Re: Countable models of ZFC

george <greeneg@xxxxxxxxxx> writes:

On Oct 1, 12:27 pm, Alan Smaill <sma...@xxxxxxxxxxxxxxxx> wrote:
A membership relation cannot be a model on its own.

I'm sorry, you're just lying.

obviously!

Or you're just moot, since NO relation EVER IS
"on its own" in ZFC; in ZFC, EVERY relation HAS
ONE domain AND ONE range.

Right -- a model (for a theory with just a single 2-place
relation) therefore consists of a set, together with
a relation over that set.

So IF you HAVE
the relation (no matter how "on its own" you may think
that may be), THEN you have its domain. You do NOT
need to SPECIFY some set as the domain

the definition of model asks for that set to be given explicitly (of
course this is just nit-picking, but let's get the definitions right).

UNLESS
THE RELATION YOU ARE STARTING WITH is itself
NOT a set, BUT RATHER, A PREDICATE, like epsilon.

I don't follow;
predicates are syntactic entities that denote relations
in the usual view of model theory, so they cannot be the
same thing.

If the outer model is non-standard then
restricting its nonstandard e to some submodel is not
likely to cure the non-standardness.

There are other ways of looking for standard models than as
substructures of non-standard ones --
look up Mostowski collapse.

That is NOT the point!

see below.

The point IS that Rupert
and AK *BOTH* alleged that it was SUFFICIENT to
create a standard model to MERELY restrict e to A!
Or did you not see where AK wrote,
"A standard model is always isomorphic to a model
form <A,e> where e is the membership relation on A."

No, that was not the allegation (Mostowski collapse was
already mentioned IIRC).

It is that given a set N and a relation R (not membership), where
<N,R> satisfies the appropriate conditions, there is a *different* set
A such that <A,e> is isomorphic to <N,R> , where e is the membership
relation.

Or it it?
Is it a theorem that a non-standard model cannot have
a non-standard submodel?

It is known that ZF cannot prove that if there is a model
of ZF, then there is a standard model of ZF

Of course it is; Rupert correctly began by saying that the latter
was a stronger assumption. But YOUR statement here
could be the result of a standard
model having only non-standard submodels.

in my terminology, any submodel of a standard model is standard.

--
Alan Smaill
.

Relevant Pages

  • Re: Countable models of ZFC
    ... standard model for ZFC, that is, a model where the membership relation ... "The membership relation" IS the model of ZFC ... Is it a theorem that a non-standard model cannot have ...
    (sci.logic)
  • Re: Countable models of ZFC
    ... "on its own" in ZFC; ... create a standard model to MERELY restrict e to A! ... Is it a theorem that a non-standard model cannot have ...
    (sci.logic)
  • Re: Countable models of ZFC
    ... "The membership relation" IS the model of ZFC ... Is it a theorem that a non-standard model cannot have ... then there is a standard model of ZF ...
    (sci.logic)
  • Re: Countable models of ZFC
    ... As I understand it, the constructible universe L is constructed in stages, ... L_kappa is a model of ZFC. ... believing that kappa must be countable, but I'm not sure how to prove ... standard model for ZFC, that is, a model where the membership relation ...
    (sci.logic)
  • Re: Continuum hypothesis
    ... The third-order language of arithmetic. ... going to be ANY DIFFERENT from the UNIQUE standard model ... At 2nd-order, however, THE AXIOMS ... No, it wouldn't, BECAUSE THE CONTEXT IS ZFC. ...
    (sci.logic)
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