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

Am Mon, 14 Oct 2024 13:35:41 +0200 schrieb WM:

On 13.10.2024 00:50, joes wrote:

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

Cardinality is nonsense. Density is numbers per places for every finite
set.

Any number div. by an inf. number ought to return density zero. OTOH,
We have an inf. number per inf. interval, so that's even more undefined.

The real line reaches until, but does not include omega, no matter your
step size.

Then double ℕ U {ω}.

That is 2N u {2w} = G u {w}. If you insist on multiplying the other way
around, that's still G u {w*2} = 2N u {w*2}. No trace of w, since there
was nothing between N and w to be doubled. Why do you want to include
the non-natural omega? It has nothing to do with N.

What value do you suppose n^2 and n^n diverge to?

More difficult to determine than 2n.

w^2 and w^w, which both still have cardinality Aleph_0.

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.

Then you are outside of basic mathematics. The double number is larger
than the doubled - in every case, even for infinite numbers.

w*2 != 2w = w
And we are only doubling natural (->finite) numbers.
The set 2N=G of even numbers obviously also includes multiples of 4.

--
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.