Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:


Wrong.. You have not understood the meaning. We have: The set of all
sets does not exist. But if all mathematical entities including all
sets do exist in a Platonist universe, how can it be that the set of
all sets does not exist?

That is NOT Fraenkel, Bar-Hillel, and Levy's point at all!

You admitted that you do not understand their philosophical remarks. So
let it be.

No, please do not misparaphrase me. I said that I admit that I don't
fully MASTER their philosophical remarks. But I said that I do know
basically what they're talking about and what they're saying. And what
they're saying is definitely NOT that sets themselves grow or are
indeterminate, but rather, as I explained for you, that we may make
choices as to which axioms to adopt as those choices will determine how
inclusive or exclusive is any universe of sets for those axioms, which
is roughly put by me, but good enough along with a recommendation that
the reader read for himself the section in the book.

So you have not made ANY point regarding your own views by quoting
Fraenkel, Bar-Hillel, and Levy.

That does not suggest that
Fraenkel, Bar-Hillel, and Levy consider that a set itself can grow or
have different members at different times or anything like that.

They do not consider it for themselves but they report that some
mathematicians could adhere to that point of view. As far as I
remember, they do *not* say hat this point of view is wrong or
illogical or silly.

And they don't advocate it at all.

They mention this point of view as a possible one, and, as far as I
remember, as the only possibility for a platonist.

WHERE? Where in that book do they mention the point of view that a set
can grow or have different members at different times and that that is
the only possible view for a platonist? You've got it completely
backwards. It is very much a platonist view that sets are NOT as just
described.

If you want a theory in which there
are growing sets, then, by all mean, go ahead and develop one.

Not necessary. It has already been there. It is the well known
mathematics without set theory.

What are the axioms, primitives, and definitions of this theory that
has "already been there" and that has growing sets?

For
such axiomatic set theories, membership in a set is definite and sets
are determined strictly by membership. And Fraenkel, Bar-Hillel, and
Levy do not dispute that.

If sets are strictly determined by membership, and if all sets do
exist, why then doesn't just that set exist the members of which are
sets with no further specification.

Because no one who works with Z set theory claimed that for any
possible description there is a set that has the properties described.

It is not important whether one claims that or not.

It is CRUCIAL whether one claims that or not.

And what do you mean "all sets exist"? Do you mean all sets that exist
do exist? (Which is of course true.)

No, it is wrong, because the set of all those sets does not exist.

That in Z set theories there is no set S that has the property that all
other sets are members of S does not entail that it is is not the case
that all sets that exist do exist.

And you didn't answer my question.

Or do you mean that for any
specification of properties, there exists a set having those
properties?

Not even all sets with a well-defined specific property do exists.

In Z set theories, we do NOT claim that for any property expressed as a
formula that there is a set that has as members all and only those
objects that have the property.

Already the definition of the natural numbers n > 0 by all sets with
cardinality n fails,

Yes, there is not such a definition in Z set theory. So?

because the set of all sets equinumerous to n does
not exist.

Yes, in Z set theories, for n not equal to 0, there is no set whose
members are all and only those sets equinumerous with n. So?

So, since you seem in the last couple of your remarks only to be
reiterating Z set theory, what is your point? What is your answer to my
question, which is: What do you mean when you say "all sets exist"?

MoeBlee

.

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