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

From: Mike Oliver (mike_lists_at_verizon.net)
Date: 02/10/05 

Date: Thu, 10 Feb 2005 09:49:51 -0600

Ralph Hartley wrote:
> Mike Oliver wrote:
>
>> OK, let me give a candidate for a property that ought to hold
>> of a non-pathological model: It ought to contain all the reals.
>> E.g. if M is non-pathological, then there shouldn't be any
>> other model N that knows about a real number that M doesn't
>> know about. Otherwise M might think there's a wellorder of
>> the reals whose every proper initial segment is countable, only
>> for the silly reason that M doesn't know about all the reals.
>
>
> I think that *might* be reasonable, but I'm not sure. I'm certainly
> willing to look at what follows from making that requirement.
>
>> (Again, we have an issue with cross-model identification, which
>> I'll address in the following way: Consider a real to be a set
>> of naturals. There's an obvious identification between the standard
>> naturals of M and those of N, and if M has any nonstandard naturals
>> it's easy to see M must leave out some reals, so that's not an issue.
>
>
> It may be easy, but I don't see it.

Remember that we're considering a real to be a set of naturals. The
only way for a model to have nonstandard naturals is for it to leave
out the set of all (standard) naturals. Otherwise the model would be
able to tell that its nonstandard naturals *were* nonstandard, so
it wouldn't think they were naturals anymore.

Thus the model is missing a real, namely the set of all its standard
naturals.

>> Then N knows about a real that M doesn't, if there's an object x that
>> N thinks is a set of naturals, such that when you look at the objects
>> N believes to be in x and match them up by this natural identification,
>> M has no object it believes to contain the corresponding naturals. Sorry
>> for the verbiage--once you unravel the concept you'll see it's actually
>> simple and natural, just hard to say concisely.)
>
>
> Perhaps so, but I had a great deal of difficulty even parsing it. Are
> you trying to prove something, or are you trying to define what you mean
> by the phrase "N knows about a real that M doesn't".

The latter. It's really just a detail.

>> Now, here's the question: Can two non-pathological models, in this
>> sense, differ on the truth value of CH?
>
>
> Lets see. Suppose one model has aleph_n reals with n>1 (CH false), and
> another only has c=aleph_1 (CH true). Then (unless they disagree about
> how big the alephs are, or on how to compare cardinals etc.) the second
> model must be pathological. So if CH is false for any model
> (pathological or not) it should be false for any non-pathological model
> (that agrees with the first model on some other stuff).

Remember the models both have *all* the reals, so they can't have different
reals. The only way they can disagree is by disagreeing on how big
the alephs are and how to compare cardinals. That is, one of them has
an object of type *higher* than the reals, that the other model lacks.

The next step up from the reals is sets of reals. Any relation on
the reals can be coded by a set of reals (because pairs of reals can
be coded by reals).

So, say one of the models, M for example, thinks that CH is true. That
means M has a wellordering of the reals whose every proper initial segment is
countable. That wellordering can be coded by a set of reals (call it S).

Now if N, or any model containing all the reals, thinks CH is *false*,
it can only be because N leaves out S. If S were an element of N,
N would be able to decode it and see that there's a wellordering of
the reals whose every proper initial segment is countable. Oh, you
have to check that N thinks S codes a wellordering and that it thinks
all the proper initial segments are countable, but this isn't hard
given that M and N each have all the reals. (Since M has all the
reals it has to be right about these two facts, and since N has
all the reals, it correctly understands that they're true of S.)

So if N thinks CH is false, it's only because N has left out a set
of reals (if this scenario is the real one). Therefore N is just wrong.

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