Does Such a Relation Exist?
Charlie-Boo
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
James Burns
Would you mind sharing why the question is interesting?
(2) and (3) imply that
R(x,y) iff x = {y}
and (1) certainly follows from that (with z = x).
So, yes to your first question; there is such a relation.
I'll leave it to you to come up an answer
to your second question, that is, a name.
Jim Burns
James Burns
> 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?
>
> (2) and (3) imply that
>
> R(x,y) iff x = {y}
>
> and (1) certainly follows from that (with z = x).
Ooops. (1) only follows from (2) and (3) if y is
of the correct type, that is, if y is a singleton
(or empty). So, in order for such a relation
to exist, its range on its first parameter needs
to be restricted to singletons (and the empty set).
This wasn't specified in the problem statement, but,
if the unspecified sets X and Y for which R is
a subset of XxY also satisfy X = Y, then
I think X = { {}, {{}}, {{{}}}, {{{{}}}}, ... }
is the unique solution (err, that and {} ).
Charlie-Boo
> 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.
> (2) and (3) imply that
>
> R(x,y) iff x = {y}
Not quite e.g. R(x,y) iff x = {y,{}} also satisfies (2) and (3).
> and (1) certainly follows from that (with z = x).
See your next post.
> So, yes to your first question; there is such a relation.
>
> I'll leave it to you to come up an answer
> to your second question, that is, a name.
How about “answer to a long standing problem”?
C-B
> Jim Burns
Charlie-Boo
> James Burns 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?
>
> > (2) and (3) imply that
>
> > R(x,y) iff x = {y}
>
> > and (1) certainly follows from that (with z = x).
>
> Ooops. (1) only follows from (2) and (3) if y is
> of the correct type, that is, if y is a singleton
> (or empty). So, in order for such a relation
> to exist, its range on its first parameter needs
> to be restricted to singletons (and the empty set).
>
> This wasn't specified in the problem statement, but,
> if the unspecified sets X and Y for which R is
> a subset of XxY also satisfy X = Y, then
> I think X = { {}, {{}}, {{{}}}, {{{{}}}}, ... }
> is the unique solution
Are you saying that X is a Quine atom (to satisfy 2)?
> (err, that and {} ).
R can’t be {} because (1) needs some tuples in R.
C-B
>
>
> > So, yes to your first question; there is such a relation.
>
> > I'll leave it to you to come up an answer
> > to your second question, that is, a name.
>
> > Jim Burns- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Charlie-Boo
> 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
Michael Stemper
>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!
Jesse F. Hughes
> On Nov 5, 10:29 am, James Burns <burns...@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
Ken Pledger
<997257e0-e214-45fd...@j19g2000vbp.googlegroups.com>,
Charlie-Boo <shyma...@gmail.com> wrote:
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.
Dan Cass
excerpt....
> On Nov 5, 10:02 am, Charlie-Boo
> <shymath...@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
>
Then your condition (1) is true, since z = x does the job.
Charlie-Boo
wrote:
> In article <997257e0-e214-45fd-bd3d-45e0344a0...@j19g2000vbp.googlegroups.com>, Charlie-Boo <shymath...@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
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 5, 10:29 am, James Burns <burns...@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?
Isn’t one enough – after all, it’s 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 -
Charlie-Boo
> In article
> <997257e0-e214-45fd-bd3d-45e0344a0...@j19g2000vbp.googlegroups.com>,
>
> > 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.
Charlie-Boo
Ok, but is that R a relation?
C-B
Jesse F. Hughes
> On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > On Nov 5, 10:29 am, James Burns <burns...@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, so I
still haven't a clue what you're going on about.
>> 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?
>> 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 up http://www.godel.com/ .)
Which relation did he discuss that satisfies these properties?
Forgive me if I won't simply take your word for it.
>> And in what context?
>
> Published material.
You realize that "published material" does not specify the context?
--
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.
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Charlie-Boo <shymath...@gmail.com> writes:
> >> > On Nov 5, 10:29 am, James Burns <burns...@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?
