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")
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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