Re: Small Set Theory,Updated.


hagman wrote:
zuhair schrieb:


Ok, let me try one more trial.

x is P_defined <-> Ay(yex<->(Py & ~y=x))

were P is a predicate in one variable, that doesn't use x.
and such that if Q is any predicate in one variable that fulfil
Ay(yex<->(Py & ~y=x))
and ~Q[x], then ~P[x], and if there doesn't exist such Q
then P[x].

I don't know how to formulate the above, here is a trial.

x is P_defined <-> (Ay(yex<->(Py & ~y=x)) &

(Px<->~EQ(Ay(yex<->(Qy & ~y=x))&~Qx)))&P&Q are predicates each in one
free variable,that

doest use the symbole x ).

The definition of Predicate in one free variable,that doesn't use the
symbole x is well known,
it is the same as how it is defined in the Axiom of separation.

In this maner as I said before "0 is (~y=y)_not embedded" is a true
statement
but "0 is (Az(~zey))_embedded" is a false statement because Qy<->~y=y
fulfills
Ay(ye0<->(Qy & ~y=0))&~Q0, therefore EQ(Ay(yex<->(Qy & ~y=x))&~Qx)).
and since Py<->(Az(~zey)) also holds for 0,i.e P0 then this violates
(Px<->~EQ(Ay(yex<->(Qy & ~y=x))&~Qx))). Therefor ~(0 is
(Az(~zey))_embedded) is the true statement.

This will quarantee uniqueness.

Complicated.
Let's say
x is P_protodefined <-> Ay(yex<->(P(y) & ~y=x))
where P is a predicate not using x (typographically).
Then x is P_defined if x is P_protodefined and
i) either P(x) is false or
ii) Q(x) is true for any Q such that x is Q_protodefined.
In case i) we say that x is P_non_embedded, in case ii) x is
P_embedded.

Note that
x is P_non_embedded
is equivalent to
~P(x) & Ay(yex<->(P(y) & ~y=x))
and thus to
Ay(yex<->P(y))

x is P_embedded
is equivalent to

From the axiom of infinity the existance of all finite ordinals is
quaranteed,givin
these new restrictions on P.

Then one can simply define what is finite and infinite.

x is finite <-> x 1-1 y , yeN.

... which requires a definition of 1-1, hence of pairs (a,b), hence of
parings {a,b}.

It looks like we can actually (somewhat, see comments) prove
PROPOSITION: Given a,b there exists a unique set c such that
Az(zec <-> (z=a v z=b)).
PROOF:
Given a,b, let P(y):<-> y=a v y=b. (Even this step is somewhat doubtful
as we suddenly
either use the symbols "a", "b" on the right hand side or presuppose
that
y=a and y=b can be resolved using the EXT axiom in a way that gets rid
of all extra symbols).
By Ax.2 there exists a set c such that c is P_defined, especially ~cec.
Either c is P_non_embedded or P_embedded
1) c is P_non_embedded:
Then Az(zec <-> P(z)) and we are done.
2) c is P_embedded:
Then P(c), i.e. c=a v c=b.
Wlog. c=a, the other case being symmetric.
It follows that c is (c=b)_protodefined.
2.1) ~a=b:
Then c is (c=b)_non_embedded and cannot be P_emebedded. --
contradiction
2.2) a=b:
2.2.1) Ez(zea):
From a=c and a=b, we have P(z) and thus z=a, i.e. cec -- contradiction
2.2.2) ~Ez(zea), i.e. a=0:
Then also c=0, but 0 is known to be non_embedded -- contradiction.
Combining all, we conclude that c is (y=a v y=b)_non_embedded and
Az(zec <-> (z=a v z=b)).
Uniqueness follows by EXT.
QED(modulo doubts mentioned)

DEFINITION: Given a,b, the set c in the above proposition is denoted as
{a,b}
DEFINITION: {a} := {a,a}.

COROLLARY: {a,b}={b,a}. ~{a,b}=a. ~{a,b}=b. ~{a}=a.
PROOF: Clear (use Ax(~xex)).


PROPOSITION:
There exists a set V such that
Az(zeV <-> ~z=V).
PROOF:
By Ax.2 there exists a (y=y)_defined set.
It follows that Az(zeV <->(z=z & ~z=V)), i.e.
Az(zeV <-> ~z=V).
QED

Unfortunately, this kills foundation:

COROLLARY:
There exist sets a,b such that aeb and bea.
PROOF:
Let W:= {V}. Then ~W=V and hence both VeW and WeV.
QED

By Axiom of Exclusion W doesn't exist.
here W is equal to x were x is (y=V)_not embedded.
But this is impossible since by Exclusion W doesn't exist.
According to axiom of exclusion
AxAy(yex & y is P_defined ->~Px)
then since V is (y=y)_defined, then W shouldn't fulfill y=y
i.e it follows that ~W=W, which violates Extensionality
therefore W doesn't exist.

COROLLARY:
There is a non-empty set that intersects each of its elements.
PROOF:
Let X={V,W}. Then (X \cap V) = W and (X \cap W) = V.
QED

Same mistake as above X doesn't exist in this theory. because it
contains V
and no set can contain V, since by then it wouldn't be equal to itself
,Thus violating
extensionality.

