On NSω1 being saturated

Definition 0.1 Let δ be a cardinal. We say that δ is Woodin with 3 iff there is some sequence (a κ : κ < δ) such that a κ ⊂ V κ for every κ < δ and for every A ⊂ V δ the set {κ < δ : A ∩ V κ = a κ ∧ κ is A–strong up to δ} is stationary in δ. Proof. Let us define ((a κ , c κ) : κ < δ) recursively as follows. If ((a κ , c κ) : κ < µ) is defined for some µ < δ, then we let (a µ , c µ) be the least (in the order of constructibility) pair (a, c) such that a ⊂ V µ , c ⊂ µ is club in µ, and {κ < µ : a ∩ V κ = a κ ∧ κ is a–strong up to µ} ∩ c = ∅ (if such a pair (a, c) exists). We claim that (a κ : κ < δ) is as desired. If not, then let (A, C) be least (in the order of constructibility) such that A ⊂ V δ , C ⊂ δ is club in δ, and {κ < δ : A ∩ V κ = a κ ∧ κ is A–strong up to δ} ∩ C = ∅. (1) As the set {κ < δ : κ is A–strong up to δ} is stationary in δ, an easy Skolem hull argument together with condensation for L[E] yields some κ ∈ C which is A–strong up to δ and (A ∩ V κ , c ∩ κ) is the least (in the order of constructibility) pair (a, c) such that a ⊂ V κ , c ⊂ κ is club in κ, and {λ < κ : a ∩ V λ = a λ ∧ λ is a–strong up to κ} ∩ c = ∅.