Download 6. Cardinals

6. Cardinals
This chapter is concerned with the basics of cardinal arithmetic.
Definition and basic properties
A cardinal, or cardinal number, is an ordinal α such that there is no smaller ordinal
which can be put in one-one correspondence with α. We generally use Greek letters κ, λ, µ
for cardinals. Obviously if κ and λ are distinct cardinals, then they cannot be put in
one-one correspondence with each other.
Proposition 6.1. For any set X there is an ordinal α which can be put in one-one
correspondence with X.
Proof. By the well-ordering principle, let < be a well-ordering of X. Then X under
< is isomorphic to an ordinal.
By this proposition, any set can be put in one-one correspondence with a cardinal—namely
the least ordinal that is in one-one correspondence with the set. This justifies the following
definition. For any set X, the cardinality, or size, or magnitude, etc. of X is the unique
cardinal |X| which can be put in one-one correspondence with X. The basic property of
this definition is given in the following theorem.
Theorem 6.2. For any sets X and Y , the following conditions are equivalent:
(i) |X| = |Y |.
(ii) There is a one-one function mapping X onto Y .
The following proposition gives obvious facts about the particular way that we have defined
the notion of cardinality.
Proposition 6.3. (i) |α| ≤ α.
(ii) |α| = α iff α is a cardinal.
Proposition 6.4. Every natural number is a cardinal.
Proof. We prove by ordinary induction on n that for every natural number n and
for every natural number m, if m < n then there is no bijection from n to m. This is
vacuously true for n = 0. Now assume it for n, but suppose that m is a natural number
less than n + 1 and f is a bijection from n + 1 onto m. Since n + 1 6= 0, obviously also
m 6= 0. So m = m′ + 1 for some natural number m′ . [An easy induction shows that every
nonzero natural number is a successor ordinal.] Let g be the bijection from m onto m
which interchanges m′ and f (n) and leaves fixed all other elements of m. Then g ◦ f is a
bijection from n + 1 onto m which takes n to m′ . Hence (g ◦ f ) ↾ n is a bijection from n
onto m′ , and m′ < n, contradicting the inductive hypothesis.
Thus the natural numbers are the first cardinals, in the ordering of cardinals determined
by the fact that they are special kinds of ordinals. A set is finite iff it can be put in one-one
correspondence with some natural number; otherwise it is infinite. The following general
lemma helps to prove that ω is the next cardinal. It is easily proved by induction.
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Lemma 6.5. If (A, <) is a simple ordering, then every finite nonempty subset of A
has a greatest element.
Theorem 6.6. ω is a cardinal.
It is harder to find larger cardinals, but they exist; in fact the collection of cardinals is so
big that, like the collection of ordinals, it does not exist as a set. We will see this a little
bit later.
Note that a cardinal is infinite iff it is greater or equal ω. The following fact will be
useful later.
Proposition 6.7. Every infinite cardinal is a limit ordinal.
Proof. Suppose not: κ is an infinite cardinal, and κ = α + 1. We define f : α → κ as
follows: f (0) = α, f (m + 1) = m for all m ∈ ω, and f (β) = β for all β ∈ α\ω. Clearly f
is one-one and maps onto κ, contradiction.
Lemma 6.8. If κ and λ are cardinals and f : κ → λ is one-one, then κ ≤ λ.
Proof. We define α ≺ β iff α, β ∈ κ and f (α) < f (β). Clearly ≺ well-orders κ. Let g
be an isomorphism from (κ, ≺) onto an ordinal γ. Thus κ ≤ γ by the definition of cardinals.
If α < β < γ, then g −1 (α) ≺ g −1 (β), hence by definition of ≺, f (g −1 (α)) < f (g −1 (β)).
Thus f ◦ g −1 : γ → λ is strictly increasing. Hence γ ≤ λ. We already know that κ ≤ γ, so
κ ≤ λ.
The purpose of this lemma is to prove the following basic theorem.
Theorem 6.9. If A ⊆ B, then |A| ≤ |B|.
Proof. Let κ = |A|, λ = |B|, and let f and g be one-one functions from κ onto A and
of λ onto B, respectively. Then g ◦ f −1 is a one-one function from κ into λ, so κ ≤ λ.
Corollary 6.10. For any sets A and B the following conditions are equivalent:
(i) |A| ≤ |B|.
(ii) There is a one-one function mapping A into B.
(iii) A = 0, or there is a function mapping B onto A.
Corollary 6.11. If there is a one-one function from A into B and a one-one function
from B into A, then there is a one-one function from A onto B.
This corollary is called the Cantor-Bernstein, or Schröder-Bernstein theorem. Our proof,
if traced back, involves the axiom of choice. It can be proved without the axiom of choice,
and this is sometimes desirable when describing a small portion of set theory to students.
Some exercises outline such a proof.
