SILVER’S THEOREM
MAXWELL LEVINE
Theorem. If κ is a singular cardinal of uncountable cofinality and {ν < κ : 2ν =
ν + } is a stationary subset of κ (and hence κ is a strong limit), then 2κ = κ+ .
The idea behind Silver’s original proof is this: Use the Levy collapse, P =
Col(ω, 2cf κ ) to move to an extension V P in which we construct an ultraproduct
V cf κ /D where P (c) is small for some c ∈ V cf κ /D with |c| = κ. P preserves κ, so
we can argue that 2κ is still small in V .
Proof. We begin by describing the construction of an ultrafilter. Let T = {ν <
κ : 2ν = ν + }, let λ := cf κ, and let h be a continuous (i.e. if α is a limit then
h(α) = supβ<α h(β)) and strictly increasing map from λ onto a cofinal subset of
κ. X := {α < λ : h(α) ∈ T } is a stationary subset of λ: If C ⊂ λ is a club, then
D = {h(α) : α ∈ C} is a club in κ, so there is some β = h(α) ∈ T ∩ D, and hence
α ∈ C ∩ X.
Let µ = 2λ (which is less than the strong limit κ) and let P = Col(ω, µ). If
U := (λλ )V , then V P |= |U | = ω. Note that V P is just a construction for our
purposes, and that the real goal is to show that V |= 2κ = κ+ . We will construct
an ultrafilter D in P (λ)V (but D ∈ V P ) such that every regressive function in U is
constant on some member of D.
Let {fi : 0 < i < ω} enumerate the regressive functions in U and construct a
descending sequence of stationary subsets X0 := X ⊇ X1 ⊇ X2 ⊇ . . . inductively
using Fodor’s Theorem:1 Given fi+1 regressive, let Xi+1 ⊂ Xi be a stationary
subset such that fi+1 is constant on Xi+1 , and then define D := {B ∈ V : B ⊂
λ, ∃i s.t. Xi ⊂ B}.
Claim. D is an ultrafilter on P (λ)V .
Proof. Showing that D is a filter is simple: If A ∈ D and A ⊂ B, then Xi ⊂ A
implies Xi ⊂ B, and thus B ∈ D. If A, B ∈ D, Xi ⊂ A, Xj ⊂ B, and i ≤ j, then
Xi ⊂ A ∩ B ∈ D.
To show that D is an ultrafilter, we need to establish the following fact: If
Y ⊂ Xi , there is some j ≥ i such that either Y ∩Xj = ∅ or Xj ⊂ Y . By construction
fi is constant on Xi+1 and takes value β > 0 (without loss of generality). Define
g such that g(α) = fi (α) if α ∈ Y and g(α) = 0 if α ∈
/ Y . g is regressive and
not constant on Xi+1 , so g = fj for some j > i (note that fj is constant on Xi if
j < i because of the nested-ness of the X’s). g is constant on Xj ⊂ Xi , so either
g 00 Xj = {β} or g 00 Xj = {0}. Hence either Xj ∩ Y = ∅ or Xj ⊂ Y .
Now if A ∈ P (λ)V \ D, then Xi 6⊂ A for all i. Apply the claim to X0 ∩ A) to
find j ≥ 0 such that Xj ∩ A = (X0 ∩ A) ∩ Xj = ∅ or (X0 ∩ A) ⊂ Xj , and hence
1Note that this is where we use the fact that λ > ω. Fodor’s Theorem only makes sense in
cardinals that contain limit points.
1
2
MAXWELL LEVINE
Xj ⊂ A. The second case is a contradiction, so the first case holds and Xj ⊂ λ \ A.
Therefore λ \ A ∈ D and we have shown that D is an ultrafilter.
D allows us to define an ultraproduct V λ /D within V P which consists of equivalence classes [f ]D of functions f : λ → V , f ∈ V , under D. (Generally we suppress
the subscript.) Note that this model is not necessarily well-founded since D is not
ω1 -complete, so if x is an ordinal in V λ /D, it may not be an ordinal in V P . There is
also a canonical injection j : V → V λ /D defined by j(x) = [cx ]D where cx (α) = x
for all α < λ.2
The next stage of the proof is to show that we can find a stand-in for κ in V λ /D.
