Re: ZFC in 4 Axioms.


hagman wrote:
zuhair schrieb:

hagman wrote:
zuhair schrieb:
I just occured to my mind, that ZFC can be reduced to four axioms as
below:

1)Extensionality 2) Comprehension 3)Infinity 4)AC.

1),2),3) are as in the traditional presentation of ZFC.

Axiom 2) (ExAy yex<->P(y))<->x=/=y

I am not sure of the above.

Then
(ExAy yex<->P(y))<->~x=y
should rather be written as either
(ExAy (yex<->P(y)) <-> ~u=v)
with free variables u and v (and whatever is free in P(y) besides x and
y)
or
Ex (Ay (yex<->P(y)) <-> ~x=v)
with free variables u and whatever is free in P(y) besides x and y
or
ExAy ((yex<->P(y)) <-> ~x=y)

Yes, this is the one.
with no free variables, except what is free in P(y) apart from x and y.
Of course, all these variants have different meaning.

Then let P(y):= y=y (about the simplest choice possible).
Then your axiom scheme claims
ExAy ((yex<->y=y) <-> ~x=y)
ExAy ((yex<->T) <-> ~x=y)
ExAy (yex <-> ~x=y)
"There is a set that contains all sets except itself". Call this set A.
Now, since you also claimed that given sets a,b there is a set (denoted
{a,b}) having exactly a and b as elements
xe{a,b} <-> (x=a v x=b)
(the proof was not conclusive, but the fact should be true anyway), and
also we should have that, given a set a, the union Ua of all its
elements exists,
xeUa <-> Ez(xez & zea)
we conclude that there is a set V,
V := U{{A,A},A}
apparently the set of all sets !? Especially, VeV which you thought you
had proved to be wrong.

you mean V=U{{A},A}

yea apparentlly this is true. but this should be checked by
comprehesion. what do I mean is that union and pairing are theorums in
this theory, while comprehension is an axiom.
so union and pairing should be designed to avoid this.

Zuhair

.

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