18. Large cardinals

18. Large cardinals
The study, or use, of large cardinals is one of the most active areas of research in set theory
currently. There are many provably different kinds of large cardinals whose descriptions
are different from one another. We restrict ourselves in this chapter to three important
kinds: Mahlo cardinals, weakly compact cardinals, and measurable cardinals. All of these
large cardinals are uncountable regular limit cardinals (which are frequently called weakly
inaccessible cardinals), and most of them are strongly inaccessible cardinals.
Mahlo cardinals
As we mentioned in the elementary part of these notes, one cannot prove in ZFC that
uncountable weakly inaccessible cardinals exist (if ZFC itself is consistent). But now
we assume that even the somewhat stronger inaccessible cardinals exist, and we want
to explore, roughly speaking, how many such there can be. We begin with some easy
propositions. A strong limit cardinal is an infinite cardinal κ such that 2 λ < κ for all
λ < κ.
Proposition 18.1. Assume that uncountable inaccessible cardinals exist, and suppose that
κ is the least such. Then every uncountable strong limit cardinal less than κ is singular.
The inaccessibles are a class of ordinals, hence form a well-ordered class, and they can be
enumerated in a strictly increasing sequence 〈ι α : α ∈ O〉. Here O is an ordinal, or On,
the class of all ordinals. The definition of Mahlo cardinal is motivated by the following
simple proposition.
Proposition 18.2. If κ = ι α with α < κ, then the set {λ < κ : λ is regular} is a
nonstationary subset of κ.
Proof. Since κ is regular and α < κ, we must have sup β

(ii) {λ < κ : λ is inaccessible} is stationary in κ.
Proof. (i)⇒(ii): Let S = {λ < κ : λ is regular}, and S ′ = {λ < κ : λ is inaccessible}.
Assume that κ is Mahlo. In particular, κ is uncountable and inaccessible. Suppose that
C is club in κ. The set D = {λ < κ : λ is strong limit} is clearly club in κ too. If
λ ∈ S ∩ C ∩ D, then λ is inaccessible, as desired.
(ii)⇒(i): obvious.
The following proposition answers a natural question one may ask after seeing Corollary
18.3.
Proposition 18.5. Suppose that κ is minimum such that ι k = κ. Then κ is not Mahlo.
Proof. Suppose to the contrary that κ is Mahlo, and let S = {λ < κ : λ is
inaccessible} For each λ ∈ S, let f(λ) be the α < κ such that λ = ι α . Then α = f(λ) < λ
by the minimality of κ. So f is regressive on the stationary set S, and hence there is an
α < κ and a stationary subset S ′ of S such that f(λ) = α for all λ ∈ S ′ . But actually f is
clearly a one-one function, contradiction.
Mahlo cardinals are in a sense larger than “ordinary” inaccessibles. Namely, below every
Mahlo cardinal κ there are κ inaccessibles. But now in principle one could enumerate all
the Mahlo cardinals, and then apply the same idea used in going from regular cardinals to
Mahlo cardinals in order to go from Mahlo cardinals to higher Mahlo cardinals. Thus we
can make the definitions
• κ is hyper-Mahlo iff κ is inaccessible and the set {λ < κ : λ is Mahlo} is stationary in κ.
• κ is hyper-hyper-Mahlo iff κ is inaccessible and the set {λ < κ : λ is hyper-Mahlo} is
stationary in κ.
Of course one can continue in this vein.
Weakly compact cardinals
• A cardinal κ is weakly compact iff κ > ω and κ → (κ, κ) 2 . There are several equivalent
definitions of weak compactness. The one which justifies the name “compact” involves
infinitary logic, and it will be discussed later. Right now we consider equivalent conditions
involving trees and linear orderings.
• A cardinal κ has the tree property iff every κ-tree has a chain of size κ.
Equivalently, κ has the tree property iff there is no κ-Aronszajn tree.
• A cardinal κ has the linear order property iff every linear order (L,

