Does Such a Relation Exist?
in [Logic]
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From: Herman Jurjus on 11 Nov 2009 04:30 Jesse F. Hughes wrote: > In the theory ZFC, certainly not (because R would not be a set). If > we amend ZFC so that it makes sense to speak of proper classes, ... Is there a difference between 'ZFC with classes' and NBG? -- Cheers, Herman Jurjus
From: Jesse F. Hughes on 11 Nov 2009 09:06 Herman Jurjus <hjmotz(a)hetnet.nl> writes: > Jesse F. Hughes wrote: >> In the theory ZFC, certainly not (because R would not be a set). If >> we amend ZFC so that it makes sense to speak of proper classes, ... > > Is there a difference between 'ZFC with classes' and NBG? I've no idea, since I'm unfamiliar with NBG. But if NBG simply extends ZFC by adding a Set predicate and appropriate axioms (a set union is a set, etc.), then that's what I have in mind. -- Jesse F. Hughes "To be honest, I don't have enough interest in math to spend the time it would take to clean up the mess that I believe has been created in the past 100 or so years." -- Curt Welch lets the world down.
From: Herman Rubin on 11 Nov 2009 20:41 In article <87ws1xnojl.fsf(a)phiwumbda.org>, Jesse F. Hughes <jesse(a)phiwumbda.org> wrote: >Herman Jurjus <hjmotz(a)hetnet.nl> writes: >> Jesse F. Hughes wrote: >>> In the theory ZFC, certainly not (because R would not be a set). If >>> we amend ZFC so that it makes sense to speak of proper classes, ... >> Is there a difference between 'ZFC with classes' and NBG? >I've no idea, since I'm unfamiliar with NBG. But if NBG simply >extends ZFC by adding a Set predicate and appropriate axioms (a set >union is a set, etc.), then that's what I have in mind. What is "ZFC with classes"? It is the case that any theorem of NBG which does not involve classes is a theorem of ZF. The same holds if choice is added. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: Aatu Koskensilta on 11 Nov 2009 20:46 hrubin(a)odds.stat.purdue.edu (Herman Rubin) writes: > What is "ZFC with classes"? It doesn't really matter. Any second-order formulation of set theory will do, with predicative or full class comprehension. > It is the case that any theorem of NBG which does not involve classes > is a theorem of ZF. The same holds if choice is added. What was at issue was global choice. To establish that NBG is conservative over ZF and NBG with choice conservative over ZFC we either use cut-elimination (for predicative second-order logic) or a simple model-theoretic construction. To show that global choice is conservative (a particularly simple form of) forcing is needed. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechen kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Charlie-Boo on 16 Nov 2009 07:54
On Nov 10, 1:37 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > I'll savor this one and give you a chance. (Also upping the ante.) > > But also maybe we're onto something big (relatively.) Someone proved > > that something doesn't imply AC? And what could we conclude from that > > little morsel? (I realized this only on my second reading.) > > The axiom of choice is independent of ZF, you know. Thus, someone > proved that ZF does not imply AC (also, that ZF does not imply ~AC). > > It has also been proved that ZF + CC is independent of AC. Thus, > ZF + CC does not prove AC (nor its negation). > > No idea what you're going to conclude from this little morsel. I just realized that if AC is true then ZF+CC (anything) proves it, so if ZF+CC does not prove AC then AC must be false. Fancy that! C-B If I can't write a book on new theorems of Logic, maybe I can at least write one on falacies (by perusing my many posts of the past.) > I conclude a few things (ZF + ~AC is equiconsistent to ZF, and so is > ZF + AC, for instance), but all of my conclusions are obvious and > well-known. > > -- > "There are people [...] who think it's socially acceptable to level > accusations of mental illness in insulting exchanges to make > points[...] [They] are rather sick [them]selves, and in reality, are > sociopathic." --- James Harris, evidently a self-described sociopath |