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

On 9/12/24 7:18 AM, WM wrote:

Le 12/09/2024 à 03:00, Richard Damon a écrit :
 

So, you can't "index" an unbounded set of unit fractions from 0, as there isn't a "first" unit fraction from that end.
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We can "address" those unit fractions with the value, but we can not "index" them from 0, only from 1/1.

 If you can index all unit fractions, then you can index them from every side.
Fact is that NUF(x) increases from 0, but at no point it can inc4rease by more than 1 because of
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Regards, WM
 

Nope, that ASSUMPTION just means you can't actually have an infinite set, as you can't get to the upper end to let you count down.
1/n - 1/(n+1) > 0 thus says that no 1/n is the smallest, as there will exist a 1/(n+1) that is smaller that that.
Your claim boils down to a claim that there exists some n in the Natural Numbers that doesn't have a next number, and thus your set of Natural Numbers is NOT the infinite set they are defined to be, but only some finite initial subsequence of the real numbers, and your logic never had a correct idea of infinity,
Sorry, you are just proving that you werte NEVER talking about the actual set of Natural Numbers, just some made up finite subset of them, and your "dark" numbers are just the actual existing Natural Numbers that you have refused to look at because you ran out of ways to count to them.