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

Am Wed, 09 Oct 2024 16:40:21 +0200 schrieb WM:

On 09.10.2024 16:13, Alan Mackenzie wrote:

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

On 08.10.2024 23:08, Alan Mackenzie wrote:

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

On 07.10.2024 18:11, Alan Mackenzie wrote:

What I should have written (WM please take note) is:

 

The idea of one countably infinite set being "bigger" than another
countably infinite set is simply nonsense.

The idea is supported by the fact that set A as a superset of set B
is bigger than B.

What do you mean by "bigger" as applied to two infinite sets when one
of them is not a subset of the other?

That is not in every case defined. But here are some rules:
Not all infinite sets can be compared by size, but we can establish
some useful rules.

Possibly.  But these rules would require proof, which you haven't
supplied.

These rules are self-evident.

I.e. can be dismissed without comment.

This breaks down in a contradiction, as shown by Richard D in another
post: To repeat his idea:
The set {0,   2,   4,  6, ...} is a subset of the natural numbers N,
        {0, 1, 2, 3, 4, 5, 6, ...}, thus is smaller than it.
We can replace the second set by one of the same "size" by multiplying
each of its members by 4.  We then get the set
         {0,       4,      8, ...}.
Now this third set is a subset of the first hence is smaller than it.

 
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.

So 2N = G u {w, w+2, w+4, ..., w+w-2}?

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.

It is astounding to see you jump from "there is no such number" to
"surely there must be one anyway".

Proof: If not, then there would be at least one endsegment with less
numbers.

... which is gobbledegook, not maths.
 

Note: The shrinking endsegments cannot acquire new numbers.

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}
{ } .
Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets. Quantifier shift: There
is a subset of three elements common to all TN-sets.

In the infinite case: All predecessors are supersets. But we are not
interested in some finitely indexed endsegment.

Then we get: All sets which have lost at most n elements have the
remainder in common. Note: All sets which are infinite have lost at most
a finite number of elements.

But what about the limit case, the intersection of all endsegments,
or the set which has lost an infinite number of elements?

--
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.