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

Am Wed, 09 Oct 2024 11:41:31 +0200 schrieb WM:

On 08.10.2024 21:17, joes wrote:

Am Tue, 08 Oct 2024 17:40:50 +0200 schrieb WM:

On 08.10.2024 15:36, joes wrote:

Am Tue, 08 Oct 2024 12:40:26 +0200 schrieb WM:

On 08.10.2024 12:04, Alan Mackenzie wrote:

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

>

Hence all must be visible including the point next to zero, but
they are not.

There is no point next to zero.

Points either are or are not. The points that are include one point
next to zero.

But not the point inbetween?

If it exists then this point is next to zero.

Ah, then the former point wasn’t the one next to zero. Same goes for
this one. There are always infinitely many points between any two
reals.

The infinite sets contain what? No natural numbers? Natural numbers
dancing around, sometimes being in a set, sometimes not? An empty
intersection requires that the infinite sets have different
elements.

These are infinite sets: {2, 3, 4, …}, {3, 4, 5, …}, {4, 5, 6, …}.
They contain all naturals larger than a given one, and nothing else.
Every natural is part of a finite number of these sets (namely, its
own value is that number). The set {n+1, n+2, …} does not contain n
and is still infinite; there are (trivially) infinitely many further
such sets. All of them differ.

All of them differ by a finite set of numbers (which is irrelevant)
but contain an infinite set of numbers in common.

Every *finite* intersection.

As long as infinitely many numbers are captivated in endsegments, only
finitely many indices are available, and the intersection is between
finitely many infinite endsegments.

WDYM, all numbers in the segments are indices.
But what about the intersection between all infinitely many segments?

Think about it this way: we are taking the limit of N\{0, 1, 2, …}.

In the limit not a single natural number remains, let alone infinitely
many.

What does this mean for the infinite intersection?

Shrinking sets which remain infinite have not lost all elements.

This goes for every single of these sets, but not for their
infinite(!) intersection.

If every single set is infinite, then the intersection is infinite
too. These sets have lost some natural numbers but have kept
infinitely many.

 

If you imagine this as potential infinity,

No, in potential infinity there are no endsegments.

Uh. So the naturals don’t have successors?

They have successors but endsegments are sets and must be complete.

What is the difference between the sef of successors and an endsegment?
Why can’t the segments be potentially infinite, or the successors
actually inf.?

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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.