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NNTP-Posting-Date: Sat, 02 Nov 2024 18:02:18 +0000
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions? (infinitary)
Newsgroups: sci.math
References: <vb4rde$22fb4$2@solani.org> <vfgil3$38pn6$1@dont-email.me>
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 <vflnoq$l02$1@news.muc.de> <vfloff$1fpr2$1@solani.org>
 <vflrt7$h190$3@dont-email.me> <vfnp4e$unin$3@dont-email.me>
 <vfns3o$3r5kq$4@i2pn2.org> <vfoq77$14lcd$5@dont-email.me>
 <vfq6rb$3v4c4$4@i2pn2.org> <vfq7hj$1fqil$3@dont-email.me>
 <c388920e-e951-4b06-a8fc-0eb3e6ea7989@att.net> <vftjeq$26ql1$1@dont-email.me>
 <f19c53f7-2509-4f84-8b1e-c42354aa06b5@att.net> <vfu2vd$29m4j$3@dont-email.me>
 <7def94cc-4a51-4305-8e62-0c5b5f5a6b0a@att.net> <vfve6i$9264$1@solani.org>
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 <01c88350-8726-4d1f-9c67-ec6dd699ef2d@att.net>
From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Sat, 2 Nov 2024 11:02:15 -0700
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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".