Does Such a Relation Exist?
in [Logic]
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From: Charlie-Boo on 10 Nov 2009 10:23 On Nov 10, 9:36 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(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 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.) C-B > 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- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 10 Nov 2009 10:41 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. A is related to B but A does not involve B? > > 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, I don't even think you should say AC and R are related but noninvolving. > but your > conditions claim that there is essentially a *global* choice function, > that is a choice function for the particular class V. I also don't think you should say but. However, I do think you should think about using Theory of Computation proofs of completeness and incompleteness to prove R exists or not, to address AOC. (As long as I get 1/2 of the prize money. (What's it up to?)) The first question could be (start with the simple stuff - substitution), what about R if we substitute YES for SE in the definition of R? Anybody? Plz excuse me for a few hours or days while I switch gears from being the first to prove (orchestrate) that AOC is impossible, to being the first to write an algorithm (as evidenced by its nonexistance on the internet) for the world's first HTML to SQL translator (speaking of formalizing and automating things.) C-B > 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.- Hide quoted text - > > - Show quoted text -
From: Aatu Koskensilta on 10 Nov 2009 11:06 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > 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. Your figgering can be backed up with a logical result. It is also a logical result, an easy and illustrative application of forcing, that any invocation of global choice in a proof of a result about sets only can be eliminated (given ordinary choice). -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechen kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Jesse F. Hughes on 10 Nov 2009 13:35 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: >> > 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. > > A is related to B but A does not involve B? Let us not quibble on such dull matters of terminology. >> > 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, > > I don't even think you should say AC and R are related but > noninvolving. > >> but your >> conditions claim that there is essentially a *global* choice function, >> that is a choice function for the particular class V. > > I also don't think you should say but. > > However, I do think you should think about using Theory of Computation > proofs of completeness and incompleteness to prove R exists or not, to > address AOC. (As long as I get 1/2 of the prize money. (What's it up > to?)) > > The first question could be (start with the simple stuff - > substitution), what about R if we substitute YES for SE in the > definition of R? Anybody? > > Plz excuse me for a few hours or days while I switch gears from being > the first to prove (orchestrate) that AOC is impossible, to being the > first to write an algorithm (as evidenced by its nonexistance on the > internet) for the world's first HTML to SQL translator (speaking of > formalizing and automating things.) No idea what you're going on about. -- Jesse F. Hughes "I am the next legend--living, breathing and solving mega problems in the here and now." -- James S. Harris
From: Jesse F. Hughes on 10 Nov 2009 13:33
Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes: > "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > >> 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. > > Your figgering can be backed up with a logical result. It is also a > logical result, an easy and illustrative application of forcing, that > any invocation of global choice in a proof of a result about sets only > can be eliminated (given ordinary choice). It is, of course, only coincidental when my figgering and the truth line up so well. -- "Tempted and tried we're oft made to wonder Why it should be thus all the day long When there are others living about us Never molested though in the wrong." -- Bad Livers, "Farther Along" |