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NNTP-Posting-Date: Tue, 17 Sep 2024 19:18:25 +0000
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions? (repleteness)
Newsgroups: sci.math
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From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Tue, 17 Sep 2024 12:18:27 -0700
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On 09/17/2024 11:57 AM, Ross Finlayson wrote:
> On 09/17/2024 10:59 AM, Jim Burns wrote:
>> On 9/16/2024 10:55 PM, Ross Finlayson wrote:
>>> On 09/16/2024 11:24 AM, Jim Burns wrote:
>>>> On 9/16/2024 2:13 PM, Jim Burns wrote:
>>>>> On 9/15/2024 9:31 PM, Ross Finlayson wrote:
>>>>>> On 09/15/2024 03:07 PM, FromTheRafters wrote:
>>
>>>>>>> What is the successor function on the reals?
>>>>>>> Give me that, and maybe we can find the
>>>>>>> 'next' number greater than Pi.
>>>>>>
>>>>>> Ah, good sir, then I'd like you to consider
>>>>>> a representation of real numbers as
>>>>>> with an integer part and a non-integer part,
>>>>>> the integer part of the integers, and
>>>>>> the non-integer part a value in [0,1],
>>>>>> where the values in [0,1], are as of
>>>>>> this model of (a finite segment of a) continuous domain,
>>>>>> these iota-values, line-reals,
>>>>>> as so established as according to the properties of
>>>>>> extent, density, completeness, and measure,
>>>>>> fulfilling implementing the Intermediate Value Theorem,
>>>>>> thus for
>>>>>> if not being the complete-ordered-field the field-reals,
>>>>>> yet being these iota-values a continuous domain [0,1]
>>>>>> these line-reals.
>>>>>
>>>>> As n → ∞, (ι=⅟n), ⟨0,ι,2⋅ι,...,n⋅ι⟩ → ℚ∩[0,1]
>>
>> Here follows the meaning of
>> my.best.guess.at.what.you.mean.
>> ⎛ I invite you to either
>> ⎜ continue implicitly accepting my guess or
>> ⎝ clarify what.you.mean.
>>
>> Define
>> f: ℕ → 𝒫(ℝ)
>> f(n) = ⟨0/n,1/n,2/n,...,n/n⟩
>>
>> Define
>> ran(f) = limⁿᐧᐧᐧ f(n)
>>
>>
>> lim.infⁿᐧᐧᐧ f(n) ⊆ limⁿᐧᐧᐧ f(n) ⊆ lim.supⁿᐧᐧᐧ f(n)
>>
>> https://en.wikipedia.org/wiki/Set-theoretic_limit
>>
>> lim.infⁿᐧᐧᐧ f(n)  =
>> ⋃⁰ᑉⁿ ⋂ⁿᑉʲ f(j)  =
>> ⋃⁰ᑉⁿ {0,1}  =
>> {0,1}
>>
>> lim.supⁿᐧᐧᐧ f(n)  =
>> ⋂⁰ᑉⁿ ⋃ⁿᑉʲ f(j)  =
>> ⋂⁰ᑉⁿ ℚ∩[0,1]  =
>> ℚ∩[0,1]
>>
>> {0,1} ⊆ limⁿᐧᐧᐧ f(n) ⊆ ℚ∩[0,1]
>>
>> ----
>> ⋂ⁿᑉʲ f(j)  =  {0,1}
>> ⋃ⁿᑉʲ f(j)  =  ℚ∩[0,1]
>>
>> ⎛ ⋂ⁿᑉʲ f(j)  ⊆
>> ⎜ ⟨0/j,1/j,...,j⋅/j⟩ ∩ ⟨0/j⁺¹,1/j⁺¹,...,j⁺¹⋅/j⁺¹⟩  =
>> ⎜ {0,1}
>> ⎜
>> ⎝ ⋂ⁿᑉʲ f(j)  ⊇  {0,1}
>>
>> ⎛ ∀ᴺj>n:  f(j) ⊆ ℚ∩[0,1]
>> ⎜ ⋃ⁿᑉʲ f(j)  ⊆  ℚ∩[0,1]
>> ⎜
>> ⎜ ∀p/q ∈ ℚ∩[0,1]:
>> ⎜  nq > n  ∧
>> ⎜  np/nq ∈ f(nq) ⊆ ⋃ⁿᑉʲ f(j)  ∧
>> ⎜  np/nq = p/q ∈ ⋃ⁿᑉʲ f(j)
>> ⎝ ℚ∩[0,1]  ⊆  ⋃ⁿᑉʲ f(j)
>>
>>> It's shewn that [0,1] has no points not in ran(f).
>>
>> Has it been shown, though?
>>
>> For ran(f) = limⁿᐧᐧᐧ ⟨0/n,1/n,2/n,...,n/n⟩
>> {0,1} ⊆ ran(f) ⊆ ℚ∩[0,1] ∌ ⅟√2 ∈ [0,1]
>>
>> If ran(f) isn't that, then what is ran(f)?
>>
>>> About 2014, ....
>>
>> Was that your last word on ran(f), in 2014?
>>
>> I had hoped you would answer the point I've made.
>>
>>
>
> Well what it is that it was arrived at that the
> only way to define the diagonal resulted being
> according to only the diagonal of this ran(f),
> here that these days and here that's the "pick
> one of anti-diagonal and only-diagonal, ha, they're
> combined, you get both or none", that the results
> were stated along the lines of that "for any x in ran(f),
> that ~x in ran(f)", then also "there is no x in [0,1]
> not in ran(d)".
>
> So, here we still have this bit of the only-diagonal
> that there exists "non-Cartesian functions", meaning
> that this range can't be re-ordered or sub-setted
> as for its definition, that just like anti-diagonal
> results "this class of functions doesn't exist, rest
> given to Cantor-Schroeder-Bernstein theorem", then
> this only-diagonal simply makes "this class of functions
> doesn't exist, not given to Cantor-Schroeder-Bernstein
> theorem", which just reflects transitivity of cardinal
> identity, why it's crossed over these "bridges" or "ponts"
> as with regards to the book-keeping, of these _structures_.
>
> Then that each of a) extent, b) density, c) completeness,
> d) measure are established, I do seem to recall that
> your account was around when it was set out, for example,
> that least-upper-bound is nice neat trivial next, while
> at least four sigma-algebras for measure 1.0 were given.
>
>

Another great discussion of Finlayson and Burns was
the "Correct Presentation of the Diagonal Argument"
thread on sci.logic.