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| Jesse F. Hughes... |
 Posted: Thu Nov 05, 2009 4:45 pm |
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Charlie-Boo <shymathguy at (no spam) gmail.com> writes:
[quote]On Nov 5, 4:02 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 2:09 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 10:29 am, James Burns <burns... at (no spam) 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?
[/quote]
I didn't say that he didn't prove them.
[quote]so I still haven't a clue what you're going on about.
Godel proved theorems in set theory as well as in logic.
[/quote]
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.
[quote]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 != {} ?
Yes, that's what I meant.
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?
What does (1) alone mean?
Ask the original question but with only (1) instead of (1), (2) and
(3).
[/quote]
(1) has a free variable x. It's not clear what you mean by (1).
[quote]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?
Isn't one enough - after all, it's Godel (yes, THE Godel - not that
Accountant who keeps showing uphttp://www.godel.com/.)
Which relation did he discuss that satisfies these properties?
Forgive me if I won't simply take your word for it.
http://www.amazon.com/Consistency-Continuum-Hypothesis-AM-3-Godel/dp/0691079277/ref=sr_1_9?ie=UTF8&s=books&qid=1257455839&sr=1-9#noop
[/quote]
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).
[quote]And in what context?
Published material.
You realize that "published material" does not specify the context?
AMAZON.COM Just like me.
[/quote]
You don't know what context means, do you?
Look which relation R do you have in mind? Just say that much.
[quote]
C-B
--
Jesse F. Hughes
"Well, you know as soon as you have a new number I will be happy to
add it to the list. Don't try those childish tit-for-tat games with
me." -- Ross Finlayson on Cantor's theorem.- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -
--[/quote]
Jesse F. Hughes
"Anything was possible last night. That was the trouble with last
nights. They were always followed by this mornings."
-- Terry Pratchett, /Small Gods/ |
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| Jesse F. Hughes... |
| Posted: Thu Nov 05, 2009 5:32 pm |
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Charlie-Boo <shymathguy at (no spam) gmail.com> writes:
[quote]On Nov 5, 4:45 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 4:02 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 2:09 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 10:29 am, James Burns <burns... at (no spam) 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"
[/quote]
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.
[quote]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.
[/quote]
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?
[quote](1) has a free variable x. It's not clear what you mean by (1).
I would say that x is universally quantified.
[/quote]
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.
[quote]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.
[/quote]
--
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 |
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| Tim Little... |
| Posted: Thu Nov 05, 2009 6:55 pm |
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On 2009-11-05, Jesse F. Hughes <jesse at (no spam) phiwumbda.org> wrote:
[quote]Right. You've quite a skill
[/quote]
Heh. Heh.
[quote](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.
[/quote]
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 |
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| Jesse F. Hughes... |
| Posted: Thu Nov 05, 2009 7:31 pm |
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Tim Little <tim at (no spam) little-possums.net> writes:
[quote]On 2009-11-05, Jesse F. Hughes <jesse at (no spam) 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.
[/quote]
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 |
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| William Elliot... |
| Posted: Fri Nov 06, 2009 2:36 am |
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On Thu, 5 Nov 2009, Charlie-Boo wrote:
[quote]Is there a two-place relation R such that:
Yes.[/quote]
[quote]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)[/quote]
[quote]2. If R(x,y) then y is an element of x.
R(x,y) ==> y in x[/quote]
R(x,y) ==> y in x ==> some z with R(x,z) ==> some z with z in x
[quote]3. If R(x,y) and R(x,z) then y=z.
R is a function[/quote]
[quote]What should it be called?
Choice function.[/quote]
What's the domain and codomain of R? |
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| Jesse F. Hughes... |
| Posted: Fri Nov 06, 2009 10:45 am |
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Charlie-Boo <shymathguy at (no spam) gmail.com> writes:
[quote]The choice function referred to in the Axiom of Choice.
[/quote]
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. |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 2:37 am |
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On Nov 5, 5:32 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 4:45 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 4:02 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 2:09 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 10:29 am, James Burns <burns... at (no spam) 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
[/quote]
*blush*
[quote]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.
[/quote]
Trim it?
[quote]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.
[/quote]
Ok. # 1 = AOC doesn’t involve my R.
[quote]The relation R that you
described would be more like what one would see in an axiom of choice
for classes.
[/quote]
Ok. # 2 = AOC involves my R when talking about classes.
But # 1 => ~(# 2).
[quote]Tell you what. Why don't you write down the axiom of choice and point
out where it involves such an R?
[/quote]
Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
[quote](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) ) ).
[/quote]
No.
C-B
[quote]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 -[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 2:40 am |
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On Nov 5, 7:31 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Tim Little <t... at (no spam) little-possums.net> writes:
On 2009-11-05, Jesse F. Hughes <je... at (no spam) 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.
[/quote]
I thought haste makes waste, not stupidity.
C-B
[quote]--
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 -[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 3:51 am |
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On Nov 6, 2:36 am, William Elliot <ma... at (no spam) rdrop.remove.com> wrote:
[quote]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?
[/quote]
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 doesn’t 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 |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 4:06 am |
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On Nov 10, 7:54 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 7:31 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Tim Little <t... at (no spam) little-possums.net> writes:
On 2009-11-05, Jesse F. Hughes <je... at (no spam) 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.
[/quote]
Good. Thanks.
[quote]I'm sure you've never made a silly blunder
yourself, so obviously you have the right to mock me.
[/quote]
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
[quote]--
"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 -[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 4:11 am |
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On Nov 10, 7:51 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) 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).
[/quote]
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
[quote]--
"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[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 4:13 am |
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On Nov 10, 7:53 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) 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.
[/quote]
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?
C-B
[quote]--
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[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 5:14 am |
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On Nov 10, 9:34 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 10, 7:51 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) 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.
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,
[/quote]
Ok.
[quote]but your conditions claim that there is essentially a *global* choice function,
that is a choice function for the particular class V.
[/quote]
How is that a but - what does it have to do with R being about classes
per se?
[quote] That's not what
AC says.
[/quote]
But that's what I say! That's what CBL says, too. (And CBL proves
all sorts of theorems very easily and amazingly short, due to several
subterfuges in use.)
Why can't we say that instead?
The question is (as you discuss) whether AC = Global AC. At the least
let us add that to the questions discussed, in the mainstream
literature (full of fraud) as well as this counter-technology that we
are now all collectively developing in a huge collaboration.
(Billions access Google.)
But 1st things 1st - I asked you first - is there such an R? (Is it
close enough to what books with pretty covers talk about?)
The first problem with ZF addressing AOC is that ZF doesn't define
what a function is - there are NO REFERENCES to them - so naturally ZF
is consistent with AOC. **
Well DUH, Mr. Godel!
C-B
(** = changes everything)
[quote]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 -[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 5:23 am |
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Guest
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On Nov 10, 9:36 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 10, 7:53 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) 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
[/quote]
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
[quote]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 -[/quote] |
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| Charlie-Boo... |
| Posted: Tue Nov 10, 2009 5:41 am |
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Guest
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On Nov 10, 9:34 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 10, 7:51 am, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
Charlie-Boo <shymath... at (no spam) 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.
[/quote]
A is related to B but A does not involve B?
[quote]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,
[/quote]
I don't even think you should say AC and R are related but
noninvolving.
[quote]but your
conditions claim that there is essentially a *global* choice function,
that is a choice function for the particular class V.
[/quote]
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
[quote] 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 -[/quote] |
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