Let d(α) = α be the constant function on λ.
Claim. V λ /D |= [d] is a cardinal.
Proof. By Los’ Theorem it is enough to show that A = {α < λ : α is a cardinal} ∈
D. Otherwise λ \ A = {α : α is not a cardinal} ∈ D. Let f (α) = |α| if α is
not a cardinal and f (α) = 0 otherwise. f is regressive, so it is constant on some
Xi ⊂ λ \ A. Xi ⊂ {α < λ : |α| = γ} ⊂ γ + for some γ. Stationary sets are
unbounded, so this is a contradiction.
Note that [d] is totally ordered, even though it may not be well-ordered.
Claim. {x ∈ V λ /D : x ∈ [d]} = {j(α) : α < λ}.
Proof. Suppose [f ] ∈ [d]; then {α < λ : f (α) < d(α) = α} ∈ D, so {α : f (α) =
γ} ∈ D. Take intersections if necessary to obtain [f ] = [cγ ] = j(γ).
Also, if γ < λ then {α : α ≥ γ} ∈ D. If not, then {α : α < γ} ∈ D, meaning
Xi ⊂ {α : α < γ} for some stationary Xi . This is a contradiction because stationary
sets are unbounded. It follows that {α : d(α) ≥ γ} ∈ D, and so [d] > j(γ).
Claim. If c := j(h)([d]) then |c| = κ.
Proof. By Los, j(h) is still continuous and strictly increasing, and we have established that [d] is a cardinal—in particular it is a limit ordinal—in V λ /D. This
means c = supβ<[d] j(h)(β). Therefore if [f ] < c, then there is j(α) < [d] with
α < λ, such that [f ] < (jh)(j(α)) < c. j(h(α)) is represented by a function in V
from λ into h(α) (changing the function on a set not in D if necessary), but there
are at most h(α)λ -many such representatives and h(α)λ < κ. Thus |[f ]| < κ. Since
initial segments of c are strictly smaller than κ, it follows that |c| ≤ κ.
On the other hand, if γ < κ then γ < h(α) for some α < λ, and so j(γ) <
j(h(α)) = j(h)(j(α)) < j(h)([d]) = c. j is injective, so {j(γ) : γ < κ} ⊂ c shows
that κ ≤ |c|.
The last stage of the proof is to examine Q := P (κ)V
to P (κ)V .
P
λ
/D
∈ V P and compare it
P
Claim. V P |= |Q| ≤ κ+ , i.e. |Q|V ≤ (κ+ )V .
2j looks like the embedding used for a measurable cardinal, but it isn’t quite as well-behaved.
Nonetheless, it still has some nice properties.
SILVER’S THEOREM
3
Proof. X0 ⊂ {α < λ : 2h(α) = h(α)+ } ∈ D, so V λ /D |= 2j(h)[d] = (j(h)[d])+ by
Los, i.e. V λ /D |= P (c) = c+ . If a ∈ V λ /D is such that V λ /D |= c+ = a, then
there is some f ∈ V λ /D such that f : Q ,→ a. But V P |= |a| ≤ κ+ because |c| = κ,
P
and f ∈ V P ⊃ V λ /D, and so it follows that |Q| ≤ (κ+ )V .
P
P
Claim. If V |= κ+ < 2κ , then V P |= κ+ < |Q|, i.e. (κ+ )V < |Q|V .
Proof. P has the µ-chain condition, so it preserves cardinals > µ; in particular it
preserves cardinals ≥ κ. If V |= 2κ > κ+ , then there is S = hCα : α < κ++ i in V
where the Cα ’s are subsets of κ.
Using preservation of cardinals, V P |= |S| = κ++ , so we can construct k :
++ V P
(κ ) → Q in V P by letting k(α) = j(Cα ) ∩ c. k is injective too: If γ ∈ Cα \ Cβ ,
then j(γ) ∈ j(Cα ) and j(γ) ∈ c, however j(γ) ∈
/ k(Cβ ). It follows that V P |= κ++ ≤
|Q|.
Silver’s Theorem follows because we get a contradiction if V 6|= 2κ = κ+ .