Download Cardinals 1. Introduction to Cardinals We work in the base theory ZF

Cardinals
1. Introduction to Cardinals
We work in the base theory ZF. The definitions and many (but not all) of the
basic theorems do not require AC; we will indicate explicitly when we are using
choice.
Definition 1.1. For sets X, Y , we write X Y to mean there is a one-to-one
function from X to Y . We write X ∼ Y to mean there is a bijection from X to Y .
Theorem 1.2. (Cantor-Schröder-Bernstein) If X Y and Y X then X ∼ Y .
Proof. Let f : X → Y and g : Y → X be one-to-one. Let X0 = X, Y0 = Y
and define recursively Xn = g(Yn−1 ), Yn = f (Xn−1 ) for n ≥ 1. Thus, X0 = X,
X1 = g(Y ), X2 = g ◦ f (X), X3 = g ◦ f ◦ g(Y ), etc., and similarly for the
T Yn .
Note that
X
⊇
X
⊇
X
⊇
.
.
.
,
and
Y
⊇
Y
⊇
Y
⊇
.
.
.
.
Let
X
=
0
1
2
0
1
2
∞
n Xn ,
T
Y∞ = n Yn . Define h : X → Y by


if x ∈ Xn − Xn+1 and n is even
f (x)
−1
h(x) = g (x) if x ∈ Xn − Xn+1 and n is odd


f (x)
if x ∈ X∞
Easily h is a bijection between Xn − Xn+2 = (Xn − Xn+1 ) ∪ (Xn+1 − Xn+2 ) and
Yn − Yn+2 = (Yn − Yn+1 ) ∪ (Yn+1 − Yn+2 ) for any even n, as f bijects (Xn − Xn+1 )
and (Yn+1 − Yn+2 ) and g −1 bijects (Xn+1 − Xn+2 ) and (Yn − Yn+1 ). Thus, h is
a bijection between X − X∞ and Y − Y∞ . It is easy to also see that f induces a
bijection from X∞ to Y∞ , and thus h is a bijection from X to Y .
Note that the above proof did not use AC.
Definition 1.3. κ is a cardinal if κ ∈ ON and ∀α < κ ¬(α ∼ κ).
Note that this definition is easily formalized into set-theory, and thus the collection of cardinals is a class. Also, note that AC is not used in the definition of
cardinal. We frequently use κ, λ, µ for cardianls.
If α ∈ ON, we let |α| denote the least ordinal β such that β ∼ α. Clearly |α| is
a cardinal. Also, |α| ≤ α. Again, AC is not needed to define |α| for α ∈ ON. We
refer to |α| as the cardinality of α.
Exercise 1. Show that every n ∈ ω, ¬(n + 1 n). Make sure you are giving a
“real” proof from within ZF; in particular, you are using only the official definition
of integer.
Exercise 2. Show that every n ∈ ω is a cardinal, and ω is a cardinal.
We define addition and multiplication on the cardinals. Note that these are
different operations that those defined on the ordinals. Thus, when we write κ · λ,
it needs to be specified (or be clear from the context) whether we are talking about
multiplication as ordinals or as cardinals.
Definition 1.4. for cardinals κ, λ:
(1) κ + λ = |(κ × {0}) ∪ (λ × {1})|
(2) κ · λ = |κ × λ|
1
2
Note that κ+λ is just the cardinality of the ordinals sum of κ and λ, and likewise
for multiplication. Thus, these operations are clearly associative (also trivial to
show directly). They are also clearly commutative, unlike the ordinal operations.
Lemma 1.5. Any infinite cardinal is a limit ordinal.
Proof. If κ is an infinite cardinal and κ = α + 1, then α is also infinite. However, if
α is an infinite ordinal then α = 1 + α ∼ α + 1 and so |κ| = |α|, a contradiction. Exercise 3. Show that for any cardinals κ, λ that κ + λ = max{κ, λ}. [hint: Say
λ ≤ κ. Prove by induction on β ∈ ON that if α ≤ β and γ is the mingling of α and
β according to xη < yη < xη+1 (here γ is the order-type of X ∪ Y , and the xη , yη
enumerate X, Y respectively), then γ ≤ β + n for some n ∈ ω if β is a successor,
and γ ≤ β if β is a limit].
Using AC, we can the notion of cardinality of an arbitrary set.
