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

On 10/1/2024 6:13 PM, Ross Finlayson wrote:

On 10/01/2024 01:13 PM, Jim Burns wrote:

On 10/1/2024 2:02 PM, Ross Finlayson wrote:

Here it's that "Eudoxus/Dedekind/Cauchy is
_insufficient_ to represent the character
of the real numbers".
>
Then, that there are line-reals and signal-reals
besides field-reals, has that of course there are
also models of line-reals and signal-reals in the
mathematics today, like Jordan measure and the ultrafilter,
and many extant examples where a simple deliberation
of continuity according to the definitions of
line-reals or signal-reals, results any contradictions
you might otherwise see as arriving their existence.
>
Then, besides noting how it's broken, then also
there's given a reasoning how it's repaired,
resulting "less insufficient", or at least making
it so that often found approaches in the applied,
and their success, make the standard linear curriculum,
unsuited.
>
Then, I think it's quite standard how I put it,
really very quite standard.

>
I hope this will help me understand you better.
Please accept or reject each claim and
-- this is important --
replace rejected claims with
what you _would_ accept.
>
⎛ ℝ, the complete ordered field, is
⎝ the consensus theory in 2024 of the continuum.
>
⎛ ℝ contains ℚ the rationals and
⎜ the least upper bound of
⎝ each bounded nonempty subset of ℚ and of ℝ
>
( The greatest lower bound of ⅟ℕ unit fractions is 0
>
⎛ A unit fraction is reciprocal to a natural>0
⎜
⎜ A set≠{} ⊆ ℕ naturals holds a minimum
⎜ A natural≠0 has a predecessor.natural.
⎜ A natural has a successor.natural.
⎜
⎜ The sum of two naturals is a natural
⎝ the product of two naturals is a natural.
>
⎛ There are no points in ℝ
⎜ between 0 and all the unit fractions
⎝ (which is what I mean here by 'infinitesimal').
>
Thank you in advance.

>
Well, first of all there's a quibble that
R is not usually said to contain Q as much as that
there's that in real-values that
there's a copy of Q embedded in R.

I take your lack of an explicit rejection of
the Dedekind.complete continuum.consensus
to be an implicit acceptance of
the Dedekind.complete continuum.consensus.
A quibble for your quibble:
A set isomorphic to ℚ is usually said to _be_ ℚ
Each model of ℚ is ℚ
A model of complete.ordered.field ℝ supersets
a model of rational.ordered.field ℚ,
which is to say, ℝ contains ℚ
I think that you (RF) are pointing to this:
⎛ Consider a model ℚ₀ of the rationals
⎜ which has only urelements.
⎜
⎜ Using ℚ₀  construct
⎜ a model of ℝ complete ordered field
⎜ in any of several known ways:
⎜ a partition of Cauchy sequences of rationals,
⎜ open.foresplits of rationals,
⎜ or something else.
⎜
⎜ The set ℝₛ of open.foresplits of ℚ₀
⎜ { S⊆ℚ₀: {}≠Sᵉᵃᶜʰ<ₑₓᵢₛₜₛSᵉᵃᶜʰ<ᵉᵃᶜʰℚ₀\S≠{} }
⎜ is a model of ℝ
⎜ It holds an open foresplit for each irrational point,
⎜ and an open foresplit for each rational point
⎜
⎜ The set ℚₛ of open.foresplits for rationals
⎜ { {q′∈ℚ₀:q′<q}: q∈ℚ₀ }
⎝ is not ℚ₀
Yes,
ℚₛ ≠ ℚ₀
However,
both ℚₛ and ℚ₀ are models of ℚ,
we say both ℚₛ and ℚ₀ we are ℚ,
Each theorem we prove for ℚ,
for example, that no element of ℚ is √2,
is true of both ℚₛ and ℚ₀,
and that's enough for (consensus) us.

The, "1/N unit fractions", what is that,
that does not have a definition.

Read a bit more and you'll see a definition.

Is that some WM-speak?
I suppose that
if it means the set 1/n for n in N
then the g-l-b is zero.

Thank you.
My motivation has been to find out if you accept that.
The rest is to make sure we're talking about
the same things.
Because
g.l.b of ⅟ℕ (⅟n for n in ℕ)  is 0
there is no positive lower bound of ⅟ℕ
A point between 0 and ⅟ℕ would be
a positive lower bound of ⅟ℕ
Such a point doesn't exist.
When I say infinitesimals don't exist,
I mean points between 0 and set ⅟ℕ
in the complete.ordered.field
don't exist.
When I say that, and then you name.check
various other systems which have infinitesimals,
it _sounds to me_ as though
you object to my claim.
All of this has been my attempt to sort out
_what you're saying_

Then otherwise what you have there appear facts
about N and R.

They're facts which identify ℕ and ℝ from among
a host of other possible things.called ℕ or ℝ
I take your lack of an explicit rejection
to be an implicit acceptance, and
I take you and I to be talking about
the same ℕ and the same ℝ

Then,
where there exists a well-ordering of R,
then to take the well-ordering it results that
first there's a well-ordering of [0,1]
for both simplicity and necessity,

Yes.
Note that a well.order of ℝ
is not the usual order of ℝ
However,
well.orders being well.orders,
a well.order for ℝ is
a well.order for each subset of ℝ,
including [0,1]
and is also not the usual order of [0,1]

and it's as the range of the function n/d
with 0 <= n < d and as d -> oo,
i.e., only in the infinite limit,
that the properties of the range of naturals,
apply to the properties of the range of [0,1].

No.
lim.infᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩  ⊆
limᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩  ⊆
lim.supᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩
⋃ᵈᐧᐧᐧ⋂ᵈᑉᵐᐧᐧᐧ⟨0/m,1/m,...,m/m⟩  ⊆
limᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩  ⊆
⋂ᵈᐧᐧᐧ⋃ᵈᑉᵐᐧᐧᐧ⟨0/m,1/m,...,m/m⟩
⋃ᵈᐧᐧᐧ{0,1}  ⊆
limᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩  ⊆
⋂ᵈᐧᐧᐧ[0,1]∩ℚ
{0,1}  ⊆
limᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩  ⊆
[0,1]∩ℚ  ⊉  [0,1]
limᵈᐧᐧᐧ⟨0/d,1/d,...,d/d⟩  ≠  [0,1]
----
I have been less bewildered by your (RF's) responses
since I have started to think of your posts as
brainstorming exercises with the prompt being
the previous post,
especially its last dozen lines.
https://en.wikipedia.org/wiki/Brainstorming
I can't guess if you'd consider that good or bad.
I thought you deserved to know how
what you're sending out
is being received.
Received by me, anyway.