The following simple theorem is very important and basic for the theory of cardinals.
It embodies in perhaps its simplest form the Cantor diagonal argument.
Theorem 6.12. For any set A we have |A| < |P(A)|.
Proof. The function given by a 7→ {a} is a one-one function from A into P(A), and
so |A| ≤ |P(A)|. [Saying that a 7→ {a} is giving the value of the function at the argument
a.] Suppose equality holds. Then there is a one-one function f mapping A onto P(A). Let
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X = {a ∈ A : a ∈
/ f (a)}. Since f maps onto P(A), choose a0 ∈ A such that f (a0 ) = X.
Then a0 ∈ X iff a0 ∈
/ X, contradiction.
By this theorem, for every ordinal α there is a larger cardinal, namely |P(α)|. Hence we
can define α+ to be the least cardinal > α. Cardinals of the form κ+ are called successor
cardinals; other infinite cardinals are called limit cardinals. Is κ+ = |P(κ)|? The statement
that this is true for every infinite cardinal κ is the famous generalized continuum hypothesis
(GCH). The weaker statement that ω + = |P(ω)| is the continuum hypothesis (CH).
It can be shown that the generalized continuum hypothesis is consistent with our
axioms. But also its negation is consistent; in fact, the negation of the weaker continuum
hypothesis is consistent. All of this under the assumption that our axioms are consistent.
(It is not possible to prove this consistency.)
S
Theorem 6.13. If Γ is a set of cardinals, then Γ is also a cardinal.
S
S
def S
Proof. We
S know already that Γ is an ordinal. Suppose that κ = | Γ| < Γ. By
definition
of , there isSa λ ∈ Γ such that κ < λ. (Membership is the same as <.) Now
S
λ ⊆ Γ. So λ = |λ| ≤ | Γ| = κ, contradiction.
We can now define the standard sequence of infinite cardinal numbers, by transfinite recursion:
ℵ0 = ω;
ℵα+1 = ℵ+
α;
[
ℵβ =
ℵα for β a limit ordinal.
α<β
For historical reasons, one sometimes writes ωα in place of ℵα . Now we get the following
two results.
Lemma 6.14. If α < β, then ℵα < ℵβ .
Lemma 6.15. α ≤ ℵα for every ordinal α.
Theorem 6.16. For every infinite cardinal κ there is an ordinal α such that κ = ℵα .
Proof. Let κ be any infinite cardinal. Then κ ≤ ℵκ < ℵκ+1 . Here κ + 1 refers to
ordinal addition. This shows that there is an ordinal α such that κ < ℵα ; choose the least
such α. Clearly α 6= 0 and α is not a limit ordinal. Say α = β + 1. Then ℵβ ≤ κ < ℵβ+1 ,
so κ = ℵβ .
We can now say a little more about the continuum hypothesis. Not only is it consistent that
it fails, but it is even consistent that |P(ω)| = ℵ2 , or |P(ω)| = ℵ17 , or |P(ω)| = ℵω+1 ;
the possibilities have been spelled out in great detail. Some impossible situations are
|P(ω)| = ℵω and |P(ω)| = ℵω+ω ; we will establish this later in this chapter.
Addition of cardinals
Let κ and λ be cardinals. We define
κ + λ = |{(α, 0) : α ∈ κ} ∪ {(β, 1) : β ∈ λ}|.
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The idea is to take disjoint copies κ × {0} and λ × {1} of κ and λ and count the number
of elements in their union.
Two immediate remarks should be made about this definition. First of all, this is not,
in general, the same as the ordinal sum κ + λ. We depend on the context to distinguish
the two notions of addition. For example, ω + 1 = ω in the cardinal sense, but not in
the ordinal sense. In fact, we know that ω < ω + 1 in the ordinal sense. To show that
ω + 1 = ω in the cardinal sense, it suffices to define a one-one function from ω onto the set
{(m, 0) : m ∈ ω} ∪ {(0, 1)}.
Let f (0) = (0, 1) and f (m + 1) = (m, 0) for any m ∈ ω.
Secondly, the definition is consistent with our definition of addition for natural numbers (as a special case of ordinal addition), and thus it does coincide with ordinal addition
when restricted to ω:
Proposition 6.17. If m and n are natural numbers, then addition in the sense of
chapter 2 and in the cardinal number sense are the same.
Proof. By induction on n.
Aside from simple facts about addition, there is the remarkable fact that κ + κ = κ for
every infinite cardinal κ. We shall prove this as a consequence of the similar result for
multiplication.
The definition of cardinal addition can be extended to infinite sums, and very elementary properties of the binary sum are then special cases of more general results; so we
proceed with the general definition. Let hκi : i ∈ Ii be a system of cardinals (this just
means that κ is a function with domain I whose values are always cardinals). Then we
define
[
X
κi = (κi × {i}) .
i∈I
i∈I
This is a generalization of summing two cardinals, as is immediate from the definitions:
Proposition 6.18. If hκi : i ∈ 2i is a system of cardinals P
(meaning that κ is a function
with domain 2 such that both κ0 and κ1 are cardinals), then i∈2 κi = κ0 + κ1 .
