Rupert
Posts:
3,797
Registered:
12/6/04
|
|
Re: another boring critisism of Cantor's Theorem
Posted:
Apr 7, 2004 12:28 AM
|
|
david_lawrence_petry@yahoo.com (David Petry) wrote in message news:<25bac3c0.0404061214.47821ef@posting.google.com>...
> hrubin@odds.stat.purdue.edu (Herman Rubin) wrote
>
> > In article <25bac3c0.0404051304.4fc07cf1@posting.google.com>,
> > David Petry <david_lawrence_petry@yahoo.com> wrote:
> > >rupertmccallum@yahoo.com (Rupert) wrote
> >
> > <> It's not
> > <> consistent with ZFC+Con(ZFC) that every model of ZFC is countable. If
> > <> ZFC is consistent, it has models of arbitrarily large cardinality by
> > <> the Loewenheim-Skolem theorem.
> >
> > >OK. But it's consistent with ZFC+Con(ZFC)+Con(ZFC+Con(ZFC))
> > >that ZFC+Con(ZFC) has a countable model. Then the models of
> > >of ZFC of arbitrarily large cardinality are objects that live
> > >in the possibly countable model of ZFC+Con(ZFC), and hence
> > >it consistent that they are really and truly countable, which
> > >is what I was getting at in the first place. I'm not sure if
> > >it's wrong to say that is consistent with ZFC itself.
> >
> > Something can be countable in a metalanguage, without being
> > countable in the language. Do not confuse them.
>
> I'm not sure that I was confusing them. But I guess a better way
> to say what I wanted to say is that for every language, there is a
> metalanguage in which every object in the language is countable.
This doesn't really mean anything. How do you define "language", for
example? Do you mean a countable first-order language?
How about this: for every consistent theory in a countable first-order
language there is a countable model for that theory. At least that's
correct.
|
|
