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

On 10/29/24 4:42 AM, WM wrote:

On 29.10.2024 01:21, Richard Damon wrote:

On 10/28/24 3:52 PM, WM wrote:

On 28.10.2024 12:21, Richard Damon wrote:

On 10/28/24 6:36 AM, WM wrote:

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NUF increases by 1 or more, but more would violate mathematics.

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No, NUF(x) jumps from 0 to Aleph_0 in the domain of finite numbers, because there is no finite x where it has the value of 0.

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It has the value 0 for all x =< 0. And it cannot jump by more than 1 at any point.

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Of course it can. just not at any finite value.

 There are no others on the real line.

Right, so your dark numbers aren't real.
NUF(x) MUST jump from 0 to Aleph_0 at all real values x, as below ANY real number x, there are Aleph_0 unit fractions.

 

No, but the first steps happen at undefinable x.

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No, it happens at an x that isn't a finite number.

 The it is dark.

And, as you said above, NOT REAL.

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If you allow your NUF to accept infintesimal numbers

 No, that is strongly forbidden.

Then it just jumps.

 

This infinity between 0 and (0, 1] is not what I can accept.

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Note, it isn't an "infinity between" it is that the "bottom" of (0, 1] doesn't exist as a definable point.

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That is true. The bottom is dark.

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No, the bottom is outside the set.

 Then it is dark bottom.

No because you just showed that you "dark" numbers are not real.

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Your "Darkness" is just your attempt to hide the problems with your logic.

 And your bottom outside the set is what?

0., since 0 is the greatest lower bound of the series: 1/1, 1/2, 1/3, 1/4, 1/5, ... 1/n, ...
That series approaches 0, but never reaches it, so its bound is 0, which isn't in the infinite set of elements,

 Regards, WM