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

On 11/2/24 6:23 AM, WM wrote:

On 02.11.2024 01:55, Richard Damon wrote:

On 11/1/24 6:07 AM, WM wrote:

On 31.10.2024 21:46, joes wrote:
>

At single points a function has a single value, not a jump.

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It jumps in case of NUF by 1 at a unit fraction with respect to the foregoing unit fraction and the many points between both.

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If it jumps a *A* point, it has two values at that point,

 No it has a larger value at that point than at the interval before that point.

Which means you are comparing points to intervals.
For there to be a point where NUF(x) increases, it must be counting the number of unit fractios less than or equal to itself.
And thus, in the interval below x, for NUF(x) to be jumping to 1, there must be NO unit fractions in the interval (0, x), but there always will be so NUF(x) can't just from 0 to 1 at any finite point.

 

In the
case of NUF, that value is infinite everywhere except at 0.

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That is impossible because there are not infinitely many unit fractions between 0 and everywhere.
>

Why not?

 If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions must be between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of uncountably many points must be between 0 and x₀. That cannot happen at x₀ = 0.

Right, NUF(x)is 0 at x - 0, and Aleph_0 at any x > 0, since as you said, for any positive finite x, there are Aleph_0 unit fractions below it.

 Regards, WM