From: Mike Oliver Subject: Re: How Does V=L Prove CH? Date: Mon, 18 Jan 1999 10:44:41 -0800 Newsgroups: sci.math,sci.logic Keywords: (Set Theory axiom) V=L implies Continuum Hypothesis Daryl McCullough wrote: > > In a book on set theory and independence proofs, it is > said that V=L implies CH (actually GCH), but I don't > understand the reason they gave. Why must every subset > of omega be in L(omega_1)? Here's a sketch; could be wrong in a detail or two. Say x is a subset of omega which is an element of L. Form the Skolem hull of {x} (L has definable Skolem functions, or you can use the "rudimentary" functions). That's a countable wellfounded model of V=L. Take its Mostowski collapse. None of the integers moved when you collapsed, so x didn't move either. Now the collapsed model is a transitive model of V=L, so by the Condensation lemma it's an actual level of L, e.g. L_alpha. But the model is countable, so alpha is countable. Therefore x \in L_alpha \subset L_{omega_1}. -- Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love") Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver. 1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9 ============================================================================== From: ikastan@sol.uucp (ilias kastanas 08-14-90) Subject: Re: How Does V=L Prove CH? Date: 19 Jan 1999 06:05:41 GMT Newsgroups: sci.math,sci.logic In article <77vka4\$kv8@edrn.newsguy.com>, Daryl McCullough wrote: >In a book on set theory and independence proofs, it is >said that V=L implies CH (actually GCH), but I don't >understand the reason they gave. Why must every subset >of omega be in L(omega_1)? One way is through this theorem (ZFC): Suppose phi(x,y) is a formula of the language of set theory that is absolute (Sigma-1), and defines a function y = F(x). Then |F(x)| <= |TransClos(x)| + aleph_0. (Proof, roughly: Find a V_a containing x, F(x), and absolute for phi (Reflection). Let C = trans. closure of {x}. By Loewenheim-Skolem V_a has a subset K satisfying the same sentences (parameters in C); |K| <= |C|+aleph_0. The Mostowski collapse of K is a transitive set Z; C is rigid, so C is in Z. Hence x is in Z. By absoluteness of phi, F(x) is in Z). For any z in L, let o(z) = order of z (least ordinal s.t. z is in L_o(z)). Then y = o(z) is absolute; applying the theorem, |o(z)| <= |TransClos(z)| + aleph_0. GCH now follows: if z is subset of aleph_a, |o(z)| <= aleph_a and hence z is in L_aleph_a+1; so P(aleph_a) is a subset of L_aleph_a+1, and |P(aleph_a)| <= |L_aleph_a+1| = aleph_a+1. Ilias