Prev: found the 10% in Geometry; Poincare dodecahedral space; 36 degree twist; matched-circles #378 Correcting Math
Next: The set theoretic content of the transfinite recursion theorem
From: Marc Alcobé García on 3 Feb 2010 16:18
> I think that's right, except in some cases, ZF proves they're not
> sets? For example, it is proven in ZF that there is no set of all the
> sets that are not members of themselves.

I do not know what's your point here and in the rest of your post.

I mean, ok, V is a proper class, the class of all sets, and we have a
variety of equivalent formulas to refer to them (x=x, or for example x
is well-founded, if we are in ZF). What can we learn from this? I
neither do understand your 'technical' V = 0. what does that mean?

Shoenfield and Levy agree in saying that set theory is but an
investigation of what is meant by the "and so on" of this particular
collection of individuals:
{}, {{}}, {{},{{}}}, and so on. This is exactly what I have in mind
when I think of V. However it is clear that it does not seem to
satisfy the needs of first order semantics to be a model of anything.


From: MoeBlee on 3 Feb 2010 16:45
On Feb 3, 3:18 pm, Marc Alcobé García <malc...(a)gmail.com> wrote:
> > I think that's right, except in some cases, ZF proves they're not
> > sets? For example, it is proven in ZF that there is no set of all the
> > sets that are not members of themselves.
>
> I do not know what's your point here

You said, "ZF cannot prove the classes M, and E, to be sets, being
such
statements undecidable."

My point is just that there may be even decidable cases.

> I mean, ok, V is a proper class, the class of all sets, and we have a
> variety of equivalent formulas to refer to them (x=x, or for example x
> is well-founded, if we are in ZF). What can we learn from this? I
> neither do understand your 'technical' V = 0. what does that mean?

When we use the Fregean method for dealing with improperly referring
terms (which results from failed instances of conditional
definitions), we usually set improperly referring terms to 0. If you
are unfamiliar with this, you can find discussion of it in set theory
books by Suppes, Quine, Bernays, and many more.

> Shoenfield and Levy agree in saying that set theory is but an
> investigation of what is meant by the "and so on" of this particular
> collection of individuals:
> {}, {{}}, {{},{{}}}, and so on. This is exactly what I have in mind
> when I think of V. However it is clear that it does not seem to
> satisfy the needs of first order semantics to be a model of anything.

Yes, and those authors never claimed it to be the universe of a
model.

There is a distinction, on the one hand, of the NOTION of a set as
being something that is in V, and on the other hand, the technical
matter of a universe for a model of first order set theory.

MoeBlee

From: Marc Alcobé García on 4 Feb 2010 10:53
> My point is just that there may be even decidable cases.

And these must be those where M, and E, are proper classes: so not a
model in the sense of
first order logic. Or not?

> Also, it might help to be clear where the relativization method is
> being used not to prove consistency itself, but rather RELATIVE
> consistency (i.e., theorems of the form "If theory T is consistent
> then theory T* is consistent").

This is interesting. How could such a distinction be made?

> When we use the Fregean method for dealing with improperly referring
> terms (which results from failed instances of conditional
> definitions), we usually set improperly referring terms to 0. If you
> are unfamiliar with this, you can find discussion of it in set theory
> books by Suppes, Quine, Bernays, and many more.

I think this is also standard, but it seems rather odd to think that
the class term {x | x = x} refers to the empty set.

> Yes, and those authors never claimed it to be the universe of a
> model.

> There is a distinction, on the one hand, of the NOTION of a set as
> being something that is in V, and on the other hand, the technical
> matter of a universe for a model of first order set theory.

The problem is that we use the same language to refer to:

i. The 'real' sets of V
ii. The sets of a model of ZF in the sense of model of a first order
theory.

and that our metatheory seems to apply only for objects of the kind of
ii.
From: MoeBlee on 4 Feb 2010 13:52
On Feb 4, 9:53 am, Marc Alcobé García <malc...(a)gmail.com> wrote:
> > My point is just that there may be even decidable cases.
>
> And these must be those where M, and E, are proper classes: so not a
> model in the sense of
> first order logic. Or not?

Yes, I thought youre point was that in ZF in certain cases we couldn't
conclude (it is undecidable) whether certain classes are proper
classes or sets. My point is that also in certain cases, ZF does
decide that certain classes are proper. (Note: ZF proves there are no
proper classes anyway, so my point just mentioned is couched in ZF in
a different way, e.g., the ZF theorem ~ExAy(yex <-> ~yey).)

> > Also, it might help to be clear where the relativization method is
> > being used not to prove consistency itself, but rather RELATIVE
> > consistency (i.e., theorems of the form "If theory T is consistent
> > then theory T* is consistent").
>
> This is interesting. How could such a distinction be made?

I'm not sure I understand your question.

For example, the following are two different kinds of statements (the
first is an assertion of consistency, and the second is an assertion
of relative consistency):

(1) ZF is consistent.

(2) If ZF is consistent then ZFC is consistent.

> > When we use the Fregean method for dealing with improperly referring
> > terms (which results from failed instances of conditional
> > definitions), we usually set improperly referring terms to 0. If you
> > are unfamiliar with this, you can find discussion of it in set theory
> > books by Suppes, Quine, Bernays, and many more.
>
> I think this is also standard, but it seems rather odd to think that
> the class term {x | x = x} refers to the empty set.

It seems odd unless one looks at the particular technical mechanics of
it.

> > Yes, and those authors never claimed it to be the universe of a
> > model.
> > There is a distinction, on the one hand, of the NOTION of a set as
> > being something that is in V, and on the other hand, the technical
> > matter of a universe for a model of first order set theory.
>
> The problem is that we use the same language to refer to:
>
> i. The 'real' sets of V
> ii. The sets of a model of ZF in the sense of model of a first order
> theory.
>
> and that our metatheory seems to apply only for objects of the kind of
> ii.

I agree that much of the informal discussion (rubric such as 'the real
universe of sets') about set theory can be difficult. But in the
meantime, I think we can be confident that writers on the subject are
usually diligent enough not to claim, e.g., that the cumulative
hierarchy is a universe for a model of the first order theory of ZF,
where 'model' is in the specific technical sense.

MoeBlee

From: Marc Alcobé García on 5 Feb 2010 07:56
> I'm not sure I understand your question.

You said:

> it might help to be clear where the relativization method is
> being used not to prove consistency itself, but rather RELATIVE
> consistency

and also gave an example of statements of consistency and relative
consistency.

My question is: when is a consistency proof using relativization not a
relative consistency proof? Can these kinds of proof be distinguished?
Is it when the relativization is made by means of a class that happens
to be provably a set?
First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4 5 6 7 8 9
Prev: found the 10% in Geometry; Poincare dodecahedral space; 36 degree twist; matched-circles #378 Correcting Math
Next: The set theoretic content of the transfinite recursion theorem