Assume the linear order property, and let (T,

• z
• y
y • • z
y • • z x • • z
x •
• x
• x
Case 1 Subcase 2.1 Subcase 2.2 Case 3
x •
y
•
z •
x •
y
•
z •
x •
y
•
z •
Subcase 4.1 Subcase 4.2 Subcase 4.3
We claim that B is a chain in T of size κ. Suppose that t 0 , t 1 ∈ B with t 0 ≠ t 1 , and choose
x 0 , x 1 ∈ L correspondingly. Say wlog x 0 < ′′ x 1 . Now t 0 ∈ B and x 0 ≤ ′′ x 1 , so t 0 ≤ x 1 .
And t 1 ∈ B and x 1 ≤ x 1 , so t 1 ≤ x 1 . So t 0 and t 1 are comparable.
Now let α < κ; we show that B has an element of height α. For each t of height α let
V t = {x ∈ L : t ≤ x}. Then
{x ∈ L : ht(x) ≥ α} = ⋃
ht(t)=α
since there are fewer than κ elements of height less than κ, this set has size κ, and so there
is a t such that ht(t) = α and |V t | = κ. We claim that t ∈ B. To prove this, take any
x ∈ V t such that t < x. Suppose that a ∈ L and x ≤ ′′ a. Choose y ∈ V t with a < ′′ y and
t < y. Then t < x, t < y, and x ≤ ′′ a < ′′ y. If x = a, then t ≤ a, as desired. If x < ′′ a,
then t < a by (*).
This finishes the case in which L has a subset of order type κ. The case of order type
κ ∗ is similar, but we give it. So, suppose that L has order type κ ∗ . Define
V t ;
B = {t ∈ T : ∃x ∈ L∀a ∈ L[a ≤ ′′ x → t ≤ a]}.
We claim that B is a chain in T of size κ. Suppose that t 0 , t 1 ∈ B with t 0 ≠ t 1 , and choose
x 0 , x 1 ∈ L correspondingly. Say wlog x 0 < ′′ x 1 . Now t 0 ∈ B and x 0 ≤ x 0 , so t 0 ≤ x 0 . and
t 1 ∈ B and x 0 ≤ ′′ x 1 , so t 1 ≤ x 0 . So t 0 and t 1 are comparable.
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Now let α < κ; we show that B has an element of height α. For each t of height α let
V t = {x ∈ L : t ≤ x}. Then
{x ∈ L : ht(x) ≥ α} = ⋃
V t ;
ht(t)=α
since there are fewer than κ elements of height less than κ, this set has size κ, and so there
is a t such that ht(t) = α and |V t | = κ. We claim that t ∈ B. To prove this, take any
x ∈ V t such that t < x. Suppose that a ∈ L and a ≤ ′′ x. Choose y ∈ V t with y < ′′ a and
t < y. Then t < x, t < y, and y < ′′ a ≤ ′′ x. If a = x, then t < a, as desired. If a < ′′ x,
then t < a by (*).
Theorem 18.7. For any uncountable cardinal κ the following conditions are equivalent:
(i) κ is weakly compact.
(ii) κ is inaccessible, and it has the linear order property.
(iii) κ is inaccessible, and it has the tree property.
(iv) For any cardinal λ such that 1 < λ < κ we have κ → (κ) 2 λ .
Proof. (i)⇒(ii): Assume that κ is weakly compact. First we need to show that κ is
inaccessible.
To show that κ is regular, suppose to the contrary that κ = ∑ α

(ii)⇒(iii): By Lemma 18.6.
(iii)⇒(iv): Assume (iii). Suppose that F : [κ] 2 → λ, where 1 < λ < κ; we want to
find a homogeneous set for F of size κ. We construct by recursion a sequence 〈t α : α < κ〉
of members of

