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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.logic,sci.math
Subject: Re: Replacement of Cardinality
Date: Sat, 24 Aug 2024 13:06:48 -0400
Organization: i2pn2 (i2pn.org)
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On 8/23/24 10:49 AM, WM wrote:
> Le 23/08/2024 à 03:23, Richard Damon a écrit :
>> On 8/22/24 6:54 AM, WM wrote:
> 
>>> In a linear order of elements which all have distances from each 
>>> other, there is necessarily a last one (if nothing follows) because 
>>> the only alternative would be more than one. Matheologial conjuring 
>>> trick is outside of mathematics.
> 
>> But that is just a false statement that tries to assume the conclusion.
> 
> NUF(x) increases from 0 to more. But it  cannot increase by more than 1 
> without shzowing a constant level.
> 
> Regards, WM

And, since there is no finite value of x that makes NUF(x) equal to 1, 
since for every finite x, their exists an x/2 that is smaller than x, 
and also must have a unit fraction below it, and thus a NUF(x) >= 1, but 
there is also a unit fraction in the interval x/2 to x, so NUF(x) > 
NUF(x/2), so NUF(x) must be > 1.

Just because you have verbally defined a function doesn't mean it exists 
and has proper values. After all, the square root function can be 
defined, but in the set of reals sqrt(-1) doesn't have a value.

So, by that same logic we see that INVNUF(n) might not be defined for 
all values of n, and it seems, it isn't defined for the range of real 
value results for ANY natural number n > 0.

Thus, for INVNUF(n) to exist, its output range must be beyond the reals, 
and extend to some post-finite domain where NUF(x) considers some 
sub-finite values to be "unit fractions" so that NUF(x) can "increment" 
from 0 to Aleph_0 as it needs to be for all "finite" x values.

But, since you have shown that you l;ogic can't even handle the "normal" 
infinity of the Natural Numbers, to expect it to handle the 
beyond-finite behavior of such a number system is incorrect.