An inconsistency between set theory and classical mathematics

[WM]

The number of unit fractions between x and 0 is, according to ZF, given by the step function NUF(x) (Number of Unit Fractions between 0 and x) which contains an infinite step at x = 0:

NUF(x) = ℵo for x > 0
NUF(x) = 0 for x ≤ 0 .

Classical mathematics however predicts a decrease of unit fractions in every point 1/n by only 1, because never two unit fractions inhabit the same point on the real line

∀∈ ℕ: 1/n - 1/(n+)1= 1/(n*(n+1)) > 0 .

Are ℵo and 1 different quantities?
If so, which result is correct?

Regards, WM

[JVR]

(1) NUF(x) = # of integers > 1/x is not a step function
(2) Card{n | n > k} = Card{n | n > n+1}
(3) Unit fractions don't inhabit points
(4) People who think they can see numbers need to see a doctor
(5) There is no such thing as 'classical mathematics'
(6) Mathematics doesn't 'predict' anything
(7) Sum(1/n^2) converges
(8) All of your 'results' are nonsense

[WM]

JVR schrieb am Montag, 13. März 2023 um 10:51:49 UTC+1:
> On Monday, March 13, 2023 at 9:30:40 AM UTC+1, WM wrote:
> > The number of unit fractions between x and 0 is, according to ZF, given by the step function NUF(x) (Number of Unit Fractions between 0 and x) which contains an infinite step at x = 0:
> >
> > NUF(x) = ℵo for x > 0
> > NUF(x) = 0 for x ≤ 0 .
> >
> > Classical mathematics however predicts a decrease of unit fractions in every point 1/n by only 1, because never two unit fractions inhabit the same point on the real line
> >
> > ∀∈ ℕ: 1/n - 1/(n+)1= 1/(n*(n+1)) > 0 .
> >
> > Are ℵo and 1 different quantities?
> > If so, which result is correct?

> (3) Unit fractions don't inhabit points

Wrong. Algebra/arithmetic and geometry are different languages for the same subject.

> (5) There is no such thing as 'classical mathematics'

There is a difference between what you know and what is reality. I'll try to help you:
The success of set theory has lead many to believe that it provides an unshakeable foundation for mathematics. It does not [...] Always remember that practically all classical mathematics was invented before modern logic and set theory. [Andrej Bauer in "Set theory and model theory", MathOverflow (30 Apr 2010)]

> (6) Mathematics doesn't 'predict' anything

There is a difference between what you know and what is reality. Mathematics predicts that never more than one unit faction inhabilts one point on the real line.

> (7) Sum(1/n^2) converges

Irrelevant for the above question. No term of the series is zero.

> (8) All of your 'results' are nonsense

As we have seen above there is a difference between what you know and what is reality.

Regards, WM

[Jim Burns]

Message-ID: <ed7ce49b-9a36-4a2a...@googlegroups.com>
On 3/12/2023 6:37 AM, WM wrote:
> Jim Burns schrieb am Samstag,
> 11. März 2023 um 19:29:31 UTC+1:

>> For each pair 1/n 1/m
>> there is an x between them.
>>
>> 1/n < (1/n + 1/m)/2 < 1/m
>
> But there are unit fractions which
> cannot be identified as individuals
> because
> they lie between zero and x
> for every x and every 1/n and
> every 1/m that you can choose.

No.

A description can exist of
something which does not exist.

What you describe,
a positive lower bound of FUFs,
does not exist in this line.

In _this line_
no all-continuous function jumps.

Thus,
a point g exists between
bounds and not-bounds at which
a function can be not-continuous.

In _this line_
arithmetic works.

Thus,
if g > 0 is a lower bound of FUFs
then g+g > 0+g is a greater lower bound
and
g is not between bounds and not-bounds.

Thus,
g = 0 is the greatest lower bound of FUFs
and
no greater, positive lower bound of FUFs
exists in this line.

It is NOT that positive lower bounds
are there but dark.
Positive lower bounds are not there.

We know that
positive lower bounds are not there
because
we know that
no all-continuous function jumps
and
arithmetic works.

