Re: Largest Set in ZFC?

On Mar 5, 8:38 pm, reaste...@xxxxxxxxx wrote:
On Mar 5, 10:20 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

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

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?

Actually, to be very precise (to avoid a possible later confusion), in
Z set theories (I'll let that be the context from now on, unless
stated otherwise), we don't derive the existence of an function that
is the successor function on all SETS. Rather, we define a 1-place
function symbol that takes any term to form a new term. Then, for any
given set, we can define the successor function on that set.

As to definitions, it seems to me that among differing approaches,
there are two basic kinds:

(1) Definitions are prescriptions in the meta-theory for
"abbreviating" formulas.

vs.

(2) A definition (of a function symbol or predicate symbol) is itself
a very special kind of axiom, called a 'definitional axiom'. Each
definitional axiom extends the language by adding a symbol to the
language and extends the theory by adding the definitional axiom
itself to the theory. However, defintional axioms satisfy the criteria
of eliminability and non-creativity, so that again definitional axioms
in effect also only allow for abbreviation. (This is made precise by
the exact theorems of mathematical logic having to do with
eliminability and non-creativity). I strongly recommend Suppes's
'Introduction To Logic' for the best introductory explanation of this
concept I've seen (it's also in shortened form in his 'Axiomatic Set
Theory'). For even more technical details you can consult just about
any of the more widely used textbooks on mathematical logic.

Personally, I prefer (2), especially as it is so widely used in
textbooks on mathematical logic and for the way it hooks up with other
concepts in mathematical logic.

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}.

Right. To be even more precise:

An instance of pairing and using extensionality:

AsE!xAy(yex <-> (y=s v y=s))
so
AsE!xAy(yex <-> y=s)

Then definitional axiom:
Asx(x={s} <-> Ay(yex <-> y=s))

Usually we take the leading universal quantifiers as tacit understood,
so we state:

Definitional axiom:
x = {s} <-> Ay(yex <-> y=s)

What prevents me from doing the following:

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

What do you mean with the '->' symbol mean there?

I think you mean, e.g.,

From { }, we construct { { } { } }.

If so, then nothing prevents you from doing what you just mentioned.

But why use '{ }' rather than '0' when '{ }' is harder to read when
it's nested.

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?

First, even if I couldn't give an example, the point stands that the
axiom itself does not RULE OUT that there are other members.

Second, example (here 'x+' stands for 'xu{x}'):

{0 1 2 ... x x+ x++ ...}

where x is any set other than a natural number. Notice, that for the
purpose of this question I don't even need to prove the existence of
such a set; all I need to do is point out that the axiom of infinity
itself does not RULE OUT that there exists a successor inductive set
with also members x, x+, etc. And in ZF, we can prove, for any x, the
existence of said set.

For example, can {0,5} be a member of X?

Sure, as long as {0 5}+ and all the rest down the line are members of
X.

If so, S({0,5}) = {0,5,{0,5}} is a member of X.

Right.

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

I don't know the motivation of the actual people who wrote it, but the
reason I personally would give is this: We don't NEED to be any more
specific, since separation does the rest of the job for us to get to
the specific set w that we want.

If X is the only set the axiom define,

Be careful. What do you mean by 'X'? It is not the case that the axiom
asserts the existence of a PARTICULAR X. Rather, the axiom asserts
that there is AT LEAST one set X that is successor inductive. So the
axiom doesn't give us a PARTICULAR X to even talk about. All we can do
is existentially instantiate to SOME set X. What gives us a PARTICULAR
X, which we then name 'w' is separation.

wouldn't
it make sense to specify exactly which sets
are members of X?

Again, there's no particular X at that step. But when we apply
separation, we do get a particular X that we call 'w', and THEN we
prove what you mentioned:

The only members of w are 0 and successors of another member of w.

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?

My own answer: Because we don't need to put in an axiom what we can
derive from it and the other axioms anyway.

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

No, the power set axiom says there is a set whose members are all and
only the subsets of X. The power set axiom doesn't itself say anything
about what subsets exist.

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.

Right.

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.

I don't know how we'd prove the existence of larger cardinalities than
that of w in ZFC-Power. Maybe it can be done; but I don't know. Some
of the other more expert posters can answer better.

I assume the Powerset axiom is independent of the other axioms.

Yes, that has been proven.

Has anyone studied systems where the Powerset axiom is false?

You mean ZFC-Power+~Power ?

I'm pretty sure you can find a fair amount on ZFC-Power. But I don't
know about ZFC-Power+~Power.

MoeBlee
.

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