Solutions to exercises in Chapter 18
E18.1 Let κ be an uncountable regular cardinal. We define S < T iff S and T are
stationary subsets of κ and the following two conditions hold:
(1) {α ∈ T : cf(α) ≤ ω} is nonstationary in κ.
(2) {α ∈ T : S ∩ α is nonstationary in α)} is nonstationary in κ.
Prove that if ω < λ < µ < κ, all these cardinals regular, then Eλκ < Eµκ , where
Eλκ = {α < κ : cf(α) = λ},
and similarly for Eµκ .
First of all, {α ∈ Eµκ : cf(α) ≤ ω} is empty, so of course it is nonstationary in κ.
For (2), let C = (µ, κ). We claim that
{α ∈ Eµκ : Eλκ ∩ α is nonstationary in α} ∩ C = ∅;
this will prove (2). In fact, suppose that α is in the indicated intersection. Let D be club
in α such that Eλκ ∩ D = ∅. Now α ∈ Eµκ , so cf(α) = µ. Define αξ for all ξ < λ as follows.
Let α0 be the least member of D. If αξ ∈ D has been S
defined, take any member αξ+1
of D greater than αξ . If ξ S
is limit less than λ, let αξ = η<ξ αη . Then αξ ∈ D because
D is closed. Now let β = ξ<λ αξ . Then β ∈ D since D is closed, and cf(β) = λ. So
β ∈ Eλκ ∩ D, contradiction.
E18.2 Continuing exercise E18.1: Assume that κ is uncountable and regular. Show that
the relation < is transitive.
Suppose that A < B < C. Then by definition
(1) {α ∈ C : cf(α) ≤ ω} is nonstationary in κ.
(2) {α ∈ B : α ∩ A is nonstationary} is nonstationary in κ.
(3) {α ∈ C : α ∩ B is nonstationary} is nonstationary in κ.
We want to show
{α ∈ C : α ∩ A is nonstationary} is nonstationary.
Our assumptions give us clubs M, N in κ such that
{α ∈ B : α ∩ A is nonstationary} ∩ M = ∅ and
{α ∈ C : α ∩ B is nonstationary} ∩ N = ∅.
Let M ′ be the set of all limits of members of M ; so also M ′ is club in κ. Now it suffices
to show that
{α ∈ C : α ∩ A is nonstationary} ∩ M ′ ∩ N = ∅.
So, suppose that α ∈ C ∩ M ′ ∩ N ; we show that α ∩ A is stationary in α. To this end, let
P be club in α, and let P ′ be the set of all of its limit points. Now α ∈ C ∩ N , so α ∩ B is
1
stationary. Since α ∈ M ′ , it follows that α ∩ M is club in α. So M ∩ P ′ is club in α, and
so we can choose β ∈ α ∩ B ∩ M ∩ P ′ . Now β ∈ B ∩ M , so β ∩ A is stationary in β. Since
β ∈ P ′ , it follows that P ∩ β is club in β. So β ∩ A ∩ P 6= ∅, hence A ∩ P 6= ∅, as desired.
E18.3 If κ is an uncountable regular cardinal and S is a stationary subset of κ, we define
Tr(S) = {α < κ : cf(α) > ω and S ∩ α is stationary in α}.
Suppose that A, B are stationary subsets of an uncountable regular cardinal κ and A < B.
Show that Tr(A) is stationary.
Assume the conditions of the exercise. Thus by definition, {α ∈ B : cf(α) ≤ ω} is
nonstationary in κ, and also {α ∈ B : A ∩ α is non-stationary in α} is non-stationary
in κ. Hence there is a club C in κ such that C ∩ {α ∈ B : cf(α) ≤ ω} = ∅ and also
C ∩ {α ∈ B : A ∩ α is non-stationary in α} = ∅. Thus B ∩ C ⊆ Tr(A), and it follows that
Tr(A) is stationary in κ.
E18.4 (Real-valued measurable cardinals) We describe a special kind of measure. A measure on a set S is a function µ : P(S) → [0, ∞) satisfying the following conditions:
(1) µ(∅) = 0 and µ(S) = 1.
(2) If µ({s}) = 0 for all s ∈ S,
(3)
P If hXi : i ∈ ωi is a system of pairwise disjoint subsets of S, then µ(
i∈ω µ(Xi ). (The Xi ’s are not necessarily nonempty.)
S
i∈ω
Xi ) =
Let κ be an infinite cardinal. Then µ is κ-additive iff for every system hXα : α < γi of
nonempty pairwise disjoint sets, wich γ < κ, we have
µ
[
Xα
α<γ
!
=
X
µ(Xα ).
