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

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Sujet : Re: How many different unit fractions are lessorequal than all unit fractions? (repleteness)
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 17. Sep 2024, 19:59:06
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <05bf9c39-b715-41d6-b349-87ddc67941ff@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Mozilla Thunderbird
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.

Date Sujet#  Auteur
21 Sep 24 o 

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