From: kramsay@aol.com (KRamsay)
Subject: Re: Zermelo set theory
Date: 19 Jan 1999 06:16:38 GMT
Newsgroups: sci.math
Keywords: Non-ZFC models of Set Theory

In article <2gzp7gul54.fsf@hera.wku.edu>,
Allen Adler <adler@hera.wku.edu> writes:
|Where can I find a construction of a model of Zermelo set theory
|which is not a model of Zermel-Frankel set theory?

I don't know a reference offhand, but the sets of rank <w+w are such
a model. One way to think of the difference between ZF and Z is that
ZF requires that the levels of the cumulative hierarchy "keep going"
and include all ordinals (considered as order types of well-ordered
sets). There's nothing in Z to require sets exist which are more than
finitely many ranks above the set of integers, which is required to be
present by the axiom of infinity. One can define a relation nRx which
holds iff n is a natural number and x is the set of sets of rank <w+n.
In Z, you can prove that for each n there exists a unique x such that
nRx, but you can't prove that {x: nRx for some n} exists.

In this model, the von Neumann ordinals, each represented as the set
of all previous ordinals, only go up to w+w, while of course one can
prove in Z the existence of well-orderings of higher order type.

Keith Ramsay
==============================================================================

From: kramsay@aol.com (KRamsay)
Subject: Re: Zermelo set theory
Date: 20 Jan 1999 21:47:10 GMT
Newsgroups: sci.math

In article <2gyamz10sj.fsf@hera.wku.edu>,
Allen Adler <adler@hera.wku.edu> writes:
|Tal Kubo and Keith Ramsay have given the example of the set of
|all sets of rank less than $\omega+\omega$, in which it is alleged
|that one can't prove that $\omega+\omega$ is well ordered.

It's a common question of a formal system to ask what's the smallest
ordinal which it can't prove exists. First-order Peano arithemetic,
for instance, can't show epsilon_0 is well-ordered, and from epsilon_0
being well-ordered is enough to show PA is formally consistent. We
wouldn't want to leave the impression that Z is weaker than PA in this
regard, however!

The problem is that while one can prove in Z that there is a well
ordering of order type w+w (e.g. nRm iff n<m and n-m is even or n is
even and m is odd defines a well-ordering on the natural numbers),
it's not necessarily representable as a von Neumann ordinal, the set
of its predecessors. The von Neumann ordinals are 0={}, 1={0}={{}},
2={0,1}={{},{{}}}, etc., and because of the axiom of infinity one can
prove in Z that w={0,1,2,3,...}, w+1={0,1,2,...,w}, and in general w+k
exist for natural numbers k. But one can't prove in Z that w+w defined
in the same way, {0,1,2,...,w,w+1,w+2...} exists. It has rank w+w, as
in general von Neumann ordinals are their own ranks. It's another
failure of replacement; the relation between k and w+k can be defined,
and one can prove that for each k there exists a unique w+k, but since
one doesn't have replacement or collection, it doesn't immediately
follow that {w+k: k is an integer} exists. And since the sets of rank
<w+w are a model of Z in which it doesn't exist, one can't prove it
exists in Z.

That's why although in ZF it's customary to identify an ordinal with
its von Neumann representation, one has to be a little careful in Z.
One can use (Scott's?) trick of representing an isomorphism class as
the set of all representatives of minimal rank, in this case the set
of all order-isomorphic well-ordered sets having the least rank. By
that definition there would be an omega+omega and much higher ordinals
too, laid out "horizontally", but not the "vertically" arranged von
Neumann version of the same thing.

|Two questions:
|(1) Where can I find a detailed account of Zermelo set theory
|    and its models? 

I don't know. What I've read of it has all been in discussions of why
one has the axiom of replacement in ZF.

|(2) Does there exist a model of Zermelo set theory which satisfies
|    the axiom of choice but which is not a model of Zermelo-Frankel
|    set theory?

If the axiom of choice is true, then the sets of rank <w+w satisfy
the axiom of choice. Without assuming the axiom of choice, take any
model of ZF which does satisfy the axiom of choice (like Goedel's L
of some model) and take the submodel of sets which have rank <w+w
relative to that model.

