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| Charlie-Boo... |
 Posted: Thu Nov 05, 2009 5:02 am |
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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 |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 8:45 am |
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On Nov 5, 10:29 am, James Burns <burns... at (no spam) osu.edu> wrote:
[quote]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?
[/quote]
A lot of people write about it.
[quote](2) and (3) imply that
R(x,y) iff x = {y}
[/quote]
Not quite e.g. R(x,y) iff x = {y,{}} also satisfies (2) and (3).
[quote]and (1) certainly follows from that (with z = x).
[/quote]
See your next post.
[quote]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.
[/quote]
How about “answer to a long standing problem”?
C-B
> Jim Burns |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 8:47 am |
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On Nov 5, 10:43 am, James Burns <burns... at (no spam) osu.edu> wrote:
[quote]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
[/quote]
Are you saying that X is a Quine atom (to satisfy 2)?
[quote](err, that and {} ).
[/quote]
R can’t be {} because (1) needs some tuples in R.
C-B
[quote]
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 -
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 8:50 am |
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On Nov 5, 10:02 am, Charlie-Boo <shymath... at (no spam) gmail.com> wrote:
[quote]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).
[/quote]
Is there a relation R that meets (1) ? (Or something that is not a
relation.)
C-B
[quote]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[/quote] |
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| Michael Stemper... |
| Posted: Thu Nov 05, 2009 9:03 am |
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In article <997257e0-e214-45fd-bd3d-45e0344a0a03 at (no spam) j19g2000vbp.googlegroups.com>, Charlie-Boo <shymathguy at (no spam) gmail.com> writes:
[quote]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.
[/quote]
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?
[quote]3. If R(x,y) and R(x,z) then y=z.
[/quote]
I guess so. At least, this is consistent with the first two.
--
Michael F. Stemper
#include <Standard_Disclaimer>
Reunite Gondwanaland! |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 10:15 am |
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On Nov 5, 2:03 pm, mstem... at (no spam) walkabout.empros.com (Michael Stemper)
wrote:
[quote]In article <997257e0-e214-45fd-bd3d-45e0344a0... at (no spam) j19g2000vbp.googlegroups..com>, Charlie-Boo <shymath... at (no spam) 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?
[/quote]
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.
[quote]3. If R(x,y) and R(x,z) then y=z.
I guess so. At least, this is consistent with the first two.
[/quote]
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
[quote]--
Michael F. Stemper
#include <Standard_Disclaimer
Reunite Gondwanaland![/quote] |
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| James Burns... |
| Posted: Thu Nov 05, 2009 10:29 am |
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Charlie-Boo wrote:
[quote]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?
[/quote]
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 |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 10:34 am |
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On Nov 5, 2:09 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]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?
[/quote]
Does the name “Godel” ring a bell?
[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).
[/quote]
Yes.
[quote]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 != {}
[/quote]
You mean x != {} ?
[quote]satisfies (1'), (2) and (3), but such an operation requires a version
of AC on classes (or so it seems to me).
[/quote]
How about (1) alone – can we satisfy that with a relation R?
[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?
[/quote]
Isn’t one enough – after all, it’s Godel (yes, THE Godel - not that
Accountant who keeps showing up http://www.godel.com/ .)
[quote]And in what context?
[/quote]
Published material.
C-B
[quote]--
"I'm a very well-educated, successful, intelligent person. This is
insane to me that I have an armed guard outside my door when I've
cooperated with everything other than the whole solitary-confinement-
in-Italy thing." --A. Speaker, on the whole T.B.-quarantined thing- Hide quoted text -
- Show quoted text -[/quote] |
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| James Burns... |
| Posted: Thu Nov 05, 2009 10:43 am |
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James Burns wrote:
[quote]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).
[/quote]
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 {} ).
[quote]
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[/quote] |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 10:51 am |
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On Nov 5, 2:49 pm, Ken Pledger <ken.pled... at (no spam) mcs.vuw.ac.nz> wrote:
[quote]In article
997257e0-e214-45fd-bd3d-45e0344a0... at (no spam) j19g2000vbp.googlegroups.com>,
Charlie-Boo <shymath... at (no spam) gmail.com> wrote:
Is there a two-place relation R such that:
1. If x is an element of y then there exists a z such that R(y,z).
2. If R(x,y) then y is an element of x.
3. If R(x,y) and R(x,z) then y=z.
What should it be called?
C-B
It looks suspiciously like a choice function.
[/quote]
Don't be too suspicious or else someone might prescribe risperdal.
[quote]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?
[/quote]
Yes!
[quote]That's a pretty ambitious attempt.
[/quote]
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. |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 11:24 am |
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On Nov 5, 4:02 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]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,
[/quote]
Why would they be theorems if he didn't prove them?
[quote]so I still haven't a clue what you're going on about.
[/quote]
Godel proved theorems in set theory as well as in logic.
[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?
[/quote]
Ask the original question but with only (1) instead of (1), (2) and
(3).
[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.
[/quote]
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]And in what context?
Published material.
You realize that "published material" does not specify the context?
[/quote]
AMAZON.COM Just like me.
C-B
[quote]--
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] |
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| Charlie-Boo... |
| Posted: Thu Nov 05, 2009 12:08 pm |
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On Nov 5, 4:45 pm, "Jesse F. Hughes" <je... at (no spam) phiwumbda.org> wrote:
[quote]Charlie-Boo <shymath... at (no spam) gmail.com> writes:
On Nov 5, 4:02 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.
[/quote]
You said, "Goedel didn't prove his famous theorems"
[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.
[/quote]
The choice function referred to in the Axiom of Choice.
[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).
(1) has a free variable x. It's not clear what you mean by (1).
[/quote]
I would say that x is universally quantified.
[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/...
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]
Godel proved that the Axiom of Choice is consistent with ZF.
C-B
[quote]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 -
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- Show quoted text -- Hide quoted text -
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| Jesse F. Hughes... |
| Posted: Thu Nov 05, 2009 2:09 pm |
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Charlie-Boo <shymathguy at (no spam) gmail.com> writes:
[quote]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.
[/quote]
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 |
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| Ken Pledger... |
| Posted: Thu Nov 05, 2009 2:49 pm |
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Guest
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In article
<997257e0-e214-45fd-bd3d-45e0344a0a03 at (no spam) j19g2000vbp.googlegroups.com>,
Charlie-Boo <shymathguy at (no spam) gmail.com> wrote:
[quote]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
[/quote]
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. |
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| Jesse F. Hughes... |
| Posted: Thu Nov 05, 2009 4:02 pm |
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Charlie-Boo <shymathguy at (no spam) gmail.com> writes:
[quote]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?
[/quote]
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.
[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 != {} ?
[/quote]
Yes, that's what I meant.
[quote]
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?
[/quote]
What does (1) alone mean?
[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 up http://www.godel.com/ .)
[/quote]
Which relation did he discuss that satisfies these properties?
Forgive me if I won't simply take your word for it.
[quote]And in what context?
Published material.
[/quote]
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. |
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