Does Such a Relation Exist?
in [Logic]
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From: Charlie-Boo on 5 Nov 2009 15:34 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? > Your clause (1) is fairly unclear to me. Do you mean: > > (1') If there is an x such that x e y, then there is a z such that > R(y,z). Yes. > or do you mean that for each x, there is a relation R_x such that > (1) - (3) hold? > > It's not clear to me whether there is a relation R satisfying > conditions (1'), (2) and (3). We could construct R if we could define > an operation F:Set -> Set such that F(y) in y if y != {} and F({}) = > (). With such an operation, the relation > > R(x,y) <-> y = F(x) & y != {} You mean x != {} ? > satisfies (1'), (2) and (3), but such an operation requires a version > of AC on classes (or so it seems to me). How about (1) alone can we satisfy that with a relation R? > If we interpret the question in terms of R_x, where x is a fixed set, > then the relation > > R(y,z) <-> z = x & x in y > > satisfies (1) - (3). > > So who are all these people writing about an R satisfying the above? Isnt one enough after all, its Godel (yes, THE Godel - not that Accountant who keeps showing up http://www.godel.com/ .) > And in what context? Published material. C-B > -- > "I'm a very well-educated, successful, intelligent person. This is > insane to me that I have an armed guard outside my door when I've > cooperated with everything other than the whole solitary-confinement- > in-Italy thing." --A. Speaker, on the whole T.B.-quarantined thing- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 5 Nov 2009 15:51 On Nov 5, 2:49 pm, Ken Pledger <ken.pled...(a)mcs.vuw.ac.nz> wrote: > In article > <997257e0-e214-45fd-bd3d-45e0344a0...(a)j19g2000vbp.googlegroups.com>, > > Charlie-Boo <shymath...(a)gmail.com> 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? > > > C-B > > It looks suspiciously like a choice function. Don't be too suspicious or else someone might prescribe risperdal. > Your condition 1 > ensures that the domain of R includes all non-empty sets. Then your > conditions 2 and 3 show that every such set x has a unique element y > such that R(x,y). So, is R attempting to be a choice function on the > proper class of all non-empty sets? Yes! > That's a pretty ambitious attempt. > :-) The undertaking is ambitious or my effort is ambitious or you are being facetious or did you mispell ambiguous? Fooling people into working on an equivalent problem with new insights is also like the risperdal user - you have the pleasure of learning about happy events over and over. (It also gives you simpler questions like (1) above that can alone answer the original bigger question.) C-B > Ken Pledger.
From: Jesse F. Hughes on 5 Nov 2009 17:32 Charlie-Boo <shymathguy(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 at quoting out of context, but you need to trim the original in order to get away with it. In any case, 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. >> >> 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. The relation R that you described would be more like what one would see in an axiom of choice for classes. Tell you what. Why don't you write down the axiom of choice and point out where it involves such an R? >> (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) ) ). 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
From: Tim Little on 5 Nov 2009 18:55 On 2009-11-05, Jesse F. Hughes <jesse(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. - Tim
From: Jesse F. Hughes on 5 Nov 2009 19:31
Tim Little <tim(a)little-possums.net> writes: > On 2009-11-05, Jesse F. Hughes <jesse(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. -- 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 |