The following is easily proved by induction on |I|:
P Proposition 6.19. If hmi : i ∈ Ii is a system of natural numbers, with I finite, then
i∈I mi is a natural number.
We mention some important but easy facts concerning the cardinalities of unions:
6.20.
S Proposition
P
i∈I Ai =
i∈I |Ai |.
If hAi : i ∈ Ii is a system of pairwise disjoint sets, then
S
P
Proposition 6.21. If hAi : i ∈ Ii is any system of sets, then i∈I Ai ≤ i∈I |Ai |.
Corollary 6.22. If hκi : i ∈ Ii is a system of cardinals, then
53
S
i∈I
κi ≤
P
i∈I
κi .
Finally, we gather together some simple arithmetic of infinite sums:
P
Proposition
6.23.
(i)
i∈I 0 = 0.
P
(ii) Pi∈0 κi = 0.P
(iii) i∈I κi = i∈I,κ
P i 6=0 κi . P
(iv) If I ⊆ J, then i∈I κi ≤ i∈J
P κi .
P
(v) IfPκi ≤ λi for all i ∈ I, then i∈I κi ≤ i∈I λi .
(vi) i∈I 1 = |I|.
(vii) If κ is infinite, then κ + 1 = κ.
Multiplication of cardinals
By definition,
κ · λ = |κ × λ|.
The following simple result can be used in verifying many simple facts concerning products.
Two sets are equipotent iff there is a bijection between them.
Proposition 6.24. If A is equipotent with C and B is equipotent with D, then A × B
is equipotent with C × D.
Proposition 6.25. (i) κ · λ = λ · κ;
(ii) κ · (λ · µ) = (κ · λ) · µ;
(iii) κ · (λ + µ) = κ · λ + κ · µ;
(iv) κ · 0 = 0;
(v) κ · 1 = κ;
(vi) κP
· 2 = κ + κ;
(vii) i∈I κ = κ · |I|;
(viii) If κ ≤ µ and λ ≤ ν, then κ · λ ≤ µ · ν.
Proposition 6.26. Multiplication of natural numbers means the same in the cardinal
number sense as in ordinal sense.
The basic theorem about multiplication of infinite cardinals is as follows.
Theorem 6.27. κ · κ = κ for every infinite cardinal κ.
Proof. Suppose not, and let κ be the least infinite cardinal such that κ · κ 6= κ. Then
κ = κ· 1 ≤ κ· κ, and so κ < κ· κ. We now define a relation ≺ on κ×κ. For all α, β, γ, δ ∈ κ,
(α, β) ≺ (γ, δ) iff max(α, β) < max(γ, δ)
or max(α, β) = max(γ, δ) and α < γ
or max(α, β) = max(γ, δ) and α = γ and β < δ.
Clearly this is a well-order. It follows that (κ × κ, ≺) is isomorphic to an ordinal α; let f
be the isomorphism. We have |α| = |κ × κ| = κ · κ > κ by the remark at the beginning of
this proof. So κ < α. Therefore there exist β, γ ∈ κ such that f (β, γ) = κ. Now
f [{(δ, ε) ∈ κ × κ : (δ, ε) ≺ (β, γ)}] = κ,
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so, with ϕ = max(β, γ)+1,
κ = |{(δ, ε) ∈ κ × κ : (δ, ε) ≺ (β, γ)}|
≤ |ϕ × ϕ| = |ϕ| · |ϕ|.
But ϕ < κ, so either ϕ is finite, and |ϕ| · |ϕ| is then also finite, or else ϕ is infinite, and
|ϕ| · |ϕ| = |ϕ| by the minimality of κ. In any case, |ϕ| · |ϕ| < κ, contradiction.
Corollary 6.28. If κ and λ are nonzero cardinals and at least one of them is infinite,
then κ + λ = κ · λ = max(κ, λ).
Corollary 6.29. If hAi : i ∈ Ii is any system of sets, then
[ [
|Ai |.
Ai ≤ |I| ·
i∈I
i∈I
A set A is countable if |A| ≤ ω. So another corollary is
Corollary 6.30. A countable union of countable sets is countable.
Proposition 6.31. If hκi : i ∈PIi is a systemS of nonzero cardinals, and either I is
infinite or some κi is infinite, then i∈I κi = |I| · i∈I κi .
By the above results, the binary operations of addition and multiplication of cardinals are
trivial when applied to infinite cardinals; and the infinite sum is also easy to calculate.
