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From: Marc Alcobé García on 2 Feb 2010 08:51
Those of you who are contributing to my posts here (for which I am
very grateful) may know by now that I am having some trouble to
obtain, let us call it a "unified picture" of the contents of Levi's
"Basic Set Theory", Kunen's "Set theory. An introduction to
independence proofs" and Shoenfield's "Mathematical Logic". There are
so many concerns I have about this that I hardly know what to ask in
order to disentangle my particular web of doubts.

I have often asked for different literature, but I believe the problem
is not in the literature: I must be thinking things are far more
complicated than they really are.

In Shoenfield, models of first-order theories are sets, but generally
models of set theory are given by means of a relative interpretation
of set theory into itself, i. e. by means of two formulas in the first
order language of set theory M(x, v) and E(x, y, v), with no free
variables other than the ones shown. The first one defining a class of
sets {x | M(x,v)} and the second one a relation {<x,y> | E(x,y,v)},
both with parameter v.

So, in what sense does the picture of models used in Shoenfield to
prove theorems about first order theories and their models fits the
picture used to buid up (for example, transitive) models of set
theory? Am I making a mess of things that have no relation to each
other?
From: LauLuna on 2 Feb 2010 18:48
On Feb 2, 2:51 pm, Marc Alcobé García <malc...(a)gmail.com> wrote:
> Those of you who are contributing to my posts here (for which I am
> very grateful) may know by now that I am having some trouble to
> obtain, let us call it a "unified picture" of the contents of Levi's
> "Basic Set Theory", Kunen's "Set theory. An introduction to
> independence proofs" and Shoenfield's "Mathematical Logic". There are
> so many concerns I have about this that I hardly know what to ask in
> order to disentangle my particular web of doubts.
>
> I have often asked for different literature, but I believe the problem
> is not in the literature: I must be thinking things are far more
> complicated than they really are.
>
> In Shoenfield, models of first-order theories are sets, but generally
> models of set theory are given by means of a relative interpretation
> of set theory into itself, i. e. by means of two formulas in the first
> order language of set theory M(x, v) and E(x, y, v), with no free
> variables other than the ones shown. The first one defining a class of
> sets {x | M(x,v)} and the second one a relation {<x,y> | E(x,y,v)},
> both with parameter v.
>
> So, in what sense does the picture of models used in Shoenfield to
> prove theorems about first order theories and their models fits the
> picture used to buid up (for example, transitive) models of set
> theory? Am I making a mess of things that have no relation to each
> other?

No, Marc, you are not entirely confused, though, of course, whatever
model theory proves for first order theories applies to first order
set theory as well.

As I see it, there are two circumstances that render the topic of
models of set theory a bit slippery.

Firstly, model theory requires models to have a domain which must be a
non empty set while the usual set theoretic universe (the cumulative
hierarchy) is no set.

Secondly, we can only construct a model for set theory by means of a
theory stronger than that for which we provide the model; otherwise
set theory would be able to prove its own consistency and then it
would have no model at all (by Gödel's second theorem). This wouldn't
be a problem if we entertained absolutely no doubt that set theory,
and whatever is required to build a model of it, is consistent; if,
let's say, set theory were as obviously a consistent theory as Peano
arithmetic is. But, as you know, many people find the case of set
theory much more problematic...

From: Marc Alcobé García on 3 Feb 2010 11:56
> [...] whatever
> model theory proves for first order theories applies to first order
> set theory as well.

Therefore, if we think models of first order theories are always sets
(please correct me if I am wrong) and assuming consistency of ZF:

1. Completeness of first order logic implies that, since the axioms of
ZF must have a model, then in the set theoretic universe V there must
exist a set m and a relation e (which is also a set, of ordered pairs,
but not necessarily the restriction of membership to m) such that the
axioms of ZF are true in m with e. Thus not entailing that V itself is
a set (which we know it is not because so ZF tells us).

2. Gödels second theorem implies that ZF cannot prove the existence of
such sets e and m. Therefore, if some classes M and E are found wich
are a relative interpretation of set theory into itself that make true
(provably within ZF) the relativized
axioms of ZF, then either:

a) M, and E, must be proper classes: so not a model in the sense of
first order logic, or
b) ZF cannot prove the classes M, and E, to be sets, being such
statements undecidable.

