Re: GCH vs. Axiom of Choice.
From: KRamsay (kramsay_at_aol.com)
Date: 12/06/04
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Date: 06 Dec 2004 05:01:04 GMT
In article <39d6e584.0412022029.767e52f2@posting.google.com>,
troubled6man@yahoo.com (J.E.) writes:
|Ah, I should have been more clear. L is a model (right?),
It's a proper class of sets, the sets satisfying a certain condition.
(Being "constructible".) Another way to describe it is that it's the
minimal model of ZF containing all the ordinals.
| ZF is an
|axiom system (right?). GCH is true for L (you said so in your post
|right?). You also said that we prove that GCH is true for L? But
|what do we need to prove it?
ZF is strong enough by itself to prove that GCH holds in L.
ZF is strong enough to prove that V=L holds in L, and that
V=L->GCH.
| For instance if we had to assume ZFC+GCH
|to prove it, then it wouldn't mean much IMO.
Heh, it's sort of amusing when to prove that some fact holds true
when relativized to a model one needs to use the same fact not
relativized to the model. I can see how that might seem underwhelming,
but it's not necessarily trivial or uninteresting. For instance, one
had to prove that the axioms of ZF hold in L, too, and obviously we
need to assume some of them in order to prove they hold in L.
| I assume that ZF can
|prove only limited things about L, since ZF is incomplete, but if I
|add GCH to ZF then can ZF+GCH prove something new about L?
I'm guessing it probably can, but I don't know of a good example.
My guess is that some assumption intermediate between ZF and ZF+V=L
is sufficient to prove (say) that aleph_1 is the same as aleph_1
relative to L. Certainly V=L implies that!
On the other hand, there aren't any first order properties of L
that can be proven in ZF+GCH but not in ZF. In fact, for an arbitrary
sentence X in the language of ZF, the following are equivalent:
(a) ZF |- (V=L->X)
(b) ZF + V=L |- X
(c) ZF |- L |= X.
(d) ZF + V=L |- L |= X
(a)->(b) is a case of the deduction theorem. (b)->(c) holds because
it's a theorem of ZF that V=L relativized to L is true Given a model
M of ZF, we can consider constructibility as relativized to M. The
elements of M satisfying this relativized constructibility form a
submodel L_M, which is another model of ZF. If we repeat this process,
to get L_{L_M}, it's possible to prove that we don't shrink the model
any; L_{L_M} = L_M. So L_M is a model of V=L, the axiom stating that
every set is in L. So anything that's a consequence of V=L holds
relative to L. If X is a theorem of ZF+V=L, then that fact can be
proven in ZF, and then applied to L.
(c)->(d) is obvious.
To prove (d)->(a), assume first that (a) is false. By Goedel's
completeness theorem, there would exist a model of ZF+V=L+~X.
Suppose M is a model of ZF+V=L+~X. Since the L relativized to M
is the same as M, we get that M |= ~(L|=X). Hence M is also a
model of ZF+V=L in which L|=X fails.
Since GCH is a consequence of V=L,
(c') ZF + GCH |- L|= X
is also equivalent.
| How about
|ZF+(~GCH), can that also prove things about L?
Since ~GCH is so weak, I don't think there are many theorems about
L in ZF+(~GCH) that aren't also theorems in ZF. We can deduce from
~GCH the existence of sets not in L, of course.
| Do the truths of L
|vary depending on what other axioms L is defined with? I'm trying to
|be more clear, but I apologize in advance if I failed again.
How would whether something is true of L depend on axioms?
Of course, what can be proven to be true of L from a set of axioms
does depend on the axioms! I'm not familiar with the details, but I
gather that some large cardinal axioms (I think the existence of a
measurable cardinal is strong enough) imply a relatively detailed
account of the structure of L. Note that the existence of a
measurable cardinal is inconsistent with V=L.
This real 0# that is believed to be outside of L encodes some of the
structure of L (if it exists).
Keith Ramsay
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