Definition 1.6. (ZFC) Let |X| denote the least ordinal α such that α ∼ X.
Using AC, this definition is well-defined, as every set X can be well-ordered, and
thus put into bijection with an ordinal. Clearly the least such ordinal, namely |X|,
is a cardinal.
Exercise 4. Using AC, give another proof of the Cantor-Schröder-Bernstein theorem. (hint: First identify X, Y with cardinals, say κ, λ. Without loss of generality,
κ < λ. Show, without using Cantor-Schröder-Bernstein, that if there were a oneto-one map from λ to κ, then κ ∼ λ, a contradiction. Use the previous exercise.)
The next result shows that for infinite cardinals κ, λ, κ · λ = max{κ, λ}, and
thus addition and multiplication on the infinite cardinals coincide.
Theorem 1.7. If κ is an infinite cardinal, then κ · κ = κ.
Proof. We prove this by induction on κ. Consider the following ordering (Gödel
ordering) on κ × κ:
(α, β)/(α0 , β 0 ) ↔ max{α, β} < max{α0 , β 0 } ∨
[(max{α, β} = max{α0 , β 0 }) ∧ (α, β) <lex (α0 , β 0 )].
This is easily a well-order of κ×κ. For each (α, β) ∈ κ×κ, the initial segment of this
order determined by (α, β) is in bijection with a subset of max{α, β} × max{α, β}
which by induction has cardinality at most max{α, β} < κ. Thus, the ordering /
has length at most κ.
Note again that the above theorem, which completely characterizes addition and
multiplication on the infinite cardinals, is proved without AC. The next lemma
generalizes theorem 1.7, but now requires the axiom of choice.
S
Theorem 1.8. (ZFC) Let κ ∈ CARD. Let X = α<κ Xα , where |Xα | ≤ κ for all
α < κ. Then |X| ≤ κ.
Proof. Using AC, let f be a function with domain κ such that for all α < κ,
f (α) : κ → Xα is a bijection. Define F : κ × κ → X by F (α, β) = f (α)(β). Clearly
F is onto, and the result follws by lemma 1.7.
Exercise 5. Show that |R| = |ω ω | = 2ω .
3
Definition 1.9. We say a set X is finite if |X| < ω, we say X is countable if
|X| ≤ ω.
Exercise 6. Show that Z, Q, and the set of algebraic numbers are countable. Show
that the set of formulas is the language of set theory is countable.
To show that there is any uncountable cardinal requires the power set axiom (this
can be stated as a precise theorem; will mention later). Armed with the power set
axiom, however, the following is a basic result.
Theorem 1.10. (Cantor) For any set X, there does not exist a map from X onto
P(X). In particular, ¬(X ∼ P(X)).
Proof. Suppose f : X → P(X) were onto. Define A ⊆ X by A = {x ∈ X : x ∈
/
f (x)}. Since A ∈ P(X), let a ∈ X be such that f (a) = A. Then a ∈ f (a) ↔ a ∈
A↔a∈
/ f (a), a contradiction.
With the axiom of choice it follows that for all cardinals κ that P(κ) has cardinality greater than κ, and thus a cardinal greater than κ exists. We would like
to get this without the axiom of choice. To do this, fix a cardinal κ, and let
Wκ = {S ∈ P(κ × κ) : S is a well-ordering }. Note the use of the power set axiom
here. Using the replacement axiom, let A = {α ∈ ON : ∃S ∈ Wκ (o.t.(S) = α)}.
Then ∪A ∈ ON, in fact, A is easily a limit ordinal so ∪A = A. First note that
is α < A then |α| = κ since by definition α = o.t.(S) where S is a well-ordering
of a subset of κ. Since any well-ordering of κ clearly has an order-type which is
bijection with κ, we get that ∀α < A (|α| ≤ κ). Next we claim that |A| > κ. For
suppose f : κ → A were a bijection. Let ≺ be the corresponding well-ordering of κ
defined by γ ≺ δ ↔ f (γ) < f (δ). F itself is an order-isomorphism between (κ, ≺)
and (A ∈), and so o.t.(≺) = A. This is a contradiction, since A is a limit (i.e., we
can easily now define a well-order ≺0 of length A + 1).
We have thus shown in ZF the following:
Lemma 1.11. For each cardinal κ there is a cardinal greater than κ.
The following definition is therefore well-defined in ZF.
