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