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- Newsgroups: sci.math
- Path: sparky!uunet!caen!zaphod.mps.ohio-state.edu!magnus.acs.ohio-state.edu!wjcastre
- From: wjcastre@magnus.acs.ohio-state.edu (W.Jose Castrellon G.)
- Subject: Re: Axioms of set theory, infinity and R. Rucker
- Message-ID: <1992Nov8.023839.16479@magnus.acs.ohio-state.edu>
- Sender: news@magnus.acs.ohio-state.edu
- Nntp-Posting-Host: top.magnus.acs.ohio-state.edu
- Organization: The Ohio State University,Math.Dept.(studnt)
- References: <1992Nov6.133138.16642@prl.philips.nl> <1992Nov6.182447.25955@infodev.cam.ac.uk> <1992Nov7.001459.7644@magnus.acs.ohio-state.edu>
- Date: Sun, 8 Nov 1992 02:38:39 GMT
- Lines: 65
-
- In article <1992Nov7.001459.7644@magnus.acs.ohio-state.edu> I write:
-
- [...]
- >Actually without using the Axiom of Choice, it is possible to prove the exis-
- >tence of the set of natural numbers, from the other axioms of set theory with
- >"There is an infinite set" in place of the Axiom of Infinity.
- >
-
- Herman Rubin has posted a proof of this, involving Hartog's function.
-
- I have received a few requests to clarify what definition of _infinite_
- I had in mind for the statement "There is an infinite set". Well, one
- can think of Dedekind's: A set is infinite if there is a function from
- it into itself that is injective but not onto; however it turns out
- that all definitions of _infinite_ will work as well, if one makes a
- small assumption on the property of being _infinite_.
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- [some blank lines, in case you want to think about it]
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- A reasonable request is that if X is _infinite_ then it remains so when we
- remove one of its elements. [and that the empty set not be infinite...]
- So let I(X) _X is infinite_ a formula that satisfies:
- For all X and all y in X, I(X) --> I (X\{y})
- Then one would like to show that the existence of a set satisfying I(X),
- implies the existence of the set of natural numbers (set of all the
- ordinals strictly smaller than any non-zero limit ordinal, or equivalently
- the smallest set containing 0 and the succesor of each of its elements).
- For this, take the range of the Hartogs function, restricted to the
- ordinals smaller than all limits (intuitively this is the set of finite
- ordinals that can be mapped injectively into X), and claim that this set H
- is the set of natural numbers (and some more to help in the proof).
- Claim: If t is an ordinal strictly smaller than all limits then t
- belongs to S (i.e. there is an injective f: t --> X), and
- I(X\f(t)) (i.e. the complement of the range of f is _infinite_)
- Suppose not, then take the least r for which it is false. Then since r
- is not a limit, the claim is true for the ordinal r-1:
- so there is an injective g: r-1 --> X and I(X\g(r-1)). By the property
- of the formula I, one can remove an element z of X\g(r-1) and still have
- I( (X\g(r-1)) \ z ). This is enough to get a contradiction, since one can
- extend g to an f: r --> X by sending the new element to z, and have
- that the claim in fact did hold for r , contrary to the assumption that
- r was a counterexample.
-
- By the way, I think the proposer's original problem is very interesting, and
- giving a correct solution is not simple at all.
-