constructing well ordered sets

A set A well-orderable if there is a well-ordering on A. For example, the empty setMathworldPlanetmath , is well-orderable, the well-ordering itself is . Similarly, any singleton has as the unique well-ordering, and is therefore well-orderable. More generally, every finite setMathworldPlanetmath is well-orderable. To construct a well-ordering on a finite set, pick any element from the set and designate the element to be the smallest element of the set. Then, pick an element from the remaining collectionMathworldPlanetmath, and make this element the next smallest element. Keep doing this until every element has been picked and assigned. Obviously, this process will not go on forever. If there are n elements in the set, there are n! well-orderings on the set (since every well-ordering is a linear orderingPlanetmathPlanetmath).

By the axiom of choiceMathworldPlanetmath, every set is well-orderable. However, this does not provide us with an explicit way to construct the well-order. For example, how does one well-order , the set of real numbers? Nevertheless, given two well-ordered sets (A,<A),(B,<B), one can explicitly construct well-orderings on the following sets, without AC.

Proposition 1.

AB is well-orderable.


We may assume AB=, since AB=A(A-B), and any subset of an well-ordered set is well-ordered. Define a binary relationMathworldPlanetmath <(A,B) on AB as follows: for x,yAB,

x<(A,B)yiff{x,yA and x<Ay, or x,yB and x<By, or xA and yB.

So <(A,B) is a (strict) linear ordering on AB. Furthermore, <(A,B) extends both <A and <B. If XAB, then the least element of X with respect to <(A,B) is the least element of X with respect to <B if XB, or the least element of XA with respect to <A, if XA. Hence (AB,<(A,B)) is a well-ordered set. ∎

Proposition 2.

A×B is well-orderable.


We may assume that A,B are both non-empty, since the empty set is well-orderable. Define <A×B on A×B as follows:

(x1,x2)<A×B(y1,y2)iff{y1<By2, or y1=y2 and x1<Ax2.

Again, it is easy to see that <A×B is a strict linear ordering on A×B. Next, let XA×B. Define X[A]:={yB(x,y)X for some xA}. Let b be the least element of X[A]. Define X-1(b):={xA(x,b)X}. Let a be the least element of X-1(b). Then (a,b)X. For any (c,d)X distinct from (a,b), since dX[A], either b<Bd or b=d. In the first case, (a,b)<A×B(c,d). In the second case, cX-1(b), either a<Bc or a=c. In the first case, (a,b)<A×B(c,d), and the second case is impossible for otherwise (a,b)=(c,d). Therefore <A×B) is a well-ordering on A×B. ∎


  • The orderingMathworldPlanetmath constructed in PropositionPlanetmathPlanetmathPlanetmath 2 is called the Hebrew lexicographic ordering. It is similar to the usual lexicographic ordering, except that the second coordinates are ordered first before the first coordinates. The reason for choosing this ordering is to make the ordering on A×2 compatible with the ordering on AA, the disjoint unionMathworldPlanetmathPlanetmath of A with itself. Had we chosen the lexicographic ordering instead, on, say, ω×2, then every initial segment of it would be finite, whereas the initial segment of (0,1) on ωω=(ω×{0})(ω×{1}) is infiniteMathworldPlanetmath.

  • The above constructions parallel the ordinal arithmetic: if α,β are ordinalsMathworldPlanetmathPlanetmath such that (A,<A)α and (B,<B)β, then

    1. (a)

      (AB,<(A,B))α+β, provided that AB=,

    2. (b)


    where denotes order isomorphism between well-ordered sets.

  • One would hope that for well-ordered sets A,B, the set BA is well-orderable without resorting to AC. However, this is not possible, and the best one can do is the following: if B is linearly ordered, and A well-ordered, then one can linearly order BA without AC:


    Let C:=BA. For f,g:AB, let [f,g]:={aAf(a)<Bg(a)}. If [f,g], it has a least element min[f,g]. Define <C on C as follows, for any f,gC:

    f<Cgiff[f,g] and f(a)=g(a) for all a<Amin[f,g].

    If fg, then [f,g][g,f] so that f<Cg or g<Cf. To show that <C is transitiveMathworldPlanetmathPlanetmathPlanetmathPlanetmath, suppose f<Cg and g<Ch. Let b=min[f,g] and c=min[g,h]. Then:

    • if bAc, then f(b)<Bg(b)Bh(b), which implies [f,h] and min[f,h]Ab.

    • if c<Ab, then f(c)=g(c)<Bh(c), which implies [f,h] and min[f,h]Ac.

    In either case, min[f,h]Amin(b,c). Now, suppose a<min[f,h]. Then f(a)=g(a)=h(a). As a result, f<Ch. ∎

  • In fact, it can be shown that the following are all equivalentMathworldPlanetmathPlanetmathPlanetmath to AC in ZF:

    1. (a)
    2. (b)

      if A,B are well-orderable, so is BA,

    3. (c)

      if A is well-orderable, so is P(A), the powerset of A.

Title constructing well ordered sets
Canonical name ConstructingWellOrderedSets
Date of creation 2013-03-22 18:49:12
Last modified on 2013-03-22 18:49:12
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 9
Author CWoo (3771)
Entry type Example
Classification msc 03E25
Classification msc 06A05
Defines well-orderable
Defines Hebrew lexicographic ordering