Re: Two results of set geometry

On Oct 3, 9:28 pm, Tony Orlow <t...@xxxxxxxxxxxxx> wrote:
MoeBlee wrote:
On Sep 29, 8:41 am, Tony Orlow <t...@xxxxxxxxxxxxx> wrote:
MoeBlee wrote:
On Sep 28, 10:24 am, Tony Orlow <t...@xxxxxxxxxxxxx> wrote:
G. Frege wrote:
WM in convinced that _assuming_ that
there are infinite sets, still /the union of finite sized elements
cannot result in an infinite sized element/.
They cannot, if each is a proper subset of some other in the set. That
is the case, is it not?
No, it's not.

Are you sure?

Yes.

w is an obvious counterexample.

Only by axiomatic/definitionistic declaration without justification.

No, we PROVE from the axioms that w is a counterexample. As to having
axioms at all, at least we have them. As to justification of them,
you'll find various justifications in the wide literature of the
subject. I won't argue that you should find such justifications to be
correct or even plausible, but at least they are there for you to
consider and critique if you like. Meanwhile, the axiom does suit the
intuitions of a great many people. That it doesn't suit your
intuitions is not ipso facto a reason it should not suit the
intuitions of others nor ipso facto a reason not to adopt it.
Moreover, you say you do endorse that there exist infinite sets, so
it's still not been said by you which axioms you do reject (and saying
that it's not the axioms you reject but rather the way their
entailments are emphasized (or however you put it) really is rank
twaddle). Moreover, you endorse mathematics (such as Robinson's non-
standard analysis) that goes beyond even the Z axioms but also
includes at least some form of a choice principle. Your very position
then in such conversations is irrational.

It's the case that each is a proper subset of some other. I'm sure you
don't dispute that. The first statement was my own, and I know your
opinion of it. However, I have mine. If no subset includes any more than
a finite number of elements, then the set cannot either.

Yes, obviously if every subset of S is finite, then S is finite.

Is N then finite?

Please don't troll me. Obviously it's not the case that every subset
of N is finite, so your question is just silliness.

Virgil just
wrote something like, "the size of N is at least as large, but not
necessarily larger, then every element in N". I thought that was a step
in the right direction.

Whatever it is, it doesn't dispute that there are sets such that every
member of the set is finite, every member is a proper subset of
another member of the set, and the union of the set is infinite.

Omega is not the union of the set? It's not, like, the limit, Man? Wow,
like, what the hell IS it? I mean, c'mon, get real.

I didn't say w is not a union. Your comments are getting even stupider
than usual for you. As to the precise definition of w, it's been given
to you about a hundred times already. Please don't insult me by asking
for it yet again.

If each is the last plus 1, then the nth is n, and there cannot be n of
them without n being in the set. That's basic logic. No non-logical (ala
Ross) axioms required. It's pre-axiomatic, and therefore has precedence.

If n ranges over only natural numbers, then what you mentioned is a
theorem of Z set theory.

That's very excellent. It's "constructive".

But if n ranges over cardinal numbers, then I

know of no system of logic that takes what you just mentioned as
"basic".

That's because of what the system of cardinal numbers misses, as far as
distinctions between numbers goes.

Like I said, you've shown no system of logic in which your principle
is "basic", let alone present at all.

But you're always welcome to point one out if it exists or

create one of your own.

Thanks. I'm trying.

Like I'm trying break DiMaggio's consecutive 56.

MoeBlee

.

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