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NNTP-Posting-Date: Mon, 28 Oct 2024 21:45:53 +0000
Subject: Re: How many different unit f [actions are lessorequal than all unit
 fractions? (infinitary)
Newsgroups: sci.math
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From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Mon, 28 Oct 2024 14:45:50 -0700
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On 10/28/2024 01:15 PM, Jim Burns wrote:
> 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
>
>

Well, the property that "a set is complete if
it contains each of its least-upper-bounds",
is completeness, is the one ascribed to line-reals,
ran(EF).

It's an upper-bound, it's least, rather trivially
as either a finite set contains its upper-bound,
or, a finite set has a next-greater upper-bound,
one or the other of those is discernible and
one or the other of those exists and
one or the other of those is an upper-bound and
one or the other of those is least,
thusly the least-upper-bound "LUB" property holds,
========== REMAINDER OF ARTICLE TRUNCATED ==========