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persres@googlemail.com
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 Posted: Fri Jul 11, 2008 10:46 pm Post subject: Pair set axiom in ZF set theory |
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I have a question about the pair set axiom in set theory. Its rather a
question about how to interpret a first order formula.
The pair set axiom says :
(for any y) (there exists x) (y E X <=> y=a V y=b).
Now, thinking about it a little I got myself confused. This seems
wrong to me (obviously it wont be wrong  ).
It seems to me that even if x had only a or only b, this would be
still true. It doesnt have to have both a and b.
say y=a , x = {a}
then
a E {a} <=> a=a V a=b.
which is true.
x doesnt have to be {a, b} which is what the pair set axiom should
establish. Isnt it true?.
So, where am I going wrong?.
Thanks |
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Dave Seaman
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| Posted: Fri Jul 11, 2008 10:58 pm Post subject: Re: Pair set axiom in ZF set theory |
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On Fri, 11 Jul 2008 15:46:43 -0700 (PDT), persres@googlemail.com wrote:
| Quote: |
I have a question about the pair set axiom in set theory. Its rather a
question about how to interpret a first order formula.
The pair set axiom says :
(for any y) (there exists x) (y E X <=> y=a V y=b).
|
No, the pairing axiom says:
(for any a) (for any b) (there exists x) (for all y)
(y e x <=> y=a V y = b).
| Quote: |
Now, thinking about it a little I got myself confused. This seems
wrong to me (obviously it wont be wrong ).
It seems to me that even if x had only a or only b, this would be
still true. It doesnt have to have both a and b.
say y=a , x = {a}
then
a E {a} <=> a=a V a=b.
which is true.
x doesnt have to be {a, b} which is what the pair set axiom should
establish. Isnt it true?.
So, where am I going wrong?.
Thanks
|
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php> |
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persres@googlemail.com
Guest
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| Posted: Fri Jul 11, 2008 11:38 pm Post subject: Re: Pair set axiom in ZF set theory |
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isn't any free variable treated as a for any.
I mean the one you have written is exactly the same as I have. Just
that yours is the universal closure. I have treated a and b as free.
(And one more thing. For any is the same as for all. Isnt it? you have
written both below, just wanted to clarify.)
In any case the form I have written is taken from the book
"Intermediate set theory" by "Drake and Singh" , page 28.
That cant be wrong. But please do someone explain also whether the
universal closure of a formula says exactly what the ones with free
variable say.
Thanks
On Jul 11, 11:58 pm, Dave Seaman <dsea...@no.such.host> wrote:
| Quote: |
On Fri, 11 Jul 2008 15:46:43 -0700 (PDT), pers...@googlemail.com wrote:
I have a question about the pair set axiom in set theory. Its rather a
question about how to interpret a first order formula.
The pair set axiom says :
(for any y) (there exists x) (y E X <=> y=a V y=b).
No, the pairing axiom says:
(for any a) (for any b) (there exists x) (for all y)
(y e x <=> y=a V y = b).
Now, thinking about it a little I got myself confused. This seems
wrong to me (obviously it wont be wrong ).
It seems to me that even if x had only a or only b, this would be
still true. It doesnt have to have both a and b.
say y=a , x = {a}
then
a E {a} <=> a=a V a=b.
which is true.
x doesnt have to be {a, b} which is what the pair set axiom should
establish. Isnt it true?.
So, where am I going wrong?.
Thanks
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
http://www.indybay.org/newsitems/2008/03/29/18489281.php |
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smn
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| Posted: Sat Jul 12, 2008 1:45 am Post subject: Re: Pair set axiom in ZF set theory |
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On Jul 11, 3:46 pm, "pers...@googlemail.com" <pers...@googlemail.com>
wrote:
| Quote: |
I have a question about the pair set axiom in set theory. Its rather a
question about how to interpret a first order formula.
|
The pair set axiom says :
(for any y) (there exists x) (y E X <=> y=a V y=b).
