Download p. 1 Math 490 Notes 4 We continue our examination of well

p. 1
Math 490 Notes 4
We continue our examination of well-ordered sets. If (X, ≤) is a w.o.set and x ∈ X, we
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define Sx = {y ∈ X ¯ y < x}. Sx is called the initial section of X determined by x.
Recall from a previous example the well-ordered set
A ∪ B = {a0 , a1 , . . . , an , . . . , b0 , b1 , . . . , bm , . . . , . . .} ,
formed from the union of two disjoint well-ordered sets A and B. For this set, we have
Sa0 = φ, Sa2 = {a0 , a1 }, San = {a0 , a1 , . . . , an−1 }, Sb0 = A, etc.
Proposition N4.1 There exists an uncountable well-ordered set SΩ such that every initial
section of SΩ is countable.
Proof : Start with any uncountable set X and use the Well Ordering Theorem to well-order
X. Let Y be another uncountable well-ordered set obtained by setting Y = X × X with the
dictionary order. If x1 is not the least element of X, then the initial segment of Y determined
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by (x1 , x1 ) is clearly uncountable, and so A = {y ∈ Y ¯ Sy is uncountable } is a non-empty
set in Y , which by assumption contains a least element; call it Ω. Note that by assumption,
SΩ ≤ Sy for all y ∈ Y .
Claim 1: Every initial segment of SΩ countable. If not, then there would exists a z < Ω such
that the initial section (SΩ )z = Yz is uncountable. By definition of A and Ω, we’d then have
z ∈ A and z ≥ Ω, which is a contradiction.
Claim 2: SΩ is uncountable. This follows immediately, since Ω ∈ A.
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This uncountable w.o.set SΩ plays an important role as a source of examples in topology, so
we’ll explore it further.
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Prop. N4.2 If z ∈ SΩ , the set U (z) of upper bounds of z is uncountable.
Proof : This follows because SΩ = U (z) ∪ (SΩ )z , and by Prop. N4.1, SΩ is uncountable,
whereas (SΩ )z is countable.
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Prop. N4.3 If A = {zn ¯ n ∈ N} ⊆ SΩ , then z = sup A ∈ SΩ , and (SΩ )z is countable.
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Proof : First note that B = A ∪ {∪(SΩ )zn ¯ n ∈ N} is a countable union of countable sets,
and is therefore countable. Thus SΩ − B is uncountable, by Prop. N4.1. But every element
of SΩ − B is an upper bound of A, so z = sup A ∈ SΩ follows by Proposition 5, and (SΩ )z is
countable by Prop. N4.1.
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The next important theorem about well-ordered sets generalizes the well-known induction
theorem for the natural numbers.
Transfinite Induction Theorem Let A be a well-ordered set, and let B ⊆ A have the
property that Aa ⊆ B =⇒ a ∈ B for all a ∈ A. Then B = A.
Proof : If B 6= A, then A − B 6= φ, and we can let y be the least element of A − B. If z < y,
then z ∈ B, and hence Ay ⊆ B. It follows by assumption that y ∈ B, a contradiction. Since
A − B 6= φ yields a contradiction, we have A = B.
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Two well-ordered sets A and B are similar (we’ll write A ≈ B) iff there is an order-preserving
bijection ψ from A to B. This is a bijection ψ : A → B such that x ≤ y in A iff ψ(x) ≤ ψ(y)
in B. Note that similar w.o.sets have the same cardinal number. If there is a b ∈ B such
that A ≈ Bb , we write A ≺ B.
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Consider the following w.o.sets, ordered from left to right:
N = {0, 1, 2, . . . , n, . . .}
E = {0, 2, 4, 6, 8, . . .}
D = {d0 , d1 , d2 , . . . , dn , . . . , e0 , e1 , e2 }
Note that N ≈ E ≈ De0 , so N ≺ D and E ≺ D. One can show that every uncountable
well-ordered set A for which every initial segment is countable is similar to SΩ as constructed
in Prop. 6.
Suppose one wishes to find an order-preserving bijection (an isomorphism) ψ between two
w.o.sets A and B. Clearly, ψ must map the least element of A to the least element of
B, the successor of the least element of A to the successor of the least element of B,
and so on. If we denote the immediate successor of a ∈ A by a+ , then in general, we
must have ψ(a+ ) = ψ(a)+ for all a ∈ A. Furthermore, if A′ ⊆ A and a = sup A′ , then
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ψ(a) = sup{ψ(x) ¯ x ∈ A′ } = sup ψ(A′ ). That is, ψ preserves sup’s. These basic facts, along
with the Transfinite Induction Theorem, lead to the following result.
Trichotomy Theorem If A and B are well-ordered sets, then exactly one of the following is true: (a) A ≺ B, (b) B ≺ A, (c) A ≈ B. Furthermore, in all three cases, the
order-preserving bijection is unique.