We now introduce infinite products which, as we shall see, are not so trivial. We need the
following standard elementary notion: for hAi : i ∈ Ii a family of sets, we define
Y
Ai = {f : f is a function, dmn(f ) = I, and ∀i ∈ I[f (i) ∈ Ai ]}.
i∈I
This is the cartesian product of the sets Ai . Now if hκi : i ∈ Ii is a system of cardinals, we
define
Y Y
κi = κi .
i∈I
i∈I
Q
Here on the right i∈I κi is the cartesian product of the cardinals κi , and on the left is the
defined product ofQthem, a certain new cardinal. We depend on the context to distinguish
these two uses of .
Some elementary properties of this notion are summarized in the following proposition.
Q
Q
Proposition 6.32. (i) i∈I Ai =
Q i∈I |Ai |.
(ii) IfQκi = 0 for some i ∈ I, then i∈I κi = 0.
(iii) Q i∈0 κi = Q
1.
(iv) Q i∈I κi = i∈I,κi 6=1 κi .
(v) i∈I 1 = 1.
Q
Q
(vi) IfQκi ≤ λi for all i ∈ I, then i∈I κi ≤ i∈I λi .
(vii) i∈2 κi = κ0 · κ1 .
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Proof. Most of these facts are very easy. We give the proof for (i). Q
According to
the
Q definition of product, it suffices to find a one-one function g mapping i∈I Ai onto
mapping Ai onto |Ai . (We are using
i∈I |Ai |. For each i ∈ I, let fi be a one-one function
Q
the axiom of choice here.) Then for each x ∈ i∈I Ai and each i ∈ I let g(x)i = fi (xi ). It
is easily checked that g is as desired.
General commutative, associative, and distributive laws hold also:
Proposition 6.33. (Commutative law) If hκi : i ∈ Ii is a system of cardinals and
f : I → I is one-one and onto, then
Y
Y
κi =
κf (i) .
i∈I
then
i∈I
Proposition 6.34. (Associative law) If hκij : (i, j) ∈ I × Ji is a system of cardinals,


Y Y
Y

κij  =
κij .
i∈I
j∈J
(i,j)∈I×J
Proposition 6.35. (Distributive law) If hλi : i ∈ Ii is a system of cardinals, then
κ·
X
λi =
i∈I
X
(κ · λi ).
i∈I
Theorem 6.36. (König) Suppose that hκi : i ∈ Ii and hλi : i ∈ Ii are systems of
cardinals such that λi < κi for all i ∈ I. Then
X
Y
λi <
κi .
i∈I
i∈I
Proof. The proofQis another P
instance of Cantor’s diagonal argument. Suppose that
this is notQtrue; thus i∈I κi ≤
i∈I λi . It follows that there is a one-one function f
mapping i∈I κi into {(α, i) : i ∈ I, α < λi }. For each i ∈ I let
Ki = {(f −1 (α, i))i : α < λi , (α, i) ∈ rng(f )}.
Clearly Ki ⊆ κi . Now |Ki | ≤ λi < κi , so we can choose xi ∈ κi \Ki (using the axiom of
choice). Say f (x) = (α, i). Then xi = (f −1 (α, i))i ∈ Ki , contradiction.
Exponentiation of cardinals
We define
κλ = |λ κ|.
The elementary arithmetic of exponentiation is summarized in the following proposition:
Proposition 6.37. (i) κ0 = 1.
(ii) If κ 6= 0, then 0κ = 0.
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(iii) κ1 = κ.
(iv) 1κ = 1.
(v) κ2 = κ · κ.
(vi) κλ · κµ = κλ+µ .
(vii) (κ · λ)µ = κµ · λµ .
(viii) (κλ )µ = κλ·µ .
(ix) Q
If κ ≤ λ 6= 0 and µ ≤ ν, then κµ ≤ λν .
|I|
(x) P
i∈I κ = κ .
Q
λ
(xi) κ i∈I i = i∈I κλi .
λ Q
Q
(xii)
κ
= i∈I κλi .
i
i∈I
Proof. The proofs are straightforward. We give the proof of (xi) as an illustration.
By the definitions of arbitrary sums and products it suffices to find a bijection f from the
set
S
{λi ×{i}:i∈I}
(∗)
κ
Q
to the set i∈I λi κ, where the product here is just the product of sets. So take any
Q
member x of (∗). We define f (x) in i∈I λi κ by defining its value (f (x))i for each i ∈ I;
and we define (f (x))i in λi κ by defining its value ((f (x))i)(α) for each α ∈ λi . Since
(α, i) ∈ λi × {i}, the pair (α, i) is in the domain of x. We define ((f (x))i)(α) = x(α, i).
f is one-one: suppose that x, y ∈ (∗) and f (x) = f (y). Take any u ∈ dmn(x); say
u = (α, i) with i ∈ I and α ∈ λi . Then x(u) = x(α, i) = ((f (x))i )(α) = ((f (y))i)(α) =
y(α, i) = y(u). Q
Q
f maps onto i∈I λi κ: take any g ∈ i∈I λi κ. Define x ∈ (∗) by setting, for any i ∈ I
and α ∈ λi , x(α, i) = (g(i))(α). Then ((f (x))i )(α) = x(α, i) = (g(i))(α), and since i and
α are arbitrary we get f (x) = g.
