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

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.