The Continuum Hypothesis - Logic at Harvard

Theorem 5.8 (Koellner and Woodin). Assume that there is a proper classof Woodin cardinals. Suppose that A is an axiom such that(i) ZFC+A is Ω-satisfiable and(ii) ZFC+A is Ω-complete for Σ 2 2 .Then there is an axiom B such that(i ′ ) ZFC+B is Ω-satisfiable and(ii ′ ) ZFC+B is Ω-complete for Σ 2 2and T A ≠ T B .This is the parallel of Theorem 5.2.To complete the parallel one would need that CH is among all of the T A .This is not known. But it is a reasonable conjecture.Conjecture 5.9. Assume large cardinal axioms.(1) There is an Σ 2 2 axiom A such that(i) ZFC+A is Ω-satisfiable and(ii) ZFC+A is Ω-complete for the Σ 2 2 .(2) Any such Σ 2 2axiom A has the feature thatZFC+A |= Ω CH.Should this conjecture hold it would provide a true analogue of Theorem 5.1.This would complete the parallel with the first step.There is also a parallel with the second step. Recall that for the secondstep in the previous subsection we had that although the various T A did notagree, they all contained ¬CH and, moreover, from among them there is onethat stands out, namely the theory given by (∗), since this theory maximizesthe Π 2 -theory of the structure 〈H(ω 2 ),∈,I NS ,A | A ∈ P(R)∩L(R)〉. In thepresent context of CHwe again(assuming theconjecture) have that althoughthe T A do not agree, they all contain CH. It turns out that once again, fromamong them there is one that stands out, namely, the maximum one. Forit is known (by a result of Woodin in 1985) that if there is a proper class31