I think you mean to say that there is a set X so that the statement
(yEX<=>y=a or y=b ) holds for all choices of y {you don't want to
allow a different X for each y}.To say this you have to reverse the
phrases in the first two parenthesis -thats the convention everybody
uses.
If a |= b and X had only the element a as a member then for the
choice y=b the statement yEX would be wrong (b is not in X) but the
statement y=a or y =b is correct (" b=a OR b=b "; is a true statement)
so the equivalence is false.
Thus X= a set with just a as its only member would not work .
The set you want is the one with members a AND b but when you write
it with equivalence statements the conjunction OR is used .This might
be what is really bothering you .
For instance the set A union B is frequently spoken of as A and B (or
A + B) but y E( A union B )<=> y E A or y E B ) Regards smn
| Quote: |
Now, thinking about it a little I got myself confused. This seems
wrong to me (obviously it wont be wrong ).
It seems to me that even if x had only a or only b, this would be
still true. It doesnt have to have both a and b.
say y=a , x = {a}
then
a E {a} <=> a=a V a=b.
which is true.
x doesnt have to be {a, b} which is what the pair set axiom should
establish. Isnt it true?.
So, where am I going wrong?.
Thanks |
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Dave Seaman
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| Posted: Sat Jul 12, 2008 2:53 am Post subject: Re: Pair set axiom in ZF set theory |
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On Fri, 11 Jul 2008 16:38:07 -0700 (PDT), persres@googlemail.com wrote:
| Quote: |
isn't any free variable treated as a for any.
I mean the one you have written is exactly the same as I have. Just
that yours is the universal closure. I have treated a and b as free.
|
Aren't you overlooking the small matter of the order of the quantifiers?
| Quote: |
(And one more thing. For any is the same as for all. Isnt it? you have
written both below, just wanted to clarify.)
In any case the form I have written is taken from the book
"Intermediate set theory" by "Drake and Singh" , page 28.
That cant be wrong. But please do someone explain also whether the
universal closure of a formula says exactly what the ones with free
variable say.
Thanks
|
If your version "can't be wrong", then where is the flaw in your own
criticism of the formula? And what happens if you try to apply the same
criticism to the version that I wrote?
| Quote: |
On Jul 11, 11:58 pm, Dave Seaman <dsea...@no.such.host> wrote:
On Fri, 11 Jul 2008 15:46:43 -0700 (PDT), pers...@googlemail.com wrote:
I have a question about the pair set axiom in set theory. Its rather a
question about how to interpret a first order formula.
The pair set axiom says :
(for any y) (there exists x) (y E X <=> y=a V y=b).
No, the pairing axiom says:
(for any a) (for any b) (there exists x) (for all y)
(y e x <=> y=a V y = b).
Now, thinking about it a little I got myself confused. This seems
wrong to me (obviously it wont be wrong ).
It seems to me that even if x had only a or only b, this would be
still true. It doesnt have to have both a and b.
say y=a , x = {a}
then
a E {a} <=> a=a V a=b.
which is true.
x doesnt have to be {a, b} which is what the pair set axiom should
establish. Isnt it true?.
So, where am I going wrong?.
Thanks
|
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php> |
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smn
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| Posted: Sun Jul 13, 2008 10:11 am Post subject: Re: Pair set axiom in ZF set theory |
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On Jul 12, 10:53 am, no comment <adler.m...@gmail.com> wrote:
| Quote: |
On Jul 12, 9:52 am, "pers...@googlemail.com" <pers...@googlemail.com
wrote:
Ok, I didnt know the order matters.
The book gives the order that you both have said it would.
So, that explains it. Thanks.
As for universal closure, any help on how some authors treat it
differently from free variables?.
For instance, as of now, my understanding is that the pair set axiom
can be written as :
(for any a,b) (there exists x) (for any y)(y E x <=> y=a V y=b).
My understanding is that the above means exactly the same as :
(there exists x) (for any y)(y E x <=> y=a V y=b).