α<γ
Here this sum (where the index set γ might be uncountable), is understood to be
sup
F ⊆γ,
F finite
X
µ(Xα ).
α∈F
We say that an uncountable cardinal κ is real-valued measurable iff there is a κ-additive
measure on κ. Show that every measurable cardinal is real-valued measurable. Hint: let µ
take on only the values 0 and 1.
Suppose that κ is measurable. Thus κ is uncountable, and there is a κ-complete nonprincipal ultrafilter U on κ. Now for any X ⊆ κ we define
µ(X) =
n
if X ∈ U ,
otherwise.
1
0
2
Conditions (1) and (2) in the definition of measure are clear. We can check (3) and κadditivity simultaneously, by assuming
that hXα : α < βi is a system of pairwise disjoint
S
subsets of κ, with β < κ. If µ( α<β Xα ) = 0, clearly µ(Xα ) = 0 for all α < β, and so
[
µ
Xα
α<γ
!
=
X
µ(Xα ).
α<γ
S
S
Suppose that µ( α<β Xα ) = 1. Thus α<β Xα ∈ U . If Xα ∈
/ U for all α < β, then
κ\Xα ∈ U for all α < β, and hence by κ-completeness,

κ\ 
[
α<β

Xα  =
\
(κ\Xα ) ∈ U,
α<β
contradiction. Hence Xα ∈ U for some α < β. There can be only one such α, since if
γ 6= α and Xγ ∈ U , then ∅ = Xα ∩ Xγ ∈ U , contradiction. Hence again
µ
[
α<γ
Xα
!
=
X
µ(Xα ).
α<γ
E18.5 Suppose that µ is a measure on a set S. A subset A of S is a µ-atom iff µ(A) > 0
and for every X ⊆ A, either µ(X) = 0 or µ(X) = µ(A). Show that if κ is a realvalued measurable cardinal, µ is a κ-additive measure on κ, and A ⊆ κ is a µ-atom, then
{X ⊆ A : µ(X) = µ(A)} is a κ-complete nonprincipal ultrafilter on A. Conclude that κ is
a measurable cardinal if there exist such µ and A.
Let F be the indicated set. Obviously A ∈ F . Suppose that X ∈ F and X ⊆ Y ⊆ A.
Then
µ(A) = µ(X ∪ (Y \X) ∪ (A\Y )) = µ(X) + µ(Y \X) + µ(A\Y ) = µ(A) + µ(Y \X) + µ(A\Y ),
and so µ(A\Y ) = 0. Hence µ(A) = µ((A\Y ∪ Y ) = µ(A\Y ) + µ(Y ) = µ(Y ). So Y ∈ F .
Now suppose that Y, Z ∈ F . Then
µ(A) = µ(Y ) = µ(Y ∩ Z) + µ(Y \Z) and
µ(A) = µ(Z) = µ(Y ∩ Z) + µ(Z\Y ).
It follows that µ(Y \Z) = µ(Z\Y ). If µ(Y \Z) = µ(A), then also µ(Z\Y ) = µ(A), and
hence
2µ(A) = µ(Y \Z) + µ(Z\Y ) = µ((Y \Z) ∪ (Z\Y )) ≤ µ(A),
contradiction. So µ(Y \Z) = 0, and hence µ(A) = µ(Y ∩ Z). It follows that Y ∩ Z ∈ F .
So, F is a filter.
Clearly ∅ ∈
/ F , so F is proper.
3
If X ⊆ A, then µ(A) = µ(X) + µ(A\X), and hence µ(X) = µ(A) or µ(A\X) = µ(A).
So X ∈ F or A\X ∈ F . Thus F is an ultrafilter.
T
<κ
Finally,
for
κ-completeness,
suppose
that
A
∈
[F
]
Suppose
that
A ∈
/ F . Then
T
A\
A ∈ F . Let hXα : α < λi be an enumeration of A . For each α < λ let Yα =
T
β<α Xβ \Xα .
[
(1)
[
Yα =
α<λ
(A\Xα )
α<λ
S
In fact, ⊆ is clear. Suppose that ξ ∈ α<λ (A\Xα ), and choose α < λ minimum such that
ξ ∈ (A\Xα ). Then ξ ∈ Yα . So (1) holds.
Clearly the Yα ’s are pairwise disjoint. So from (1) we get
\ µ(A) = µ A\ A
!
[
=µ
(A\Xα )
α<λ
=µ
[
α<λ
=
X
Yα
!
µ(Yα ),
α<λ
and hence there is a α < λ such that µ(Yα ) = 1. Hence µ(A\Xα ) = µ(A) also, contradiction.
