Re: Largest Set in ZFC?

On Mar 5, 10:20 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Mar 5, 12:15 am, reaste...@xxxxxxxxx wrote:

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:

I agree that the successor function can bederived using the
other axioms, but the other axioms don't define successor.
Are definitions considered to be different from axioms?

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..

Thanks. I was curious how that worked.
If S is a set then pairing says {S,S} is a set
and extensionality says {S,S} = {S}.

What prevents me from doing the following:

{} -> {{},{}} = {{}}
{{}} -> {{{}},{{}}} = {{{}}}
{{{}}} -> {{{{}}},{{{}}}} = {{{{}}}}
etc.

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.

What other members?
Can you give an example of a member of X
that is not the empty set or the successor
of some other element of X?

For example, can {0,5} be a member of X?
If so, S({0,5}) = {0,5,{0,5}} is a member of X.

It seems AoI is deliberately vague about what
sets are members of X. Why is that?

If X is the only set the axiom define, wouldn't
it make sense to specify exactly which sets
are members 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).

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.

Why deosn't AoI just do that to begin with?

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.

if the powerset of X is uncuontable, there must be
such subsets.

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?

Yes.

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.

I assume the Powerset axiom is independent of the other axioms.
Has anyone studied systems where the Powerset axiom is false?


Russell
- 2 many 2 count
.

Relevant Pages

  • 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 ... My answer to this is that 1 is not the union of 0 and 0'. ...
    (sci.math)
  • Re: infinity
    ... It is just the union of two disjoint sets, ... and with a successor relation that does ... I don't see any axiom that says 0 is not ...
    (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: Galileos Paradox and the Project of the Reals
    ... too many times already exactly what the axiom of infinity is. ... There is no general definition of 'one after' in set theory, ... That's how successor rests on "<". ...
    (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)