Re: Small Set theory:Revised.


zuhair wrote:
hagman wrote:
zuhair schrieb:

zuhair wrote:
Below is the last version of this theory:-

-Small Set Theory-


Primitive e


Definitions:-
x is P_defined <-> Ay (((yex ->P[y])&( (~yex&~y=x)->~P[y] )).
x is P_embedded <->( (x is P_defined)&P[x] ).
x is P_not embedded<->( (x is P_defined)&~P[x] )

x is P_defined<->( x is P_embedded v x is P_not embedded).


were P is a formula in one free variable.




Axioms:-
Ax.0: Extensionality: AxAy(x=y <-> (Az(zex<->zey))
Ax.1: Description:AxEP ( x is P_defined )

Note: There may be several such P for a given x.
No it cannot.
If x is P_defined and Q_defined, we know that
P(y)<->Q(y) for all y except possibly for y=x.
We know nothing about P(x) and Q(x).

But by then neither P nor Q is the formula I meant in Ax.1.

Suppose that for set x there exist formulas P0 , P1, P2,.....
such that Pi <-> Pj . here the formula of definition of x (i.e P in x
is P_defined) would be

P<-> ( P0 v P1 v P2 v P3 v.............. ).


Ax.2: Exclusion:AxAy ( (yex & y is P_defined) -> ~P[x] )

Especially xex must be false.
Otherwise, assume xex and let P be such that x is P_defined (Ax.1).
Then (xex & x is P_defined) implies ~P[x] by Ax.2.
But xex also implies P[x] by definition of P_defined.
Contradiction, the assumption xex must be false.

Ok, that was my goal of developing this axom. However as an extention
to what you already nicely presented, if yex then x!ey. this is another
result of Ax.1 and Ax.2.

Ax.3: Comprehension:ExAy ( (yex ->P[y])&( (~yex & ~y=x)->~P[y] ) )

Or shorter: Ex x is P_defined

No that is not correct. you forgot Ay.

Ex Ay x is P_defined such that y is the variable of P. this is an
alternative.

Again, this direction is only almost unique:
If x and z are both P_defined,

if x and z are both P_defined then x=z.

then for all y except possibly y=x or
y=z, we have
yex <->P(y) <-> yez.

No there is an error here. this doesn't follow from my theory. I don't
know from were you infered it.
To show that x=z, it still must be shown that
xex <-> xez and zex<->zez.

In reality there is no need for that, in this theory. two sets
are equivalent only if they have the same members.
Both xex and zez are false (see above).
So x and z are either equal or at least one of xez, zex holds.
by extensionality we know that x=z.
Assume xez. Then ~P[z] (Ax.2) and P[x] (z is P_defined).
Similarly, the assumption zex would lead to ~P[x] and P[z].
Thes cases are mutually exclusive, hence we have:
For a given P,
either there exists a unique P_defined set x

Yes.
or there exist exactly two P_defined sets x,z and one is contained in
the other.

This is incorrect. You analysis is wrong.
Notation (inspiration: both are approximations to {y|P(y)} ):
Let
{y$P(y)}
stand for the uniquely determined set in the first case and for the
bigger of the two sets in the second case.
Let
{y§P(y)}
stand for the uniquely determined set in the first case and for the
smaller of the two sets in the second case.
Hence we have either {y§P(y)} = {y$P(y)} or {y§P(y)} e {y$P(y)}.
If x is P_defined then x={y§P(y)} or x={y$P(y)}. wrong.

Ax.4: Infinity: EN:0eN & (Ax(xeN->S(x)eN))

According to Ax.1 there should exist a P such that N is P_defined.
I have no idea, what such a P might look like.

P(y)<->( ( y is a subset of Power y and every member m of y is a subset
of Power m ) & y is not infinite).

so that we do not have a misunderstanding here let me use the following
definitions.

y is an ordinal <-> ( y is transitive and Am(mey -> m is transitive)).

In reality this is equivalent to my version which is:

y is an ordinal <-> ( y is a subset of P(y) & Am(mey -> m is a subset
of P(m)) ).

Anyhow you can use either.

y is recursive <-> Uy e y. were Uy refers to Union y.

y is a recursive ordinal <-> ( y is an ordinal & y is recursive ).

y is a recursive ordinal <-> y is a last membered ordinal.

y is a finite ordinal <-> As( ((s is a subset of y)& (s is ordinal))->
s is last membered ).

y is an infinite ordinal <-> ~ [As( ((s is a subset of y)& (s is
ordinal))-> s is last membered )].

Now let me define N in Ax.4.

here N is P_defined.

P(y)<-> y is a finite ordinal.

it is clear here that N is itself an ordinal that is not last membered,
since otherwise it will be in itself and thus violates axiom 2.

just to extend the defintions of finite and infinite to all sets.

x is finite <-> x 1-1 y, were yeN .
x is infinite <-> ~(x 1-1 y , were yeN).

Zuhair

This only makes me think that Ax.4 is a theorum in this theory. rather
than an axiom.

+/- AC.

In zuhair set theory, the Axiom Of Choice is probably not needed.

---------------------------------------------------------------------------­­-



Now I think this axiomatic system is clear. Theorums in this thoery
are: 1) Universe 2) Pairing 3) Union 4) Separation 5) Replacement
6) Power.

Theorems neeed proofs.
For some of these claims, however, a proof will definiitely fail.
For example you will have a hard time with the power set of N (see
below).


Of course all these spring from Ax.3 and should be modefied after
Ax.2.



This theory avoids Russell's paradox, and at the same time has more
models
than do ZFC, or NBG have.

According to Ax.1, there exists a (meta-theoretic) mapping from the
zuhair-sets to formulae (i.e. finite strings over a finite alphabet)
that maps a given zuhair-set x to a formula P such that x is P_defined
(if there are several formulas, you may want to select the lexically
first).
For a given formula P there are at most two zuhair-sets that are mapped
to P by this.

There exist at most ONE zuhair-sets that are mapped to P by this.
Hence there are at most twice as many zuhair-sets as formulae.
There are only countably many formulae.

What? I disagree with this.

Example: let us take the constant formula. ( I call P a constant
formula if P(y)<-> y = x , were x is a specific set, example P(y)<-> y
= 0 ).

For example Let xi be a member of a P_embedded set G. then let
Pi(y)<->y=xi. for all xi in G.

Now it is clear that any P_embedded set is an uncountable set. so it
has uncountable number of members. so we can have uncountable number of
constant Pi for the example above.

so your argument that we have a countable number of formulas is
erronous.

Zuhair


Hence there are only countably many zuhair sets.






just an idea occured to me that if Ax.3 if even further modefied to the
following to grant what I want from it to do.

Ax.3: Comprehension:ExAy ( (yex ->(P[y]&~y=x))&((~yex & ~y=x)->~P[y]) )

Makes no difference.
I have shown above that xex is false, i.e. yex implies ~y=x.

Zuhair
-hagman

.

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