Sketch of proof : If a0 is the least element of A and b0 is the least element of B, define
ψ(a0 ) = b0 . Next, for a ∈ A, assume the “induction hypothesis” that ψ : Aa → Bψ(a) is an
order-preserving bijection, and for all x ∈ Aa , there exists a b ∈ B such that b > ψ(x). If a
has an immediate predecessor y (i.e. y + = a), then ψ(y) is already defined by hypothesis,
and we let ψ(a) = ψ(y)+ (note that we know ψ(y)+ exists, because ψ(y) can not be the
greatest element of B by assumption). If a has no immediate predecessor, then let
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ψ(a) = sup{ψ(x) ¯ x ∈ Aa }. It should be relatively easy to see that a < a′ =⇒ ψ(a) < ψ(a′ ),
so the function ψ constructed in this way is both one-to-one and order-preserving. This
construction ends when either all elements of A or all elements of B (or both) have been
exhausted.
Suppose the elements of A are exhausted first. In other words, after ψ(a) defined for all
a ∈ A, there exists a b ∈ B such that b > ψ(a) for all a ∈ A. If A has a greatest element z,
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then A ≈ Bψ(z)+ . If A has no greatest element, then A ≈ Bω , where ω = sup{ψ(a) ¯ a ∈ A}.
Either way, A ≺ B. If the elements of B are exhausted first, then we may apply the same
reasoning to ψ −1 to show B ≺ A. Finally, if the elements of A and B are exhausted simultaneously, then A ≈ B.
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Trichotomy Corollary 1 If α and β are cardinal numbers, then either α ≤ β or β ≤ α.
Proof : Choose A ∈ α, B ∈ β, and well-order both A and B. By the preceding theorem,
exactly one of the following holds: A ≺ B, B ≺ A, or A ≈ B. If A ≺ B or A ≈ B, then
α ≤ β. Otherwise, β ≤ α.
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Trichotomy Corollary 2 A well-ordered set is not similar to any of its initial sections.
Proof : If A is well-ordered and a ∈ A, the identity injection from Aa to A is the unique
order-preserving map which establishes Aa ≺ A. Thus Aa ≈ A is impossible by the Trichotomy theorem.
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Ordinal Numbers
Recall that cardinal numbers are defined by partitioning the class S of all sets into equivalence classes, with A ∼ B iff there is a bijection from A to B. Similarly, we define ordinal
numbers to be the equivalence classes obtained when the class W of all well-ordered sets is
partitioned into equivalence classes relative to the equivalence relation similarity. If L ∈ W,
let hLi denote the ordinal number of the equivalence class containing L. As before, we denote
by |L| the cardinal number for L. For each ordinal number λ, there is an associated cardinal
number |λ|, which is the cardinality of any set L ∈ λ.
Let O denote the class of ordinal numbers. A partial order on O is defined by λ ≤ µ
iff there exists L ∈ λ and M ∈ µ such that L ≺ M or L ≈ M , with λ < µ corresponding
to L ≺ M . Note that this definition is independent of the representatives L and M chosen
from the given equivalence classes. It follows by the Trichotomy Theorem that ≤ is a simple
order on O, and also λ ≤ µ =⇒ |λ| ≤ |µ|.
The finite ordinal numbers are those ordinals λ for which |λ| is finite. Note that the
empty set φ is a well-ordered set (vacuously), and the ordinal containing φ is naturally denoted 0 (zero). Now consider all well-ordered sets with exactly n elements for some n ∈ N.
It should be easy to see that all such well-ordered sets are similar to each other, and thus
they all belong to the same equivalence class in W. So there is a unique ordinal number
corresponding to an n-element set, which it is natural to denote by n. Since there is an
obvious order-preserving bijection between finite ordinals and finte cardinals, it is customary
to make no distinction between finite ordinal numbers and finite cardinal numbers.
When we consider the infinite ordinals (those ordinals λ for which |λ| ≥ ℵ0 ), the situation
is drastically different. We define denumerable ordinals (resp., countable ordinals) to
be those of cardinality ℵ0 (resp., ≤ ℵ0 ). We can consider the denumerable ordinals to
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correspond to the distinct ways to well-order a denumerable set. For instance, N is already
well-ordered in its natural ordering. Some other w.o.sets defining denumerable ordinals are
listed here:
ω = h{a0 , a1 , . . . , an , . . .}i = hNi
ω + 1 = h{a0 , a1 , . . . , an , . . . , b0 }i
ω + ω = h{a0 , a1 , . . . , an , . . . , b0 , b1 , . . . , bm , . . .}i
ω × ω = hN × N, with dictionary orderi
How many distinct denumerable ordinals are there? Recall the uncountable w.o.set SΩ such
that each of its initial sections is countable. Note that (by Corollary 2 to the Trichotomy Theorem) no two distinct sections of the same w.o.set are similar, so each section of SΩ defines a
distinct, countable ordinal. Among these are the finite ordinals, but the cardinal number of
the set of all finite ordinals is ℵ0 , so the cardinal number of the set of denumerable ordinals
is uncountable. In fact, it is customary to refer to SΩ as the set of countable ordinals, and
Ω = hSΩ i is called the first uncountable ordinal. ω = hNi is the smallest infinite ordinal,
whereas Ω = hSΩ i is the least uncountable ordinal. An important fact about ordinal numbers which we shall use but not prove is the following.
