Re: Largest Set in ZFC?

On Mar 8, 12:22 am, reaste...@xxxxxxxxx wrote:
On Mar 7, 2:55 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

On Mar 7, 12:35 pm, reaste...@xxxxxxxxx wrote:
I would be interested in ZF - Infinity - Powerset with these two
axioms:

All sets are finite.
(not AoI)

Which definition of 'finite'?  "1-1 with a natural number" or "not 1-1
with a proper subset"?

Good question.
I forgot there are many definitions of infinite.

"All sets have a bijection with a finite ordinal"
would be circular.

That would be an axiom and not a defintion. It's quite fine. We can
define 'finite' and 'ordinal'. And we would take it as an axiom that
every set bijects with some finite ordinal. Circularity is not a
problem here.

"All non-empty sets have an bijection with
a successor of the empty set" might work.

As an axiom I take it. But what is the definition of 'a successor of
the empty set"?

I'm not sure "No set has an bijection with
a proper subset of itself" really prevents
infinite sets.

That depends on whether 'infinite' is taken as 'not finite (not
bijectable with some finite ordinal) or as Dedekind infinite. Anyway,
as an axiom, "all sets are finite" is stronger than "all sets are
Dedekind finite (no set is bijectable with a proper subset of
itself)".

What about "uncountable" sets?
Since I am negating the Powerset axiom,
I can have sets where I can't prove all
of the subsets actually exist.

Please get rid of this "actually exist" talk. I've explained already
why it is ill-premised. Please look again at my explanation as to what
we prove as regards existence. Briefly, we prove that there exist
objects that have certain properties. We don't prove or disprove that
something exists or not. There is no thing that does not exist. If
it's an object, if it's a thing, then it exists. If it's a subset,
then it exists. What doesn't exist is an object that has a certain
PROPERTY. Talk of this or that "actually existing" is confusion.

I think I would prefer:
"All non-empty sets have a bijection with a successor of the empty
set".

You haven't defined "is a successor of the empty set".

Anyway, the following is quite fine:

AxEn(n is a finite ordinal & x is equinumerous with n)

where 'x is equinumerous with n' is defined as 'there exists a
bijection between x and n'.

You think ZF |- all sets are finite  <-> ~ axiom of infinity ?

If so, and depending on your definition of 'finite', how did you
arrive at that conclusion?

No finite set can be closed under the successor function?

Yes. But that doesn't show that ZF proves "all sets are finite <-> ~
axiom of infinity".

Yes, I think we can show:

Ax(x is finite -> ~x is closed under successor)

so we have:

all sets are finite -> ~ axiom of infinity

But you need to prove also:

~ axiom of infinity -> Ax x is finite.

I guess I could say:
"There is no set closed under successor."

That's not enough. See above.

There exists a set, X, such that no set contains every subset of X.
(not Powerset)
I assume I don't need Choice if all sets are finite.

I think I could prove there exists a finite ordinal, X,
such that the complete powerset of X doesn't exist
as a set in this system.

Okay, I take it you mean:

Ex(x is a finite ordinal & ~EsAz(zes <-> z subset of x)).

Your proof?

Let x be a set that does not have a powerset.

Ex(~EsAz(zes <-> z subset of x)).

Since every non-empty set has an bijection with a
successor ordinal, there exists finite ordinal, N,
that has a bijection with X.

Okay, but you mean 'x' not 'X'. And let's use 'n' rather than 'N',
since 'N' is usually taken as standing for the set of natural numbers.

The powerset of N can't exist.

Means, there is no set that has as members all susets of n.

Proof: Assume P contains all of the subsets of N.
Let B be a bijection between X and N.

Okay. But to make it easier (see below), let the bijection be from N
to X.

For every member of P, replace the elements of N
inside this subset with the corresponding element of X
(with respect to B).

Maybe you mean this?:

Let f be function on P such that for y in P, f(y) = {B(z) | zey}.

If P contains every subset of N, this new set formed
by replacement

The "new set" would be the range of f.

But the problem for me is that I would have to go back to check if and
where the power set axiom was used to establish that there always
exist such functions and ranges as descibed above. The Cartesian
product operation is established by way of the power set axiom, so I'd
want to check whether we can assert the existence of functions as
described above without taking them as subsets of a Cartesian product.

must contain every subset of X.
Since no set contains every subset of X,
the set P can not exist.

Cantor's diagonal argument could be used to show
that for any set, S, there is a subset of X that is not
an element of S.

You mean to prove the previous assertion?

I can use an argument similar to the one above.
I think the diagonal argument works too.

Let's see it then. And also keep in mind what I mentioned about the
possibility that the power set axiom is needed even to prove the
existence of whatever functions as you describe them. If you so
radically alter the axioms, you may need to go back to some very early
stages to verify what your theory can still prove, and not just assume
that there exists such described functions.

MoeBlee

MoeBlee
.

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