Hence F is κ-complete.
Since all members of F have size κ by κ-completeness and nonprincipality, it follows
that |A| = κ. So κ is a measurable cardinal.
E18.6 Prove that if κ is real-valued measurable then either κ is measurable or κ ≤ 2ω .
Hint: if there do not exist any µ-atoms, construct a binary tree of height at most ω1 .
Let µ be a κ-additive measure on κ. By exercise E18.5, if there is a µ-atom, then κ is
measurable. So, suppose that there does not exist any µ-atoms. We construct a tree
under ⊃ by constructing the levels Lα , as follows. L0 = {κ}. Suppose that Lα has been
constructed, and that it is a nonempty collection of subsets of κ each of positive measure.
For each X ∈ Lα let YX be a subset of X such that 0 < µ(YX ) < µ(X); such a set exists
since X is not a µ-atom. Then we define
Lα+1 = {YX , X\YX : X ∈ Lα }.
If α is a limit ordinal and Lβ has been constructed for every β < α, then we define


(
)
\
\
Lα =
Zβ : Zβ ∈ Lβ for all β < α and µ 
Zβ  > 0 ,
β<α
β<α
4
except that if Lα = ∅ the construction stops.
Clearly this gives a tree. Let α be the least ordinal such that Lα is not defined. So α
is a limit ordinal.
(1) α ≤ ω1 , and in fact, if hZβ : α < γi is a branch of the tree, thus with Zβ ⊂ Zδ if
δ < β < γ, then γ is countable.
In fact, we have µ(Zβ \Zβ+1 ) > 0 for every β < γ, and the sets Zβ \Zβ+1 are pairwise
disjoint. If γ ≥ ω1 , then
γ=
[
n∈ω
1
β < γ : µ(Zβ \Zβ+1 ) >
n+1
and hence there would be an n ∈ ω such that
β < γ : µ(Zβ \Zβ+1 ) >
1
n+1
,
is uncountable, which is not possible. So (1) holds.
Similarly each level of our tree is countable. It follows that the tree has at most 2ω
branches.
for each B ∈ B let WB =
T Let B be the collection of all branches in this tree, and
ω
X∈B X. Let C = {WB : B ∈ B}\{∅}. Now clearly |C | ≤ 2 , and C consists of measure
0 sets.
S
(2) κ = C .
In fact, if α ∈ κ, then B = {X ∈ T : α ∈ X} is a branch, and so α ∈ WB .
From (2) it follows that κ ≤ 2ω , since the measure µ is κ-additive and µ(κ) = 1. In
fact, 2ω < κ would imply by (2) that µ(κ) = 0, contradiction.
E18.7 Let κ be a regular uncountable cardinal. Show that the diagonal intersection of the
system h(α + 1, κ) : α < κi is the set of all limit ordinals less than κ.
For any β ∈ κ,
β ∈ △α<κ (α + 1, κ) iff
iff
∀α < β[β ∈ (α + 1, κ)]
∀α < β[α + 1 < β]
iff
β is a limit ordinal.
E18.8 Let F be a filter on a regular uncountable cardinal κ. We say that F is normal
iff it is closed under diagonal intersections. Suppose that F is normal, and (α, κ) ∈ F for
every α < κ. Show that every club of κ is in F . Hint: use exercise E18.7.
Let C be a club, and let hαξ : ξ < κi be the strictly increasing enumeration of C, and let
D be the set of all limit ordinals less than κ. By exercise E18.7 suffices to show that
D ∩ △ξ<κ (αξ , κ) ⊆ C.
5
So, take any β ∈ D ∩ △ξ<κ (αξ , κ). Thus β is a limit ordinal, and ∀ξ < β[β ∈ (αξ , κ)], i.e.,
∀ξ < β[αξ < β]. Now ξ ≤ αξ for all ξ, so C ∩ β is unbounded in β. Hence β ∈ C.
E18.9 Let F be a proper filter on a regular uncountable cardinal κ. Show that the following
conditions are equivalent.
(i) F is normal
(ii) For any S0 ⊆ κ, if κ\S0 ∈
/ F and f is a regressive function defined on S0 , then
there is an S ⊆ S0 with κ\S ∈
/ F and f is constant on S.
(i)⇒(ii): Assume (i), and suppose that S0 ⊆ κ, κ\S0 ∈
/ F , and f is a regressive function
on S0 . Suppose that the conclusion fails. Then for every γ < κ we have κ\f −1 [{γ}] ∈ F ,
as otherwise we could take S = f −1 [{γ}]. By (i), take β ∈ △γ<κ (κ\f −1 [{γ}]). Then
∀γ < β[β ∈ κ\f −1 [{γ}]; in particular, β ∈ κ\f −1 [{f (β)}], contradiction.