Theorem N4.1 Any set of ordinals in their natural order is a well-ordered set.
This theorem has important consequences. For example, it implies that every ordinal has an
immediate successor, but not necessarily an immediate predecessor. Ω and ω, for example,
are ordinals with no immediate predecessor.
As we have seen, each infinite cardinal corresponds to a large set of infinite ordinals. Indeed,
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if α is an infinte cardinal, let Oα = {λ ∈ O ¯ |λ| = α}. Then one may generalize our previous
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results to show that |Oα | > α. But by the preceding theorem, Oα contains a least ordinal
such that |λ| = α, denoted by λα . We’ll call λα the minimal ordinal corresponding to α. The
correspondence α ↔ λα is one-to-one and order-preserving between cardinals and minimal
ordinals, which gives us the following corollary.
Corollary Any set of cardinal numbers in their natural order is well-ordered.
For instance, ℵ1 = |Ω| is the immediate cardinal successor of ℵ0 . Where does |R| = 2ℵ0
fit in? Since |R| is known to be uncountable, it follows that |R| ≥ ℵ1 . Using the usual set
theory axioms that we’ve been using up to this point, along with the Axiom of Choice, it
can neither be proved nor disproved that ℵ1 = |R|. The assertion that ℵ1 = |R| is called
the Continuum Hypothesis. This hypothesis, like the Axiom of Choice, can be accepted
or rejected as an axiom of set theory. The Generalized Continuum Hypothesis is the
assertion that for any infinite cardinal α, there are no cardinal numbers properly between α
and 2α . To say it another way, 2α is the immediate successor of α. Of course, this assertion
is clearly not true for finite cardinals.
Ordinal Arithmetic
We’ll discuss only ordinal addition and multiplication here, since ordinal exponentiation is
more technical and of no interest in this course. Let λ, µ ∈ O and choose L ∈ λ, M ∈ µ
such that L ∩ M = φ. Then:
and
λ+µ = hL∪M i, where x ∈ L, y ∈ M =⇒ x < y and the internal orderings within
L and M remain unchanged
µ+λ = hM ∪Li, where x ∈ M, y ∈ L =⇒ x < y and the internal orderings within
L and M remain unchanged.
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It turns out that ordinal addition is associative but not commutative. If λ = ω, β = 1,
L = N ∈ λ, M = {b} ∈ β, then
ω + 1 = h{0, 1, 2, . . . , n, . . . , b}i, whereas
1 + ω = h{b, 0, 1, 2, . . . , n, . . .}i = ω.
So ω + 1 > 1 + ω.
Returning to arbitrary ordinals λ, µ with L ∈ λ and M ∈ µ, we define the ordinal products
λµ = hL × M, dictionary orderi
µλ = hM × L, dictionary orderi .
To see that ordinal multiplication is also not commutative, consider α = ω, β = 2, L = N ∈
α, M = {a, b} ∈ β. When the dictionary orderings for these two product sets are written in
an “increasing left-to-right” format, we get
2ω = h{(a, 0), (a, 1), (a, 2), . . . , (b, 0), (b, 1), . . .}i = ω + ω,
ω2 = h{(0, a), (0, b), (1, a), (1, b), (2, a), (2, b), . . .}i = ω .
One use of ordinal arithmetic is to provide a notation for ordinal numbers. For example,
among the denumerable ordinals listed in increasing order are:
ω, ω + 1, ω + 2, . . . , 2ω, 2ω + 1, 2ω + 2, . . . , nω, nω + 1, . . . , ω 2 = ωω, ω 2 + 1, . . . , ω 2 + ω, . . . , ω n , . . .
.
Among the ordinals of cardinality ℵ1 are found:
Ω, Ω + 1, Ω + 2, . . . , 2Ω, 2Ω + 1, 2Ω + 2, . . . , nΩ, nΩ + 1, . . . , Ω2 = ΩΩ, Ω2 + 1, . . . , Ω2 + Ω, . . . , Ωn , . . .
.
An ordinal λ is called a limit ordinal if each set in λ has no greatest element, or, in other
words, λ has no immediate predecessor. The ordinals with an immediate predecessor (along
with 0) are called non-limit ordinals. In the above list of denumerable ordinals, ω, 2ω,
nω, ω 2 , ω 2 + ω, ω n are all limit ordinals, whereas ω + 1, ω + 2, 2ω + 1, 2ω + 2, nω + 1, ω 2 + 1
are non-limit ordinals. The distinction between limit ordinals and non-limit ordinals turns
out to be important in topological examples involving ordinal numbers.