> so I still haven't a clue what you're going on about.
Godel proved theorems in set theory as well as in logic.
> >> 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).
> >> 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.
> >> And in what context?
>
> > Published material.
>
> You realize that "published material" does not specify the context?
AMAZON.COM Just like me.
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 -
Jesse F. Hughes
> On Nov 5, 4:02 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Charlie-Boo <shymath...@gmail.com> writes:
>> >> > On Nov 5, 10:29 am, James Burns <burns...@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.
>> 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.
>> >> 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).
(1) has a free variable x. It's not clear what you mean by (1).
>> >> 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
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).
>> >> And in what context?
>>
>> > Published material.
>>
>> You realize that "published material" does not specify the context?
>
> AMAZON.COM Just like me.
You don't know what context means, do you?
Look which relation R do you have in mind? Just say that much.
>
> 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 -
>
--
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/
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 5, 4:02 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Charlie-Boo <shymath...@gmail.com> writes:
> >> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> >> Charlie-Boo <shymath...@gmail.com> writes:
> >> >> > On Nov 5, 10:29 am, James Burns <burns...@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"
> >> 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.
> >> >> 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).
>
> (1) has a free variable x. It's not clear what you mean by (1).
I would say that x is universally quantified.
> >> >> 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/...
>
> 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.
C-B
> >> >> And in what context?
>
> >> > Published material.
>
> >> You realize that "published material" does not specify the context?
>
> > AMAZON.COM Just like me.
>
> You don't know what context means, do you?
>
> Look which relation R do you have in mind? Just say that much.
>
> > 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 -
>
> --
> 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/- Hide quoted text -
Jesse F. Hughes
> On Nov 5, 4:45 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > On Nov 5, 4:02 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Charlie-Boo <shymath...@gmail.com> writes:
>> >> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> >> Charlie-Boo <shymath...@gmail.com> writes:
>> >> >> > On Nov 5, 10:29 am, James Burns <burns...@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
Tim Little
> 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
Jesse F. Hughes
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
William Elliot
> 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?
Dan Cass
If a relation R is defined as a subset of a product of two sets, then yes.
Jesse F. Hughes
> 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.
Jesse F. Hughes
No, not the way that Charlie specified it! He didn't say that for
every set w, there is a relation R such that for all y in w.... .
He said there is a relation R such that for all y ... . Thus, his R
is a proper *class* of ordered pairs, not a subset of a product of two
sets.
Seems to me that he's talking about a global choice function, which is
not what AC alleges exists.
--
Jesse F. Hughes
"If you really think there's a bug you should report a bug. Maybe
you're not using it properly... It turns out Luddites don't know how
to use software properly, so you should look into that." -- Bill Gates
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 5, 4:45 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Charlie-Boo <shymath...@gmail.com> writes:
> >> > On Nov 5, 4:02 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> >> Charlie-Boo <shymath...@gmail.com> writes:
> >> >> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> >> >> Charlie-Boo <shymath...@gmail.com> writes:
> >> >> >> > On Nov 5, 10:29 am, James Burns <burns...@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 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).
> 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 -
Charlie-Boo
> Tim Little <t...@little-possums.net> writes:
> > On 2009-11-05, Jesse F. Hughes <je...@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 -
Jesse F. Hughes
> On Nov 5, 7:31 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Tim Little <t...@little-possums.net> writes:
>> > On 2009-11-05, Jesse F. Hughes <je...@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.
Jesse F. Hughes
>> 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
Jesse F. Hughes
>> 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
Charlie-Boo
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
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 5, 7:31 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Tim Little <t...@little-possums.net> writes:
> >> > On 2009-11-05, Jesse F. Hughes <je...@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 -
Charlie-Boo
> Charlie-Boo <shymath...@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.
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
Charlie-Boo
> Charlie-Boo <shymath...@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?
C-B
Charlie-Boo
And is there a set (or superset) of all nonempty sets?
