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

Chris M. Thomasson submitted this idea :

On 11/3/2024 6:34 AM, joes wrote:

Am Sat, 02 Nov 2024 22:39:36 +0100 schrieb WM:

On 02.11.2024 21:34, Richard Damon wrote:

On 11/2/24 1:42 PM, WM wrote:

On 02.11.2024 14:50, Moebius wrote:

Am 02.11.2024 um 14:21 schrieb joes:

Am Fri, 01 Nov 2024 18:03:26 +0100 schrieb WM:

>

If an invariable set of numbers is there, then there is a smallest
and a largest number of those which are existing.

or each and every n e IN there is an n' e IN (say n' = n+1)

Actual infinity is not based on claims for each and every, but
concerns all.

But if it applies to ALL, it must apply to ANY, so a property of ANY
must apply to each on of the ALL.
So, for ALL the Natural Numbers, there can't be a highest, because for
ANY Natural Number there is a following one

That cannot be true for all dark numbers.

And that is why "dark" numbers are not natural (or the naturals are all
not dark).
>

>
Well, what if the dark numbers are natural wrt:
>
1, 2, 3, 4, ...
>
Oh shit! WM says ... is dark. I say 5, then WM says well okay 5 is not dark now. Shit like that? Its rather hilarious to me.

 It may make more sense to go higher. Consider the 123^456^789th prime number. I would be pairing the natural number 123^456^789 with some really large prime number. Which prime? I don't know. WM eschews such a pairing because the value of that prime is unknown, or 'dark' and he thinks it cannot be used in a pairing.
 Had I been matching instead of pairing, he would have a point because it would likely be impossible for me to check that the 123^456^789th prime has the correct natural number index. For pairing, I don't need the values, only the countability, but for matching I might.