Does Such a Relation Exist?
in [Logic]
|
From: Jesse F. Hughes on 10 Nov 2009 07:51 Charlie-Boo <shymathguy(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). -- "There's lots of things in this old world to take a poor boy down. If you leave them be, you can save yourself some pain. You don't have to live in fear, but you best have some respect, For rattlesnakes, painted ladies and cocaine." -- Bob Childers
From: Jesse F. Hughes on 10 Nov 2009 07:54 Charlie-Boo <shymathguy(a)gmail.com> writes: > 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. It was a stupid mistake. I'm sure you've never made a silly blunder yourself, so obviously you have the right to mock me. -- "It's one of the easiest tickets to true fame--not this silly stuff where people cheer you for a few years and then forget about you--but the kind of fame where school kids have to read your biography and do reports on you." -- Another reason to support James S. Harris.
From: Charlie-Boo on 10 Nov 2009 08:51 On Nov 6, 2:36 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > 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? Anything as long as R satisfies 1-3. The idea is to, 1. Formalize the Axiom of Choice using well-known (well-understood) primitives: Predicate Calculus. 2. Use logic to develop simpler requirements that if impossible make AOC impossible. This simplifies the question of whether AOC is true or not. (1) is an example. 3. Apply incompleteness proofs in other domains e.g. Computability to this formalization. Define, YES(x,y) iff Turing Machine x halts yes on input y. SE(x,y) iff y is an element of x. M defines r.e. set YES(M,x) and (general) set SE(M,x). There is no M that defines an r.e. set ~YES(x,x). There is no M that defines a (general) set ~SE(x,x). Thus we show there is no set of sets that contain themselves. With a little bit of logic we can likewise say there is no r.e. set (exists y)~YES(x,y) and similarly with other wffs (theorems) of Computability. If we substitute YES for SE in the definition of AOC or its necessary conditions 1-3, we can very directly manipulate that wff as referring to Turing Machines and e.g. appeal to known theorems. Then we apply that same manipulation to SE. BTW: If there is an R that meets (1) it doesnt necessarily meet (2) or (3). However, does the existence of such an R mean there is some R that meets (1) and (2)? (1) and (3)? Which of the 8 subsets of 1-3 are equivalent to which others in this sense? This would even more directly reduce AOC to simpler questions. C-B
From: Charlie-Boo on 10 Nov 2009 09:06 On Nov 10, 7:54 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > 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. > > It was a stupid mistake. Good. Thanks. > I'm sure you've never made a silly blunder > yourself, so obviously you have the right to mock me. Hmmm . . . Doesn't everyone have the right to mock anyone? Or at least the same rights? My point is that attributing a mistake to haste leaves something to be desired. (1) Why bother - what's the point? Shouldn't we ALWAYS not judge something someone did by judging something else that they did? So it has no relevance to anything. (2) It is a little suspiocious when someone says they said something due to haste. I would agree that we can attribute it to not taking the time to think about it. Is that what you meant? But that occurs when someone posts an easy problem because they thought it was neat (and very well may be) but didn't then check that it is actually difficult before posting it. But then again, now we are poking a hole in a defense mechanism that needn't be used anyway, so that is a waste. (3) What you did wasn't bad. Bad is using ad hominem. Or worse, defending the use of ad hominem. So it also not worth defending. (4) In general, let's all be big boys and not waste time tending to foolish pride (the root of all evil to many.) Sorry if it offended you! C-B > -- > "It's one of the easiest tickets to true fame--not this silly stuff > where people cheer you for a few years and then forget about you--but > the kind of fame where school kids have to read your biography and do > reports on you." -- Another reason to support James S. Harris.- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 10 Nov 2009 09:11
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 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). > > 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. 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. C-B > -- > "There's lots of things in this old world to take a poor boy down. > If you leave them be, you can save yourself some pain. > You don't have to live in fear, but you best have some respect, > For rattlesnakes, painted ladies and cocaine." -- Bob Childers |