Definition 1.12. For κ a cardinal, we let κ+ denote the least cardinal greater than
κ.
Lemma 1.13. If A ⊆ CARD, then sup(A) ∈ CARD.
Proof. Let µ = sup(A). Let α < µ. Let κ ∈ A be such that alpha < κ. Since κ is a
cardinal, there does not exist a one-to-one map from κ into α. Since κ ≤ µ, there
does not therefore exist a one-to-one map from µ into α. So, ∀α < µ ¬(|α| = µ)
and µ is thus a cardinal.
We can now describe the general picture of the cardinals. We define by transfinite
recursion on the ordinals for each α ∈ ON the cardinal ℵα which is also denoted
ωα .
Definition 1.14. ωα is defined by transfinite recursion by:
(1) ω0 = ω
(2) ωα+1 = (ωα )+
(3) For α limit, ωα = supβ<α ωβ .
4
Theorem 1.15. ωα is defined for all α ∈ ON and is a cardinal. Furthermore,
every cardinal is of the form ωα for some α ∈ ON.
Proof. That ωα is well-defined and a cardinal follows from the transfinite recursion
theorem and lemmas 1.11, 1.13. Suppose κ ∈ CARD. Let α ∈ ON be least such
that ωα ≥ κ. This is well-defined since a trivial induction gives ωβ ≥ β for all
β ∈ ON. If α = β + 1 is a successor, then ωβ < κ and so ωα = (ωβ )+ ≤ κ, and thus
ωα = κ. If α is a limit, then ωα = supβ<α ωβ ≤ κ as each ωβ for β < α is < κ. So
again ωα = κ.
Finally in this section we introduce cardinal exponentiation.
Definition 1.16. For α, β ∈ ON, let α β = {f : f is a function from α to β}. For
κ, λ cardinals, let κλ = |λ κ|. Let κ<λ = supα<λ κα .
Note that we need AC to discuss κλ , and we henceforth assume AC in discussions
where cardinal exponentiation is involved.
The following easy lemma says that the “familiar” exponent rules apply to cardinal exponentiation.
Lemma 1.17. For κ, λ, σ cardinals we have κλ+σ = κλ · κσ , (κλ )σ = κλ·σ .
Proof. This follows from the following facts about sets A, B, C:
1) If B ∩C = ∅, then there is an obvious bijection between (B∪C) A and B A× C A.
2.) There is an obvious bijection between C (B A) and (C×B) A.
Here is one easy computation:
Lemma 1.18. If 2 ≤ κ ≤ λ are cardinals, then κλ = 2λ .
Proof. Clearly κλ ≥ 2λ . Also, 2λ ≤ (2κ )λ = 2κ·λ = 2λ .
We also have the following easy basic fact.
Lemma 1.19. For all cardinals κ, 2κ = |P(κ)|.
Proof. This follows immediately from the
( fact that we may identify any A ∈ P(κ)
1 if α ∈ A
with its characteristic function fA (α) =
.
0 if α ∈
/A
So by Cantor’s theorem we have that 2κ > κ for all cardinals κ. Exactly how
big is 2κ ? More generally, how do we compute κλ ? We will turn to these questions
shortly, but first we introduce some terminology.
Definition 1.20. The continuum hypothesis, CH, is the assertion that 2ω = ω1 .
The generalized continuum hypothesis, GCH, is the assertion that 2ωα = ωα+1 for
all α ∈ ON.
ω
Exercise 7. Show that there are 22 functions from R to R. Show that there are
2ω many continuous functions from R to R.
Exercise 8. Let X be a second countable space. Show that there are 2ω open
(likewise for closed) subsets of X. Show that there are 2ω many Borel subsets of
X.
Exercise 9. Let f : ω1 → R be an ω1 sequence of reals. Show that for some
α0 < α1 < α2 < . . . we have f (α0 ) > f (α1 ) > f (α2 ) > . . . (here we mean in the
usual ordering on the reals).
5
2. Cofinality
The following is a basic, important definition.
Definition 2.1. Let α be a limit ordinal. Then cof(α) is the least ordinal β such
that there is a function f : β → α which is unbounded (also said to be cofinal) in
α. That is, ∀α0 < α ∃β 0 < β (f (β) ≥ α0 ).
Clearly cof(α) ≤ α and is an infinite limit ordinal.