C-B
Jesse F. Hughes
> On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@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
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
Jesse F. Hughes
> On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@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, but your
conditions claim that there is essentially a *global* choice function,
that is a choice function for the particular class V. 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.
Jesse F. Hughes
> Charlie-Boo <shyma...@gmail.com> writes:
>
>> On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>> Charlie-Boo <shymath...@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
> gives a hint as to how one would show Global AC does not imply AC,
^^^^^^^^^^^^^^^^^^^^^^^^^^^
> but I'm not familiar with that argument.
Sorry, I meant to say "AC does not imply Global AC".
--
Jesse F. Hughes
"My baby don't allow me in the kitchen
and I've come to love her decision."
-- Bad Livers
Charlie-Boo
Ok.
> but your conditions claim that there is essentially a *global* choice function,
> that is a choice function for the particular class V.
How is that a but - what does it have to do with R being about classes
per se?
> That's not what
> AC says.
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)
> 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 -
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Charlie-Boo <shymath...@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
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
> 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 -
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Charlie-Boo <shymath...@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.
A is related to B but A does not involve B?
> > 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,
I don't even think you should say AC and R are related but
noninvolving.
> but your
> conditions claim that there is essentially a *global* choice function,
> that is a choice function for the particular class V.
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
> Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.- Hide quoted text -
Aatu Koskensilta
> 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.
Your figgering can be backed up with a logical result. It is also a
logical result, an easy and illustrative application of forcing, that
any invocation of global choice in a proof of a result about sets only
can be eliminated (given ordinary choice).
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Jesse F. Hughes
> On Nov 10, 9:34 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Charlie-Boo <shymath...@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.
>
> A is related to B but A does not involve B?
Let us not quibble on such dull matters of terminology.
>> > 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,
>
> I don't even think you should say AC and R are related but
> noninvolving.
>
>> but your
>> conditions claim that there is essentially a *global* choice function,
>> that is a choice function for the particular class V.
>
> 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.)
No idea what you're going on about.
--
Jesse F. Hughes
"I am the next legend--living, breathing and solving mega problems in
the here and now." -- James S. Harris
Jesse F. Hughes
> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>
>> 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.
>
> Your figgering can be backed up with a logical result. It is also a
> logical result, an easy and illustrative application of forcing, that
> any invocation of global choice in a proof of a result about sets only
> can be eliminated (given ordinary choice).
It is, of course, only coincidental when my figgering and the truth
line up so well.
--
"Tempted and tried we're oft made to wonder
Why it should be thus all the day long
When there are others living about us
Never molested though in the wrong." -- Bad Livers, "Farther Along"
Jesse F. Hughes
> On Nov 10, 9:34 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > 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,
>
> Ok.
>
>> but your conditions claim that there is essentially a *global* choice function,
>> that is a choice function for the particular class V.
>
> How is that a but - what does it have to do with R being about classes
> per se?
Your conditions of R amount to the axiom of choice for a particular
class, not for all classes of sets.
>> That's not what
>> AC says.
>
> 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.)
But who cares what you say? You said that lots of people are writing
about relations R satisfying those three conditions. That's just not
true.
> 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.)
Well, you can certainly ask that question. Seems to me that the
answer is almost certainly "no", but I haven't a proof of that fact.
Do you have any argument why the answer may be "yes"?
> 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?)
In the theory ZFC, certainly not (because R would not be a set). If
we amend ZFC so that it makes sense to speak of proper classes, then I
have no proof that there is no class R satisfying your conditions.
Nor do I have a proof that there is such a class. Moreover, I
sincerely doubt that the latter claim is provable (though I haven't an
argument to that effect).
> 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. **
I'm not sure what you're going on about. The axioms of ZF do not
define function, but it is easy enough to introduce such a
definition. Here it is:
Let f, X and Y be sets. Then f is a function with domain X and
codomain Y (written f:X -> Y) iff the following hold:
(1) f c X x Y (f is a subset of X x Y)
(2) (Ax in X)(Ey in Y)( <x,y> in f )
(3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) ->
y = y' )
What's the issue?
