Re: Largest Set in ZFC?

On Mar 5, 12:15 am, reaste...@xxxxxxxxx wrote:
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.

In ZF, the existence of an empty set is derivable from the theorem
schema of separation (which is derivable from the strong form of the
axiom schema of replacement). Uniqueness is derived from the axiom of
extensionality.

Its probably simpler to note that AoI says the
empty set exists and is a member of a set
closed under the successor function.

As the axiom of infinity is actually stated, it does not itself
declare that there exists a set that has no members. Rather, it uses
the symbol '0'. That that symbol designates a unique set that has no
members is derivable from the definition of '0' as based on the
existence and uniqueness theorem derived from the theorem schema of
separation and the axiom of extensionality.

AoI also "defines" successor, since none of the
other axioms actually define successor.

Wrong. The pairing theorem (derivable from the strong form of axiom
schema of replacement and the power set axiom) and the union axiom
provide for a successor of any set, and the axiom of extensionality
for uniqueness. The definition of 'successor of S' is as you give it
below. That definition is not itself part of the axiom of infinity:

The successor of set S is defined as S U {S}.

AoI assumes the union of two sets is a set.

Pairing and union do that. And extensionality for uniqueness.

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}}.

No, the way to do it this:

xuy = U{x y}

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.

The definition is:

{S} = {S S}

and, of course {S S} we get from pairing..

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.

No. X might have OTHER members.

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).

X is not yet a DEFINED set. The variable 'X' at this point is just an
existential instantiation.

The we prove there is a unique X such that X satisfies the axiom of
infinity and is a subset of any set that satisfies the axiom of
infinity. Then THAT UNIQUE X is defined as w.

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.

Not functions, but rather formulas. The subsets are defined by
formulas, and there are only countably many formulas.

The Powerset axiom says all the subsets of X exist,
even the "inseparable" ones.

I don't know what you mean by "inseparable' in this context. Anyway,
the power set axiom makes no mention of any such thing as
'inseparable'

These would be subsets
of X that can't be defined in a finite number of symbols.

If there are such subsets of the given set, yes.

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

Larger in the sense of greater cardinality?

And by "can there be?" do you mean "is it consistent that there are?"
or do you mean "is it proven that there are?" (Ordinarily, we woud
take the former sense.) Anyway, as far as I can tell(?), the answer is
that it's consistent but not provable.

MoeBlee

.

Relevant Pages

  • Re: Small Set theory:Revised.
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  • Re: Small Set theory:Revised.
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  • Re: Why Regularity?
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    (sci.math)
  • Re: Why Regularity?
    ... set is a member of itself it will be either x or x'. ... First he said the successor of U was ... By axiom of Pairing if x is a set, y is a set, so is ... Do you want or x union? ...
    (sci.math)
  • Re: Why Regularity?
    ... set is a member of itself it will be either x or x'. ... First he said the successor of U was ... By axiom of Pairing if x is a set, y is a set, so is ... Do you want or x union? ...
    (sci.math)