Definition 2.2. An ordinal α is said to be regular if cof(α) = α. Otherwise α is
said to be singular.
Sometimes the convention is made that for α a successor that cof(α) = 1, but
we generally only refer to cof(α) when α is a limit.
Lemma 2.3. Let α be a limit ordinal. Then there is an increasing map f : cof(α) →
α which is cofinal.
Proof. Let g : cof(α) → α be cofinal. Define f by
f (β) = max{g(β), sup f (γ) + 1}.
γ<β
Thus, cofinalities are always witnessed by strictly increasing maps. The next
fact is a sort of converse to this.
Lemma 2.4. If α, β are limit ordinals and f : α → β is cofinal and monotonically
increasing (i.e., if α1 < α2 then f (α1 ) ≤ f (α2 )), then cof(α) = cof(β).
Proof. If g : cof(α) → α is cofinal, then f ◦ g : cof(α) → β is cofinal, so cof(β) ≤
cof(α). Suppose h : cof(β) → β is increasing, cofinal. Define k : cof(β) → α by
k(γ) = least η such that f (η) > g(γ). This is well-defined, and easily cofinal into
α. Thus, cof(α) ≤ cof(β).
Lemma 2.5. If α is a limit ordinal, then cof(α) is a regular cardinal.
Proof. We have already observed that cof(α) is an infinite limit ordinal. We have
cof(α) regular, that is, cof(cof(α)) = cof(α) as otherwise by composing functions
(as in lemma 2.4) we would have a cofinal map from cof(cof(α)) to α. It remains
to observe that a regular ordinal is necessarily a cardinal. Suppose β is a regular
limit ordinal, but κ = |β| < β. Let f : κ → β be a bijection. In particular, f is
cofinal, so cof(β) ≤ κ < β, a contradiction.
Thus, cof(cof(α)) = cof(α) for all (limit) α.
The above lemmas did not use AC. The next lemma does use AC, and can fail
without it.
Lemma 2.6. (ZFC) For every cardinal κ, κ+ is regular.
Proof. If not, then there would be a cofinal map f : λ → κ, where λ ≤ κ. This
would write κ+ as a κ union of sets, each of which size ≤ κ. This contradicts
theorem 1.8.
Thus, in ZFC all successor cardinals are regular. What about limit cardinals?
(note: by “limit cardinal” we mean a cardinal not of the form κ+ . Of course, all
infinite cardinals are limit ordinals.) The next lemma computes the cofinality of a
limit cardinal.
6
Lemma 2.7. Let ωα be a limit cardinal. Then α is a limit ordinal and cof(ωα ) =
cof(α).
Proof. Clearly α must be a limit ordinal as otherwise ωα would be a successor
cardianl. The map β 7→ ωβ is increasing and cofinal from α to ωα , so by lemma 2.4
it follows that cof(ωα ) = cof(α).
Examples. ω is regular. Assuming ZFC, so are ω1 , ω2 , . . . , ωn , . . . . ωω is
singular of cofinality ω. All limit cardinals below ωω1 are singular of cofinality
ω. ωω1 is singular of cofinality ω1 . ω(ωω2 +ω1 ) is singular of cofinality ω1 , while
cof(ωωω2 ) = ω2 .
All the limit cardinals mentioned above are singular. Are there any regular
limit cardinals? We will see that this question is independent of ZFC (in fact, the
existence of such cardinals has a stronger consistency strength than ZFC). For now,
we make this into a definition.
Definition 2.8. κ is weakly inaccessible if κ is a regular limit cardinal. κ is strongly
inaccessible if κ is regular and ∀λ < κ (2λ < κ). κ is a strong limit cardinal if
∀λ < κ (2λ < κ).
Exercise 10. Assuming GCH, identify the strong limit cardinals, and show that
κ is weakly inaccessible iff κ is strongly inaccessible.
We end this section with a basic result on cardinal arithmetic.
Theorem 2.9. (König) For every infinite cardinal κ, κcof(κ) > κ.
Proof. Let f : cof(κ) → κ be cofinal. Suppose {gα }α<κ enumerated (cof(κ)) κ. Define g ∈ (cof(κ)) κ by g(β) = the least element of κ − {gα (β) : β < f (β)}. Clearly
g 6= gα for any α, a contradiction.
Corollary 2.10. For every infinite cardinal κ, cof(2κ ) > κ. More generally,
cof(κλ ) > λ.
