Re: ZFC in 4 Axioms.
- From: "zuhair" <zaljohar@xxxxxxxxx>
- Date: 12 Dec 2006 18:42:56 -0800
hagman wrote:
zuhair schrieb:
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}
I had not yet defined {A}. But let us define {A} to be the set
havingexactly A as element
xe{a} <-> x=a
By Extension, this is the same as {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.
How will you define union and pairing in such a way that
1) they mostly look the way they naively should
2) their existence follows from your axioms
3) the contradiction is avoided?
Here is a trial.
Primitive e
Ax.1) Extensionality: As in ZFC
Ax.2) Universe: E!V Ay yeV
Ax.3) Comprehesnion: Ex Ay ( yex<->P(y) /\ y!=x /\ y!=V )
Ax.4) Infinity: As in ZFC
Ax.5) Choice: As in ZFC.
1) Pairing: Ay Az Ex Am ( mex<-> ( m=y v m=z ) /\ m!=x /\ m!=V )From this the following theorums arise:
2) Union: Ax Ey Az (zey<->(Eu:zeu /\ uex) /\ z!=y /\ z!=V)
3) Separation: Ax Ey Az ( zey<->zex/\P(z) /\ z!=y /\ z!=V)
4) Replacement:(AxE!y:P(x,y))->AaEbAy:yeb<->Exea:P(x,y) /\ y!=b /\
y!=V.
5) Power:Ax EP(x) Az ( zeP(x)<->(Ay:yez->yex) /\ z!=P(x) /\ z!=V)
All these theories can be derived simply from the above 5 axiomatic
system.
I think this will distroy your argument about V being in itself.
Zuhair
Also note that I needed union and pairing only to come from the "set
that contains all sets except itself" to the "set of all sets" and from
here to a contradiction to the Foundation axiom of ZFC.
Personally, I don't want the "set that contains all sets except itself"
to exist either.
.
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