Re: Replacement of Cardinality

Am 15.08.2024 um 19:34 schrieb Moebius:

Am 15.08.2024 um 19:01 schrieb Moebius:

Am 13.08.2024 um 19:02 schrieb Jim Burns:

On 8/13/2024 10:21 AM, WM wrote:

>

[There is] a real [number] x with NUF(x) = 1.

>
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
>
NUF(INVNUF(1)) > 1
Contradiction.

>
Well, no, this just isn't a proof, sorry about that, Jim.
>
(Hint: The term "INVNUF(1)" is not defined. Hence you may not even use it in a proof by contradiction. Actually, you didn't state an assumption in your "proof".)
>
Proof by contradiction:
>
Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such that NUF(x0) = 1. This means that there is exactly one unit fraction u such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 + 1) is an unit fraction which is smaller than u0 and hence smaller than x0. Hence NUF(x0) > 1. Contradiction!
>
(Of course, it's clear that I'm using the same "proof idea" as you used in your attempt of a proof.)

 Ok, I'll give it a try.
 Assume that there is an x e IR such that NUF(x) = 1.
 May we (by using this assumption) define INVNUF in such a way that 1 is in its domain?
 I mean can we define
         INVNUF(n) = _first_ x with NUF(x) = n     for n e {1, ...}
 now? (This would allow to use the term INVNUF(1) in our proof.)
 For this we would have to show/prove that there is a _first_ x with NUF(x) = 1. And for this we would have to show that (a) there is an x e IR such that NUF(x) = 1 (which is our assumption, so nothing to do here) and (b) that there is no smaller real number x' such that NUF(x') = 1. But I DOUBT that we will be able to prove/show that (even with our assumption).
 So I consider this a dead end.
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 Oh, wait! From our assumtion we can derive a contradiction (see my proof above)! Hence we can even prove (b) now (in the context of this proof by contradiction)! For that we assume ~(b) and immediately get ~~(b) and hence (b) from the assumption ~(b) and our original assumption, by contradiction. (The assumption ~(b) is now "discharged").
 Now we can define INVNUF such that 1 is in its domain. And hence now we may use your line of thought:
 

INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
>
NUF(INVNUF(1)) > 1
Contradiction.

 And hence we finally get: there is no x e IR such that NUF(x) = 1.  qed
 Well...

Im mean, in this case we would need (the mayor part of) my original proof
[...] Let x0 e IR such that NUF(x0) = 1. This means that there is exactly one unit fraction u such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 + 1) is an unit fraction which is smaller than u0 and hence smaller than x0. Hence NUF(x0) > 1. Contradiction!
as a "subproof". You see... (unnecessarily complicated).