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Topic: Don't get Axiom of Choice?
Replies: 55   Last Post: Oct 1, 2009 1:52 PM

 Messages: [ Previous | Next ]
 Keith Ramsay Posts: 1,745 Registered: 12/6/04
Re: Don't get Axiom of Choice?
Posted: Sep 29, 2009 4:24 AM

On Sep 27, 6:03 pm, Arturo Magidin <magi...@member.ams.org> wrote:
|While Goedel was explicit in his model construction (from the
|assumption that ZF is consistent, he takes a model for ZF and
|constructs an "internal" model in which ZF+C holds), forcing works
|differently; I'm not sure it would be accurate to say Cohen "found an
|interpretation in which AC is false" in the same sense as Goedel's
|work does.

Goedel's result does give an obvious interpretation
(being true in his class L) which makes AC true.
Cohen's proof is not so obvious about it, but could
be construed as also giving an interpretation in
which it is false.

The tricky thing about the operation in his proof is
that he adjoins a "generic" set to a specific model
(the minimal model, assuming there is one), which
doesn't give us an explicit such set. But the
relation of "forcing" which he defines gives us an
interpretation of sentences (making them true if
they are true for any such model, whatever the
generic set was). It's less of an interpretation in
the sense that ~A being forced is not the same as
A not being forced, while ~A being true in L is just
the same as A not being true in L.

There's a more recent style of forcing proof, which
uses what are called Boolean-valued models. Given
a Boolean-valued model, and a sentence, the sentence
has a value which is some element of a Boolean
algebra (and can be other than 1=true or 0=false).
Cohen's construction in effect supplies a value to
each sentence.

I've wondered now and then whether there was a
construction that worked in the opposite direction.
Despite what I wrote in another message, the way
AC was proved consistent was by trimming down the
universe to something small enough to make AC true.
model in which it is true an element that breaks it.
But our usual intuition is that AC is made true by
allowing the universe to be big enough to include
all those choice sets and so on. I've wondered if
AC fails and somehow fluff it up into one in which
AC holds.

Cohen mentions in relation to his proof the issue
that in order to be able to add a set, in the way
that he does, one has to have a universe that doesn't
already contain all the possibilities. (What he
adds is in principle an actual set that happened
not to be in the minimal mode.) Perhaps one wants
to trim down the universe in which AC fails maybe
to some equivalent countable universe first, or
maybe just define some idealized interpretation
of set existence which makes AC true anyway.

My vague thought was to make the sets of the new
universe be something like imagined maximum elements
of partially ordered sets satisfying the conditions
of Zorn's lemma. But I don't know if such a
construction works out at all.

Keith Ramsay

Date Subject Author
9/26/09 anon5874@yahoo.com
9/26/09 William Elliot
9/26/09 Christopher J. Henrich
9/26/09 fishfry
9/26/09 Peter Webb
9/26/09 Martin Eisenberg
9/27/09 fishfry
9/27/09 Virgil
9/27/09 Stephen J. Herschkorn
9/27/09 Virgil
9/27/09 Arturo Magidin
9/27/09 Arturo Magidin
9/27/09 Virgil
9/27/09 Peter Webb
9/27/09 Tim Little
9/26/09 David Bernier
9/26/09 anon5874@yahoo.com
9/26/09 Tim Little
9/26/09 Ted Hwa
9/26/09 Brian Chandler
9/26/09 Stuart M Newberger
9/26/09 Peter Webb
9/26/09 anon5874@yahoo.com
9/26/09 Arturo Magidin
9/26/09 Arturo Magidin
9/26/09 Arturo Magidin
9/26/09 markwh04@yahoo.com
9/26/09 Arturo Magidin
9/27/09 LauLuna
9/27/09 Arturo Magidin
9/27/09 markwh04@yahoo.com
9/27/09 Arturo Magidin
9/28/09 Jon Slaughter
9/28/09 Peter Webb
9/28/09 Tim Little
9/28/09 Frederick Williams
9/28/09 Arturo Magidin
9/29/09 Keith Ramsay
9/29/09 T.H. Ray
9/29/09 Keith Ramsay
9/29/09 Peter Webb
9/29/09 Aatu Koskensilta
9/29/09 Alan Morgan
9/30/09 Michael Stemper
9/30/09 A N Niel
9/30/09 Michael Stemper
9/30/09 FredJeffries@gmail.com
10/1/09 Peter Webb
10/1/09 David Bernier
10/1/09 Michael Stemper
9/30/09 Tim Little
9/30/09 Stephen J. Herschkorn
9/30/09 Dave L. Renfro
9/30/09 Robert Israel
10/1/09 Dave L. Renfro
9/29/09 Keith Ramsay