so since |B| = κ it follows that |H ξ | = κ for some ξ < λ, as desired.
(iv)⇒(i): obvious.
Now we go into the connection of weakly compact cardinals with logic, thereby justifying
the name “weakly compact”. This is optional material.
Let κ and λ be infinite cardinals. The language L κλ is an extension of ordinary first
order logic as follows. The notion of a model is unchanged. In the logic, we have a sequence
of λ distinct individual variables, and we allow quantification over any one-one sequence
of fewer than λ variables. We also allow conjunctions and disjunctions of fewer than κ
formulas. It should be clear what it means for an assignment of values to the variables to
satisfy a formula in this extended language. We say that an infinite cardinal κ is logically
weakly compact iff the following condition holds:
(*) For any language L κκ with at most κ basic symbols, if Γ is a set of sentences of the
language and if every subset of Γ of size less than κ has a model, then also Γ has a model.
Notice here the somewhat unnatural restriction that there are at most κ basic symbols.
If we drop this restriction, we obtain the notion of a strongly compact cardinal. These
cardinals are much larger than even the measurable cardinals discussed later. We will not
go into the theory of such cardinals.
Theorem 18.8. An infinite cardinal is logically weakly compact iff it is weakly compact.
Proof. Suppose that κ is logically weakly compact.
(1) κ is regular.
Suppose not; say X ⊆ κ is unbounded but |X| < κ. Take the language with individual
constants c α for α < κ and also one more individual constant d. Consider the following
set Γ of sentences in this language:
⎧
⎫
⎨ ∨ ∨ ⎬
{d ≠ c α : α < κ} ∪ (d = c α )
⎩
⎭ .
β∈X α

Suppose that ∆ ∈ [Γ]

By induction, |F α | ≤ κ for all α ≤ κ, and F κ is the set of all formulas. (One uses that κ is
inaccessible.)
Expand L to L ′ by adjoining a set C of new individual constants, with |C| = κ. Let
Θ be the set of all subformulas of the sentences in Γ. Let 〈ϕ α : α < κ〉 list all sentences
of L ′ which are of the form ∃xψ α (x) and are obtained from a member of Θ by replacing
variables by members of C. Here x is a one-one sequence of variables of length less than
κ; say that x has length β α . Now we define a sequence 〈d α : α < κ〉; each d α will be a
sequence of members of C of length less than κ. If d β has been defined for all β < α, then
⋃
β

Clearly f is an element of T with height α. So (6) holds.
Thus T is clearly a κ-tree, so by the tree property we can let B be a branch in T of
size κ. Let Ξ = {f(α) : α < κ, f ∈ B, f has height α + 1}. Clearly Γ ∪ Ω κ ⊆ Ξ and for
every α < κ, ψ α ∈ Ξ or ¬ψ α ∈ Ξ.
(7) If ϕ, ϕ → χ ∈ Ξ, then χ ∈ Ξ.
In fact, say ϕ = f(α) and ϕ → χ = f(β). Choose γ > α, β so that ψ γ is χ. We may
assume that dmn(f) ≥ γ + 1. Since rng(f) has a model, it follows that f(γ) = χ. So (7)
holds.
Let S be the set of all terms with no variables in them. We define σ ≡ τ iff σ, τ ∈ S
and (σ = τ) ∈ Ξ. Then ≡ is an equivalence relation on S. In fact, let σ ∈ S. Say that
σ = σ is ψ α . Since ψ α holds in every model, it holds in any model of {f(β) : β ≤ α}, and
hence f(α) = (σ = σ). So (σ = σ) ∈ Ξ and so σ ≡ σ. Symmetry and transitivity follow
by (7).
Let M be the collection of all equivalence classes. Using (7) it is easy to see that the
function and relation symbols can be defined on M so that the following conditions hold:
(8) If F is an m-ary function symbol, then
F M (σ 0 / ≡, . . ., σ m−1 / ≡) = F(σ 0 , . . ., σ m−1 )/ ≡ .
(9) If R is an m-ary relation symbol, then
〈σ 0 / ≡, . . ., σ m−1 / ≡〉 ∈ R M iff R(σ 0 , . . ., σ m−1 ) ∈ Ξ.
Now the final claim is as follows:
(10) If ϕ is a sentence of L ′ , then M |= ϕ iff ϕ ∈ Ξ.
Clearly this will finish the proof. We prove (10) by induction on ϕ. It is clear for atomic
sentences by (8) and (9). If it holds for ϕ, it clearly holds for ¬ϕ. Now suppose that Q is
a set of sentences of size less than κ, and (10) holds for each member of Q. Suppose that
M |= ∧ Q. Then M |= ϕ for each ϕ ∈ Q, and so Q ⊆ Ξ. Hence there is a ∆ ∈ [κ]