>> 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.
--
"It's my belief that when religion and pseudoscience achieve an
official status within a culture [...], then genocide, war,
oppression, injustice, and economic stagnation are sure to follow."
-- David Petry, on why |X| < |P(X)| is bad, bad, bad.
Jesse F. Hughes
> 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.)
The axiom of choice is independent of ZF, you know. Thus, someone
proved that ZF does not imply AC (also, that ZF does not imply ~AC).
It has also been proved that ZF + CC is independent of AC. Thus,
ZF + CC does not prove AC (nor its negation).
No idea what you're going to conclude from this little morsel. I
conclude a few things (ZF + ~AC is equiconsistent to ZF, and so is
ZF + AC, for instance), but all of my conclusions are obvious and
well-known.
--
"There are people [...] who think it's socially acceptable to level
accusations of mental illness in insulting exchanges to make
points[...] [They] are rather sick [them]selves, and in reality, are
sociopathic." --- James Harris, evidently a self-described sociopath
David Hartley
<je...@phiwumbda.org> writes
>In the theory ZFC, certainly not (because R would not be a set). If we
>amend ZFC so that it makes sense to speak of proper classes, then I
>have no proof that there is no class R satisfying your conditions. Nor
>do I have a proof that there is such a class. Moreover, I sincerely
>doubt that the latter claim is provable (though I haven't an argument
>to that effect).
"V=L" provides a universal well-ordering and so a universal choice
function. So it is consistent with ZFC (+ classes) that such an R
exists.
--
David Hartley
Jesse F. Hughes
Thanks for the clarification. Thus, there is no proof that the class
R does not exist (in ZFC + classes). And, if I understood Aatu's
post, there is similarly no proof that the class R exists. Hence,
Global AC is independent of ZFC + classes.
I hope I got that right.
If so, surely, that is the answer to Charlie's question (though not,
I'd wager, the answer he wanted to receive).
--
Jesse F. Hughes
"The sole cause of all human misery is the inability of people
to sit quietly in their rooms." -- Blaise Pascal
Aatu Koskensilta
> Hence, Global AC is independent of ZFC + classes.
Yep.
Herman Jurjus
> In the theory ZFC, certainly not (because R would not be a set). If
Is there a difference between 'ZFC with classes' and NBG?
--
Cheers,
Herman Jurjus
Jesse F. Hughes
> Jesse F. Hughes wrote:
>> In the theory ZFC, certainly not (because R would not be a set). If
>> we amend ZFC so that it makes sense to speak of proper classes, ...
>
> Is there a difference between 'ZFC with classes' and NBG?
I've no idea, since I'm unfamiliar with NBG. But if NBG simply
extends ZFC by adding a Set predicate and appropriate axioms (a set
union is a set, etc.), then that's what I have in mind.
--
Jesse F. Hughes
"To be honest, I don't have enough interest in math to spend the time
it would take to clean up the mess that I believe has been created in
the past 100 or so years." -- Curt Welch lets the world down.
Herman Rubin
>> Jesse F. Hughes wrote:
>>> In the theory ZFC, certainly not (because R would not be a set). If
>>> we amend ZFC so that it makes sense to speak of proper classes, ...
>> Is there a difference between 'ZFC with classes' and NBG?
>I've no idea, since I'm unfamiliar with NBG. But if NBG simply
>extends ZFC by adding a Set predicate and appropriate axioms (a set
>union is a set, etc.), then that's what I have in mind.
What is "ZFC with classes"?
It is the case that any theorem of NBG which does not
involve classes is a theorem of ZF. The same holds if
choice is added.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Aatu Koskensilta
> What is "ZFC with classes"?
It doesn't really matter. Any second-order formulation of set theory
will do, with predicative or full class comprehension.
> It is the case that any theorem of NBG which does not involve classes
> is a theorem of ZF. The same holds if choice is added.
What was at issue was global choice. To establish that NBG is
conservative over ZF and NBG with choice conservative over ZFC we either
use cut-elimination (for predicative second-order logic) or a simple
model-theoretic construction. To show that global choice is conservative
(a particularly simple form of) forcing is needed.
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > 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.)
>
> The axiom of choice is independent of ZF, you know. Thus, someone
> proved that ZF does not imply AC (also, that ZF does not imply ~AC).
>
> It has also been proved that ZF + CC is independent of AC. Thus,
> ZF + CC does not prove AC (nor its negation).
>
> No idea what you're going to conclude from this little morsel.
I just realized that if AC is true then ZF+CC (anything) proves it, so
if ZF+CC does not prove AC then AC must be false.
Fancy that!
C-B
If I can't write a book on new theorems of Logic, maybe I can at least
write one on falacies (by perusing my many posts of the past.)
Charlie-Boo
Set Theory is all about what can be formally proven using a specific
set of axioms. If you add more (axioms, definitions) then you are not
addressing the same question.
C-B
> -- David Petry, on why |X| < |P(X)| is bad, bad, bad.- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
Jesse F. Hughes
> On Nov 10, 1:37 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > 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.)
>>
>> The axiom of choice is independent of ZF, you know. Thus, someone
>> proved that ZF does not imply AC (also, that ZF does not imply ~AC).
>>
>> It has also been proved that ZF + CC is independent of AC. Thus,
>> ZF + CC does not prove AC (nor its negation).
>>
>> No idea what you're going to conclude from this little morsel.
>
> I just realized that if AC is true then ZF+CC (anything) proves it, so
> if ZF+CC does not prove AC then AC must be false.
Well, then it also follows that the axiom of extensionality is false,
too. After all, the axiom of extensionality is independent of
ZF-extensionality.
Similarly, the axiom that zero is not a successor of any natural
number is false, since it is not proved by PA-(zero is not a
successor). And so on.
And finally, it also follows (by the exact same reasoning) that ~AC is
false, too. Thus, you've managed to prove that *both* AC and its
negation is false! All before breakfast, too! Congrats!
> Fancy that!
Yes, fancy that. You have a remarkable insight into mathematical
logic.
--
"Flowers in the Attic" was based on a true story. [...] HOW is it OK
to just butcher such an awesome piece of work? It's like passing
Pokemon off as the Mona-Lisa; sick and entirely wrong.
-- An Amazon reviewer pissed that the movie didn't include incest
Jesse F. Hughes
> On Nov 10, 1:32 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> I'm not sure what you're going on about. The axioms of ZF do not
>> define function, but it is easy enough to introduce such a
>> definition. Here it is:
>>
>> Let f, X and Y be sets. Then f is a function with domain X and
>> codomain Y (written f:X -> Y) iff the following hold:
>>
>> (1) f c X x Y (f is a subset of X x Y)
>> (2) (Ax in X)(Ey in Y)( <x,y> in f )
>> (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) ->
>> y = y' )
>>
>> What's the issue?
>
> Set Theory is all about what can be formally proven using a specific
> set of axioms. If you add more (axioms, definitions) then you are not
> addressing the same question.
The introduction of definitions for function (and ordered pair and so
on) are simple niceties, done in order to shorten presentations and
make theorems and their proofs more readable. You seem to think that
something more substantial is going on.
Any statement involving the predicate "f is a function" can be
rewritten without that predicate by inserting the conjunction of
(1)-(3). To be perfectly pedantic, we would have to rewrite (1)-(3)
to eliminate the cross-product and ordered pair notations, but this is
certainly doable.
--
Jesse F. Hughes
"But you have to support spyware if you're going to have free
file-sharing applications. Fair's fair."
-- NYU student Keith Caron in Wired
Charlie-Boo
> Charlie-Boo <shymath...@gmail.com> writes:
> > On Nov 10, 1:32 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> I'm not sure what you're going on about. The axioms of ZF do not
> >> define function, but it is easy enough to introduce such a
> >> definition. Here it is:
>
> >> Let f, X and Y be sets. Then f is a function with domain X and
> >> codomain Y (written f:X -> Y) iff the following hold:
>
> >> (1) f c X x Y (f is a subset of X x Y)
> >> (2) (Ax in X)(Ey in Y)( <x,y> in f )
> >> (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) ->
> >> y = y' )
>
> >> What's the issue?
>
> > Set Theory is all about what can be formally proven using a specific
> > set of axioms. If you add more (axioms, definitions) then you are not
> > addressing the same question.
>
> The introduction of definitions for function (and ordered pair and so
> on) are simple niceties, done in order to shorten presentations and
> make theorems and their proofs more readable. You seem to think that
> something more substantial is going on.
But the "definition" is subjective. Might not two different
definitions of function using relations lead to different results?
That's why an axiomatic system requires you to specify everything,
which ZF unfortunately does not.
C-B
> Any statement involving the predicate "f is a function" can be
> rewritten without that predicate by inserting the conjunction of
> (1)-(3). To be perfectly pedantic, we would have to rewrite (1)-(3)
> to eliminate the cross-product and ordered pair notations, but this is
> certainly doable.
>
> --
> Jesse F. Hughes
> "But you have to support spyware if you're going to have free
> file-sharing applications. Fair's fair."
> -- NYU student Keith Caron in Wired- Hide quoted text -
Herman Rubin
Charlie-Boo <shyma...@gmail.com> wrote:
>On Nov 16, 8:25=A0am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> >> I'm not sure what you're going on about. =A0The axioms of ZF do not
>> >> define function, but it is easy enough to introduce such a
>> >> =A0 Let f, X and Y be sets. =A0Then f is a function with domain X and
>> >> =A0 codomain Y (written f:X -> Y) iff the following hold:
>> >> =A0 (1) f c X x Y =A0 =A0(f is a subset of X x Y)
>> >> =A0 (2) (Ax in X)(Ey in Y)( <x,y> in f )
>> >> =A0 (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) ->
>> >> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =
>=A0 y =3D y' )
>> >> What's the issue?
>> > Set Theory is all about what can be formally proven using a specific
>> > set of axioms. =A0If you add more (axioms, definitions) then you are no=
>t
>> > addressing the same question.
>> The introduction of definitions for function (and ordered pair and so
>> on) are simple niceties, done in order to shorten presentations and
>> make theorems and their proofs more readable. =A0You seem to think that
>> something more substantial is going on.
>But the "definition" is subjective. Might not two different
>definitions of function using relations lead to different results?
>That's why an axiomatic system requires you to specify everything,
>which ZF unfortunately does not.
The definition of ordered pair is irrelevant. Whatever
definition is used can be shown equivalent to the usual
one, and this one is not the original one.
In fact, the von Neumann axiomatization of set theory,
the first one with a finite number of axioms (not
axiom schemata), has function as a primitive; it is
equivalent to the NBG system (von Neumann-Bernays-Godel)
in which the usual definition of ordered pair is used.
Charlie-Boo
> In article <77840a3f-bde8-4f61-a35f-31554ffea...@h10g2000vbm.googlegroups.com>,
Some definitions have been shown to be inconsistent. Some authors
have writen "if this form of definition is legitimate" (I believe
Kleene was one.) People debated for years whether definitions can
refer to themselves. Whether to allow certain axioms has never been
settled.
Something is missing from ZF: a definition of function. Yet you say
it's irrelevant???
You cannot dismiss an omission in such a cavalier manner.
> Whatever
> definition is used can be shown equivalent to the usual
> one,
How do you know that? You cannot systematically consider every
possible definition that anyone might propose. That makes no sense.
"It doesn't matter. They're all the same." Is there any logic that
supports such a broad statement as that?
> and this one is not the original one.
>
> In fact, the von Neumann axiomatization of set theory,
> the first one with a finite number of axioms (not
> axiom schemata), has function as a primitive;
That's right - they were smarter. But ZF did not.
