Does Such a Relation Exist?
in [Logic]
|
From: Jesse F. Hughes on 10 Nov 2009 13:32 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Nov 10, 9:34 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> > Anyway, how is R about classes and not sets? You may be getting to >> > the point, actually. Classes are for things that are not sets >> > (=relations) so AOC is really about whether R is a set. Russell >> > proved that some things aren't sets and I am trying to apply >> > additional logic to address R being a set or not. >> >> I shouldn't have said that R was about classes per se, > > Ok. > >> but your conditions claim that there is essentially a *global* choice function, >> that is a choice function for the particular class V. > > How is that a but - what does it have to do with R being about classes > per se? Your conditions of R amount to the axiom of choice for a particular class, not for all classes of sets. >> That's not what >> AC says. > > But that's what I say! That's what CBL says, too. (And CBL proves > all sorts of theorems very easily and amazingly short, due to several > subterfuges in use.) But who cares what you say? You said that lots of people are writing about relations R satisfying those three conditions. That's just not true. > Why can't we say that instead? > > The question is (as you discuss) whether AC = Global AC. At the least > let us add that to the questions discussed, in the mainstream > literature (full of fraud) as well as this counter-technology that we > are now all collectively developing in a huge collaboration. > (Billions access Google.) Well, you can certainly ask that question. Seems to me that the answer is almost certainly "no", but I haven't a proof of that fact. Do you have any argument why the answer may be "yes"? > But 1st things 1st - I asked you first - is there such an R? (Is it > close enough to what books with pretty covers talk about?) 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, then I have no proof that there is no class R satisfying your conditions. Nor do I have a proof that there is such a class. Moreover, I sincerely doubt that the latter claim is provable (though I haven't an argument to that effect). > The first problem with ZF addressing AOC is that ZF doesn't define > what a function is - there are NO REFERENCES to them - so naturally ZF > is consistent with AOC. ** I'm not sure what you're going on about. The axioms of ZF do not define function, but it is easy enough to introduce such a definition. Here it is: Let f, X and Y be sets. Then f is a function with domain X and codomain Y (written f:X -> Y) iff the following hold: (1) f c X x Y (f is a subset of X x Y) (2) (Ax in X)(Ey in Y)( <x,y> in f ) (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) -> y = y' ) What's the issue? >> Compare the following: >> >> AC: >> >> For all w, there is an R c w x Uw such that the following three >> conditions hold: >> >> (Ay in w)( (Ex)( x in y ) -> (Ez) R(y,z) ) >> (Ay)(Az)( R(y,z) -> z in y ) >> (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' ) >> >> Global AC: >> >> There is an R such that the following three conditions hold: >> >> (Ay)( (Ex)( x in y ) -> (Ez) R(y,z) ) >> (Ay)(Az)( R(y,z) -> z in y ) >> (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' ) >> >> Those are two different claims. In Global AC, it is clear that R >> cannot be a set at all. It must be a proper class of ordered pairs. >> >> Clearly, Global AC implies AC. Suppose w is a set and let R be given >> as in Global AC. Define >> >> R' = { (y,z) in w x Uw | R(y,z) } >> >> Then R' satisfies the three conditions for AC. However, AC does not >> imply Global AC. The fact that we have choice functions for each set >> does not entail, near as I can figger, a choice function for the class >> of all sets. -- "It's my belief that when religion and pseudoscience achieve an official status within a culture [...], then genocide, war, oppression, injustice, and economic stagnation are sure to follow." -- David Petry, on why |X| < |P(X)| is bad, bad, bad.
From: Jesse F. Hughes on 10 Nov 2009 13:37 Charlie-Boo <shymathguy(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 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
From: David Hartley on 10 Nov 2009 14:15 In message <87ocnap6x6.fsf(a)phiwumbda.org>, Jesse F. Hughes <jesse(a)phiwumbda.org> writes >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, then I >have no proof that there is no class R satisfying your conditions. Nor >do I have a proof that there is such a class. Moreover, I sincerely >doubt that the latter claim is provable (though I haven't an argument >to that effect). "V=L" provides a universal well-ordering and so a universal choice function. So it is consistent with ZFC (+ classes) that such an R exists. -- David Hartley
From: Jesse F. Hughes on 10 Nov 2009 14:42 David Hartley <me9(a)privacy.net> writes: > In message <87ocnap6x6.fsf(a)phiwumbda.org>, Jesse F. Hughes > <jesse(a)phiwumbda.org> writes >>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, then I >>have no proof that there is no class R satisfying your conditions. Nor >>do I have a proof that there is such a class. Moreover, I sincerely >>doubt that the latter claim is provable (though I haven't an argument >>to that effect). > > "V=L" provides a universal well-ordering and so a universal choice > function. So it is consistent with ZFC (+ classes) that such an R > exists. Thanks for the clarification. Thus, there is no proof that the class R does not exist (in ZFC + classes). And, if I understood Aatu's post, there is similarly no proof that the class R exists. Hence, Global AC is independent of ZFC + classes. I hope I got that right. If so, surely, that is the answer to Charlie's question (though not, I'd wager, the answer he wanted to receive). -- Jesse F. Hughes "The sole cause of all human misery is the inability of people to sit quietly in their rooms." -- Blaise Pascal
From: Aatu Koskensilta on 10 Nov 2009 14:49
"Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > Hence, Global AC is independent of ZFC + classes. Yep. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechen kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |