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 Posted: Wed Jul 23, 2008 12:41 am Post subject: Re: A set theory with the class of all classes |
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On Jul 22, 5:19 pm, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]On Jul 22, 3:29 pm, Zaljo...@gmail.com wrote:
On Jul 22, 10:48 am, MoeBlee <jazzm...@hotmail.com> wrote:
Don>t forget though, that a sufficiently formulated schema of
replacement entails the schema of separation.
Of course, thats why I said ...It appears to me that the other axioms
that do not entail separation nor regularity nor choice .....
Okay, but that we have different forms of replacement and that one of
them entails separation and others don>t is a subtle point. Yet your
assertion about consistency depends crucially on that distinction,
which the reader will not likely get just from the "the other axioms"
bit.
Let me first clarify what I mean by relativized to V.
But then you only give examples, not a definition of relativization.
Here>s the ordinary definition:
For a formula P, we>ll say P* is the relativization of P to V, based
on recursion:
If P is atomic, then P* is P.
If P is ~Q, then P* is ~(Q*)
If P is Q -> R, then P* is Q* -> R*.
If P is AxQ, then P* is Ax(xeV -> Q*).
And that entails:
If P is Q & R, then P* is Q* & R*
If P is Q v R, then P* is Q* v R*
If P is Q <-> R, then P* is Q* <-> R*
If P is ExQ, then P* is Ex(xeV & Q*)
Is that also what you mean?
I mean the following: The unordered pair of two sets is a set
In symbols: forall X in V, forall Y in V we have {X,Y} in V
More radically writtin
forall A in V, forall B in V, Exists X in V, forall y ( y in X <-
(y=A or y=B)).
Wait a minute:
Here>s the unrelativized pairing axiom:
(1) AxyEzAk(kez <-> (k=x v k=y)).
The relativization of that is:
(2) AxeV AyeV EzeV AkeV(kez <-> (k=x v k=y)).
Here>s what you have:
(3) AxeV AyeV EzeV Ak(kez <-> (k=x v k=y)).
At least prima facie, (2) and (3) are not equivalent.
(Note: In ordinary class theory, they>re equivalent, because kez -
keV and since xeV and yeV we also have (k=x v k=y) so relativizing to
Ak to AkeV is redundant. But it>s not clear (at least not so very
IMMEDIATELY clear that you>re clear without mentioning) that your
theory yields kez -> keV.)
[/quote]
Yes, with the axiom of reflexisivness of V, that I have added lately,
it is clear that they are equivalent.
But generally I agree with you, I need to state the formulas.
thanks Moe
[quote]
I don>t know whether they end up equivalent in your theory; but if
they are, then you should at least mention the main proof idea.
Or, maybe you don>t claim that they>re equivalent. But then your
definition of 'relativization' is not clear. It>s not predictable what
quantifiers you>re going to relativize, as indeed you did not
relativize 'k'.
Similarily Replacement is relativized and choice and regularity all to
V.
Since you didn>t relativize all the quantifiers in pairing, it would
help for you to explicitly state your relativized (or partially
relativized, or whatever) replacement schema and relativized (or
partially relativized, or whatever) axioms of choice and regularity.
Better just to state your formulas than leave it to guess work as to
what quantifiers you>re relativizing (since, as mentioned, you didn>t
relativize the 'k' quantifier in pairing).
Now what you said is that if we have pairing,union,power and infinity
unrelativized to V, then this mean that they are relativized to V?
how?
Actually, sorry, forget what I said about that. What I said is too
general. What I had in mind are cases where you have as a theorem
Ax1...AxnP where P has no quantifiers in it, then you have the
relativization as a theorem. But not all your axioms are of that form,
so nevermind.
And please specify your exact formulation of the replacement schema.
You don>t want one that entails separation.
Why not? actually I want it to entail separation, since it is
relativized to V.
Best then to give the formula itself.
For me, your description in which some things are relativized and not
others and some variables are relativized but not others - and with
schemata of such things - is confusing at best.
In such an instance, you>d be much clearer just to write the formulas.
MoeBlee[/quote] |
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MoeBlee
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| Posted: Wed Jul 23, 2008 12:48 am Post subject: Re: A set theory with the class of all classes |
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On Jul 22, 5:38 pm, Zaljo...@gmail.com wrote:
[quote]On Jul 22, 5:21 pm, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 22, 3:16 pm, Zaljo...@gmail.com wrote:
what I mean was
forall r in V' Exists X: forall y ( y in X <-> ( y in r and P(y) ).
that was what I meant.
Then, again, I have no way of predicting which quantifiers you want
relativized and which ones you don>t want relativized.
No Moe, here I made a dam mistake when I said that I am relativizing
to V', that was a mistake.
[/quote]
No, I understand that you corrected to what you wanted. But your
corrected formula is one in which you didn>t relativize all the
quantifiers. So when you say 'the relativization' of an arbitrary
formula, it is unpredictable what that formula is; that is, it is
unpredicaable which quantifiers you>ll relativize and which you won>t.
MoeBlee |
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| Posted: Wed Jul 23, 2008 4:30 am Post subject: Re: A set theory with the class of all classes |
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On Jul 22, 5:48 pm, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]On Jul 22, 5:38 pm, Zaljo...@gmail.com wrote:
On Jul 22, 5:21 pm, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 22, 3:16 pm, Zaljo...@gmail.com wrote:
what I mean was
forall r in V' Exists X: forall y ( y in X <-> ( y in r and P(y) ).
that was what I meant.
Then, again, I have no way of predicting which quantifiers you want
relativized and which ones you don>t want relativized.
No Moe, here I made a dam mistake when I said that I am relativizing
to V', that was a mistake.
No, I understand that you corrected to what you wanted. But your
corrected formula is one in which you didn>t relativize all the
quantifiers. So when you say 'the relativization' of an arbitrary
formula, it is unpredictable what that formula is; that is, it is
unpredicaable which quantifiers you>ll relativize and which you won>t.
MoeBlee
[/quote]
Ok, let me restate my theory in an easier manner:
Theory /\ is the set of all sentences entailed by (using FOL with
identity and the primitives "in" and "V" were V is a constant) the
following non logical axioms:
Definition: x is a set iff x in V
Definition: x is a class iff [ x in V or ~ x in V ]
So we have: for all x : x is a class.
1) Axiom of Extensionality:
forall z (z in A <-> z in B) -> A=B
2) Axiom of complementary class.
forall A, Exists X : forall y ( y in X <-> ~ y in A )
Define: X=A' iff forall y ( y in X <-> ~ y in A )
A' is read as 'the complementary class of A'
3) Axiom of Intersection:
forall A, Exists X : forall y ( y in X <-> forall z ( z in A -> y in
z ) )
Define: X= /\ A iff forall y ( y in X <-> forall z ( z in A -> y in
z ) )
/\ A is read as 'the intersectional class of A'.
4) All axioms of Z except regularity and sepration.
Define: X = { A,B } iff forall y ( y in X <-> (y=A or y=B) )
{A,B} is read as the unordered pair of A and B.
Define: X= Union A iff forall y ( y in X <-> Exists z ( z in A and y
in z ) ).
Union A is read as "the class union of A".
Define: X= Power A iff forall y ( y in X <-> forall z ( z in y -> z
in A ) ).
Power A is read as "the power class of A".
5) Axiom of set existence: Exists X : X in V
6) Axiom of reflexiseness of V: forall X in V , forall y ( y in X -> y
in V ).
7) Axiom of pairing of sets: The unordered pair of two sets is a set.
8) Axiom of set union: The class union of a set is a set.
9) Axiom of power set: The power class of a set is a set.
10) Axiom of Infinity:
Exists N in V ( 0 in N and forall y ( y in N -> (y in V and yU{y} in
N) ) ).
11) Axiom schema of separation: If P is a formula in which X is not
free, then all closures of
forall A in V, Exists X in V, forall y ( y in X <-> ( y in A and
P(y) ) )
are axioms.
12) Axiom schema of replacement: If P is a formula in which B is not
free, then all closures of
[forall x in V, Exists! y in V , P(x,y) , forall x in V ( P(x,y) -> y
in V )]
-> forall A in V ,Exists B in V, forall y ( y in B <-> Exists x in A:
P(x,y) ).
are axioms.
13) Axiom of Regularity for sets: For every non empty set there should
exist an element that is disjoint of it.
14) Axiom of choice for sets: V is well orderable.
15) Axiom of class comprehension over V: if P is a formula in which X
is not free, then all closure of
Exists X: forall y ( y in X <-> ( y in V and P(y) ) ).
are axioms.
/Theory definition finished
However I am contemplating adding axiom schema of infinity of classes:
If F is a once place function symbole, then all closures of
forall c, Exists X ( c in X and forall y ( y in X -> F(y) in X ) ).
are axioms.
Zuhair |
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MoeBlee
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| Posted: Wed Jul 23, 2008 5:56 pm Post subject: Re: A set theory with the class of all classes |
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On Jul 22, 9:30 pm, Zaljo...@gmail.com wrote:
[quote]Ok, let me restate my theory in an easier manner:
[/quote]
Indeed, it>s much better now. I have a fe comments:
[quote]Theory /\ is the set of all sentences entailed by (using FOL with
identity and the primitives "in" and "V" were V is a constant) the
following non logical axioms:
Definition: x is a set iff x in V
Definition: x is a class iff [ x in V or ~ x in V ]
So we have: for all x : x is a class.
1) Axiom of Extensionality:
forall z (z in A <-> z in B) -> A=B
2) Axiom of complementary class.
forall A, Exists X : forall y ( y in X <-> ~ y in A )
Define: X=A' iff forall y ( y in X <-> ~ y in A )
A' is read as 'the complementary class of A'
3) Axiom of Intersection:
forall A, Exists X : forall y ( y in X <-> forall z ( z in A -> y in
z ) )
Define: X= /\ A iff forall y ( y in X <-> forall z ( z in A -> y in
z ) )
/\ A is read as 'the intersectional class of A'.
4) All axioms of Z except regularity and sepration.
Define: X = { A,B } iff forall y ( y in X <-> (y=A or y=B) )
{A,B} is read as the unordered pair of A and B.
Define: X= Union A iff forall y ( y in X <-> Exists z ( z in A and y
in z ) ).
Union A is read as "the class union of A".
Define: X= Power A iff forall y ( y in X <-> forall z ( z in y -> z
in A ) ).
Power A is read as "the power class of A".
5) Axiom of set existence: Exists X : X in V
6) Axiom of reflexiseness of V: forall X in V , forall y ( y in X -> y
in V ).
[/quote]
Or, you could just say "AX X is a subclass of V".
[quote]7) Axiom of pairing of sets: The unordered pair of two sets is a set.
8) Axiom of set union: The class union of a set is a set.
9) Axiom of power set: The power class of a set is a set.
10) Axiom of Infinity:
Exists N in V ( 0 in N and forall y ( y in N -> (y in V and yU{y} in
N) ) ).
[/quote]
You don>t need the 'y in V' clause, since NeV so if yeN, by your
earlier axioim we have yeV.
[quote]11) Axiom schema of separation: If P is a formula in which X is not
free, then all closures of
forall A in V, Exists X in V, forall y ( y in X <-> ( y in A and
P(y) ) )
are axioms.
12) Axiom schema of replacement: If P is a formula in which B is not
free, then all closures of
[forall x in V, Exists! y in V , P(x,y) , forall x in V ( P(x,y) -> y
in V )]
[/quote]
That>s the antecedent, but it>s not well-formed. You have:
AxeV E!yeV P(x y)
which is okay and is a formula, but then you just concatenate ANOTHER
formula onto it, yet the concatenation of two formulas is not a
formula.
Did you mean this:
AxeV E!yeV P(x y) & AxeV(P(x y) -> yeV)
[quote]-> forall A in V ,Exists B in V, forall y ( y in B <-> Exists x in A:
P(x,y) ).
are axioms.
13) Axiom of Regularity for sets: For every non empty set there should
exist an element that is disjoint of it.
14) Axiom of choice for sets: V is well orderable.
15) Axiom of class comprehension over V: if P is a formula in which X
is not free, then all closure of
Exists X: forall y ( y in X <-> ( y in V and P(y) ) ).
are axioms.
/Theory definition finished
However I am contemplating adding axiom schema of infinity of classes:
If F is a once place function symbole, then all closures of
forall c, Exists X ( c in X and forall y ( y in X -> F(y) in X ) ).
are axioms.
[/quote]
Except for replacement, it looks pretty clear now.
MoeBlee |
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| Posted: Wed Jul 23, 2008 9:31 pm Post subject: Re: A set theory with the class of all classes |
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On Jul 23, 10:56 am, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]On Jul 22, 9:30 pm, Zaljo...@gmail.com wrote:
Ok, let me restate my theory in an easier manner:
Indeed, it>s much better now. I have a fe comments:
Theory /\ is the set of all sentences entailed by (using FOL with
identity and the primitives "in" and "V" were V is a constant) the
following non logical axioms:
Definition: x is a set iff x in V
Definition: x is a class iff [ x in V or ~ x in V ]
So we have: for all x : x is a class.
1) Axiom of Extensionality:
forall z (z in A <-> z in B) -> A=B
2) Axiom of complementary class.
forall A, Exists X : forall y ( y in X <-> ~ y in A )
Define: X=A' iff forall y ( y in X <-> ~ y in A )
A' is read as 'the complementary class of A'
3) Axiom of Intersection:
forall A, Exists X : forall y ( y in X <-> forall z ( z in A -> y in
z ) )
Define: X= /\ A iff forall y ( y in X <-> forall z ( z in A -> y in
z ) )
/\ A is read as 'the intersectional class of A'.
4) All axioms of Z except regularity and sepration.
Define: X = { A,B } iff forall y ( y in X <-> (y=A or y=B) )
{A,B} is read as the unordered pair of A and B.
Define: X= Union A iff forall y ( y in X <-> Exists z ( z in A and y
in z ) ).
Union A is read as "the class union of A".
Define: X= Power A iff forall y ( y in X <-> forall z ( z in y -> z
in A ) ).
Power A is read as "the power class of A".
5) Axiom of set existence: Exists X : X in V
6) Axiom of reflexiseness of V: forall X in V , forall y ( y in X -> y
in V ).
Or, you could just say "AX X is a subclass of V".
[/quote]
I think you mean AX in V X is a subclass of V.
which mean : forall X in V X is a subclass of V.
Becuase you know in this theory it is not the case that every class is
a subclass of V. But every set is a subclass of V. as in example you
have U (the class of all classes which is /\0 , this is not a subclass
of V).
One can rephrase my axiom by saying: Every set is a subclass of V.
Or by saying : any member of a set is a set.
[quote]
7) Axiom of pairing of sets: The unordered pair of two sets is a set.
8) Axiom of set union: The class union of a set is a set.
9) Axiom of power set: The power class of a set is a set.
10) Axiom of Infinity:
Exists N in V ( 0 in N and forall y ( y in N -> (y in V and yU{y} in
N) ) ).
You don>t need the 'y in V' clause, since NeV so if yeN, by your
earlier axioim we have yeV.
[/quote]
Agreed.
[quote]
11) Axiom schema of separation: If P is a formula in which X is not
free, then all closures of
forall A in V, Exists X in V, forall y ( y in X <-> ( y in A and
P(y) ) )
are axioms.
12) Axiom schema of replacement: If P is a formula in which B is not
free, then all closures of
[forall x in V, Exists! y in V , P(x,y) , forall x in V ( P(x,y) -> y
in V )]
That>s the antecedent, but it>s not well-formed. You have:
AxeV E!yeV P(x y)
which is okay and is a formula, but then you just concatenate ANOTHER
formula onto it, yet the concatenation of two formulas is not a
formula.
Did you mean this:
AxeV E!yeV P(x y) & AxeV(P(x y) -> yeV)
[/quote]
Yep.
[quote]
-> forall A in V ,Exists B in V, forall y ( y in B <-> Exists x in A:
P(x,y) ).
are axioms.
13) Axiom of Regularity for sets: For every non empty set there should
exist an element that is disjoint of it.
14) Axiom of choice for sets: V is well orderable.
15) Axiom of class comprehension over V: if P is a formula in which X
is not free, then all closure of
Exists X: forall y ( y in X <-> ( y in V and P(y) ) ).
are axioms.
/Theory definition finished
However I am contemplating adding axiom schema of infinity of classes:
If F is a once place function symbole, then all closures of
forall c, Exists X ( c in X and forall y ( y in X -> F(y) in X ) ).
are axioms.
Except for replacement, it looks pretty clear now.
MoeBlee- Hide quoted text -
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- Show quoted text -[/quote] |
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MoeBlee
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| Posted: Wed Jul 23, 2008 10:55 pm Post subject: Re: A set theory with the class of all classes |
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On Jul 23, 2:31 pm, Zaljo...@gmail.com wrote:
[quote]On Jul 23, 10:56 am, MoeBlee <jazzm...@hotmail.com> wrote:
Or, you could just say "AX X is a subclass of V".
I think you mean AX in V X is a subclass of V.
[/quote]
Right.
MoeBlee |
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| Posted: Thu Jul 24, 2008 12:13 am Post subject: Re: A set theory with the class of all classes |
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On Jul 22, 4:14 pm, Rupert <rupertmccal...@yahoo.com> wrote:
[quote]On Jul 23, 6:46 am, Zaljo...@gmail.com wrote:
On Jul 22, 7:14 am, Rupert <rupertmccal...@yahoo.com> wrote:
On Jul 22, 10:52 am, Zaljo...@gmail.com wrote:
Hi all,
What prevents the existence of a set of all sets in ZFC, is actually
separation. It appears to me that the other axioms that do not entail
separation nor regularity nor choice, will not be contradictive with
the existence of the set of all sets. Two axioms that of intersection
and that of complementary class, do not exist in ZFC, which is in my
opinion a point of weakness, since these axioms seems pritty
intuitive.
In the following theory, all axioms of Z except separation and
regularity are presented in unrestricted manner. In addition all
axioms of ZFC are presented relativized to V ( the class of all
sets ), in addition you have axiom of intersection and axiom of
complementary sets in unrestricted manner. This of course will lead to
the existence of the class of all classes.
Theory /\ is the set of all sentences entailed by (using FOL with
identity and the primitives "in" and "V" were V is a constant) the
following non logical axioms:
Definition: x is a set iff x in V
Definition: x is a class iff [ x in V or ~ x in V ]
So we have: for all x : x is a class.
1) Axiom of Extensionality:
forall z (z in A <-> z in B) -> A=B
2) Axiom of complementary class.
forall A, Exists X : forall y ( y in X <-> ~ y in A )
Define: X=A' iff forall y ( y in X <-> ~ y in A )
A' is read as 'the complementary class of A'
3) Axiom of Intersection:
forall A, Exists X : forall y ( y in X <-> forall z ( z in A -> y in
z ) )
Define: X= /\ A iff forall y ( y in X <-> forall z ( z in A -> y in
z ) )
/\ A is read as 'the intersectional class of A'.
4) All axioms of Z except regularity and sepration.
5) Axiom of set existence: Exists X : X in V
6) All axioms of ZFC relativised to V.
7) Axiom of class comprehension over V: if F is a formula in which X
is not free, then all closure of
Exists X: forall y ( y in X <-> ( y in V and P(y) ) ).
are axioms.
/Theory definition finished
Now this theory will have the class of all classes U
Proof: 0 is a class
Exists /\0 which is the class of all classes. Axiom 3
Define U=/\0
of course /\U=0
also we have the theorem unionU=U
and the theorem powerU=U
of course U in U.
So we can cary all operations of
union,pairing,powering,complementation and intersection. However
Replacement, regularity and choice are restricted within V.
Now is this theory inconsistent?
Zuhair
I am pretty sure that this theory is consistent provided it is
consistent with ZFC that an inaccessible cardinal exists.
I may write up and post a proof of this later.
what if I added two axioms to this theory.
Axiom of relfexiveness of V.
forall X in V , forall y ( y in X -> y in V ).
Axiom schema of Infinity of classes:
if F is a one place funtion symbole then all closures of
forall z, Exists A (z in A and forall x ( x in A -> F(x) in A))
are axioms.
Would the theory remain consistent?
The aim beyond this schema is to find classes like
{U, {U} , {U,{U}} , {U,{U},{U,{U}}},.............}
or like
{U , {U} , {{U}} , {{{U}}} , ..........}
By this schema we can construct such classes.
