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

Am Sat, 12 Oct 2024 19:49:23 +0200 schrieb WM:

On 10.10.2024 21:54, joes wrote:

Am Thu, 10 Oct 2024 20:53:07 +0200 schrieb WM:

On 10.10.2024 20:45, Alan Mackenzie wrote:

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

>

If all natnumbers are there and if 2n is greater than n, then the
doubled numbers do not fit into ℕ.

For any finite n greater than zero, 2n is greater than n.  The same
does not hold for infinite n.

There are no infinite n = natural numbers.

Exactly! There are furthermore no infinite doubles of naturals (2n).

But the doubles are larger. Hence after doubling the set has a smaller
density and therefore a larger extension on the real line. Hence not all
natural numbers have been doubled.

Taking "density" here to mean cardinality div. by size(?) makes it pretty
much undefined, since those are both infinite - with N just as with 2N=G
(what's the difference anyway?). The real line reaches until, but does not
include omega, no matter your step size. What value do you suppose n^2
and n^n diverge to?

Numbers multiplied by 2 do not remain unchanged.

Either doubling creates new natural numbers. Then not all have been
doubled. Or all have been doubled, then some products fall outside
of ℕ.

No.  Not even close.

Deplorable. But note that all natural numbers are finite and follow
this law: When doubled then 2n > n. If a set of natural numbers is
doubled, then the results cover a larger set than before..

Additionally: if n is finite, so is 2n. It cannot go beyond w.

Then there is no complete set. The doubling can be repeated and
repeated. Always new numbers are created. Potential infinity.

No! Actual infinity already includes all doubles of all numbers.

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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.