Does Such a Relation Exist?
in [Logic]
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From: Charlie-Boo on 5 Nov 2009 13:50 On Nov 5, 10:02 am, 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). Is there a relation R that meets (1) ? (Or something that is not a relation.) C-B > 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
From: Michael Stemper on 5 Nov 2009 14:03 In article <997257e0-e214-45fd-bd3d-45e0344a0a03(a)j19g2000vbp.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> writes: >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. Putting the first two together gives me: If x is an element of y, then there exists a z such that z is an element of y. Did I do this correctly? >3. If R(x,y) and R(x,z) then y=z. I guess so. At least, this is consistent with the first two. -- Michael F. Stemper #include <Standard_Disclaimer> Reunite Gondwanaland!
From: Jesse F. Hughes on 5 Nov 2009 14:09 Charlie-Boo <shymathguy(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? 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). 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 != {} satisfies (1'), (2) and (3), but such an operation requires a version of AC on classes (or so it seems to me). 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? And in what context? -- "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
From: Ken Pledger on 5 Nov 2009 14:49 In article <997257e0-e214-45fd-bd3d-45e0344a0a03(a)j19g2000vbp.googlegroups.com>, Charlie-Boo <shymathguy(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. 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? That's a pretty ambitious attempt. :-) Ken Pledger.
From: Charlie-Boo on 5 Nov 2009 15:15
On Nov 5, 2:03 pm, mstem...(a)walkabout.empros.com (Michael Stemper) wrote: > In article <997257e0-e214-45fd-bd3d-45e0344a0...(a)j19g2000vbp.googlegroups..com>, Charlie-Boo <shymath...(a)gmail.com> writes: > > >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. > > Putting the first two together gives me: If x is an element of y, > then there exists a z such that z is an element of y. Did I do > this correctly? Well, you did successfully use transitivity. However, using my DEF: P (a), (eA)P(A)^EQ(A,a) we have that something is an element of a set iff there is an element of that set equal to that thing, which gives you that conclusion already. > >3. If R(x,y) and R(x,z) then y=z. > > I guess so. At least, this is consistent with the first two. Then are all 3 consistent - is there such an R? Interesting. If they are consistent, then does that mean there is an R? A relation R? C-B > -- > Michael F. Stemper > #include <Standard_Disclaimer> > Reunite Gondwanaland! |