(ii)⇒(i): Assume (ii), and suppose that haα : α < κi is a system of members of
F . Suppose that △α<κ aα ∈
/ F . Now ∀α ∈ κ\△α<κ aα ∃β < α[α ∈
/ aβ ]. This gives us a
regressive function f defined on κ\△α<κ aα such that for every α in that set, α ∈
/ af (α) .
Hence by (ii) choose S ⊆ κ\△α<κ aα such that f is constant on S, say with value γ, with
κ\S ∈
/ F . Since aγ ∈ F , we have aγ 6⊆ κ\S. Choose β ∈ aγ ∩ S. Then β ∈
/ af (β) gives a
contradiction.
E18.10 A probability measure on a set S is a real-valued function µ with domain P(S)
having the following properties:
(i) µ(∅) = 0 and µ(S) = 1.
(ii) If X ⊆ Y , then µ(X) ≤ µ(Y ).
(iii) µ({a}) = 0 for all a ∈ S.
S
(iv)
If
hX
:
n
∈
ωi
is
a
system
of
pairwise
disjoint
sets,
then
µ(
n
n∈ω Xn ) =
P
n∈ω µ(Xn ). (Some of the sets Xn might be empty.)
Prove that there does not exist a probability measure on ω1 . Hint: consider an Ulam
matrix.
Suppose that µ is a probability measure on ω1 . Let f = hfρ : ρ < ω1 i be a family of
injections fρ : ρ → ω. Define the function A : ω × ω1 → P(ω1 ) by setting, for any ξ < ω
and α < ω1 ,
Aξα = {ρ ∈ ω1 \(α + 1) : fρ (α) = ξ}.
S
Take any α < ω1 . Since n∈ω Anα = ω1 \(α+1), Anα ∩Am
α = ∅ for α 6= β, and µ(ω1 \(α+1)) =
n(α)
1, choose n(α) ∈ ω such that ϕ(Aα ) > 0. Then there exist M ∈ [ω1 ]ω1 and m ∈ ω such
that n(α) = m for every α ∈ M . Then hAm
α : α ∈ M i is a system of pairwise disjoint sets
each of positive measure, contradiction.
E18.11 Show that if κ is a measurable cardinal, then there is a normal κ-complete nonprincipal ultrafilter on κ. Hint: Let D be a κ-complete nonprincipal ultrafilter on κ. Define
f ≡ g iff f, g ∈ κ κ and {α < κ : f (α) = g(α)} ∈ D. Show that ≡ is an equivalence relation
on κ κ. Show that there is a relation ≺ on the collection of all ≡-classes such that for all
f, g ∈ κ κ, [f ] ≺ [g] iff {α < κ : f (α) < g(α)} ∈ D. Here for any function h ∈ κ κ we
use [h] for the equivalence class of h under ≡. Show that ≺ makes the collection of all
equivalence classes into a well-order. Show that there is a ≺ smallest equivalence class x
6
such that ∀f ∈ x∀γ < κ[{α < κ : γ < f (α)} ∈ D. Let E = {X ⊆ κ : f −1 [X] ∈ D}. Show
that E satisfies the requirements of the exercise.
≡ is reflexive: κ = {α < κ : f (α) = f (α)}, hence f ≡ f .
≡ is symmetric: Assume that f ≡ g. Thus {α < κ : f (α) = g(α)} ∈ D. Hence
{α < κ : g(α) = f (α)} = {α < κ : f (α) = g(α)} ∈ D. Hence g ≡ f .
≡ is transitive: Assume that f ≡ g ≡ h. Thus {α < κ : f (α) = g(α)} ∈ D and
{α < κ : g(α) = h(α)} ∈ D. Hence
{α < κ : f (α) = g(α)} ∩ {α < κ : g(α) = h(α)} ∈ D;
since
{α < κ : f (α) = g(α)} ∩ {α < κ : g(α) = h(α)} ⊆ {α < κ : f (α) = h(α)},
we get {α < κ : f (α) = h(α)} ∈ D, so f ≡ h.
Now define
x≺y
iff
∃f, g[x = [f ] and y = [g] and {α < κ : f (α) < g(α)} ∈ D].
(1) ∀f, g ∈ κ κ[[f ] ≺ [g] iff {α < κ : f (ϕ) < g(α)} ∈ D].
