Ordinal Numbers, Axiom of Choice, Zorn’s
Lemma and Well-Ordering Principle
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Ordinal Numbers
A linearly ordered set (X, ≤) is called well-ordered if every nonempty subset of
X has a minimum element. For example, N , under the usual order, is wellordered, but Z is not.
Two well-ordered sets (X, ≤) and (Y, ≤) are isomorphic if there exists a bijection
f : X → Y such that for all x1 , x2 ∈ X, f (x1 ) ≤ f (x2 ) if and only if x1 ≤ x2 .
Let (X, ≤) be a nonempty well-ordered set. Then X has a minimum element
x0 , and X\{x0 } has a minimum element x1 , and so on. So if X has n elements,
it is isomorphic to {0, 1, 2, . . . , n − 1} with the usual order. If X is infinite, the
above process continues and we obtain a sequence x1 , x2 , . . . , xn , . . . such that
x0 ≤ x1 ≤ . . . ≤ xn ≤ . . .. If X\{x0 , x1 , . . . , xn , . . .} is nonempty, then it has a
minimum element xω , which is larger than each xn for all n = 1, 2, . . ..
”Being isomorphic” is an equivalence relation on the class of well-ordered sets,
so the class partitions into equivalence classes of isomorphic well-ordered sets.
Each equivalence class is assigned a number called ordinal number (or, simply,
ordinal). The empty set is assigned the number 0, the one point set the number
1, the set {0, 1} with the usual order the number 2, and so on. The set of natural
numbers N (with the usual order) is assigned the number ω. If a well-ordered
set X is in an equivalence class that is assigned an ordinal number α, we say X
has ordinality α.
Let (X, ≤) be a well-ordered set. A subset S ⊂ X is called an initial segment
of X if S = X or there exists a ∈ X such that S = Xa = {x ∈ X : x < a} (as
usual, x < a means x ≤ a and x 6= a); in the latter case, we say S is a proper
initial segment of X. Observe that for any a ∈ X the set {x : x ≤ a} is also an
initial segment (to prove, consider whether a is or is not the largest element in
X).
Exercise
1. Prove that a subset A of a well-ordered set (X, ≤) is an initial segment if and
only if each point of X which precedes (i.e. ≤) an element of A is in A.
Let α, β be two ordinal numbers with representatives (X, ≤), (Y, ≤), respectively. We say α ≤ β if X is isomorphic to an initial segment of Y . As usual,
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α < β means α ≤ β and α 6= β.
Proposition 1 For any two ordinals α, β, either α ≤ β or β ≤ α.
Proof.
Consider the family F of bijections such that each f ∈ F is an isomorphism
with domain an initial segment of X and range an initial segment of Y .
Note that F is nonempty since the empty map with domain and range both
empty sets belongs to F. Order F as follows:
For f, g ∈ F, we put f ≤ g if domf ⊂ domg and g|domf = f . If G is a chain in
F, then ∪{g : g ∈ G} is an upper bound for G.
By Zorn’s lemma, F has a maximal element f . If domf = X or rangef = Y ,
then we have α ≤ β or β ≤ α, respectively. Otherwise, if domf = Xa and
rangef = Yb , we can extend f to the initial segment {x : x ≤ a} by defining
f (a) = b, contradicting the maximality of f .
[One can avoid Zorn’s lemma in the proof using Proposition 3 below, which
does not depend on Proposition 1. For any ordinal α, denote W (α) the set of
ordinals < α. By Proposition 3, W (α) is a well-order set of ordinality α. Now
let ordinals α, β be given. Let C = W (α) ∩ W (β). By Exercise 1, C is an initial
segment of both W (α) and W (β). So C is well-ordered and we let γ be its
ordinality, i.e. C = W (γ), in particular, γ 6∈ C. Since C is an initial segment
of both W (α) and W (β), we have by definition γ ≤ α and γ ≤ β. Note that γ
cannot be < both α and β; otherwise γ ∈ A ∩ B = C, contradicting γ 6∈ C. So
γ = α or γ = β, from which it follows that α ≤ β or β ≤ α.]
Lemma 1 Let X be a well-ordered set. X cannot be isomorphic to any of its
proper initial segment.
Proof.
Suppose f : X → Xa is an isomorphism. Then f (a) < a whence f 2 (a) < f (a),
f 3 (a) < f 2 (a), . . .. So the set {f n (a) : n = 0, 1, . . .} has no least element, a
contradiction.
Proposition 2 For any two ordinals α, β, if α ≤ β and β ≤ α, then α = β.
Proof.
If f : X → A and g : Y → B, are isomorphisms, where A, B are initial segments
of Y and X respectively, then g ◦ f : X → g(A) is an isomorphism of X into
its initial segment g(A). By lemma, this is impossible unless g(A) = X, from
which it follows A = Y and B = X, i.e. X and Y are isomorphic.
