Re: Countable models of ZFC

On Oct 3, 9:34 pm, george <gree...@xxxxxxxxxx> wrote:
On Sep 26, 5:35 am, Rupert <rupertmccal...@xxxxxxxxx> wrote:

I'm sorry you had so much trouble understanding it. It seems pretty
straightforward to me. Given a set M,

Well, WHICH are we being given?? A SET?? OR A MODEL??

the standard membership relation
on M is {(x,y):x, y in M and x in y}.

Up to this point, M is a set.

So a model (M,

And THERE is the problem.
Now, SUDDENLY, M is a model INSTEAD OF a set.
Or rather, M is the domain of a model.


Of course every domain of a model is a set. That is not a problem.

If M is both a set AND the domain of a model
then IN ADDITION to the model (M,whatever),
THERE MUST EXIST ALSO
the model OF SOME SET THEORY in which M is a set,
and that model must have a domain of which M is a member.
THAT MODEL ALSO has a membership relation,
and THAT MODEL ALSO might be standard OR NONstandard.

No. This is wrong.

For me, models are sets. I don't like talking about proper classes.
When I say "M is a set", I don't mean it's a set living in some model,
I just mean it's a set, period.

Some people might want to use a theory which allows proper classes and
then the universe counts as a model too. Of course, this model is
standard (just as the standard kilogram weighs one kilogram).

You have to be able to make sense of "x is a member of y" all by
itself, not just "x is a member of y in such-and-such a model". There
has to be such a notion available, or your semantics will never get
off the ground.

IF THAT model was nonstandard then merely restricting its membership
relation to some submodel IS NOT going to guarantee that that submodel
(in this case, (M,E)) is standard. SOME nonstandard models DO have
nonstandard submodels.

If you're going to take the view that you can never understand any
sentence unless you've specified what model it's relativized to,
you're going to tie yourself in knots.

.

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