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- Newsgroups: sci.math
- Path: sparky!uunet!gatech!utkcs2!utkux1.utk.edu!utkvx1.utk.edu!dchatham
- From: dchatham@utkvx1.utk.edu (Chatham, Doug)
- Subject: Re: Foundational issues
- Message-ID: <12SEP199214515040@utkvx1.utk.edu>
- News-Software: VAX/VMS VNEWS 1.41
- Sender: usenet@utkux1.utk.edu (USENET News System)
- Organization: University of Tennessee Computing Center
- References: <92Sep12.101940edt.47512@neat.cs.toronto.edu>
- Distribution: na
- Date: Sat, 12 Sep 1992 19:51:00 GMT
- Lines: 31
-
- In article <92Sep12.101940edt.47512@neat.cs.toronto.edu>,
- arnold@cs.toronto.edu (Arnold Rosenbloom) writes...
- [Introduction and signature deleted]
-
- >First the number theory text assumes the Peano Axioms, now if we
- >view these axioms as the specification of a particular set, how
- >do we know that we have a well defined system. For example, the Peano
- >axioms assume the existence of a successor function that behaves in a certain
- >way. How do we know there is a subset of NxN which has the required properties.
- >
- >The other side (set theory) of the coin also presents some problems.
- >THat is, the natural
- >numbers are 'defined' as 0={}, 1={0}, 2={0,1}, ... , n+1={0,...,n}.
- >But this looks like an inductive definition. So how can you make an inductive
- >definition (and hope it makes sense) without the natural number system already
- >there to prove that you have actually defined something?
-
- I'm not sure what you mean by "well defined system." However, the
- reason we "know" the existence of a successor function is that we ASSUMED it.
- Unfortunately, according to Godel we cannot prove the consistency of a
- number theory including the Peano axioms [without bringing in some larger
- formal system whose consistency would be unprovable without . . .ad nauseaum,
- ad infinitum]. So we can't prove the successor function won't cause a
- contradiction somehow.
- Secondly, to make inductive definitions we need to ASSUME the
- existence of at least one inductive set. ZF (Zermelo-Frankel?) set theory
- makes this assumption an explicit axiom.
- I hope this helps.
-
- Sincerely,
- Doug Chatham
-