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

Am Wed, 09 Oct 2024 17:39:36 +0200 schrieb WM:

On 09.10.2024 17:11, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers but only half of them are smaller than ω,
i.e., are natural numbers. The other half is larger than ω.

 
Ha ha ha ha!  This is garbage.  If you think doubling some numbers
gives results which are "larger than ω" you'd better be prepared to
give an example of such a number.  But you're surely going to tell me
that these are "dark numbers

 
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Should all places ω+2, ω+4, ω+6, ... remain empty?

You haven't even written them down on the right. Are they included?
The ellipsis commonly only goes up to w.

Should the even numbers in spite of doubling remain below ω?

What else would you expect?

Then they must occupy places not existing before. That means the
original set had not contained all natural numbers. That mans no actual
or complete infinity.

Of course not. For every n, also n+1, n*2 and n^n are already included.

Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.

That is simply false.  You cannot specify a single number which is in
all endsegments.

True. This proves dark numbers.

Dark numbers don't exist, or at least they're not natural numbers.
There is no number in each and every end segment of N.

True. But those endsegments which have lost only finitely many numbers
and yet contain infinitely many, have an infinite intersection.

Not what we are discussing.

Note: The shrinking endsegments cannot acquire new numbers.

Why should it.

An end segment is what it is.  It doesn't change.

But the terms of the sequence do. Here is a simple finite example:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .

Example of what?  The reasoning you might do on finite sets mostly
isn't applicable to infinite sets.

Why not? The essence is that only finitely many numbers have been lost
and the rest is remaining.

So, for an infinite number of infinite sets infinitely many numbers
have been "lost", leaving nothing.

Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.

Which "these"?

Quantifier shift: There is a subset of three elements common to all
TN-sets. Understood?

Yes, I understand completely.

No.

LOL

--
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.