C-B
> it is
> equivalent to the NBG system (von Neumann-Bernays-Godel)
> in which the usual definition of ordered pair is used.
> --
> This address is for information only. I do not claim that these views
> are those of the Statistics Department or of Purdue University.
> Herman Rubin, Department of Statistics, Purdue University
> hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558- Hide quoted text -
Herman Rubin
Charlie-Boo <shyma...@gmail.com> wrote:
>On Dec 2, 1:39=A0pm, hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
>> In article <77840a3f-bde8-4f61-a35f-31554ffea...@h10g2000vbm.googlegroups=
>.com>,
>> Charlie-Boo =A0<shymath...@gmail.com> wrote:
>> >On Nov 16, 8:25am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Charlie-Boo <shymath...@gmail.com> writes:
>> >> > On Nov 10, 1:32pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrot=
>e:
>> >> >> I'm not sure what you're going on about. The axioms of ZF do n=
>ot
>> >> >> define function, but it is easy enough to introduce such a
>> >> >> definition. Here it is:
>> >> >> Let f, X and Y be sets. Then f is a function with domain=
> X and
>> >> >> codomain Y (written f:X -> Y) iff the following hold:
>> >> >> (1) f c X x Y (f is a subset of X x Y)
>> >> >> (2) (Ax in X)(Ey in Y)( <x,y> in f )
>> >> >> (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f =
>) -> y = y' )
>> >> >> What's the issue?
>> >> > Set Theory is all about what can be formally proven using a specific
>> >> > set of axioms. If you add more (axioms, definitions) then you a=
>re not
>> >> > addressing the same question.
>> >> The introduction of definitions for function (and ordered pair and so
>> >> on) are simple niceties, done in order to shorten presentations and
>> >> make theorems and their proofs more readable. You seem to think t=
>hat
>> >> something more substantial is going on.
>> >But the "definition" is subjective. Might not two different
>> >definitions of function using relations lead to different results?
>> >That's why an axiomatic system requires you to specify everything,
>> >which ZF unfortunately does not.
>> The definition of ordered pair is irrelevant.
>Some definitions have been shown to be inconsistent. Some authors
>have writen "if this form of definition is legitimate" (I believe
>Kleene was one.) People debated for years whether definitions can
>refer to themselves. Whether to allow certain axioms has never been
>settled.
>Something is missing from ZF: a definition of function. Yet you say
>it's irrelevant???
>You cannot dismiss an omission in such a cavalier manner.
I can and do. In von Neumann's thesis, function is primitive,
and a class is defined as a function which takes on two designated
values only. Yours assumes that X x Y exists; this is not needed
if you replace (1) and (2) by
(Au in f)(Ex in X)(Ey in Y)(u = <x,y>).
Now in this form, <x,y> can be replaced by <<1,x><2,y>>, or
other formulations. The essence is preserved.
There have been many "definitions" of ordered pair; the one
now used is the simplest. What is usually called a function
should really be called a representation of that function,
and this can be useful. If a different representation was
used, one could still refer to this representation if needed
to prove theorems, but nothing important would be changed.
>> Whatever
>> definition is used can be shown equivalent to the usual
>> one,
>How do you know that? You cannot systematically consider every
>possible definition that anyone might propose. That makes no sense.
>"It doesn't matter. They're all the same." Is there any logic that
>supports such a broad statement as that?
No, but you could introduce the current on and show that
the current representation could be produced from whatever
other one was used, by showing that a class with the
desired properties can be produced therefrom.
>> and this one is not the original one.
>> In fact, the von Neumann axiomatization of set theory,
>> the first one with a finite number of axioms (not
>> axiom schemata), has function as a primitive;
>That's right - they were smarter. But ZF did not.
Since NBG is a conservative extension of ZF allowing
proper classes, this is not the case. In NBG, one
usually uses the current definition, which was the
result of several papers by Kuratowski and Wiener.
Bernays replaced the von Neumann approach with the
current use of ordered pairs, and it was easily seen
to be equivalent. Godel equated classes and sets
with the same elements.