Son of George and V = L

This paper has three parts. In this first part, we formulate and prove from V= L a new combinatorial principle, O ++. In the second part, we discuss the topological problem which led to the formulation of O +. Finally, we use O ++ to construct a space solving the topological problem. ?1. 0 ++: Formulation and proof from V = L. The combinatorial principles 0 * and 0 + can be thought of as giving an c1-sequence of countable approximations to the set a(cwl). To construct the space of this paper, this is not enough; we need filters which approximate the club filter and which define an accurate notion of stationary. Specifically what we need is 0 ++, which asserts There are A and C such that: 1(a) A is a function with domain Wi; for all ax E w1, A(ac) E [2(oc)]'. (b) C is a function from 9(wi) to the family of club subsets of oi1. (c) For all X E 9(w1), if r E C(X), then X n r E A(T) and C(X) n r E A(r). Part I is simply a statement of 0 +. Given A and C as above, define for s E c W, Va = {c E A(d): c is club in ct}, and for X EA 9(cw1), define S(X) = e c1: for all c ,ac c n X# 0}. 0 ++, continued. Additionally, there is D such that: 2(a) D is a stationary subset of w1. (b) For all s e D, Wj is afilter. (c) If f is a countable family of stationary subsets of Wi, then n {S(X): X E a} n D is stationary. In outline, the proof of 0 ++ from V = L is the same as that of 0 +. However, to get part 2 of 0 ++, we need a few definitions and a lemma. For c E c 1, set Sa = {e E cw,: a= cxLw and L, l= ZF-}. Set D = {ox: Sa 7 0 and Sa has no last element}. LEMMA 1. D is not empty. Moreover, for any a E L,,2, there is ca E D with a cofinal subset, N, of Sa such that if v E N, then there is an elementary embedding of L, to a transitive model of ZF+ V = L containing a in the range. PROOF. We will prove the second assertion. Let a E L,,2. Define P,,, n E c, so that Received November 10, 1980. AMS(MOS) subject classifications (1970). Primary 03E45; Secondary 54E30.