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- Newsgroups: sci.logic
- Path: sparky!uunet!pmafire!mica.inel.gov!guinness!garnet.idbsu.edu!holmes
- From: holmes@garnet.idbsu.edu (Randall Holmes)
- Subject: Re: Impredicativity - was: Russell's Paradox
- Message-ID: <1992Nov5.164251.29649@guinness.idbsu.edu>
- Sender: usenet@guinness.idbsu.edu (Usenet News mail)
- Nntp-Posting-Host: garnet
- Organization: Boise State University
- References: <Bx693z.H37@cantua.canterbury.ac.nz> <1992Nov4.073717.22625@CSD-NewsHost.Stanford.EDU>
- Date: Thu, 5 Nov 1992 16:42:51 GMT
- Lines: 98
-
- In article <1992Nov4.073717.22625@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
- >In article <Bx693z.H37@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
- >>Indeed, ZF intrinsically suggests it
- >>*without* the bother of a foundation axiom, by limiting the set formation
- >>possibilities so that self-membership is at least not immediately achieved
- >
- >What is the reasoning here? That a theory T "intrinsically suggests"
- >not-P for any P not in T? If neither P nor not-P are in T, which of
- >not-P or not-not-P (= P) does T "intrinsically suggest?"
- >
- >>The slick form of AF, that every nonempty set has a member disjoint from it,
- >
- >Where is this form given? And why doesn't it rule out Omega = {Omega}?
- >(Just asking, I'm not expert on this stuff.)
-
- Thomas Jech, _Set Theory_, first page. It does rule out Omega =
- {Omega}; it is a form of the axiom of foundation (AF, not AFA).
-
- >
- >>But I'm open on well-foundedness: for all I know, there may well be sets like
- >>A = { 1 , { 2 , { 3 , { 4 , {.....}}}}} . Such a set seems very well defined on
- >>the face of it, so it *could* exist, for all I know. No self-membership there.
- >>
- >>REQUEST: If someone can produce a way of getting a self-membered set from the
- >>set A above, using the other ZF constructions, I would *love* to see it !!!!
- >
- >If you are comfortable with A, you should not mind
- >B = { 1 , { 1 , { 1 , { 1 , {.....}}}}} .
- >
- >But by AFA, B = {1, B}, making B a member of itself.
-
- This uses a very strong extensionality property which follows from
- AFA; there is no reason why there cannot be a sequence of distinct
- sets xi such that xi = {1,x[i+1]} in ZFC without Foundation. Of
- course, this makes the symbol above ambiguous (the original expression
- with 1,2,3,4... also fails to specify a unique set in ZFC- -- there
- could be many such sets).
-
- >
- >As soon as you weaken FA to AFA, the possibility of self-membership
- >arises.
-
- You do not "weaken" FA to AFA; ZFC- + FA and ZFC- + AFA are both
- stronger than ZFC-, but neither is stronger than the other.
-
- >
- >>That is another reason I went to the trouble of posting my first article; it
- >>seems to me to make a very clear criterion for locating the essential
- >>impredicativity of Russell, which shows up most clearly in the *set* form:
- >> { x | P(x) } ,
- >>the essential impredicativity comes from the bad range of x, not the form of P.
- >>
- >>Doesn't *anyone* else think so ? :-(
- >
- >I do. But I forgot to say in my earlier message that the main point of
- >my positive reformulation of Russell's paradox was to remove this
- >impredicativity from the argument. Witness Y was constructed as the
- >set of all *elements of X* not members of themselves. This established
- >a positive result without any impredicativity. Russell's theorem is
- >not so much a corollary of this theorem as an interpretation of it, in
- >that "there is no set of all sets" formalizes as "every set is
- >nonuniversal", or "every set has a nonmember".
- >
- >>That's partly why I don't think that Aczel's
- >>set theory is really basic, delightful though it is. It can't look like
- >>fundamental set theory if a convincing show of consistency has to route
- >>through ZF's consistency.
-
- It doesn't; its consistency can be proven in ZFC- (or ZF-) (just like
- that of full ZFC). Foundation or the lack thereof has nothing to do
- with consistency strength.
-
- >
- >ZF+FA is even stronger than ZF+AFA. So by that reasoning ZF+FA can't
- >look like like a fundamental set theory either. Yet you seem to prefer
- >ZF+FA to ZF+AFA.
-
- No, it isn't. They have exactly the same strength (and both describe
- hierarchically structured universes).
-
- >
- >>I hope you are taking your own rubbishing
- >>(concerning Z vs ZF) like a man ! It's a tough life on the net !
- >
- >Being mistaken at the top of your voice does have its drawbacks.
- >Yesterday definitely wasn't my day, I got something really wrong on
- >sci.math involving measure theory, much more wrong than Z vs. ZF, which
- >I maintain is a highly debatable matter, the near-universal acceptance
- >of the cumulative hierarchy as a model of ZF notwithstanding.
- >--
- >Vaughan Pratt There's no truth in logic, son.
-
-
- --
- The opinions expressed | --Sincerely,
- above are not the "official" | M. Randall Holmes
- opinions of any person | Math. Dept., Boise State Univ.
- or institution. | holmes@opal.idbsu.edu
-