Lemma 18.9. Let A be a set of infinite cardinals such that for every regular cardinal
κ, the set A ∩ κ is non-stationary in κ. Then there is a one-one regressive function with
domain A.
Proof. We proceed by induction on γ def
= ⋃ A. Note that γ is a cardinal; it is 0 if
A = ∅. The cases γ = 0 and γ = ω are trivial, since then A = ∅ or A = {ω} respectively.
Next, suppose that γ is a successor cardinal κ + . Then A = A ′ ∪ {κ + } for some set A ′
of infinite cardinals less than κ + . Then ⋃ A ′ < κ + , so by the inductive hypothesis there
is a one-one regressive function f on A ′ . We can extend f to A by setting f(κ + ) = κ, and
so we get a one-one regressive function defined on A.
Suppose that γ is singular. Let 〈µ ξ : ξ < cf(γ)〉 be a strictly increasing continuous
sequence of infinite cardinals with supremum γ, with cf(γ) < µ 0 . Note then that for every
cardinal λ < γ, either λ < µ 0 or else there is a unique ξ < cf(γ) such that µ ξ ≤ λ < µ ξ+1 .
For every ξ < cf(γ) we can apply the inductive hypothesis to A ∩ µ ξ to get a one-one
regressive function g ξ with domain A ∩ µ ξ . We now define f with domain A. In case
cf(γ) = ω we define, for each λ ∈ A,
⎧
g 0 (λ) + 2 if λ < µ 0 ,
⎪⎨ µ ξ + g ξ+1 (λ) + 1 if µ ξ < λ < µ ξ+1 ,
f(λ) = µ ξ if λ = µ ξ+1 ,
⎪⎩ 1 if λ = µ 0 ,
0 if λ = γ ∈ A.
Here the addition is ordinal addition. Clearly f is as desired in this case. If cf(γ) > ω, let
〈ν ξ : ξ < cf(γ)〉 be a strictly increasing sequence of limit ordinals with supremum cf(γ).
Then we define, for each λ ∈ A,
⎧
g ⎪⎨ 0 (λ) + 1 if λ < µ 0 ,
µ
f(λ) = ξ + g ξ+1 (λ) + 1 if µ ξ < λ < µ ξ+1 ,
⎪⎩ ν ξ if λ = µ ξ ,
0 if λ = γ ∈ A.
Clearly f works in this case too.
Finally, suppose that γ is a regular limit cardinal. By assumption, there is a club C
in γ such that C ∩ γ ∩ A = ∅. We may assume that C ∩ ω = ∅. Let 〈µ ξ : ξ < γ〉 be the
strictly increasing enumeration of C. Then we define, for each λ ∈ A,
⎧
⎨g 0 (λ) + 1 if λ < µ 0 ,
f(λ) = µ
⎩ ξ + g ξ+1 (λ) + 1 if µ ξ < λ < µ ξ+1 ,
0 if λ = γ ∈ A.
Clearly f works in this case too.
Lemma 18.10. Suppose that κ is weakly compact, and S is a stationary subset of κ. Then
there is a regular λ < κ such that S ∩ λ is stationary in λ.
Proof. Suppose not. Thus for all regular λ < κ, the set S ∩ λ is non-stationary in
λ. Let C be the collection of all infinite cardinals less than κ. Clearly C is club in κ, so
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S ∩ C is stationary in κ. Clearly still S ∩ C ∩ λ is non-stationary in λ for every regular
λ < κ. So we may assume from the beginning that S is a set of infinite cardinals.
Let 〈λ ξ : ξ < κ〉 be the strictly increasing enumeration of S. Let
⎧ ⎡
⎤⎫
⎨
T =
⎩ s : ∃ξ < κ ⎣s ∈ ∏ ⎬
λ η and s is one-one⎦
⎭ .
η

