Obections to Cantor's Theory (Wikipedia article)
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From: Han de Bruijn on 27 Jul 2005 06:44 Torkel Franzen wrote: > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > >>The infinitary ones. It seems that the "first five" are sufficient for >>defining set theory for _finite_ sets. > > ZFC without the axiom of infinity is formally equivalent to PA. > > However, you haven't explained what you mean by saying "the finitary > result still exists", or what is puzzling you. So if you replace ZFC by the PA system then you can still formulate the theorem, but not "prove" it. Is that right? >>But what is ACA ? > > See my earlier response. Got it. Han de Bruijn
From: Torkel Franzen on 27 Jul 2005 06:46 Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > So if you replace ZFC by the PA system then you can still formulate the > theorem, but not "prove" it. Is that right? Goodstein's theorem is formalizable in the language of arithmetic and is provable in ZFC but not in PA, right. What is it you find puzzling about this?
From: Han de Bruijn on 27 Jul 2005 07:27 MoeBlee wrote: > Meanwhile, I'm still fascinated by your inconsistent theory, posted at > your web site, which is: > > Z, without axiom of infinity, bu with your axiom: Ax x = {x}. > > Would you say what we are to gain from this inconsistent theory? No. But first we repeat the mantra: A little bit of Physics would be NO Idleness in Mathematics It is physically correct that every member of a set is at the same time a _part_ of the set, meaning that a e A ==> a c A , where e stands for membership and c for being a subset. The above axiom that a member of a set cannot physically distinguished from a set which contains only the member - the envelope {} means nothing, physically - is only a weaker form of this idea. There has been a thread on this topic in 'sci.math' as well, called "Set inclusion and membership". It is noted that a "set theory without the membership" actually exists. Google it up as "mereology" and you will find quite some clues. Now I simply add this axiom to ZFC and see what happens. Nothing else. If the consequence may be that ZFC scrumbles into nothingness, then, unfortunately, we are not going to gain anything. But I consider that not to be my problem, because I'm just doing .. the mantra. No kidding: it IS a problem, but I find that first things come first. Han de Bruijn
From: David Kastrup on 27 Jul 2005 07:50 Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > MoeBlee wrote: > >> Meanwhile, I'm still fascinated by your inconsistent theory, posted at >> your web site, which is: >> Z, without axiom of infinity, bu with your axiom: Ax x = {x}. >> Would you say what we are to gain from this inconsistent theory? > > No. But first we repeat the mantra: > > A little bit of Physics would be NO Idleness in Mathematics Oh, it is there alright, like a little bit of real life is in a novel worth reading. > It is physically correct that every member of a set is at the same > time You blew it. A "set" is not a physical entity, so there is nothing that can be "physically correct" about a set. > a _part_ of the set, meaning that a e A ==> a c A , where e stands > for membership and c for being a subset. The above axiom that a > member of a set cannot physically distinguished from a set which > contains only the member - the envelope {} means nothing, physically The whole set means nothing, physically. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Robert Low on 27 Jul 2005 08:05
Torkel Franzen wrote: > Robert Low <mtx014(a)coventry.ac.uk> writes: >>OK: without trying too hard to think about it, I can persuade myself >>that's enough to let one build the sequence of decreasing ordinals >>used in the proof of Goodstein's theorem. (Unless I'm mistaken >>about what one needs to prove the theorem, of course.) > The ordinals smaller than epsilon_zero kan be represented by a > primitive recursive set of notations, and the relation "the ordinal > [represented by] n is smaller than the ordinal [represented by] m" is > also primitive recursive. The proof of Goodstein's theorem proceeds by > well-founded induction on this primitive recursive relation. The > non-arithmetical part of ACA enters into the proof that the relation > is indeed well-founded. That's much more precise than what I was groping towards---but it's pleasant to see that I can claim I was thinking a woolly version of what was correct. One day when time permits, I should get hold of your Inexhaustibility book; maybe even read it :-) |