Does Such a Relation Exist?
in [Logic]
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From: William Elliot on 6 Nov 2009 02:36 On Thu, 5 Nov 2009, Charlie-Boo wrote: > Is there a two-place relation R such that: > Yes. > 1. If x is an element of y then there exists a z such that R(y,z). > x in y ==> some z with R(y,z) > 2. If R(x,y) then y is an element of x. > R(x,y) ==> y in x R(x,y) ==> y in x ==> some z with R(x,z) ==> some z with z in x > 3. If R(x,y) and R(x,z) then y=z. > R is a function > What should it be called? > Choice function. What's the domain and codomain of R?
From: Jesse F. Hughes on 6 Nov 2009 10:45 Charlie-Boo <shymathguy(a)gmail.com> writes: > The choice function referred to in the Axiom of Choice. Then your conditions are not quite correct. The axiom of choice says: For every set w, there is a relation R c w x Uw such that the following hold: (1) (A y in w)(A x)( x in y -> (Ez) R(y,z) ) (2) (A y in w)(A z)( R(y,z) -> z in y ) (3) (A y in w)(A z)(A z')( ( R(y,z) & R(y,z') ) -> z = z' ) Your relation R was not relativized to a set w, so it wasn't the axiom of choice. It was more like an axiom of choice for the class V of all sets. That is, you wrote: there is a relation R (evidently not a set, but a class of ordered pairs) such that (1) (A y)(A x)( x in y -> (Ez) R(y,z) ) (2) (A y)(A z)( R(y,z) -> z in y ) (3) (A y)(A z)(A z')( ( R(y,z) & R(y,z') ) -> z = z' ) This isn't what AC claims. (My apologies for the freshman error in predicate logic in a previous post. I guess my first-order logic skills are rustier than I realized.) -- "So why are mathematicians NOT what most people suppose? Why are they not these brilliant and wonderful people who act in favor of humanity instead of against it?" -- James S. Harris, on public confusion about mathematicians and superheroes.
From: Charlie-Boo on 10 Nov 2009 07:37 On Nov 5, 5:32 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Nov 5, 4:45 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Charlie-Boo <shymath...(a)gmail.com> writes: > >> > On Nov 5, 4:02 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> >> Charlie-Boo <shymath...(a)gmail.com> writes: > >> >> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> >> >> Charlie-Boo <shymath...(a)gmail.com> writes: > >> >> >> > On Nov 5, 10:29 am, James Burns <burns...(a)osu.edu> wrote: > >> >> >> >> Charlie-Boo wrote: > >> >> >> >> > Is there a two-place relation R such that: > > >> >> >> >> > 1. If x is an element of y then there exists a z such that R(y,z). > > >> >> >> >> > 2. If R(x,y) then y is an element of x. > > >> >> >> >> > 3. If R(x,y) and R(x,z) then y=z. > > >> >> >> >> > What should it be called? > > >> >> >> >> Would you mind sharing why the question is interesting? > > >> >> >> > A lot of people write about it. > > >> >> >> How about some context then? Who writes about it? > > >> >> > Does the name ``Godel" ring a bell? > > >> >> Yes, but Goedel didn't prove his famous theorems in set theory, > > >> > Why would they be theorems if he didn't prove them? > > > > I didn't say that he didn't prove them. > > > You said, "Goedel didn't prove his famous theorems" > > Right. You've quite a skill *blush* > I thought you were going on about the incompleteness > theorems, but you were talking about the independence of CH, so never > mind what I said above. Trim it? > >> >> so I still haven't a clue what you're going on about. > > >> > Godel proved theorems in set theory as well as in logic. > > >> Yes, that's true. But where did he assume the existence of this > >> relation R that you're going on about? > > >> Don't be coy. Just spell it out. > > > The choice function referred to in the Axiom of Choice. > > Sorry, still not clear on what you mean. The axiom of choice does not > involve a relation R that you described. Ok. # 1 = AOC doesnt 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). > 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). > >> (1) has a free variable x. It's not clear what you mean by (1). > > > I would say that x is universally quantified. > > Okay, so you mean to ask: is there a relation R such that > > (Ax)(Ay)( x e y -> (Ez)( R(y,z) ) )? > > This is equivalent to > > (Ay)( ( (Ax) x e y ) -> (Ez)( R(y,z) ) ). No. C-B > The antecedent is always false and hence the conditional is true, > regardless of what R is -- unless I'm making some silly error as I > toss this off. So, I doubt this is really what you meant after all. > > >> You think I'll purchase and read this book to figure out what you're > >> talking about? I have a better idea. You can just explicitly state > >> which relation you mean (and, perhaps, where Goedel introduces this > >> relation). > > > Godel proved that the Axiom of Choice is consistent with ZF. > > -- > Jesse F. Hughes > "You do know that after the get done with [outlawing] cigarettes, > they're gonna come after guns, right?" > -- AM talk radio host Mike Gallagher- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 10 Nov 2009 07:40 On Nov 5, 7:31 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tim Little <t...(a)little-possums.net> writes: > > On 2009-11-05, Jesse F. Hughes <je...(a)phiwumbda.org> wrote: > >> Right. You've quite a skill > > > Heh. Heh. > > >> (Ax)(Ay)( x e y -> (Ez)( R(y,z) ) )? > > >> This is equivalent to > > >> (Ay)( ( (Ax) x e y ) -> (Ez)( R(y,z) ) ). > > >> The antecedent is always false and hence the conditional is true, > >> regardless of what R is -- unless I'm making some silly error as I > >> toss this off. > > > Yes, unfortunately. Quantifiers do not distribute over implication > > like that. The first statement asserts something about all nonempty > > sets y, while the second asserts something about a universal set y. > > D'oh! Eh, that's what I get for trying to toss off a reply while I'm > heading out the door. I thought haste makes waste, not stupidity. C-B > -- > Jesse F. Hughes > "Now 'pure math' makes sense as well as clearly it's a peacock game, > where some of you see it as a way to show you as being highly > intelligent and thus more desirable to women." -- James S. Harris- Hide quoted text - > > - Show quoted text -
From: Jesse F. Hughes on 10 Nov 2009 07:53
Charlie-Boo <shymathguy(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. -- 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 |