Proof. Suppose not, and let X = {b ∈ B : G(b) ≠ b}. Since we are assuming that X
is a nonempty subclass of A, choose b ∈ X such that y ∈ A and yRb imply that y /∈ X.
Then
contradiction.
G(b) = {G(y) : y ∈ A and yRb}
= {G(y) : y ∈ B and yRb}
= {y : y ∈ B and yRb}
= {y : y ∈ B and y ∈ b}
= {y : y ∈ b}
= b,
Lemma 18.13. Let κ be weakly compact. Then for every U ⊆ V κ , the structure (V κ , ∈, U)
has a transitive elementary extension (M, ∈, U ′ ) such that κ ∈ M.
(This means that V κ ⊆ M and a sentence holds in the structure (V κ , ∈, U, x) x∈Vκ iff it
holds in (M, ∈, U ′ , x) x∈Vκ .)
Proof. Let Γ be the set of all L κκ -sentences true in the structure (V κ , ∈, U, x) x∈Vκ ,
together with the sentences
c is an ordinal,
α < c (for all α < κ),
where c is a new individual constant. The language here clearly has κ many symbols. Every
subset of Γ of size less than κ has a model; namely we can take (V κ , ∈, U, x, β) x∈Vκ , choosing
β greater than each α appearing in the sentences of Γ. Hence by weak compactness, Γ has
a model (M, E, W, k x , y) x∈Vκ . This model is well-founded, since the sentence
¬∃v 0 v 1 . . .
[ ∧
(v n+1 ∈ v n )
n∈ω
]
holds in (V κ , ∈, U, x) x∈Vκ , and hence in (M, E, W, k x , y) x∈Vκ .
Note that k is an injection of V κ into M. Let F be a bijection from M\rng(k) onto
{(V κ , u) : u ∈ M\rng(k)}. Then G def
= k −1 ∪ F −1 is one-one, mapping M onto some set
N such that V κ ⊆ N. We define, for x, z ∈ N, xE ′ z iff G −1 (x)EG −1 (z). Then G is an
isomorphism from (M, E, W, k x , y) x∈Vκ onto N def
= (N, E ′ , G[W], x, G(y)) x∈Vκ . Of course
N is still well-founded. It is also extensional, since the extensionality axiom holds in (V κ , ∈)
and hence in (M, E) and (N, E ′ ). Let H, P be the Mostowski collapse of (N, E ′ ). Thus P
is a transitive set, and
(1) H is an isomorphism from (N, E ′ ) onto (P, ∈).
(2) ∀a, b ∈ N[aE ′ b ∈ V κ → a ∈ b].
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In fact, suppose that a, b ∈ N and aE ′ b ∈ V κ . Let the individual constants used in the
expansion of (V κ , ∈, U) to (V κ , ∈, U, x) a∈Vκ be 〈c x : x ∈ V κ 〉. Then
(V κ , ∈, U, x) a∈Vκ |= ∀z
[
z ∈ k b → ∨ w∈b(z = k w )
]
,
and hence this sentence holds in (N, E ′ , G[W], x, G(y)) x∈Vκ as well, and so there is a w ∈ b
such that a = w, i.e., a ∈ b. So (2) holds.
(3) ∀a, b ∈ V κ [a ∈ b → aE ′ b]
In fact, suppose that a, b ∈ V κ and a ∈ b. Then the sentence k a ∈ k b holds in (V κ , ∈
, U, x) x∈Vκ , so it also holds in (N, E ′ , G[W], x, G(y)) x∈Vκ , so that aE ′ b.
We have now verified the hypotheses of Lemma 18.12. It follows that H ↾ V κ is
the identity. In particular, V κ ⊆ P. Now take any sentence σ in the language of (V κ , ∈
, U, x) x∈Vκ . Then
(V κ , ∈, U, x) x∈Vκ |= σ iff (M, E, W, k x ) x∈Vκ |= σ
iff
(N, E ′ , G[W], x) x∈Vκ |= σ
iff (P, ∈, H[G[W]], x) x∈Vκ |= σ.
Thus (P, ∈, H[G[W]]) is an elementary extension of (V κ , ∈, U).
Now for α < κ we have
(M, E, W, k x , y) x∈Vκ |= [y is an ordinal and k α Ey],
hence
(N, E ′ , G[W], x, G(y)) x∈Vκ |= [G(y) is an ordinal and αE ′ G(y)],
hence
(P, ∈, H[G[W]], x, H(G(y))) x∈Vκ |= [H(G(y)) is an ordinal and α ∈ H(G(y))].
Thus H(G(y)) is an ordinal in P greater than each α < κ, so since P is transitive,
