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- From: torkel@sics.se (Torkel Franzen)
- Newsgroups: sci.logic
- Subject: Re: Impredicativity - was: Russell's Paradox
- Message-ID: <TORKEL.92Nov5103134@isis.sics.se>
- Date: 5 Nov 92 09:31:34 GMT
- References: <Bx81Ho.9HL@cantua.canterbury.ac.nz>
- Sender: news@sics.se
- Organization: Swedish Institute of Computer Science, Kista
- Lines: 24
- In-Reply-To: wft@math.canterbury.ac.nz's message of Thu, 5 Nov 1992 02:27:23 GMT
-
- In article <Bx81Ho.9HL@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz
- (Bill Taylor) writes:
-
- >A = { 1 , { 2 , { 3 , { 4 , {.....}}}}}
- >B = { 1 , { 1 , { 1 , { 1 , {.....}}}}}
- >Can B above, be obtained from A above, by ZF methods.
-
- The question is vague in several respects, but the most likely
- answer is no. A, to put this into formal ZF terms, is, let us say, the
- range of a function f defined on the natural numbers, where
- f(n)={n,f(n+1)} for all n. Assuming the existence of such a function
- in ZF (without foundation) does not in any obvious way (and hence not
- at all, by the argument from experience) imply the existence of a set
- x such that x={1,x}. In particular, Replacement by itself is no help
- since to get x using replacement, we need to define a function h by
- h(n)=G(f(n)), where G({a,b})={1,G(b)}, and there is no way of defining
- G in ZF (without foundation) so that this is provable. In other words:
- Replacement allows us to make infinite "substitutions" in sets, but to
- make infinite substitutions in non-well-founded sets we need a
- corresponding non-well-founded recursion principle.
-
- Of course, even if this is correct it doesn't follow that accepting
- A and rejecting B makes any sense. Or in other words: it doesn't help
- explain what mathematical objects, if any, you have in mind.
-