I do not know if these ideas are correct (for example, I do not know
if the constructible universe L satisfies any of these conditions) but
they seem to give a sort of "solution" to your points:

> Firstly, model theory requires models to have a domain which must be a
> non empty set while the usual set theoretic universe (the cumulative
> hierarchy) is no set.

> Secondly, we can only construct a model for set theory by means of a
> theory stronger than that for which we provide the model; otherwise
> set theory would be able to prove its own consistency and then it
> would have no model at all (by Gödel's second theorem).

However, some questions remain open, the main one being: What is
exactly V?
From: MoeBlee on 3 Feb 2010 12:21
On Feb 2, 5:48 pm, LauLuna <laureanol...(a)yahoo.es> wrote:

> Firstly, model theory requires models to have a domain which must be a
> non empty set while the usual set theoretic universe (the cumulative
> hierarchy) is no set.

That is often a conceptual universe for set theory, but, as far as I
know (and I welcome being corrected or further informed), still to
prove consistency by means of proving the existence of a model does
require proving the existence of model in the sense of one with a
universe that is a set. As far as I know (and I welcome being
corrected or further informed), actual technical proofs of consistency
that are couched in terms of proper classes can be reformulated to
show that there is indeed a model in the ordinary sense of having a
universe that is a set.

> Secondly, we can only construct a model for set theory by means of a
> theory stronger [...]

Stronger or not comparable in strength. I.e., if T is a consistent
theory that is prone to the second incompleteness theorem, then
neither T nor any weaker theory proves the consistency of T. But there
are stronger theories and ALSO theories that are neither stronger,
weaker, nor the same as T that do prove the consistency of T.

MoeBlee
From: MoeBlee on 3 Feb 2010 12:51
On Feb 3, 10:56 am, Marc Alcobé García <malc...(a)gmail.com> wrote:

> [if ZF is consistent then]
> ZF must have a model, then in the set theoretic universe V there must
> exist a set m and a relation e (which is also a set, of ordered pairs,
> but not necessarily the restriction of membership to m) such that the
> axioms of ZF are true in m with e. Thus not entailing that V itself is
> a set (which we know it is not because so ZF tells us).

I think that is correct.

> 2. Gödels second theorem implies that ZF cannot prove the existence of
> such sets e and m. Therefore, if some classes M and E are found wich
> are a relative interpretation of set theory into itself that make true
> (provably within ZF) the relativized
> axioms of ZF, then either:
>
>   a) M, and E, must be proper classes: so not a model in the sense of
> first order logic, or
>   b) ZF cannot prove the classes M, and E, to be sets, being such
> statements undecidable.

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.

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").

>What is exactly V?

In Z (and its offshoots such as ZF), we have a theorem:

~ExAy(y=y -> yex).

Now, if we then use abstraction notation to write:

{y | y=y}

and define a constant 'V' by

V = {y | y=y}

then we find that the set abstraction "term" does not properly refer
and so the definition of 'V' is not proper.

Therefore, how one takes the term 'V' depends on ones handling of
improperly referring terms. I (and a good number of authors) prefer to
use the Fregean method, so that, TECHNICALLY, V = 0.

However, we also recognize that we may refer to such notation as 'V'
and {y | y=y} as either shorthand (cf., e.g., Suppes's 'Axiomatic Set
Theory'; also, if I recall correctly, Quine; Bernays too, though with
some difference in context), or (by more intricate syntactical
methods, cf., Takeuti and Zaring; also, basically related in this way
is Levy's 'Basic Set Theory') as defined terminology that reduces to
META-language about FORMULAS of Z set theory. In that sense,
discussion pertaining to Z set theory about, e.g., 'V' reduces (either
by convention or by syntactical manipulation) to disscussion about a
formula, such as 'y=y'.

But in, say Bernays class theory (and its variants), we prove:

E!xAy(yex <-> y is a set)

And so in such a theory, we are completely justified in defining:

V = x <-> Ay(yex <-> y is a set).

And so, in such theories V is indeed the class of all sets.

MoeBlee


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