CLUB SHOOTING FORCES DIAMOND
MONROE ESKEW
Definition Let κ be a regular cardinal. A stationary set S ⊆ κ is called fat when
for all club C ⊆ κ and all α < κ, there is a closed d ⊆ C ∩ S such that ot(d) = α.
Theorem 1 (Stavi). Suppose κ = µ+ , µ<µ = µ, and S ⊆ κ is fat. Then there is a
κ-distributive forcing PS that adds a club C ⊆ S. The forcing PS consists of closed,
bounded subsets of S, ordered by end-extension.
Theorem 2. Suppose κ = µ+ , µ<µ = µ, and S ⊆ κ is fat. Then PS forces, “For
all stationary T ⊆ κ, ♦κ (T ).”
Proof. Let S, µ, κ be as hypothesized, and let C be PS -generic over V . Let Ṫ be a
name for a stationary subset of κ. We will define a ♦κ (T ) sequence by:
(1) Working in V , for each α < κ, choose some closed cα ⊆ S \ α of order type
α + 1.
(2) Working in V [C], for α ∈ T , let βα be the least ordinal ≥ α such that C ∩ (βα +
1) VPS α̌ ∈ Ṫ .
(3) In V [C], let Aα = {ξ < α : the ξ + 1st element of cβα is in C} for α ∈ T .
(The idea is to code information into the choice of successor points of C.) Now
let Ḋ be a name for a club, and Ẋ a name for a subset of κ. Let Ė be a name for
{α ∈ D : V |= “C ∩ (α + 1) decides Ẋ ∩ α and forces that α ∈ Ḋ”}. E is easily
seen to be closed. E is unbounded because by distributivity, for every α, X ∩ α is
decided by some initial segment of the filter. So a catch-your-tail argument shows
E is unbounded.
Key observation: α ∈ E iff C ∩(α+1) V α̌ ∈ Ė. This follows from the definition
of E. Now forget about this particular C; we will make a density argument.
Note that for any p ∈ PS , there is q ≤ p such that q max(q) ∈ Ė and
q 1 max(q) ∈
/ Ṫ . For if C ⊇ p is generic, then there is an α ∈ T ∩ E above max(p)
since T is forced to be stationary. So q = C ∩ (α + 1) α ∈ Ė, and q cannot force
α∈
/ Ṫ .
Now let p ∈ PS be arbitrary. Let q ≤ p be such that q max(q) ∈ Ė and
q 1 max(q) ∈
/ Ṫ . Let α = max(q), and let q 0 ≤ q have minimal supremum β ≥ α
0
such that q α ∈ Ṫ . Let q 00 ≤ q 0 be q 0 ∪ d, where
d = {γ ∈ cβ : γ is the least point or a limit point of cβ , or
(∃ξ < α)γ is the ξ + 1st element of cβ and q ξ ∈ Ẋ ∩ α}.
00
Clearly q α ∈ Ḋ ∩ Ṫ . By the definition of Aα , q 00 Aα = Ẋ ∩ α. Since p was
arbitrary, it is forced that there is α ∈ T ∩ D such that Aα = X ∩ α.
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