From holmes@diamond.idbsu.edu Thu Oct 17 09:51 MDT 1996
Received: by diamond.idbsu.edu
(1.37.109.16/16.2) id AA034087416; Thu, 17 Oct 1996 09:50:16 -0600
Date: Thu, 17 Oct 1996 09:50:16 -0600
From: Randall Holmes
Return-Path:
To: holmes@diamond.idbsu.edu, john.harrison@cl.cam.ac.uk
Subject: Re: (no subject)
Cc: John.Harrison@cl.cam.ac.uk
Status: RO
I haven't seen a big treatment. My source for Ackermann's set theory
is Azriel Levy's article referenced at the very end of the review of the
book on Finsler set theory which is found on my Web page. That article
has further references to articles in which the equiconsistency with
ZF is shown.
My (somewhat informed) guess is that the approach would be almost
identical to an approach in ZF, except initially, where one would get
to prove things that one can't prove in ZF. The universe of sets
apparently looks exactly as in ZF, and I think the universe of classes
looks very much like more ZF!
I can't remember what I wrote to you about Ackermann's theory.
One thing about Ackermann's theory is that it might appeal to category
theorists; it has a set/class distinction, but one has a much greater
ability to do mathematical constructions with classes than in the
usual extensions of ZFC with classes. It is also elegant!
Since I can't remember what I wrote, I'll recap a little, just for fun :-)
The objects of our world are classes, on which there is a membership
relation. Some classes are sets.
Axiom I: Classes with the same elements are equal.
Axiom II: Elements of sets are sets.
Axiom III: Subclasses of sets are sets.
Axiom IV: Any condition on sets defines a class.
Axiom V: Any condition on classes which does not mention the sethood
predicate or have any non-set parameters, and which turns out to be
true only of sets, defines a set.
Axiom VI: Any class is disjoint from at least one of its elements
(foundation).
Theorem I: The empty set exists.
Proof: The condition of not existing is definable without mention of
sethood or non-set parameters and is true only of sets so defines a
set :-)
Theorem II: If X is a set, X \cup {X} is a set.
Proof: The condition "Y = X or Y \in X" does not mention sethood, has
no non-set parameter by hypothesis and is true only of sets (recall
that elements of sets are sets), so defines a set.
Theorem III: There is an infinite set.
Proof: We prove the existence of the von Neumann ordinal \omega. This
is defined by the condition "X belongs to all classes A such that A
contains the empty set and A is closed under the operation X |-> X
\cup {X}". This condition is defined without reference to sethood or
any non-set parameter. The class of all sets is closed under the
indicated operation, so the condition is true only of sets, and so
defines a set!
Theorem IV: If X is a set, the power set P(X) of X is a set.
Proof: The condition "Y is a subclass of X" does not mention sethood,
has only a set parameter, and is satisfied only by sets (since
subclasses of sets are sets by axiom).
Theorem V: The union U[A] of a set A exists.
Proof: The condition "X is an element of an element of A" does not
mention sethood or have any non-set parameter and is true only of sets
by axiom.
So we have at least Zermelo set theory. Showing that full replacement
holds seems to be a little tricky because of conditions containing
quantifiers over sets, which seem to mention sethood. Here is a
limited form of Replacement (actually Collection!), though:
Meta-Theorem: Suppose that we have a theorem "for each x in set A
there is a set y such that Pxy" where P is a condition which does not
mention sethood or have non-set parameters. Then there is a set B
such that for each x in A there is y in B such that Pxy.
Proof: The set of all pairs (x,y) such that Pxy and y is of the least
possible rank meets the usual conditions to be a set (after one does
the work to show that a class of the same ordinal rank as a set is a
set; show that the transitive closure of a set is a set, then show
that the ordinal which is its rank is a set, then show that the entire
rank is a set). One can then construct B as the range of this
set relation.
Probably there is then a nice argument to show that quantifiers over
classes in the condition P can safely be replaced by quantifiers over
sets.
(Foundation was not part of Ackermann's original theory, but one could
then relativize everything to the well-founded classes and the
well-founded sets as usual).
Examples of unexpected results:
Theorem VI: There is a non-set which is an element of a class.
Proof: Suppose otherwise. The condition of being an element does not
mention sethood or non-set parameters and by hypothesis would be
satisfied only by sets. It would thus define a set, in fact the set
of all sets. By the axiom of class comprehension, the class of all
sets which are not elements of themselves exists. It would be a
subclass of the set of all sets and so a set, which is absurd.
Definition: \Omega is the class of all set ordinals. (we mean von
Neumann ordinals here).
Observation: \Omega is a class ordinal.
Theorem VII: \Omega + 1 exists.
Proof: It is sufficient to show that \Omega is an element of some
ordinal. If \Omega were not an element of some ordinal, we could
define the class of set ordinals (\Omega itself!) by "is an ordinal
and is an element", which would clearly define a set. \Omega would
then be a set ordinal and so an element of itself, which is absurd!
And so forth. One can show the existence of lots of non-set
structure.
To interpret Ackermann set theory in ZFC, interpret the class of sets
as a particular (nonstandard) set $M$ which happens to contain all of
its elements of its elements and subsets of its elements and to be
closed under all of the instances of replacement and comprehension
that do not mention $M$ itself which occur in your arguments. Classes
are just the objects, standard and nonstandard, of your ambient ZFC.
The axioms of Ackermann set theory will clearly be satisfied.
Notice that this interpretation shows that we can't expect to prove
anything about classes in Ackermann set theory which we can't prove
about sets in ZFC. We cannot, for example, prove that \Omega is
inaccessible, though we strongly suspect this :-)
I'm curious as to whether there are ways to strengthen Ackermann set
theory in its own spirit (so to say) rather than by adjoining the
usual strong infinity axioms.
I think that an extended development of set theory in Ackermann's
system would look very familiar to a student of the usual set theory,
except that the more robust class system might be exploited in some
ways (though it clearly does not provide additional strength). The
question probably is whether a development of category theory would be
improved by the availability of Ackermann's classes.
--Randall