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

On 10/26/2024 12:22 PM, Ross Finlayson wrote:

On 10/25/2024 12:44 PM, Jim Burns wrote:

On 10/23/2024 1:38 PM, Ross Finlayson wrote:

[...] that the most direct mapping between
discrete domain and continuous range is
this totally simple continuum limit of n/d
for natural integers as only d is not finite
and furthermore
is constant monotone strictly increasing
with a bounded range in [a,b], an infinite domain.

>
The continuum limit is not the continuum.
I know:
it sounds like it should be, but it isn't.
>
The continuum limit is
the spacing of a lattice approaching 0.
>
If we are _already_ working in the continuum,
the lattice points _in the limit_
are sufficient to
uniquely determine a _continuous_ function.
For many purposes,
uniquely determining a continuous function
is sufficient for that purpose.
>
But that isn't the continuum.
In a continuum,
each split has a point at the split,
either one which ends the foresplit
or one which begins the hindsplit
_which is different_

>
Do you yet recall that these properties:
 extent density completeness measure,
would establish that ran(f) that being ran(EF)
is a continuous domain?

I still recall
you claiming that
EF(ℕ) is Dedekind.complete [0,1]ᴿ
You establishing that, not so much.
Do you recall that the continuum limit
is not the continuum?
The continuum limit is
letting the spacing of a lattice approach 0.

Then that completeness is as simply trivial
that it's defined that
the least-upper-bound of the set is
an element of the set, that
for f(...m) that f(m+1) is this?

Consider your
  n/d n->d d->oo
Is that complete real interval [0,1]ᴿ ?
If
  n/d n->d d->oo
means
  limᵈ⁻ᐣⁱⁿᶠlimⁿ⁻ᐣᵈn/d
then no.
  limᵈ⁻ᐣⁱⁿᶠlimⁿ⁻ᐣᵈn/d = limᵈ⁻ᐣⁱⁿᶠd/d = 1
If
  n/d n->d d->oo
means
  limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d
  (integer.interval [0,d]ᴺ ᵉᵃᶜʰ/d)
then also no.
⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ is
the infinite sequence of sets [0,d]ᴺ/d
E([0,c]ᴺ/c) is an end.segment of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
E([0,c]ᴺ/c) = { [0,c]ᴺ/c [0,c+1]ᴺ/(c+1) [0,c+2]ᴺ/(c+2) ... }
⋃E([0,c]ᴺ/c) is the supremum of end.segment E([0,c]ᴺ/c)
Each end.segment.supremum ⋃E([0,c]ᴺ/c) is
a superset of any set.limit of E([0,c]ᴺ/c)
-- if that set.limit exists.
⟨ ⋃E([0,c]ᴺ/c) ⟩ᶜ⁼¹ᐧᐧᐧⁱⁿᶠ is
an infinite sequence of supersets of
any set.limit of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
-- if that set.limit exists.
⋂⁰ᑉᶜ⋃E([0,c]ᴺ/c) is
also a superset of
any set.limit of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
-- if that set.limit exists.
However,
⋂⁰ᑉᶜ⋃E([0,c]ᴺ/c)  =  rational interval [0,1]ꟴ
[0,1]ꟴ is not Dedekind.complete.
Each subset of [0,1]ꟴ is not Dedekind.complete.
Any set.limit of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
  is not Dedekind.complete
-- if that set.limit exists.
Either
  limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d ≠ [0,1]ᴿ
  because complete [0,1]ᴿ ⊈ rational [0,1]ꟴ
or
  limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d ≠ [0,1]ᴿ
  because  limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d isn't anything.
If
  n/d n->d d->oo
means
  [0,1]ᴿ
  _by definition_
then who cares?
You have drawn a conclusion
no more sure.footed than
whatever that intuition was which
led you to make that definition.
And, anyway, a bare intuition is not shareable.
That's why we make proofs.

Then, about the "anti" and "only", and there being
this way that this ultimately tenuous continuum
limit (I'm glad at least we've arrived at that
being a word, "continuum-limit"),

[...]

makes for that
its range is a "continuous domain" itself

No.
That's not what the continuum limit is.
https://en.wikipedia.org/wiki/Continuum_limit