I have a theory that Avery FINIT set is P_not embedded, and Avery x is
P_embedded
is an infinite set.

At least informally, any finite set ought to be (y=a1 v y=a2 v ... v
y=an)_not_embedded.

Further I have the blind intuition that:-

EP(Ay(yex<->(Py & ~y=x)) & Px) & x is infinite -> x is P_embedded.

I doubt this (anyway it should rather be written as
EP(Ay(yex<->(Py & ~y=x)) & Px) & x is infinite -> EP(x is
P_embedded).
)

EP(Ay(yex<->(Py & ~y=x)) & ~Px) & x is finite -> x is P_not embedded.

This is clear, at least when mroe correctly written as
EP(Ay(yex<->(Py & ~y=x)) & ~Px) & x is finite -> EP(x is P_not
embedded).
since this is trivially true even without assumption on finiteness.

I think I made wrong sentences.

so EP(Ay(yex<->(Py & ~y=x)) & Px) & x is infinite -> x is P_embedded.

this is wrong.

one example against it is that w is (yez)_proto_embedded were z= wU{w}.
and at the same time it is w is (finite ordinal)_proto_not embedded.


what I meant. is the following.

1) x is finite ->EP(x is P_not embedded)
2) x is P_embedded -> x is infinite.

See hagman, to me it is Axiom of Exclusion that is the most important
of all axioms, I tried to construct this theory with this axiom as its
heart. It is this axiom that is supposed to give this theory a
different way of understanding sets.

To give an intuitive idea about this axiom I will address it and give
few examples:

Axiom of Exclusion: AxAy(yex & y is P_defined -> ~Px).

Examples: Let the set of all ordinals be J.

Now J is (y is an ordinal)_defined.

J is (y is an ordinal)_embedded.

Now can there be a set W ={J}?

the answer is yes. since since P[y]<->y=J is a predicate.
and {J} is not an ordinal.

Now can their be the following J U{J} ?

The answer is NO. because JU{J} would have J as a member, and it would
be an ordinal also, violating exclusion.( Solving Burali-forti
paradox).

Another example.

Let T be the set of all ordinals no bijectable to {0} , i.e not
bijectable to 1.

Then T is (|y|<>1)_embedded.

Can there be a set such that T is a member?

Then anser is Yes. Example: {T}.

Also sets like { { T } } , {{{ T }}} , etc... do exist.

But any non singlton set containing T then it is not allowed, since
that violates Exclusion.

You see that T here doesn't have a power set.

While the set of all ordinals J do have a power set.

Also V clearly do not have a power set( solving Cantor's paradox).

In order to Re-write this theory so that its axioms are not forgotten:

-Small Set Theory_


Primitives, e , =

Definition Schemas:


x is P_protodefined <-> Ay(yex<->(P(y) & ~y=x))
where P is a predicate not using x (typographically).
Then x is P_defined if x is P_protodefined and
i) either P(x) is false or
ii) Q(x) is true for any Q such that x is Q_protodefined.
In case i) we say that x is P_non_embedded,
in case ii) x is P_embedded.

Note that
x is P_non_embedded
is equivalent to
~P(x) & Ay(yex<->(P(y) & ~y=x))
and thus to
Ay(yex<->P(y))

x is P_embedded
is equivalent to
P(x) & Ay(yex<->(P(y) & ~y=x))
and thus to
Ay((yex v y=x)<->P(y)).

Axioms:

Ax0) Extensionality:As in ZFC.
Ax1) Regularity:As in ZFC.
Ax2) Exclusion: AxAy((yex & y is P_defined)->~P(x)).
Ax3) Comprehension:(Ex x is P_defined)<->(AQ((yex & x is P_defined &
y is Q_defined)->~Q(x))).
Ax4) Infinity: as in ZFC
Ax5) Choice: as in ZFC.

I don't know if I need the axiom of concordance here. which is

AxAz( (x is P_defined & z is Q_defined & (P<->Q))->(P(x)<->P(z)) ).

Though I feel that this axiom is needed to quarentee uniqueness!

Any suggestion on that?

About comprehension perhaps Its enough to state it as Ex x is P_defined
since the rest is excluded by Ax.2. I don't know really. I added what I
added
to Ax3 to quarentee non contradiction between it and Ax.2. since
if I state Ax.3 as Ex x is P_defined, then this would imply the
existance
of x for every P, so for example if y=V, then by this axiom W={V} would
be existing in violation to Ax.2. Therefore I added
"<->(AQ((yex & x is P_defined & y is Q_defined)->~Q(x)))"
to"Ex x is P_defined". in order to resolve this contradiction.

There is no need to mention that it is the addition of ~y=x in the
definition
of x is P_defined that solved Russell's paradox.

Theorums are the same as posted earlier at the beginning of this
thread.

The important theorums of (pairing,union,separation,replacement, and
power) are all
influenced by Ax.3. we only need to translate them to the suitable
predicates.I think
that can be done easily. The only problem here is that these theorums
are cumbersome
somehow, since they are better termed Conditional pairing, Conditional
union, ..., conditional power. However Perhpas and only Perhpas
Separation is not conditional.

Zuhair

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