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

On 11/02/2024 07:10 AM, Jim Burns wrote:

On 11/1/2024 1:24 PM, WM wrote:

On 01.11.2024 14:21, Jim Burns wrote:

>

⎛  ∀ᴿx > 0:
⎜  ∀n ∈ ℕ:
⎜ x > 0
⎜ ⅟x > 0
⎜ n+⅟x ≥ ⅟x > 0
⎜ ⌈n+⅟x⌉ ≥ ⅟x > 0
⎜ ⌈n+⅟x⌉ ∈ ℕ
⎜ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ
⎜ x⋅⌈n+⅟x⌉ ≥ x⋅⅟x > 0
⎜ x⋅⌈n+⅟x⌉⋅⅟⌈n+⅟x⌉ ≥ 1⋅⅟⌈n+⅟x⌉ > 0
⎜ x ≥ ⅟⌈n+⅟x⌉ > 0
⎜ ⅟⌈n+⅟x⌉ ∈ (0,x]  ∧  ⅟⌈n+⅟x⌉ ∈ ⅟ℕ
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
>
⎛  ∀n ∈ ℕ⁺:
⎜ n > 0
⎜ n+1 > n > 0
⎜ ⅟n⋅(n+1)⋅⅟(n+1) > ⅟n⋅n⋅⅟(n+1)
⎜ ⅟n > ⅟(n+1)
⎜ ⅟n - ⅟(n+1) > ⅟(n+1) - ⅟(n+1)
⎝ ⅟n - ⅟(n+1) > 0

>

Will you (WM) also be ignoring
the work which shows
NUF(x) grows at 0 to ℵ₀ ?

>
Of course.

>

If NUF(x) has grown to ℵ₀ at x₀,
then ℵ₀ unit fractions must be between 0 and x₀.

>
Yes, and ℵ₀.many are there,
because x₀ ≥  ⅟⌈n+⅟x₀⌉ > 0
>
NUF(0) = 0  ∧
  ∀ᴿx₀ > 0:
NUF(x₀)=ℵ₀  ⇐  ∀n∈ℕ: (0,x₀] ∋ ⅟⌈n+⅟x₀⌉ ∈ ⅟ℕ
>

Hence
at least ℵ₀ points with
ℵ₀ intervals of uncountably many points
must be between 0 and x₀.

>

That cannot happen at x₀ = 0.

>
No.
There is no x₀ > 0 which,
  between it and 0,
   that does not happen.
NUF(0) = 0  ∧
  ¬∃ᴿx₀ > 0:
NUF(x₀)<ℵ₀  ∧  ∀n∈ℕ: (0,x₀] ∋ ⅟⌈n+⅟x₀⌉ ∈ ⅟ℕ
>

Therefore
not even a fool could support your claim.

>
E pur si muove:
x₀ ≥  ⅟⌈n+⅟x₀⌉ > 0
>
>

The delta-epsilonics of course, or some put it
"delta-epsilontics", with little d and smaller e,
of often for induction arbitrary m and larger n,
is well-known to all students of calculus.
"The infinitesimal analysis", ....
Mixing transfinite cardinals and scalar quantities
about the infinite-divisibility of standard quantities
as with regards to the infinite limit and that going
to zero in delta-epsilonics and 1 for the geometric series,
has that they aren't exactly commensurate, if qualitative.
I.e., in a manner of speaking, the infinite transfinite
cardinals don't exist in delta-epsilonics any more
than plain manner-of-speaking "infinity".