- Hide quoted text -
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The axiom of reflexiveness is not a problem. But your other axiom
schema mucks up my current consistency proof and I would need to find
a new consistency proof (if one exists). Have to keep thinking about
it.- Hide quoted text -
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[/quote]
This theory has the geography of having sets that exactly the ones in
ZFC, have proper classes that are subclasses of V, that are exactly
the ones in MK and NBG, but in addition to those have the class of all
classes U, in addition to proper classes outside V. however the world
outside V or simply V' is not concured by comprehension, the rules of
union,pairing, powering, complementation and interesection are active
all over U, however there is something missing in V' which is
comprehension.
I have the general plan of trying to concure V', but starting from V.
I have the idea of Seeing V' through V.
Contemplate the following two Axiom schemata of replacement:
1) If P is a formula in which B is not free, then all closures of:
[forall x in V, Exists! y : P(x,y)] ->
forall A in V, Exists B, forall y ( y in B <-> Exists x in A:P(x,y) ).
are axioms.
2) If P is a formula in which B is not free, then all closures of:
[forall x in V, Exists! y : P(x,y)] ->
forall A subset_of V, Exists B, forall y ( y in B <-> Exists x in
A:P(x,y) ).
are axioms.
I wounder if these two replacement schemata would be consistent with
this theory.
In this way we can construct any class as far as it is equal in size
to a subset of V.
If any of these two axioms proves to be consistent with the original
theory, then there will be no need for the axiom schema of infinity of
classes, since it looks redundant to me.
If any of these two axioms proves to be consistent with the original
theory, and all the theory proves to be consistent if ZFC is
consistent, then to be it appears as if we began to concure the vague
world of V'.
Regards
Zuhair |
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george
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| Posted: Thu Jul 24, 2008 2:26 pm Post subject: Re: A set theory with the class of all classes |
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On Jul 21, 10:52 pm, Zaljo...@gmail.com wrote:
[quote]. Two axioms that of intersection
and that of complementary class, do not exist in ZFC, which is in my
opinion a point of weakness, since these axioms seems pritty
intuitive.
[/quote]
Your axiom of intersection follows as a theorem from separation (so
ZFC
*does* have it), except when the set you are "intersecting" is empty.
In that case, you probably DO get a contradiction, because you wind up
defining
the "intersection of the empty set" as the class that contains every
class,
INCLUDING ITSELF. |
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george
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| Posted: Thu Jul 24, 2008 3:05 pm Post subject: Re: A set theory with the class of all classes |
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On Jul 22, 8:41 pm, Zaljo...@gmail.com wrote:
[quote]On Jul 22, 5:19 pm, MoeBlee <jazzm...@hotmail.com> wrote:
And please specify your exact formulation of the replacement schema..
You don>t want one that entails separation.
Why not? actually I want it to entail separation, since it is
relativized to V.
Best then to give the formula itself.
For me, your description in which some things are relativized and not
others and some variables are relativized but not others - and with
schemata of such things - is confusing at best.
[/quote]
MoeBlee is very right about that.
Originally, you talked about axioms of Z being unrelativized and
the remaining axioms of ZFC being relativized. This is bad BECAUSE
once you get to Z*F*C, you suddenly don>t NEED pairing or separation
from Z
any more. Yet you can>t ditch both of them, because you want pairing
UNrestricted, while you want separation restricted. It seems to me
that pairing is the problem. |
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| Posted: Thu Jul 24, 2008 8:41 pm Post subject: Re: A set theory with the class of all classes |
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On Jul 24, 8:05 am, george <gree...@cs.unc.edu> wrote:
[quote]On Jul 22, 8:41 pm, Zaljo...@gmail.com wrote:
On Jul 22, 5:19 pm, MoeBlee <jazzm...@hotmail.com> wrote:
And please specify your exact formulation of the replacement schema.
You don>t want one that entails separation.
Why not? actually I want it to entail separation, since it is
relativized to V.
Best then to give the formula itself.
For me, your description in which some things are relativized and not
others and some variables are relativized but not others - and with
schemata of such things - is confusing at best.
MoeBlee is very right about that.
Originally, you talked about axioms of Z being unrelativized and
the remaining axioms of ZFC being relativized. This is bad BECAUSE
once you get to Z*F*C, you suddenly don>t NEED pairing or separation
from Z
any more. Yet you can>t ditch both of them, because you want pairing
UNrestricted, while you want separation restricted. It seems to me
that pairing is the problem.
[/quote]
I am not seeing any problem. See the last formulation of the theory.
Zuhair |
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