In fact, ⇐ is immediate from the definition. Now suppose that [f ] ≺ [g]. Choose f ′ , g ′ ∈ κ κ
such that [f ] = [f ′ ], [g] = [g ′ ], and {α < κ : f ′ (α) < g ′ (α)} ∈ D. Then
{α < κ : f (α) = f ′ (α)}∩{α < κ : f ′ (α) < g ′ (α)} ∩ {α < κ : g(α) = g ′ (α)}
⊆ {α < κ : f (α) < g(α)};
the left side is in D, hence also the right side is in D, so {α < κ : f (α) < g(α)} ∈ D. Thus
(1) holds.
≺ is irreflexive: {α < κ : f (α) < f (α)} = ∅ ∈
/ D, so [f ] 6≺ [f ].
≺ is transitive: Assume that [f ] ≺ [g] ≺ [h]. Then
{α < κ : f (α) < g(α)} ∩ {α < κ : g(α) < h(α)} ⊆ {α < κ : f (α) < h(α)};
the left side is in D, hence also the right side is in D, so [f ] ≺ [h].
≺ is a linear order: Suppose that f, g ∈ κ κ are such that [f ] 6= [g] and [f ] 6≺ [g]. Now
κ = {α < κ : f (α) < g(α)} ∪ {α < κ : f (α) = g(α} ∪ {α < κ : g(α) < f (α)};
The first two sets are not in D, so the third one is in D, and hence [g] ≺ [f ].
≺ is a well-order: Suppose not. Then we get a sequence hf m : m ∈ ωi of members of
κ
κ such that [fm+1 ] ≺ [fm ] for all m ∈ ω. Thus {α < κ : fm+1 (α) < fm (α)} ∈ D for all
m ∈ ω. It follows that
\
{α < κ : fm+1 (α) < fm (α)} ∈ D;
m∈ω
7
taking any element α in this intersection, we get . . . fm+1 (α) < fm (α) . . ., contradiction.
Now let k(α) = α for all α < κ. Then for any γ < κ we have
{α < κ : γ < k(α)} = {α < κ : γ < α} = κ\(γ + 1) ∈ D.
It follows that we can take the smallest equivalence class [f ] such that for any γ < κ we
have {α < κ : γ < f (α)} ∈ D. Now we let E = {X ⊆ κ : f −1 [X] ∈ D}. We claim that E
is as desired in the exercise.
∅∈
/ E: This is true since f −1 [∅] = ∅ ∈
/ D.
If X ⊆ Y ⊆ κ and X ∈ E, then Y ∈ E: In fact, assume that X ⊆ Y ⊆ κ and X ∈ E.
Then f −1 [X] ⊆ f −1 [Y ] and f −1 [X] ∈ D, so f −1 [y] ∈ D, so that Y ∈ E.
If X, Y ∈ E, then X ∩ Y ∈ E: In fact, f −1 [X ∩ Y ] = f −1 [X] ∩ f −1 [y], so this is clear.
If X ⊆ κ, then X ∈ E or (κ\X) ∈ E: For, suppose that X ∈
/ E. Then f −1 [X] ∈
/ D,
−1
−1
so f [κ\X] = (κ\f [X]) ∈ D, and hence (κ\X) ∈ E.
E is nonprincipal: for any α < κ we have {β < κ : α < f (β)} ∈ D, and {β < κ : α <
f (β)} ⊆ {β < κ : α 6= f (β)}, so {β < κ : α 6= f (β)} ∈ D, hence {β < κ : α = f (β)} ∈
/ D,
hence f −1 [{α}] ∈
/ D and so {α} ∈
/ E.
E is κ-complete: Suppose that hXα : α < βi is a system of subsets of κ, with
β <h κ and with
[Xα ] ∈ E for all α < β. Thus f −1 [Xα ] ∈ D for all α < β. Since
i
T
T
T
−1
f −1
[Xα ] ∈ D, it follows that α<β Xα ∈ E.
α<β Xα =
α<β f
E is normal: We apply exercise 18.9. Suppose that S0 ∈ E and g is regressive on S0 .
Note that f −1 [S0 ] ∈ D. Let h = g ◦ f . Then for any α ∈ f −1 [S0 ] we have h(α) < f (α),
so that [h] ≺ [f ]. By the definition of f it then follows that there is a γ < κ such that
{α < κ : γ < h(α)} ∈
/ D. Hence {α < κ : h(α) ≤ γ} ∈ D. Now
{α < κ : h(α) ≤ γ} =
[
{α < κ : h(α) = δ},
δ≤γ
and so there is a δ ≤ γ such that {α < κ : h(α) = δ} ∈ D. Now {α < κ : h(α) = δ} =
h−1 [{δ}] = f −1 [g −1 [{δ}], so g −1 [{δ}] ∈ E. This checks the condition of exercise 18.9.
8