Keith Ramsay

==============================================================================

From: ikastan@sol.uucp (ilias kastanas 08-14-90)
Subject: Re: Zermelo set theory
Date: 21 Jan 1999 17:11:55 GMT
Newsgroups: sci.math

In article <2giue1ckka.fsf@hera.wku.edu>,
Allen Adler  <adler@hera.wku.edu> wrote:
@kramsay@aol.com (KRamsay) writes:
@
@> In article <2gyamz10sj.fsf@hera.wku.edu>,
@> Allen Adler <adler@hera.wku.edu> writes:
@
@> |(1) Where can I find a detailed account of Zermelo set theory
@> |    and its models? 
@> 
@> I don't know. What I've read of it has all been in discussions of why
@> one has the axiom of replacement in ZF.
@
@After thinking about it a little, it occurred to me that people who
@work with "subsystems of analysis" (something I know very little about)
@work with models of set theory with very restricted versions of the
@replacement axiom, e.g. the replacement axiom schema holds for formulas
@satisfying some restricted condition with respect to the Levy hierarchy,
@e.g. Delta_1^2 or something like that. So maybe they talk about Zermelo
@set theory somewhere? I just tried math-sci by looking under titles
@containing "subsystems of analysis" but didn't find anything that looked
@like it might discuss Zermelo set theory systematically. Maybe someone
@else can do better?



	Subsystems of analysis restrict Comprehension (and Choice)... yes,
to Pi-1-1, Delta-1-2 or so.  The other axioms (on top of PA) are Extensionality
and full (set-variable) Induction...  all in all rather weak as a set theory!
(Which it wasn't meant to be; its aim is to study not "sets" but "sets of
reals".)  Note that it's a two-sorted theory; "model" means an  M  (|= PA)
_and_ a choice of some family of subsets of  M  to interpret the set variables.

	A typical "weakness" of  Z  is:  it cannot prove P,  P = "every
wellordering is isomorphic to an ordinal"  (because P is false in V_w+w --
all wellorderings of  w  are there, including those of type  w+w, but  w+w
is not -- and  V_w+w |= Z).  On the other hand, poor Analysis cannot even
formulate the notion "ordinal"!

	Anyway...  I posted a more detailed discussion in sci.logic a couple
of months ago  (thread subject had "second-order arithmetic" in it, I think). 


@> |(2) Does there exist a model of Zermelo set theory which satisfies
@> |    the axiom of choice but which is not a model of Zermelo-Frankel
@> |    set theory?
@> 
@> If the axiom of choice is true, then the sets of rank <w+w satisfy
@> the axiom of choice. Without assuming the axiom of choice, take any
@> model of ZF which does satisfy the axiom of choice (like Goedel's L
@> of some model) and take the submodel of sets which have rank <w+w
@> relative to that model.
@
@This looks neat, but let me make sure I understand it. Suppose one has
@an element A of L_{omega+omega} whose elements are pairwise disjoint.
@Let B be an element of L which meets each element of A in a singleton.
@Let C be the union of A. Since the union axiom holds in Zermelo set theory,
@C also belongs to L_{omega+omega}, hence to L_alpha for some
@alpha < omega+omega. Therefore every constructible subset of C, in
@particular B, belongs to L_{alpha+1}, hence to L_{omega+omega}.
@So L_{omega+omega} satisfies choice. Is this argument correct?
@
@Naively, it looks as though the same argument shows that if V is a model
@of ZFC, V_lambda is a model of Zermelo set theory with choice for every
@limit ordinal lambda greater than omega. Is that correct? Or does one
@have to make additional assumptions, such as that the cardinailty of
@lambda is a regular cardinal?



	Lambda limit and > w is correct.

	E.g.  ZF (or any consistent extension thereof) is not finitely
axiomatizable over  Z;  an easy proof is to show that for any statement  Q
consistent with  Z, there is a Q' that is a theorem of  ZF + Q  but not of
Z + Q.  Sure enough,  Q' is  "there is a limit ordinal lambda, > w, such
that  Q^V_lambda"   (Q relativized to V_lambda).

	(Q' = "Consis(Z + Q)" would also do...  but the above seems simpler).


	Roughly speaking,  PA (essentially, Z(F) - Inf) "is about"  V_w, the
hereditarily finite sets;  Analysis is about the hereditarily countable sets;
Z is about  V_w+w.



						Ilias