Largest Set in ZFC?

I was looking at the axioms of ZFC on Wikipedia:
http://en.wikipedia.org/wiki/ZFC

I was wondering which sets can be defined in ZFC.

The axioms given on Wikipedia only postulate
one set, the set guaranteed to exist by
the Axiom of Inifinity.

According to Wikipedia, the empty set can
derived from the other axioms if any set exists.
Its probably simpler to note that AoI says the
empty set exists and is a member of a set
closed under the successor function.

AoI also "defines" successor, since none of the
other axioms actually define successor.
The successor of set S is defined as S U {S}.

AoI assumes the union of two sets is a set.
The Axiom of Union only says the elements
of a set can be in a union. I guess with Pairing
and Union I can define the union of set {S,{S}}.

AoI also seems to assume if S is a set then {S}
is also a set. I would be interested to know if this
can be proven with axioms other than AoI.

Wikipedia's AoI says:
There exists a set X such that the empty set
is a member of X and whenever y is in X, so is S(y).
(where S(y) is the successor of y.)

Can I say the following about X?

If x is an element of X then
x is the empty set or
x is the successor of y
where y is a member of X.

If so, then it is easy to prove omega is
not a member of X since omega is
not the successor of any member of X.
(and omega is not the empty set).

The Axiom Schema of Separation allows me
to define subsets of X, but only a countable number
of such subsets. This follows because a
function must be finitely defined and there can
only be a countable number of finite definitions.

The Powerset axiom says all the subsets of X exist,
even the "inseparable" ones. These would be subsets
of X that can't be defined in a finite number of symbols.

WIthout the Powerset axiom, can there be any set
in ZFC larger than the set guaranteed to exist by AoI?


Russell
- 2 many 2 count
.

Relevant Pages

  • Re: Set Theory: Should you believe?
    ... with a successor function, and with a successor relationship where ... one assumes in the axioms that the successor relationship is a function ... formulate the Peano Axioms which are non-equivalent. ... Bringing in these other treatments where "number" ...
    (sci.logic)
  • Re: Set Theory: Should you believe?
    ... with a successor function, and with a successor relationship where ... of formulating the Peano Axioms are essentially the same. ... logicians commonly refer to as Peano Arithmetic, is the "first order projection" ... infinitely many naturals and others are not. ...
    (sci.logic)
  • Re: Largest Set in ZFC?
    ... I was wondering which sets can be defined in ZFC. ... derived from the other axioms if any set exists. ... AoI also "defines" successor, since none of the ...
    (sci.logic)
  • Re: infinity
    ... Not while discussing the consequences of the axioms as they actually are! ... > You never commented on my adjustment of the Peano axioms, ... IF you claim to have generated an infinite set with your ... I am applying only finitely many successor operations at ...
    (sci.math)
  • Re: Godel proved maths inconsistent not incompleteness theorem
    ... axiom replaces Peano's 5 axioms.) ... I prove theorems of Godel, Rosser, Turing, Smullyan and you say I ... that you trumpet it to be, not simply its application to recursion ... Every number has a successor number: For every N printed in line 2, ...
    (sci.logic)