κ ∈ P.
An infinite cardinal κ is first-order describable iff there is a U ⊆ V κ and a sentence σ in
the language for (V κ , ∈, U) such that (V κ , ∈, U) |= σ, while there is no α < κ such that
(V α , ∈, U ∩ V α ) |= σ.
Theorem 18.14. If κ is infinite but not inaccessible, then it is first-order describable.
Proof. ω is describable by the sentence that says that κ is the first limit ordinal;
absoluteness is used. The subset U is not needed for this. Now suppose that κ is singular.
Let λ = cf(κ), and let f be a function whose domain is some ordinal γ < κ with
rng(f) cofinal in κ. Let U = {(λ, β, f(β)) : β < λ}. Let σ be the sentence expressing the
following:
For every ordinal γ there is an ordinal δ with γ < δ, U is nonempty, and there is an
ordinal µ and a function g with domain µ such that U consists of all triples (µ, β, g(β))
with β < µ.
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Clearly (V κ , ∈, U) |= σ. Suppose that α < κ and (V α , ∈, V α ∩ U) |= σ. Then α is a limit
ordinal, and there is an ordinal γ < α and a function g with domain γ such that V α ∩ U
consists of all triples (γ, β, g(β)) with β < γ. (Some absoluteness is used.) Now V α ∩ U
is nonempty; choose (γ, β, g(β)) in it. Then γ = λ since it is in U. It follows that g = f.
Choose β < λ such that α < f(β). Then (λ, β, f(β)) ∈ U ∩ V α . Since α < f(β), it follows
that α has rank less than α, contradiction.
Now suppose that λ < κ ≤ 2 λ . A contradiction is reached similarly, as follows. Let f
be a function whose domain is P(λ) with range κ. Let U = {(λ, B, f(B)) : B ⊆ λ}. Let
σ be the sentence expressing the following:
For every ordinal γ there is an ordinal δ with γ < δ, U is nonempty, and there is an ordinal
µ and a function g with domain P(µ) such that U consists of all triples (µ, B, g(B)) with
B ⊆ µ.
Clearly (V κ , ∈, U) |= σ. Suppose that α < κ and (V α , ∈, V α ∩ U) |= σ. Then α is a limit
ordinal, and there is an ordinal γ < α and a function g with domain P(γ) such that
V α ∩U consists of all triples (γ, B, g(B)) with B ⊆ γ. (Some absoluteness is used.) Clearly
γ = λ; otherwise U ∩ V α would be empty. Note that g = f. Choose B ⊆ λ such that
α = f(B). Then (λ, B, f(B)) ∈ U ∩ V α . Again this implies that α has rank less than α,
contradiction.
The new equivalent of weak compactness involves second-order logic. We augment first
order logic by adding a new variable S ranging over subsets rather than elements. There
is one new kind of atomic formula: Sv with v a first-order variable. This is interpreted as
saying that v is a member of S.
Now an infinite cardinal κ is Π 1 1 -indescribable iff for every U ⊆ V κ and every secondorder
sentence σ of the form ∀Sϕ, with no quantifiers on S within ϕ, if (V κ , ∈, U) |= σ,
then there is an α < κ such that (V α , ∈, U ∩ V α ) |= σ. Note that if κ is Π 1 1-indescribable
then it is not first-order describable.
Theorem 18.15. An infinite cardinal κ is weakly compact iff it is Π 1 1 -indescribable.
Proof. First suppose that κ is Π 1 1-indescribable. By Theorem 18.14 it is inaccessible.
So it suffices to show that it has the tree property. By the proof of Theorem 18.7(iii)⇒(iv)
it suffices to check the tree property for a tree T ⊆

