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

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?

A shrinking infinite set which remains infinite has an infinite
core.

In the limit, it is empty.

Again, no.  There is no such thing as a "core", here.  Each of these
sets has an infinitude of elements.  No element is in all of these
sets.

Try to think better. A function of sets which are losing some elements
but remain infinite, have the same infinite core.

That is untrue.  For any element which you assert is in the "core", I
can give one of these sets which does not contain that element.

Of course, the core is dark.

 The "core" is thus empty.

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.

That argument is absolutely definite, a logical necessity. If you
cannot understand it, ....

It wasn't an argument, it was a bare statement, devoid of any
supporting argument.

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 you imagine this as potential infinity, you are
really thinking about stopping at some finite number. This „process”
is not allowed to be stopped.

I understand it full well, and I understand that it's mistaken.

Impossible. You don't understand that all sets are infinite and cannot
have lost all elements.

That is not very difficult to understand. We are not talking about any
set of successors of a natural number, though.

--
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.