Proof. Since (κλ )λ = κλ , theorem 2.9 gives that cof(κλ ) must be greater than
λ.
We can also state König’s theorem in a slightly more general form. First, we
define the infinitary sum and products operations.
P
Definition 2.11. If {κα }α<δ is a sequence of cardinals, we define α<δ κα to be
the
Q cardinality of the disjoint union of sets Xα , α < δ, where |Xα | = κα . We define
α<δ κα to the cardinality of the functions f with domain δ such that f (α) ∈ κα
for all α < δ.
P
Exercise 11. Show that α<δ κα = δ · supα<δ (κα ).
The same diagonal argument used in theorem 2.9 also shows the following.
P
Q
Theorem 2.12. (König) Let κα < λα for all α < δ. Then α<δ κα < α<δ λα .
Q
Proof.
Let X
α<δ λα , for α < δ, and |Xα | = κα . We must show that
S
Qα ⊆
λα which is not in
α<δ Xα 6=
α<δ λα . For each α < δ, let f (α) be an element of S
{g(α) : g ∈ Xα }. This exists since |Xα | = κα < λα . Clearly, g ∈
/ α<δ Xα .
Note that theorem 2.12 implies theorem 2.9 by taking κα = κ for all α < δ =
cof(κ).
7
3. Cardinal Arithmetic Under the GCH
Assuming ZFC + GCH we can readily compute κλ for all κ and λ.
Theorem 3.1. Assume ZFC + GCH. Let κ be an infinite cardinal. Then:
(1) If λ < cof(κ), then κλ = κ.
(2) If cof(κ) ≤ λ ≤ κ, then κλ = κ+ .
(3) If λ > κ then κλ = λ+ .
Proof.SIf λ < cof(κ), then any function f : λ → κ has bounded range. Thus,
κλ = α<κ αλ . Also, αλ ≤ 2max(α,λ) ≤ κ since α, λ < κ, using the GCH.
If cof(κ) ≤ λ ≤ κ, then κλ ≥ κ by theorem 2.9. Also, κλ ≤ κκ = κ+ by GCH.
If λ > κ, then κλ ≤ 2λ = λ+ . Also, clearly κλ ≥ 2λ = λ+ .
4. Cardinal Arithmetic in General
We discuss now some properties of cardinal arithmetic assuming only ZFC. We
only scratch the surface here, aiming toward one specific result. We first introduce
two cardinal functions.
Definition 4.1. The beth function iα is defined for α ∈ ON by recursion on α as
follows: i0 = ω, iα+1 = 2iα , and for α limit, iα = supβ<α iβ .
Definition 4.2. The gimel function ‫ג‬α is defined by ‫(ג‬κ) = κcof(κ) , for all cardinals
κ.
Exercise 12. Show that ‫(ג‬κ) > κ and cof(‫(ג‬κ)) > cof(κ).
We now show that cardinal exponentiation is determined etirely from the gimel
function and the cofinality function. From these two functions we compute κλ by
induction on κ as follows.
Theorem 4.3. For all infinite cardinals κ, λ we have:
(1) If λ ≥ κ then κλ = 2λ .
(2) If for some µ < κ we have µλ ≥ κ, then κλ = µλ .
(3) If for all µ < κ we have µλ < κ, then:
• If λ < cof(κ) then κλ = κ.
• If cof(κ) ≤ λ < κ, then κλ = ‫(ג‬κ).
Proof. (1) is clear, and (2) follows from κλ ≤ (µλ )λ = µλ . The first case of (3)
follows from the fact that κλ = supµ<κ µλ = κ from the assumption of the case.
For the second case of (3), clearly we have κλ ≥ κcof(κ) = ‫(ג‬κ). The upper bound
follows from the following lemma.
Lemma 4.4. If λ ≥ cof(κ), then κλ = (supα<κ αλ )cof(κ) .
Proof. Fix a cofinal map h : cof(κ) → κ. To each f : λ → κ, and each β < cof(κ)
we associate a partial function fβ : λ → h(β). Namely, let fβ = f ∩ (λ × h(β)). The
map f → (fβ )β<cof(κ) is clearly a one-to-one map from κλ to (supα<κ αλ )cof(κ) . The next lemma gives somes information about the continuum function at singular cardinals.