Proposition 6.38. If m, n ∈ ω, then mn ∈ ω
Proposition 6.39. |P(A)| = 2|A| .
For each X ⊆ A define χX ∈ A 2 by setting
n
1
χX (a) =
0
if a ∈ X,
otherwise.
[This is the characteristic function of X.] It is easy to check that χ is a bijection from
P(A) onto A 2.
The calculation of exponentiation is not as simple as that for addition and multiplication.
The following result gives one of the most useful facts about exponentiation, however.
Theorem 6.40. If 2 ≤ κ ≤ λ ≥ ω, then κλ = 2λ .
Proof. Note that each function f : λ → λ is a subset of λ × λ. Hence λ λ ⊆ P(λ × λ),
and so λλ ≤ |P(λ × λ). Therefore,
2λ ≤ κλ ≤ λλ ≤ |P(λ × λ)| = 2λ·λ = 2λ ;
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so all the entries in this string of inequalities are equal, and this gives κλ = 2λ .
Cofinality, and
regular and singular cardinals
Further cardinal arithmetic depends on the notion of cofinality. For later purposes we
define a rather general version of this notion. Let (P, <) be a partially ordered set. A
subset X of P is dominating iff for every p ∈ P there is an x ∈ X such that p ≤ x. The
cofinality of P is the smallest cardinality of a dominating subset of P . We denote this
cardinal by cf(P ).
A subset X of P is unbounded iff there does not exist a p ∈ P such that x ≤ p for all
x ∈ X. If P is simply ordered without largest element, then these notions—dominating
and unbounded—coincide. In fact, suppose that X is dominating but not unbounded.
Since X is not unbounded, choose p ∈ P such that x ≤ p for all x ∈ X. Since P does
not have a largest element, choose q ∈ P such that p < q. Then because X is dominating,
choose x ∈ X such that q ≤ x. Then q ≤ x ≤ p < q, contradiction. Thus X dominating
implies that X is unbounded. Now suppose that Y is unbounded but not dominating.
Since Y is not dominating, there is a p ∈ P such that p 6≤ x, for all x ∈ Y . Since P is a
simple order, it follows that x < p for all x ∈ Y . This contradicts Y being unbounded.
A cardinal κ is regular iff κ is infinite and cf(κ) = κ. An infinite cardinal that is not
regular is called singular.
Theorem 6.41. For every infinite cardinal κ, the cardinal κ+ is regular.
Proof. Suppose that Γ ⊆ κ+ , Γ is unbounded in κ+ , and |Γ| < κ+ . Hence
X
[
X
+
|γ| ≤
κ = κ · κ = κ,
γ ≤
κ =
γ∈Γ γ∈Γ
γ∈Γ
contradiction. The first equality here holds because Γ is unbounded in κ+ and κ+ is a
limit ordinal.
This theorem almost tells the full story about when a cardinal is regular. Examples of
singular cardinals are ℵω+ω and ℵω1 . But it is conceivable that there are regular cardinals
not covered by Theorem 6.41. A regular limit cardinal is said to be weakly inaccessible. A
cardinal κ is said to be inaccessible if it is regular and has the property that for any cardinal
λ < κ, also 2λ < κ. Clearly every inaccessible cardinal is also weakly inaccessible. Under
GCH, the two notions coincide. It is consistent with ZFC that 2ω is weakly inaccessible;
but of course it definitely is not inaccessible. It is consistent with ZFC that there are no
uncountable weak inaccessibles at all. But it is reasonable to postulate their existence,
and they are useful in some situations. In fact, the subject of large cardinals is one of the
most studied in contemporary set theory, with many spectacular results.
Theorem 6.42. Suppose that (A, <) is a simple ordering with no largest element.
Then there is a strictly increasing function f : cf(A) → A such that rng(f ) is unbounded
in A.
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Proof. Let X be a dominating subset of A of size cf(A), and let g be a bijection from
cf(A) onto X. We define a function f : cf(A) → X by recursion, as follows. If f (α) ∈ X
has been defined for all β < α, where α < cf(A), then {f (β) : β < α} has size less than
cf(A), and hence it is not dominating. Hence there is an a ∈ A such that f (β) < a for all
β < α. We let f (α) be an element of X such that a, g(α) ≤ f (α).
Clearly f is strictly increasing. If a ∈ A, choose α < cf(A) such that a ≤ g(α). Then
a ≤ f (α).