Now since κ ∈ M and (M, ∈) is a model of ZFC, Vκ
M
to V κ . Hence by (1) we get
exists, and by absoluteness it is equal
Hence
(M, ∈, U ′ ) |= ∀X ⊆ V κ ϕ V κ
(U ′ ∩ V κ ).
(M, ∈, U ′ ) |= ∃α∀X ⊆ V α ϕ V α
(U ′ ∩ V α ),
so by the elementary extension property we get
(V κ , ∈, U) |= ∃α∀X ⊆ V α ϕ V α
(U ′ ∩ V α ).
We choose such an α. Since V κ ∩On = κ, it follows that α < κ. Hence (V α , ∈, U ′ ∩V α ) |= σ,
as desired.
Measurable cardinals
Our third kind of large cardinal is the class of measurable cardinals. Although, as the
name suggests, this notion comes from measure theory, the definition and results we give
are purely set-theoretical. Moreover, similarly to weakly compact cardinals, it is not
obvious from the definition that we are dealing with large cardinals.
The definition is given in terms of the notion of an ultrafilter on a set.
• Let X be a nonempty set. A filter on X is a family F of subsets of X satisfying the
following conditions:
(i) X ∈ F.
(ii) If Y, Z ∈ F, then Y ∩ Z ∈ F.
(iii) If Y ∈ F and Y ⊆ Z ⊆ X, then Z ∈ F.
• A filter F on a set X is proper or nontrivial iff ∅ /∈ F.
• An ultrafilter on a set X is a nontrivial filter F on X such that for every Y ⊆ X, either
Y ∈ F or X\Y ∈ F.
• A family A of subsets of X has the finite intersection property, fip, iff for every finite
subset B of A we have ⋂ B ≠ ∅.
• If A is a family of subsets of X, then the filter generated by A is the set
{Y ⊆ X : ⋂ B ⊆ Y for some finite B ⊆ A }.
[Clearly this is a filter on X, and it contains A .]
Proposition 18.16. If x ∈ X, then {Y ⊆ X : x ∈ Y } is an ultrafilter on X.
An ultrafilter of the kind given in this proposition is called a principal ultrafilter. There
are nonprincipal ultrafilters on any infinite set, as we will see shortly.
Proposition 18.17. Let F be a proper filter on a set X. Then the following are equivalent:
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(i) F is an ultrafilter.
(ii) F is maximal in the partially ordered set of all proper filters (under ⊆).
Proof. (i)⇒(ii): Assume (i), and suppose that G is a filter with F ⊂ G . Choose
Y ∈ G \F. Since Y /∈ F, we must have X\Y ∈ F ⊆ G . So Y, X\Y ∈ G , hence
∅ = Y ∩ (X\Y ) ∈ G , and so G is not proper.
(ii)⇒(i): Assume (ii), and suppose that Y ⊆ X, with Y /∈ F; we want to show that
X\Y ∈ F. Let
G = {Z ⊆ X : Y ∩ W ⊆ Z for some W ∈ F }.
Clearly G is a filter on X, and F ⊆ G . Moreover, Y ∈ G \F. It follows that G is not
proper, and so ∅ ∈ G . Thus there is a W ∈ F such that Y ∩ W = ∅. Hence W ⊆ X\Y ,
and hence X\Y ∈ F, as desired.
Theorem 18.18. For any infinite set X there is a nonprincipal ultrafilter on X. Moreover,
if A is any collection of subsets of X with fip, then A can be extended to an ultrafilter.
Proof. First we show that the first assertion follows from the second. Let A be the
collection of all cofinite subsets of X—the subsets whose complements are finite. A has
fip, since if B is a finite subset of A , then X\ ⋂ B = ⋃ Y ∈B
(X\B) is finite. By the second
assertion, A can be extended to an ultrafilter F. Clearly F is nonprincipal.
To prove the second assertion, let A be a collection of subsets of X with fip, and let
C be the collection of all proper filters on X which contain A . Clearly the filter generated
by A is proper, so C ≠ ∅. We consider C as a partially ordered set under inclusion.
Any subset D of C which is a chain has an upper bound in C , namely ⋃ D, as is easily
checked. So by Zorn’s lemma C has a maximal member F. By Proposition 18.16, F is an
ultrafilter.
• Let X be an infinite set, and let κ be an infinite cardinal. An ultrafilter F on X is κ-
complete iff for any A ∈ [F] 0 let Z α =
( ⋂ β