Lemma 4.5. Let κ be a singular cardinal and suppose the continuum function
below κ is eventually constant, that is, there is a γ < κ such that 2<κ = 2γ . Then
2κ = 2γ .
8
Proof. An argument similar to lemma 4.4 shows that for any κ we have 2κ ≤
(2<κ )cof(κ) , and hence easily 2κ = (2<κ )cof(κ) . Thus, 2κ = (2<κ )cof(κ) = (2γ )cof(κ) =
(2δ )cof(κ) = 2δ = 2γ , where γ < δ < κ and δ ≥ cof(κ).
We thus have the following theorem, which shows that the continuum function
can be computed entirely from the gimel function.
Theorem 4.6. For any cardinal κ:
(1) If κ is a successor cardinal, then 2κ = ‫(ג‬κ).
(2) If κ is a limit cardinal and the continuum below κ is eventually constant,
then 2κ = 2<κ · ‫(ג‬κ).
(3) If κ is a limit cardinal and the continuum below κ is not eventually constant,
then 2κ = ‫(ג‬2<κ ).
Proof. The first part is clear as successor cardinals are regular. Suppose that κ is
a limit cardinal, and the continuum function is eventually constant below κ, say
2<κ = 2γ . If κ is singular, then 2κ = 2γ by the previous lemma. If κ is regular,
then 2κ = κκ = κcof(κ) = ‫(ג‬κ). In the singular case, note that 2γ = 2<κ ≥ κ. Thus,
‫(ג‬κ) = κcof(κ) ≤ (2γ )cof(κ) = (2δ )cof(κ) = 2δ = 2γ (where cof(κ) < δ < κ). Thus,
2<κ · ‫(ג‬κ) = 2<κ . If κ is regular, then 2<κ ≤ 2κ = κκ = ‫(ג‬κ). So, 2<κ · ‫(ג‬κ) = ‫(ג‬κ).
In either case, 2κ = 2<κ · ‫(ג‬κ). Finally, suppose κ is a limit cardinal and the
continuum function below κ is not eventually constant. Then cof(2<κ ) = cof(κ).
Then 2κ = (2<κ )cof(κ) = ‫(ג‬2<κ ).
Note that if κ is a strong limit cardinal (i.e., 2<κ = κ) then 2κ = ‫(ג‬κ), so the
continuum function and gimel function coincide in the main case of interest.
What can be said about the value of 2κ or ‫(ג‬κ) in general? We shall see that
if κ is regular, then nothing more can be said that 2κ > κ and cof(2κ ) > κ (and
of course if κ1 < κ2 then 2κ1 ≤ 2κ2 ). So, for κ regular, 2κ = ‫(ג‬κ) is basically
arbitrary subject to Cantor’s and König’s theorems.
If κ is a singular limit cardinal, the situation is much more difficult. Here there
are restrictions on 2κ . For now, let’s just observe one more fact about the gimel
function. Note that if there is a λ < κ such that λcof(κ) ≥ κ, then ‫(ג‬κ) = λcof(κ) ,
and so is computed by exponentiation on smaller cardinals.
Exercise 13. Suppose κ is a singular limit cardinal. Suppose there is a λ < κ such
that λcof(κ) ≥ κ, and let λ be the least such. Show that cof(λ) ≤ cof(κ).
More generally, if there is a λ < κ with cof(λ) ≥ cof(κ) and ‫(ג‬λ) ≥ κ, then
‫(ג‬κ) = κcof(κ) ≤ λcof(λ)·cof(κ) = λcof(λ) = ‫(ג‬λ).
Finally, suppose κ is a singular limit cardinal and there is no λ < κ such that
λcof(κ) ≥ κ. We claim that cof(‫(ג‬κ)) ≥ the the least α such that λα ≥ κ for some
λ < κ. For suppose λα < κ for all λ < κ, we show that cof(‫(ג‬κ)) > α. Then
‫(ג‬κ)α = κcof(κ)·α . We may assume α ≥ cof(κ) as we know that cof(‫(ג‬κ)) ≥ cof(κ).
Thus, ‫(ג‬κ)α = κα = (supλ<κ λα )cof(κ) = κcof(κ) = ‫(ג‬κ). Thus, cof(‫(ג‬κ)) > α by
König.
We have thus shown:
Lemma 4.7. Suppose κ is a singular strong limit cardinal. Then cof(‫(ג‬κ)) > κ.