Proposition 6.43. Suppose that (A, <) is a simple ordering with no largest element.
Then cf(cf(A)) = cf(A).
Proof. Clearly cf(α) ≤ α for any ordinal α; in particular, cf(cf(A)) ≤ cf(A). Now by
Theorem 6.42, let f : cf(A) → A be strictly increasing with rng(f ) unbounded in A, and
let g : cf(cf(A)) → cf(A) be strictly increasing with rng(g) unbounded in cf(A). Clearly
f ◦ g : cf(cf(A)) → A is strictly increasing. We claim that rng(f ◦ g) is unbounded in A.
For, given a ∈ A, choose α < cf(A) such that a ≤ f (α), and then choose β < cf(cf(A))
such that α ≤ g(β). Then a ≤ f (α) ≤ f (g(β)), proving the claim. It follows that
cf(A) ≤ cf(cf(A)).
S
Proposition 6.44. If κ is a regular cardinal, Γ ⊆ κ, and |Γ| < κ, then Γ < κ.
Proof. Since cf(κ) = κ, from the definition of cf it follows
that Γ is bounded in κ.
S
Hence there is an α < κ such that γ ≤ α for all γ ∈ Γ. So Γ ≤ α < κ.
Proposition 6.45. If A is a linearly ordered set with no greatest element, κ is a
regular cardinal, and f : κ → A is strictly increasing with rng(f ) unbounded in A, then
κ = cf(A).
Proof. By the definition of cf we have cf(A) ≤ κ. Suppose that cf(A) < κ. By
Theorem 6.42 let g : cf(A) → A be strictly increasing with rng(g) unbounded in A. For
each α < cf(α) choose βα < κ such S
that g(α) ≤ f (βα ). Then {βα : α < cf(A)} ⊆ κ and
|{βα : α < cf(A)}| < κ, so by 6.44, α<cf(A) βα < κ. Let γ < κ be such that βα < γ for
all α < cf(A). Then f (γ) is a bound for rng(g), contradiction.
Proposition 6.46. A cardinal κ isP
regular iff for every system hλi : i ∈ Ii of cardinals
less than κ, with |I| < κ, one also has i∈I λi < κ.
P
S
S
Proof.
⇒:
λ
≤
|I|
·
λ
<
κ.
⇐:
if
Γ
⊆
κ
and
|Γ|
<
κ,
then
|
Γ| ≤
i
i
i∈I
i∈I
P
S
|λ|
<
κ,
so
also
Γ
<
κ.
Thus
κ
is
regular.
λ∈Γ
Proposition 6.47. If κ is an infinite singular cardinal, then there is a P
strictly increasing sequence hλα : α < cf(κ)i of infinite successor cardinals such that κ = α<cf(κ) λα .
Proof. Choose Γ ⊆ κ unbounded and of order type cf(κ); let hγα : α < cf(κ)i
enumerate Γ in strictly increasing order. Since each γα is merely an ordinal, this does not
complete the proof. We define the desired sequence bySrecursion. Suppose that λβ < κ
has been defined for all β < α, with α < cf(κ). Then β<α λβ < κ by the definition of
cofinality. So also


+
[
max γα ,
λβ  < κ,
β<α
59
and we define λP
α to be this cardinal.
Now γδ ≤ α<cf(κ) λα for each δ < cf(κ), so
κ=
[
Γ≤
X
λα ≤ κ · cf(κ) = κ.
α<cf(κ)
The main theorem of cardinal arithmetic
Now we return to the general treatment of cardinal arithmetic.
Theorem 6.48. (König) If κ is infinite and cf(κ) ≤ λ, then κλ > κ.
Proof. If κ is regular, then κλ ≥ κκ = 2κ > κ. So, assume that κ is singular. Then
therePis a system hµα : α < cf(κ)i of nonzero cardinals such that each µα is less than κ,
and α<cf(κ) µα = κ. Hence
κ=
X
Y
µα <
α<cf(κ)
κ = κcf(κ) ≤ κλ .
α<cf(κ)
Corollary 6.49. For λ infinite we have cf(2λ ) > λ.
Proof. Suppose that cf(2λ ) ≤ λ. Then (2λ )λ > 2λ . But (2λ )λ = 2λ·λ = 2λ ,
contradiction.
We can now verify a statement made earlier about possibilities for |P(ω)|. Since |P(ω)| =
2ω , the corollary says that cf(2ω ) > ω. So this implies that |P(ω)| cannot be ℵω or ℵω+ω .
Here ω + ω is the ordinal sum of ω with ω. It rules out many other possibilities of this
sort.
We now prove a lemma needed for the last major theorem of this subsection, which
says how to compute exponents (in a way).
Lemma 6.50. If κ is a limit cardinal and λ ≥ cf(κ), then


κλ = 
[
µ<κ
µ a cardinal
cf(κ)

µλ 
.
