Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

Am Thu, 10 Oct 2024 19:56:26 +0200 schrieb WM:

On 09.10.2024 19:08, joes wrote:

Am Wed, 09 Oct 2024 14:48:17 +0200 schrieb WM:

How do you compare finite sets?

By their numbers of elements.

How is that defined?

Nonsense. Only potential infinity is used. Never the main body is
applied.

What "main body"?

The actually infinite numbers of dark elements.

Those are of course also bijected, even though you cannot see them.

For Cantor's enumeration of all fractions I have given a simple
disproof.

Your "proofs" tend to be nonsense.

Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with less
numbers.

I struggle to follow this illogic. Why should one segment have less
numbers?

All have the same numbers, namely ℕ. Some of the first numbers are
transformed from contents to indices and than lost. But almost all
numbers, namely ℵo, remain (because after every definable natnumber n ℵo
numbers follow). If the intersection is less than ℵo, at least one
endsegment must have fewer than ℵo numbers.

Yes, all of them have |N|=Aleph_0 numbers, which is also the amount
of segments (duh). Remember that we are intersecting inf. many sets:
it is only required that there be a set which does not contain that
number, for every number - there are even infinitely many successor-
segments that don't!

Note: The shrinking endsegments cannot acquire new numbers.

Not necessary, they already contain as many as needed.

It would be nessessary if all are infinite but their intersection is
empty. Then the infinitely many numbers cannot be the same in all
endsegments.

Right. Even though all sets are infinite, no two are the same.

Satz: Jede Menge, die mindestens drei Elemente enthält (nennen wir diese
Mengen DE), enthält drei Elemente gemeinsam mit allen DE-Mengen. Es gibt
also drei Elemente, die in allen DE-Mengen enthalten sind. Das ist ein
erlaubter Quantorentausch.

A quantifier shift is never valid.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.