Does Such a Relation Exist?
in [Logic]
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From: Charlie-Boo on 16 Nov 2009 08:00 On Nov 10, 1:32 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(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? Set Theory is all about what can be formally proven using a specific set of axioms. If you add more (axioms, definitions) then you are not addressing the same question. C-B > > > > >> 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.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Jesse F. Hughes on 16 Nov 2009 08:25 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Nov 10, 1:32 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> 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? > > Set Theory is all about what can be formally proven using a specific > set of axioms. If you add more (axioms, definitions) then you are not > addressing the same question. The introduction of definitions for function (and ordered pair and so on) are simple niceties, done in order to shorten presentations and make theorems and their proofs more readable. You seem to think that something more substantial is going on. Any statement involving the predicate "f is a function" can be rewritten without that predicate by inserting the conjunction of (1)-(3). To be perfectly pedantic, we would have to rewrite (1)-(3) to eliminate the cross-product and ordered pair notations, but this is certainly doable. -- Jesse F. Hughes "But you have to support spyware if you're going to have free file-sharing applications. Fair's fair." -- NYU student Keith Caron in Wired
From: Charlie-Boo on 2 Dec 2009 11:21 On Nov 16, 8:25 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Nov 10, 1:32 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> 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? > > > Set Theory is all about what can be formally proven using a specific > > set of axioms. If you add more (axioms, definitions) then you are not > > addressing the same question. > > The introduction of definitions for function (and ordered pair and so > on) are simple niceties, done in order to shorten presentations and > make theorems and their proofs more readable. You seem to think that > something more substantial is going on. But the "definition" is subjective. Might not two different definitions of function using relations lead to different results? That's why an axiomatic system requires you to specify everything, which ZF unfortunately does not. C-B > Any statement involving the predicate "f is a function" can be > rewritten without that predicate by inserting the conjunction of > (1)-(3). To be perfectly pedantic, we would have to rewrite (1)-(3) > to eliminate the cross-product and ordered pair notations, but this is > certainly doable. > > -- > Jesse F. Hughes > "But you have to support spyware if you're going to have free > file-sharing applications. Fair's fair." > -- NYU student Keith Caron in Wired- Hide quoted text - > > - Show quoted text -
From: Herman Rubin on 2 Dec 2009 13:39 In article <77840a3f-bde8-4f61-a35f-31554ffea21a(a)h10g2000vbm.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> wrote: >On Nov 16, 8:25=A0am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> > On Nov 10, 1:32=A0pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> I'm not sure what you're going on about. =A0The axioms of ZF do not >> >> define function, but it is easy enough to introduce such a >> >> definition. =A0Here it is: >> >> =A0 Let f, X and Y be sets. =A0Then f is a function with domain X and >> >> =A0 codomain Y (written f:X -> Y) iff the following hold: >> >> =A0 (1) f c X x Y =A0 =A0(f is a subset of X x Y) >> >> =A0 (2) (Ax in X)(Ey in Y)( <x,y> in f ) >> >> =A0 (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) -> >> >> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = >=A0 y =3D y' ) >> >> What's the issue? >> > Set Theory is all about what can be formally proven using a specific >> > set of axioms. =A0If you add more (axioms, definitions) then you are no= >t >> > addressing the same question. >> The introduction of definitions for function (and ordered pair and so >> on) are simple niceties, done in order to shorten presentations and >> make theorems and their proofs more readable. =A0You seem to think that >> something more substantial is going on. >But the "definition" is subjective. Might not two different >definitions of function using relations lead to different results? >That's why an axiomatic system requires you to specify everything, >which ZF unfortunately does not. The definition of ordered pair is irrelevant. Whatever definition is used can be shown equivalent to the usual one, and this one is not the original one. In fact, the von Neumann axiomatization of set theory, the first one with a finite number of axioms (not axiom schemata), has function as a primitive; it is equivalent to the NBG system (von Neumann-Bernays-Godel) in which the usual definition of ordered pair is used. -- 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: Charlie-Boo on 2 Dec 2009 16:00
On Dec 2, 1:39 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <77840a3f-bde8-4f61-a35f-31554ffea...(a)h10g2000vbm.googlegroups..com>, > > > > > > Charlie-Boo <shymath...(a)gmail.com> wrote: > >On Nov 16, 8:25=A0am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Charlie-Boo <shymath...(a)gmail.com> writes: > >> > On Nov 10, 1:32=A0pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> >> I'm not sure what you're going on about. =A0The axioms of ZF do not > >> >> define function, but it is easy enough to introduce such a > >> >> definition. =A0Here it is: > >> >> =A0 Let f, X and Y be sets. =A0Then f is a function with domain X and > >> >> =A0 codomain Y (written f:X -> Y) iff the following hold: > >> >> =A0 (1) f c X x Y =A0 =A0(f is a subset of X x Y) > >> >> =A0 (2) (Ax in X)(Ey in Y)( <x,y> in f ) > >> >> =A0 (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) -> > >> >> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = > >=A0 y =3D y' ) > >> >> What's the issue? > >> > Set Theory is all about what can be formally proven using a specific > >> > set of axioms. =A0If you add more (axioms, definitions) then you are no= > >t > >> > addressing the same question. > >> The introduction of definitions for function (and ordered pair and so > >> on) are simple niceties, done in order to shorten presentations and > >> make theorems and their proofs more readable. =A0You seem to think that > >> something more substantial is going on. > >But the "definition" is subjective. Might not two different > >definitions of function using relations lead to different results? > >That's why an axiomatic system requires you to specify everything, > >which ZF unfortunately does not. > > The definition of ordered pair is irrelevant. Some definitions have been shown to be inconsistent. Some authors have writen "if this form of definition is legitimate" (I believe Kleene was one.) People debated for years whether definitions can refer to themselves. Whether to allow certain axioms has never been settled. Something is missing from ZF: a definition of function. Yet you say it's irrelevant??? You cannot dismiss an omission in such a cavalier manner. > Whatever > definition is used can be shown equivalent to the usual > one, How do you know that? You cannot systematically consider every possible definition that anyone might propose. That makes no sense. "It doesn't matter. They're all the same." Is there any logic that supports such a broad statement as that? > and this one is not the original one. > > In fact, the von Neumann axiomatization of set theory, > the first one with a finite number of axioms (not > axiom schemata), has function as a primitive; That's right - they were smarter. But ZF did not. C-B > it is > equivalent to the NBG system (von Neumann-Bernays-Godel) > in which the usual definition of ordered pair is used. > -- > 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 > hru...(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558- Hide quoted text - > > - Show quoted text - |