Proof. LetQγ : cf(κ) → κ be strictly increasing with rng(γ) unbounded in κ. We
define F : λ κ → α<cf(κ) λ γα as follows. If f ∈ λ κ, α < cf(κ), and β < λ, then
((F (f ))α )β =
n
f (β) if f (β) < γα ,
0
otherwise.
60
Now F is a one-one function. For, if f, g ∈ λ κ and f 6= g, say β < λ and f (β) 6= g(β).
Choose α < cfκ such that f (β) and g(β) are both less than γα . Then ((F (f ))α )β = f (β) 6=
g(β) = ((F (g))α )β , from which it follows that F (f ) 6= F (g). Since F is one-one,
Y
λ
λ
λ
κ = | κ| ≤ γα α<cf(κ)


Y

 [
λ
µ
≤

α<cf(κ) µ a µ<κ
cardinal
cf(κ)


 [
µλ 
=
µ<κ
µ a cardinal
≤ (κλ )cf(κ) = κλ·cf(κ) = κλ ,
and the lemma follows.
The following theorem is not needed for the main result, but it is a classical result about
exponentiation.
Theorem 6.51. (Hausdorff) If κ and λ are infinite cardinals, then (κ+ )λ = κλ · κ+ .
Proof. If κ+ ≤ λ, then both sides are equal to 2λ . Suppose that λ < κ+ . Then
[
+ λ
λ +
λ (κ ) = | (κ )| = α
α<κ+
X
≤
|α|λ ≤ κλ · κ+ ≤ (κ+ )λ ,
α<κ+
as desired.
Here is the promised theorem giving computation rules for exponentiation. It essentially
reduces the computation of κλ to two special cases: 2λ , and κcf (κ) . Generalizations of the
results mentioned about the continuum hypothesis give a pretty good picture of what can
happen to 2λ . The case of κcf(κ is more complicated, and there is still work being done on
what the possibilities here are. Recent deep work of Shelah on pcf theory has shed some
light on this. For example, he showed that ℵℵω0 ≤ 2ℵ0 + ℵω4 . The role of ω4 here is still
unclear.
Theorem 6.52. (main theorem of cardinal arithmetic) Let κ and λ be cardinals with
2 ≤ κ and λ ≥ ω. Then
(i) If κ ≤ λ, then κλ = 2λ .
(ii) If κ is infinite and there is a µ < κ such that µλ ≥ κ, then κλ = µλ .
(iii) Assume that κ is infinite and µλ < κ for all µ < κ. Then λ < κ, and:
61
(a) if cf(κ) > λ, then κλ = κ;
(b) if cf(κ) ≤ λ, then κλ = κcf (κ) .
Proof. (i) has already been noted. Under the hypothesis of (ii),
κλ ≤ (µλ )λ = µλ ≤ κλ ,
as desired.
Now assume the hypothesis of (iii). In particular, 2λ < κ, so of course λ < κ. Next,
assume the hypothesis of (iii)(a): cf(κ) > λ. Then
[
λ κλ = |λ κ| = α
α<κ
X
|α|λ ≤ κ,
≤
(since λ < cf(κ))
α<κ
giving the desired result.
Finally, assume the hypothesis of (iii)(b): cf(κ) ≤ λ. Since λ < κ, it follows that κ is
singular, so in particular it is a limit cardinal. Then


κλ = 
[
µ<κ
µ a cardinal
cf(κ)

µλ 
≤ κcf(κ) ≤ κλ ,
finishing the proof.
In theory one can now compute κλ for infinite κ, λ as follows. If κ ≤ λ, then κλ = 2λ .
Suppose that κ > λ. Let κ′ be minimum such that (κ′ )λ = κλ . Then ∀µ < κ′ [µλ < κ′ ]. In
fact, if µ < κ′ and µλ ≥ κ′ , then (κ′ )λ ≤ (µλ )λ = µλ·λ = µλ < κλ = (κ′ )λ , contradiction.
Now (κ′ )λ is computed by 6.52(iii).
Under the generalized continuum hypothesis the computation of exponents is very
simple:
Corollary 6.53. Assume GCH, and suppose that κ and λ are cardinals with 2 ≤ κ
and λ infinite. Then:
(i) If κ ≤ λ, then κλ = λ+ .
(ii) If cf(κ) ≤ λ < κ, then κλ = κ+ .
(iii) If λ < cf(κ), then κλ = κ.
Proof. (i) is immediate from 6.51(i). For (ii), assume that cf(κ) ≤ λ < κ. Then
κ is a limit cardinal, and so for each µ < κ we have µλ ≤ (max(µ, λ))+ < κ; hence by
6.51(iii)(b) we have κλ = κcf(κ) = κ+ . For (iii), assume that λ < cf(κ). If there is a µ < κ
such that µλ ≥ κ, then by 6.51(ii), κλ = µλ ≤ (max(λ, µ))+ ≤ κ, as desired. If µλ < κ for
all µ < κ, then κλ = κ by 6.51(iii)(a).
62
EXERCISES
The first four exercises outline a proof of the Cantor-Schröder-Bernstein theorem without
using the axiom of choice. This theorem says that if there is an injection from A into B
and one from B into A, then there is a bijection from A to B. In the development in the
text, using the axiom of choice, the hypothesis implies that |A| ≤ |B| ≤ |A|, and hence
|A| = |B|. But it is of some interest that it can be proved in an elementary way, without
using the axiom of choice or anything about ordinals and cardinals. Of course, the axiom
of choice should not be used in these four exercises.
E6.1. Let F : P(A) → P(A), and assume that for all X, Y ⊆ A, ifS X ⊆ Y , then
F (X) ⊆ F (Y ). Let A = {X : X ⊆ A and X ⊆ F (X)}, and set X0 = X∈A X. Then
X0 ⊆ F (X0 ).
E6.2. Under the assumptions of exercise 1 we actually have X0 = F (X0 ).
E6.3. Suppose that f : A → B is one-one and g : B → A is also one-one. For every X ⊆ A
let F (X) = A\g[B\f [X]]. Show that for all X, Y ⊆ A, if X ⊆ Y then F (X) ⊆ F (Y ).
E6.4. Prove the Cantor-Schröder-Bernstein theorem as follows. Assume that f and g are
as in exercise E6.3, and choose F as in that exercise. Let X0 be as in exercise E6.1. Show
that A\X0 ⊆ rng(g). Then define h : A → B by setting, for any a ∈ A,
h(a) =
f (a)
if a ∈ X0 ,
−1
g (a) if a ∈ A\X0 .
Show that h is one-one and maps onto B.
E6.5. Show that if α and β are ordinals, then |α ∔ β| = |α| + |β|, where ∔ is ordinal
addition and + is cardinal addition.
E6.6. Show that if α and β are ordinals, then |α ⊙ β| = |α| · |β|, where ⊙ is ordinal
multiplication and · is cardinal multiplication.
E6.7. Show that if α and β are ordinals, 2 ≤ α, and ω ≤ β, then |· αβ | = |α| · |β|. Here
the dot to the left of the first exponent indicates that ordinal exponentiation is involved.
[This is a good exercise to keep in mind. For example, · 2ω is a countable set,
but 2ω is not.]
E6.8. Prove that if |A| ≤ |B| then |P(A)| ≤ |P(B)|.
E6.9. Prove the following general distributive law:
YX
i∈I j∈Ji
where P =
Q
i∈I
κij =
XY
κi,f (i) ,
f ∈P i∈I
Ji .
E6.10. Show that for any cardinal κ we have κ+ = {α : α is an ordinal and |α| ≤ κ}.
E6.11. For every infinite cardinal λ there is a cardinal κ > λ such that κλ = κ.
63
E6.12. For every infinite cardinal λ there is a cardinal κ > λ such that κλ > κ.
E6.13. Prove that for every n ∈ ω, and every infinite cardinal κ, ℵκn = 2κ · ℵn .
E6.14. Prove that ℵℵω1 = 2ℵ1 · ℵℵω0 .
Q
E6.15. Prove that ℵℵω0 = n∈ω ℵn .
E6.16. Prove that for any infinite cardinal κ, (κ+ )κ = 2κ .
E6.17. Show that if κ is an infinite cardinal and C is the collection of all cardinals less
than κ, then |C| ≤ κ.
E6.18. Show that if κ is an infinite cardinal and C is the collection of all cardinals less
than κ, then
!cf(κ)
X
.
2κ =
2ν
ν∈C
E6.19. Prove that for any limit ordinal τ ,
Q
ξ<τ
2ℵξ = 2ℵτ .
E6.20.
Show that if κ is an infinite cardinal and C is the set of all cardinals ≤ κ, then
P
µ
κ
= 2κ .
µ∈C
E6.21.PShow that if λ is an infinite cardinal, κ = λ+ , and C is the set of all cardinals < κ,
then µ∈C κµ = 2λ .
E6.22. Assume that κ is an infinite cardinal, and 2λ < κ for every cardinal λ < κ. Show
that 2κ = κcf(κ) .
E6.23. Suppose that λ is a singular cardinal, cf(λ) = ω, and 2κ < λ for every κ < λ. Prove
that 2λ = λω .
References
Abraham, U.; Magidor, M. Cardinal arithmetic. Chapter in Handbook of Set Theory.
Springer 2010, 2197pp.
Holz, M.; Steffens, K.; Weitz, E. Introduction to Cardinal Arithmetic. Birkhäuser
1999, 304pp.
64