contradiction. So |P| = κ.
Theorem 18.20. Suppose that κ is the least infinite cardinal such that there is a nonprincipal
σ-complete ultrafilter F on κ. Then F is κ-complete.
Proof. Assume the hypothesis, but suppose that F is not κ-complete. So there is a
A ∈ [F]

{α < κ : f α (β) = 0} and {α < κ : f α (β) = 1} is in U, so we can let ε(β) ∈ 2 be such that
{α < κ : f α (β) = ε(β)} ∈ U. Then
⋂
{α < κ : f α (β) = ε(β)} ∈ U;
β

means that for every function f : ⋃ n∈ω [κ]n → m there is a subset H ⊆ κ of order type α
such that for each n ∈ ω, f ↾ [H] n is constant.
7. 0 ♯ exists. This means that there is a non-identity elementary embedding of L into L.
Thus no actual cardinal is referred to. But 0 ♯ implies the existence of some large cardinals,
and the existence of some large cardinals implies that 0 ♯ exists.
8. Jónsson κ is a Jónsson cardinal iff every model of size κ has a proper elementary
substructure of size κ.
9. Rowbottom κ is a Rowbottom cardinal iff for every uncountable λ < κ, every model
of type (κ, λ) has an elementary submodel of type (κ, ω).
10. Ramsey κ → (κ) ν, V ν ⊆ M, A ∩ V ν = j(A) ∩ V ν ]
14. superstrong κ is superstrong iff there is a nontrivial elementary embedding j : V →
M with κ the first ordinal moved, such that V j(κ) ⊆ M.
15. strongly compact κ is strongly compact iff for any L κκ -language, if Γ is a set of
sentences and every subset of Γ of size less than κ has a model, then Γ itself has a model.
16. supercompact κ is supercompact iff for every A with |A| ≥ κ there is normal measure
on P κ (A).
17. extendible For an ordinal η, we say that k is η-extendible iff there exist ζ and a
nontrivial elementary embedding j : V κ+η → V ζ with κ first ordinal moved, with η < j(κ).
κ is extendible iff it is η-extendible for every η > 0.
18. Vopěnka’s principle If C is a proper class of models in a given first-order language,
then there exist two distinct members A, B ∈ C such that A can be elementarily embedded
in B.
19. huge A cardinal κ is huge iff there is a nontrivial elementary embedding j : V → M
with κ the first ordinal moved, such that M j(κ) ⊆ M.
20. I0. There is an ordinal δ and a proper elementary embedding j of L(V δ+1 ) into L(V δ+1 )
such that the first ordinal moved is less than δ.
In the diagram on the next page, a line indicates that (the consistency of the) existence of
the cardinal above implies (the consistency of the) existence of the one below.
215

• I0
• huge
• Vopěnka
• extendible
supercompact
•
super strong •
• strongly compact
•
Woodin
• strong
• measurable
Ramsey
•
κ → (ω 1 )

EXERCISES
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 E κ λ < Eκ µ , where
and similarly for E κ µ .
E κ λ = {α < κ : cf(α) = λ},
E18.2. Continuing exercise E18.1: Assume that κ is uncountable and regular. Show that
the relation < is transitive.
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.
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) If 〈X i : i ∈ ω〉 is a system of pairwise disjoint subsets of S, then µ( ⋃ i∈ω X i) =
∑
i∈ω µ(X i). (The X i ’s are not necessarily nonempty.)
Let κ be an infinite cardinal. Then µ is κ-additive iff for every system 〈X α : α < γ〉 of
nonempty pairwise disjoint sets, with γ < κ, we have
( ) ⋃
X α
α

{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.
E18.6. Prove that if κ is real-valued measurable then either κ is measurable or κ ≤ 2 ω .
Hint: if there do not exist any µ-atoms, construct a binary tree of height at most ω 1 .
E18.7. Let κ be a regular uncountable cardinal. Show that the diagonal intersection of
the system 〈(α + 1, κ) : α < κ〉 is the set of all limit ordinals less than κ.
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.
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 S 0 ⊆ κ, if κ\S 0 /∈ F and f is a regressive function defined on S 0 , then
there is an S ⊆ S 0 with κ\S /∈ F and f is constant on S.
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.
(iv) If 〈X n : n ∈ ω〉 is a system of pairwise disjoint sets, then µ( ⋃ n∈ω X n) =
∑
n∈ω µ(X n). (Some of the sets X n might be empty.)
Prove that there does not exist a probability measure on ω 1 . Hint: consider an Ulam
matrix.
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 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.
Reference
Kanamori, A. The higher infinite. Springer 2005, 536pp.
218