Re: THIS STATEMENT HAS NO PROOF IN ANY SYSTEM = true or false?

From: Mike Oliver (mike_lists_at_verizon.net)
Date: 01/26/05 

Date: Tue, 25 Jan 2005 20:43:35 -0600

poopdeville@gmail.com wrote:
> Mike Oliver wrote:
> It's lazily phrased and not particularly important, but I'll get to it
> below. (OK, another attempt at re-wording: "There demonstrably exists
> models of ZF (whose elements are sets, but I suspect this will be
> contentious) in which AC fails.")

This looks promising; I may be able to get my point across just from
here.

Let's take a model of ZF whose elements are uncontroversially
sets, in which AC fails. The model I have in mind is L(R). It's
a proper class model, but that's easily gotten around if you like
(at the cost of weakening ZF slightly, or strengthening the metatheory
slightly). L(R) is the class of all sets that you get by starting
with R at the bottom, and then doing the L construction. L(R)
is a *transitive* model, so the epsilon relation means exactly what
it does out in V--that saves us from doing all these translations
("M-extensions") that I had to deal with in a previous post.

Now (assuming large cardinals, but again we can get around that if
you want to), there is no wellorder of R in L(R), so L(R) satisfies
~AC. (Equivalently, there is no choice function in L(R) for P(R)\{0}.)

But that doesn't mean there really *isn't* a wellorder of R! It just
never showed up in L(R), which is missing some sets. Specifically,
any relation on R, including a wellorder, can be coded by a set
of reals (work this out for yourself), so L(R) is missing any set
of reals that codes a wellorder of R.

Now, the statement "AC is true" just means that this is what's
always going to happen--whenever you have a model whose elements
are sets, in which some set cannot be wellordered, it simply
means that the relevant wellorder is not an element of the
model.

I believe this should take care of all the points you raised in the
rest of your post, so I'm snipping. Let me know if you feel it
doesn't answer some of them.

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