

    B>:={aABba for all bB},

    is either empty or has a least element, then A is well-ordered.

  • Every subset of a well-ordered set is well-ordered.

  • If A is well-ordered and B is a poset such that there is abijectiveMathworldPlanetmathPlanetmath order morphism φ:AB, then B iswell-ordered.

Now we define an important ingredient for understanding the structureMathworldPlanetmathof well-ordered sets.

Definition 1 (section).

Let A be well-ordered. Then for every aA we define thesectionPlanetmathPlanetmathPlanetmath of a:

a^:={bAba}.

A section is also known as an initial segment. We denote the set of all sections of A by A^. This set isordered by inclusion.

Theorem 1.

Let A be a well-ordered set. Then the mapping ^:AA^ defined by aa^ is a bijective order morphism.In particular, A^ is well-ordered.

Proof.

Let a,bA with ab. Then a^b^, so^ is an order morphism. Now assume that a^=b^. Ifa didn’t equal b, we would have ba^, leading to acontradictionMathworldPlanetmathPlanetmath. Therefore ^ is injectivePlanetmathPlanetmath. Now letCA^, then there exists a cA such that c^=C, so^ is surjective.∎

Theorem 2.

Let A and B be well-ordered sets and φ:AB abijective order morphism. Then there exists a bijective order morphismφ^:A^B^ such that for all aA

φ^(a^)=φ(a)^.
Proof.

Setting φ^(a^):=φ(a)^ is well-defined byTheoremMathworldPlanetmath 1. The rest of the theorem followssince ^ and φ are bijective order morphisms.∎

Theorem 3.

Let A be a well-ordered set and aA such that there is aninjective order morphism φ:Aa^. Then A=a^.

Proof.

The image of a section of A under φ has a maximal elementMathworldPlanetmathwhich in turn defines a smaller section of A. We may thereforedefine the following two monotonically decreasing sequences of sets:

B0:=φ(a^),A0:=section defined by maximal element of B0,
Bn:=φ(An-1),An:=section defined by maximal elementof Bn.

Now A^ is well-ordered, so the set defined by the elements of thesequence (An) has a minimal element, that is AN=AN+1 andhence BN=BN+1 for some sufficiently large N. Applyingφ-1 N+1 times to the latter equation yields a^=A0,that is a is the maximal element of B, and thus of A.∎

Theorem 4.

Let A and B be well-ordered sets. Then there exists at most onebijective order morphism φ:AB.

Proof.

Let φ,ψ:AB be two bijective order morphisms. LetaA and set b:=φ(a) and c:=ψ(a). Then therestrictionsPlanetmathPlanetmath φ|a^:a^b^ andψ|a^:a^c^ are bijective ordermorphisms, so the restriction of ψφ-1 to b^ is abijective order morphism to c^. Now either b^c^or c^b^, so by Theorem 3b^=c^, hence b=c and thus φ=ψ.∎

Theorem 5.

Let A and B be well-ordered sets such that for every sectiona^A^ there is a bijective order morphism to a sectionb^B^ and vice-versa, then there is a bijective ordermorphism φ:AB.

Proof.

Let aA and let b^B^ be a section such that there is abijective order morphism ψa:a^b^. ByTheorem 3, b^ is unique, and so isψa by Theorem 4. Definingφ:AB by setting φ(a)=b gives therefore awell-defined (by Theorem 1) and injective ordermorphism. But φ is also surjective, since any bB mapsuniquely to A via b^a^, and back again by φ.∎

Theorem 6.

Let A and B be well-ordered sets. Then there is an injective ordermorphism ι:AB or ι:BA. If ιcannot be chosen bijective, then it can at least be chosen such thatits image is a section.

Proof.

Let A^0 be the set of sections of A from which there is aninjective order morphism to B. If A^0 is the empty setMathworldPlanetmath, thenB must be empty, since otherwise we could map the least element ofA to B. If A^0 is not empty, we may consider the setA0:=A^0. If A0=A, nothing remains to be shown. Otherwisethe set AA0 is nonempty an hence has a least elementa0A. By construction, there is no injective order morphism froma^0 to B, but there is an injective order morphism fromφa:a^B for every element aA which isstrictly smaller than a0. Now assume there is an element bBsuch that there is no injective order morphism from b^A. Then we can similarly construct a least element b0B forwhich there is no injective order morphism b^0A. Surely,b0 is greater than all the elements from the images of thefunctions φa, but then there is a bijective ordermorphism from a^0 to b^0 byTheorem 5 which is acontradiction. Therefore, all sections of B and B itselfmap injectively and order-preserving to A.∎

Theorem 7.

Let A be a well-ordered set and BA a nonemptysubset. Then there is a bijective order morphism from B to one ofthe sets in A^{A}.

Proof.

The set B is well-ordered with respect to the order induced byA. 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 ι:AB whoseimage is a section b^B^. The element b defines a sectionin A^ which is identical to A byTheorem 3. Thus ι is surjective which isa contradiction.∎

References

  • C G. Cantor, Beiträge zur Begründung dertransfiniten Mengenlehre (Zweiter Artikel), Math. Ann. 49,207–246 (1897).
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