Re: Largest Set in ZFC?

On Mar 8, 2:08 pm, reaste...@xxxxxxxxx wrote:

It is not obvious to me that separation is powerful enough.
What if X is uncountable? We can only specify a countable
number of subsets of X. What if the subset we "want" can't
be specified with a finite formula?
And exactly which set do we want?

That's all confusion. Separation is adequate for the task. Just look
at the ordinary proofs in a textbook. After the axiom of infinity, we
apply existential instantiation and then separation, then
extensionality, to arrive at our desired set.

Saying "we don't need to be more specific" still sounds
like we are being deliberately vague. I think the reason
for vagueness is so we can "assume" the AoI set
contains limit ordinals (other than 0).

Please! You're jabbering nonsense that rattles in your head because
you leave too much space there unoccupied by the correct information
you'd get if you just read a book on the subject. I EXPLAINED to you
why we don't need to be more specific. PROVABLY, separation gets the
job done. And there are NO limit ordinals in w.

We can derive omega from any set that satisfies AoI.
AoI says a set exists. It doesn't say more than one
such set exists. Since omega is defined as the "smallest"
set satisfying AoI, I see no reason to assume any
other set satisfies AoI.

CORRECT this time! We have to PROVE that there are sets other than w
that are successor inductive.

Your argument suggests omega is a model of ZF.

It's nonsense. w is a universe of a model of ZF, but just saying w is
a model of ZF is not enough to have any bearing on this matter. There
is no consistency issue since simply proving in ZF that there is a
least inductive set is not a a proof IN ZF that w is the universe of a
model of ZF.

Can't we prove omega exists in ZF?

In Z (don't even need ZF) we prove there is a unique least successor
inductive set. Then we name it 'w'.

I don't see how ZF can have a set with greater
cardinality than omega (without using Powerset).

I don't either. But that doesn't preclude that there isn't another
successor inductive set.

MoeBlee

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