Does Such a Relation Exist?
in [Logic]
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From: Charlie-Boo on 10 Nov 2009 09:13 On Nov 10, 7:53 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > >> Tell you what. Why don't you write down the axiom of choice and point > >> out where it involves such an R? > > > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y). > > Wow. What an utter failure to write down the axiom of choice. Want > to try again? > > You speak, after all, as if there is a single choice function. Tain't > so. I know. That's good of you to understand the set axioms so well. I like the simpler version. So, do we know that they aren't equivalent? C-B > -- > Jesse F. Hughes > "This Trojan appears to utilize a function of the Windows Media DRM > designed to enable license delivery scenarios as part of a social > engineering attack." -- MS candidly explains the role of DRM licenses
From: Jesse F. Hughes on 10 Nov 2009 09:36 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Nov 10, 7:53 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> >> Tell you what. Why don't you write down the axiom of choice and point >> >> out where it involves such an R? >> >> > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y). >> >> Wow. What an utter failure to write down the axiom of choice. Want >> to try again? >> >> You speak, after all, as if there is a single choice function. Tain't >> so. > > I know. That's good of you to understand the set axioms so well. I > like the simpler version. So, do we know that they aren't > equivalent? As I just posted Global AC (your simpler conditions) imply AC, but there is no reason to think that AC implies Global AC as far as I know. I'd imagine that the proof that countable choice does not imply AC gives a hint as to how one would show Global AC does not imply AC, but I'm not familiar with that argument. -- Jesse F. Hughes "Yes, I'm one of those arrogant people who tries to be quotable. There is actually at least one person who quotes me often." -- James Harris
From: Jesse F. Hughes on 10 Nov 2009 09:34 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> >> Sorry, still not clear on what you mean. The axiom of choice does not >> >> involve a relation R that you described. >> >> > Ok. # 1 = AOC doesn't involve my R. >> >> >> The relation R that you >> >> described would be more like what one would see in an axiom of choice >> >> for classes. >> >> > Ok. # 2 = AOC involves my R when talking about classes. >> >> > But # 1 => ~(# 2). >> >> You're talking nonsense. The axiom of choice refers to a particular >> (equivalence class of) axiom(s). It is an axiom about sets. The >> class-based axiom of choice that I mentioned (which I've never seen in >> the literature) is a different axiom. >> >> Thus #1 does not entail ~(#2). > > I guess it depends on your definition of "involves". It is a very > broad word to me. They are related, but I wouldn't say that AC involves your R. > 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, but your conditions claim that there is essentially a *global* choice function, that is a choice function for the particular class V. That's not what AC says. 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. -- Jesse F. Hughes Baba: Spell checkers are bad. Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.
From: Jesse F. Hughes on 10 Nov 2009 09:51 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > Charlie-Boo <shymathguy(a)gmail.com> writes: > >> On Nov 10, 7:53 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >>> Charlie-Boo <shymath...(a)gmail.com> writes: >>> >> Tell you what. Why don't you write down the axiom of choice and point >>> >> out where it involves such an R? >>> >>> > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y). >>> >>> Wow. What an utter failure to write down the axiom of choice. Want >>> to try again? >>> >>> You speak, after all, as if there is a single choice function. Tain't >>> so. >> >> I know. That's good of you to understand the set axioms so well. I >> like the simpler version. So, do we know that they aren't >> equivalent? > > As I just posted Global AC (your simpler conditions) imply AC, but > there is no reason to think that AC implies Global AC as far as I > know. > > I'd imagine that the proof that countable choice does not imply AC > gives a hint as to how one would show Global AC does not imply AC, ^^^^^^^^^^^^^^^^^^^^^^^^^^^ > but I'm not familiar with that argument. Sorry, I meant to say "AC does not imply Global AC". -- Jesse F. Hughes "My baby don't allow me in the kitchen and I've come to love her decision." -- Bad Livers
From: Charlie-Boo on 10 Nov 2009 10:14
On Nov 10, 9:34 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Charlie-Boo <shymath...(a)gmail.com> writes: > >> >> Sorry, still not clear on what you mean. The axiom of choice does not > >> >> involve a relation R that you described. > > >> > Ok. # 1 = AOC doesn't involve my R. > > >> >> The relation R that you > >> >> described would be more like what one would see in an axiom of choice > >> >> for classes. > > >> > Ok. # 2 = AOC involves my R when talking about classes. > > >> > But # 1 => ~(# 2). > > >> You're talking nonsense. The axiom of choice refers to a particular > >> (equivalence class of) axiom(s). It is an axiom about sets. The > >> class-based axiom of choice that I mentioned (which I've never seen in > >> the literature) is a different axiom. > > >> Thus #1 does not entail ~(#2). > > > I guess it depends on your definition of "involves". It is a very > > broad word to me. > > They are related, but I wouldn't say that AC involves your R. > > > 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? > 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.) 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.) 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?) 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. ** Well DUH, Mr. Godel! C-B (** = changes everything) > 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. > > -- > Jesse F. Hughes > > Baba: Spell checkers are bad. > Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.- Hide quoted text - > > - Show quoted text - |