Proposition 3 Let β be an ordinal number. The set S of of all ordinals less
than β is a well-ordered set isomorphic to any well-ordered set with ordinality
β.
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Proof.
Let X be a well-ordered set with ordinality β and α < β. If Y is a well-ordered
set with ordinality α, then Y is isomorphic to a unique proper initial segment
Xa of X. This defines a correspondence α 7→ a which is an order preserving
bijection. It follows that S is well-ordered and isomorphic to X.
Proposition 4 Let α and β be two ordinals. Then exactly one of the following
is true: α < β, α = β, β < α.
Proof.
By Proposition 1, either α < β or α = β or β < α. By Lemma 1, these three
possibilities are mutually exclusive.
Theorem 1 (Principle of Transfinite Induction) Let β be an ordinal number and S = {α : α < β}. Let A be a subset of S satisfying the following property:
(i) 0 ∈ A;
(ii) if Sa ⊂ A, then a ∈ A. Then A = S.
Proof.
Suppose A 6= S. Let a ∈ S be the least element in S that does not belong to A.
Then Sa ⊂ A. But then a ∈ A by (ii), a contradiction. So A = S.
Remark 1 1. Since S0 = φ ⊂ A , (ii) implies that 0 ∈ A. So condition (i)
above can be dropped.
2. Since {α : α < β} is a well-ordered set with ordinality β, we may identified
it with the ordinal number β.
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Arithmetic of Ordinal Numbers
Let α, β be ordinal numbers. Let X, Y be disjoint well-ordered sets with ordinality α, β respectively. Define a well-order on X ∪ Y such that x ≤ y for all
x ∈ X, y ∈ Y , and that the ordering among the elements of X and of Y remains
the same. Then it is easy to check that X ∪Y is well-ordered under this ordering
and its ordinality is defined as α + β.
Note that 1 + ω = ω, but that ω + 1 > ω, so generally, α + β 6= β + α, i.e + is
not commuatative.
If X × Y is ordered with antilexicographic ordering, i.e. (x1 , y1 ) ≤ (x2 , y2 ) if
and only if y1 < y2 or (y1 = y2 and x1 ≤ x2 ), the result is an well-ordered set
whose ordinality is defined as α × β or α · β or αβ. (It will be clear later why
antilexicographic, but not lexicographic, ordering is used for the definition.)
Note that 2ω = ω, while ω2 = ω + ω 6= ω, so generally, ordinal multiplication is
not commutative.
Now we attempt to define αβ . If β = 2, then either the lexicographic or antilexicographic ordering on X 2 yields a well-ordering with ordinality α2 = α × α.
The same applies if β is finite. However if β is infnite, lexicographic ordering
on X Y may not be a well-ordering, and the antilexicographic ordering may not
be a linear ordering.
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For example, if α = 2 and β = ω, we can consider X Y as the set of sequences of
0 and 1’s. The sequence ei , i = 1, 2, . . . does not have a least element in the lexicographic order, while antilexicographic ordering cannot compare two sequences
like a, b where ai = 1 if i is even and 0 if i is odd, and bi = 0 if i is even and 1 if
i is odd. If, instead, we consider the subset F of X Y consisting of all sequences
of 0 and 1’s having only finitely many 1’s, then on F , antilexicographic ordering
is a well-ordering. This generalizes to the general case as follows:
Let X, Y be well-ordered sets with ordinality α, β respectively. Let a be the least
element of X. Let F be the subset of X Y consisting of functions f : Y → X
such that f (y) = a for all but finitely many y’s. The antilexicographic ordering
on F is a well-ordering and the ordinality of F is defined as αβ .
Exercises
1. Prove that β = α + 1 is the least ordinal greater than α. (β is called the
successor of α and α is called the predecessor of β. The predecessor of β is also
written as β − 1.)
2. Let S be a set of ordinals. An ordinal α is called an upper bound of S if
β ≤ α for all β ∈ S. A least upper bound of S, if it exists, is an upper bound
which is ≤ any upper bound of S. Prove that any set of ordinals has a unique
least upper bound. The least upper bound of S is denoted by sup S.(For each
a ∈ S, let Sa be the set of all ordinals less than a and let T be the union of
Sa , a ∈ S. Then T is a well-ordered set and let β be its ordinality. Since for
each a ∈ S, Sa ⊂ T implies that a < β, we see that β is an upper bound. The
set of upper bounds of S has a least element, which is the least upper bound.)
3. Let β be a fixed ordinal. Prove that (i) β · 0 = 0, (ii)β · (α + 1) = βα + β, (iii)
βα = sup{βγ : γ < α} if α is a limit ordinal. (Note: one can take this excercise
as a definition of ordinal multiplication by transfinite induction.)
4. Let β be a fixed ordinal. Prove that (i) β 0 = 0, (ii)β α+1 = β α · β, (iii)
β α = sup{β γ : γ < α} if α is a limit ordinal. (Note: one can take this excercise
as a definition of ordinal exponentiation by transfinite induction.)
5. Find the least ordinal α such that ω + α = α. (Ans: ω 2 )
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Axiom of Choice and its Equivalences
Our next objective is to prove that the following are equivalent:
1. (Axiom of Choice) Let I be a set and Ai , i ∈ I be a family of disjoint
nonempty sets. Then there exists a set S = {xi : i ∈ I} such that xi ∈ Ai for
each i.
2. (Zorn’s Lemma) Let (X, ≤) be a partially ordered set. Suppose every linearly
ordered subset of X has an upper bound, then X has a maxmial element.
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3. (Well-Ordering Principle) Every set can be well-ordered.
First we prove that the sets in the statement of Axiom of Choice need not
be disjoint for the conclusion to hold. In other words, Axiom of Choice is equivalent to the following:
1’. Let I be a set and Ai , i ∈ I be a family of nonempty sets. Then there exists
a set S = {xi : i ∈ I} such that xi ∈ Ai for each i.
To prove, consider the disjoint family Ti = {(i, x) : x ∈ Ai }, i ∈ I. By 1, there
is a set T = {(i, xi ) : i ∈ I}, where xi ∈ Ai for all i ∈ I. It follows that
S = {xi : i ∈ I} is a set. Therefore 1 implies 1’. The converse is obvious.
Let us prove 1 implies 3. Since 1 is equivalent to 1’, we may assume that
1’ holds. Let X be a set. By 1’, we may choose a point c(S) ∈ S for each
nonempty subset S of X. Consider the family F consisting of all well-ordered
relations R such that, with D = dom(R), (i) D ⊂ X, (ii) R is a well-ordering on
D, and (iii) for any initial segment A of D with D\A 6= φ, c(X\A) is the least
element of D\A. F is nonempty since empty relation, as well as the singleton
relation {(c(X), c(X))}, is in it. Let R and T be members of F with domain
D and E respectively. Let A be the set of all points x ∈ D ∩ E such that the
sets {y : yRx} and {y : yT x} are identical and such that on these sets the two
orderings agree. Then A is a segment relative to both R and T . If both D\A
and E\A are nonempty , then c(X\A) is the first point of both of them; but
then c(X\A) ∈ A in view of the definition of A, a contradiction. It follows that
A = D or A = E. Thus any two members of F are related as follows: the
domain of one member is a segment relative to the other, and the two orderings
agree on this segment. Using this fact it is not hard to see that the union U of
the members of F is itsef a member of F; it is the largest element of F. If F is
the domain of U , then F = X, for otherwise the point c(X\F ) may be adjoined
at the end of the ordering (more precisely, U ∪ (F × {c(X\F )}) is a member of
F which properly contains U ).
Let us now prove 3 implies 2. Let (A, ≤) be a partially ordered set in which
every chain has an upper bound. The assumption (3) guarantees that there
exists a one-to-one transfinite sequence {aα : α < θ} whose range is A. Suppose on the contrary that (A, ≤) does not have a maximal element. We define
{αη : η < θ} by transfinite induction as follows: Let α0 = 0. Suppose αη have
been defined for η < β so that αη is the least ordinal < θ with the property
that aαη > aαγ for all γ < η.
Case (i): β has a predecessor β − 1. If the set S = {aα : aα > aβ−1 } is empty,
then aβ−1 is a maximal element of A (Exercise), contradicting the assumption.
So S is nonempty and we define αβ to be the least ordinal such that aαβ > aαβ−1 .
Case (ii): β is a limit ordinal. Then we let αβ to be the least ordinal such that
aαβ is an upper bound of the chain {aαη : η < β}. (The upper bound exists by
the assumption (3).)
It follows that αη ≥ η for all η < θ (Exercise). The chain {aαη : η < θ} has an
upper bound b = aγ for some γ < θ. Since γ < αγ+1 , aγ cannot be ≥ aαγ+1 due
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to the minimality in the definition of αγ+1 . This contradicts that b is an upper
bound.
Therefore A must have a maximal element.
Let us prove (2) implies (1). Let F be the family {f : domf ⊂ I : f (i) ∈ Xi }.
F is nonempty since the empty function is in it. Define ≤ in F such that f ≤ g
means g is an extension of f . Every chain in F has an upper bound which is
the union of the functions in the chain. So by (2), F has a maximal element
h. It follows that the domain of h must be the whole set I and hence the set
{h(i) : i ∈ I} satisfies the requirement of (1).
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