However, you have mentioned that some authors have different
conventions. Could either of you please explain the conventions.
I am not able to see how any other convention could give them a
differen meaning.
Thanks
On Jul 12, 1:37 pm, David C. Ullrich <dullr...@sprynet.com> wrote:
On Fri, 11 Jul 2008 16:38:07 -0700 (PDT), "pers...@googlemail.com"
pers...@googlemail.com> wrote:
isn't any free variable treated as a for any.
I mean the one you have written is exactly the same as I have.
No it's not. "for any x there exists y" is very different
from "(there exists y)(for any x)".
Just
that yours is the universal closure. I have treated a and b as free.
(And one more thing. For any is the same as for all. Isnt it? you have
written both below, just wanted to clarify.)
In any case the form I have written is taken from the book
"Intermediate set theory" by "Drake and Singh" , page 28.
I doubt that.
That cant be wrong. But please do someone explain also whether the
universal closure of a formula says exactly what the ones with free
variable say.
There are different conventions on that - yes for some authors,
no for others.
But that's not the only difference between the two versions
that you say are the same.
Thanks
On Jul 11, 11:58 pm, Dave Seaman <dsea...@no.such.host> wrote:
On Fri, 11 Jul 2008 15:46:43 -0700 (PDT), pers...@googlemail.com wrote:
I have a question about the pair set axiom in set theory. Its rather a
question about how to interpret a first order formula.
The pair set axiom says :
(for any y) (there exists x) (y E X <=> y=a V y=b)..
No, the pairing axiom says:
(for any a) (for any b) (there exists x) (for all y)
(y e x <=> y=a V y = b).
Now, thinking about it a little I got myself confused. This seems
wrong to me (obviously it wont be wrong ).
It seems to me that even if x had only a or only b, this would be
still true. It doesnt have to have both a and b.
say y=a , x = {a}
then
a E {a} <=> a=a V a=b.
which is true.
x doesnt have to be {a, b} which is what the pair set axiom should
establish. Isnt it true?.
So, where am I going wrong?.
Thanks
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
http://www.indybay.org/newsitems/2008/03/29/18489281.php
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)- Hide quoted text -
- Show quoted text -
Hello,
In a sense, there is only one convention. It comes down to the
definition of what it means for a formula in a first order language L
to be true in an L-structure M.
Some people define truth ONLY for sentences, that is, for formulas
that have no free occurrences of any variable. If that sort of
definition is used, your original formula in which the occurrences of
a and b are free does not have a truth value in M.
Others define truth for all formulas, essentially by saying that a
formula with free occurrences of variables x, y, z and so on is true
in M precisely if the universally quantified version is true. The
details of the definitions tend to differ, but it always comes down to
this.- Hide quoted text -
- Show quoted text -
|
Yes ,except the problem comes when you when you are writing a proof
(even informal,but rigorous in mathematics-like set theory) and state
a premise P(x) with free variable x .A mathematician does not think he
is assuming (x) P(x) .In this convention universal generalization -for
every x ie (x) -cannot be used if x is free in a premise .Then one
somehow proves Q(x) and then by the so called deduction theorem ,the
statement :if P(x) then Q(x) ; is then valid (or has a proof in a
deductive system) without premise or in other words for every x .
Of course a mathematician cares that in an L structure and a value of
x making P true that his proof shows thath Q is true so" if P then Q
" is true and if P is false ,then "if P then Q" is defined as true .So
the if ,then statement holds for every x .I am sure you know all
this.
I noticed recently that in the formal deductive system for first
order logic of Rosser as described in the worthy book of I. Copi -
symbolic logic,that he allows universal generalization from Premises
in derivations but he has safeguards in the deduction theorem .I don't
like that but it does no harm in the end.It makes an axiom that P(x)
implies (x)P(x) so its related to your having P(x) have a truth
value and does simplify the rules of inference slightly. Interesting
smn |
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