Abstract. We show that “saturation ” of the universe with respect to forcing over L with partial orders on ω1 is equivalent to the existence of 0 #. If P is a constructible forcing notion, then we say that G ⊆ P is P-generic iff G is P-generic over L. The statement that all countable constructible forcings have generics is rather weak, and holds for example in L[R], where R is a Cohen real over L. But it is not possible that all constructible forcings have generics: consider the forcing that collapses ω1 to ω with finite conditions. Definition. V is L-saturated for ω1-forcings iff whenever P is a constructible forcing of L-cardinality ω1 such that for any p ∈ P there is a P-generic containing p in some ω1-preserving extension of V, then there is a P-generic in V. Theorem 1. The following are equivalent: (a) V is L-saturated for ω1-forcings. (b) 0 # exists. Proof. (a) → (b) The existence of 0 # is equivalent to the statement that every stationary constructible subset of ω1 contains a CUB subset (see ). Now use the following: Fact (Baumgartner; see ). If X is a stationary constructible subset of ω1, then there is a forcing P ∈ L of L-cardinality ω1 which preserves cardinals over V and adds a CUB subset to X. (P adds a CUB subset of X using “finite conditions”.) (b) → (a) Assume that 0 # exists, and suppose that P is a constructible forcing of L-cardinality ω1 such that every condition in P belongs to a generic in an ω1preserving extension of V. We will show that there is a P-generic in V. Assume that the universe of P is exactly ω1. LetPbeof the form t(i, ω1, ∞), where i <ω1 <∞ is a finite increasing sequence of indiscernibles and t is an L-term. We claim that if i <k0 <k1 are countable indiscernibles and Gk0 is Pk0-generic over L, then there is Gk1 containing Gk0 which is Pk1-generic over L, wherePk = t(i, k, ∞). If not, then player I wins the open game G(k0,k1,Gk0), where I chooses constructible dense subsets of Pk1 and II responds with increasingly strong conditions meeting Received by the editors January 4, 2005
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