[Jim Burns]

On 3/13/2023 4:30 AM, WM wrote:

> Classical mathematics however predicts
> a decrease of unit fractions in every point 1/n
> by only 1,
> because never two unit fractions inhabit
> the same point on the real line

Classical compass-and-straightedge geometry
shows that,
for each point between 0 and x
there are NO FEWER points between 0 and x
than there are between 0 and 1

https://en.wikipedia.org/wiki/Infinity

There are points 0 and 1 and
the line segment (0,1) between them.

There is the midpoint M of (0,1)
There is the circle centered at M of radius 1/2
which runs through points 0 and 1

For each point C on the circle other than 0 1
the angle 0C1 is a right angle.
The triangle 0C1 is a right triangle.

X is a point in (0,1)
The perpendicular to (0,1) at X intersects
the circle at points C and C'

Consider the main line through 0 and 1
and the branch line through 0 and C

Each perpendicular from the main line to
the branch line and from the branch line
to the main line forms, with those two lines,
a triangle similar to every other such triangle.

Right triangle 0C1 is similar to
right triangle 0XC
|01|/|0C| = |0C|/|0X|

Thus,
|0C|² = |01|*|0X|
|0C| = x⁰ᐧ⁵

Point C is better-named X⁰ᐧ⁵

There is an endless sequence of points,
alternating on the main line and branch line,
⟨ 1 X⁰ᐧ⁵ X X¹ᐧ⁵ X² X²ᐧ⁵ X³ X³ᐧ⁵ X⁴ ... ⟩

Each point is not the end of the sequence.

Each point is distinct from point 0

The set of points on the main line
between 0 and each of ⟨ 1 X X² X³ X⁴ ... ⟩
is the empty set,

because,
for any point G hypothetically
between 0 and each of ⟨ 1 X X² X³ X⁴ ... ⟩
G+G would also be
between 0 and each of ⟨ 1 X X² X³ X⁴ ... ⟩

No point can be the greatest lower bound of
⟨ 1 X X² X³ X⁴ ... ⟩

Therefore,
either the positive lower bounds are the empty set
or some all-continuous functions jump.

[WM]

Jim Burns schrieb am Montag, 13. März 2023 um 15:37:17 UTC+1:
> Message-ID: <ed7ce49b-9a36-4a2a...@googlegroups.com>
> On 3/12/2023 6:37 AM, WM wrote:
> > Jim Burns schrieb am Samstag,
> > 11. März 2023 um 19:29:31 UTC+1:
>
> >> For each pair 1/n 1/m
> >> there is an x between them.
> >>
> >> 1/n < (1/n + 1/m)/2 < 1/m
> >
> > But there are unit fractions which
> > cannot be identified as individuals
> > because
> > they lie between zero and x
> > for every x and every 1/n and
> > every 1/m that you can choose.
>
> No.
>
> A description can exist of
> something which does not exist.

Here you need a description of something that exists as individual, but you can't get it. How would you describe and remove an individual from a collective if you cannot reduce it?

>
> What you describe,
> a positive lower bound of FUFs,
> does not exist in this line.
>
> In _this line_
> no all-continuous function jumps.

Here is the limit of my concessions. If any idea violates
∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
then this idea has to be banned.
That means NUF(x) cannot increase from NUF(0) = 0 to NUF(x) > 0 without passing the value 1.

Regards, WM

[Jim Burns]

On 3/13/2023 5:58 PM, WM wrote:
> Jim Burns schrieb am Montag,
> 13. März 2023 um 15:37:17 UTC+1:
>> Message-ID: <ed7ce49b-9a36-4a2a...@googlegroups.com>
>> On 3/12/2023 6:37 AM, WM wrote:
>>> Jim Burns schrieb am Samstag,
>>> 11. März 2023 um 19:29:31 UTC+1:

>>>> For each pair 1/n 1/m
>>>> there is an x between them.
>>>> 1/n < (1/n + 1/m)/2 < 1/m
>>>
>>> But there are unit fractions which
>>> cannot be identified as individuals
>>> because
>>> they lie between zero and x
>>> for every x and every 1/n and
>>> every 1/m that you can choose.
>>
>> No.
>> A description can exist of
>> something which does not exist.
>
> Here you need a description of
> something that exists as individual,
> but you can't get it.

No, here I need a claim that is true of
each one of the ones we're talking about.

For example, this claim:
|
| It ends a sequence from 1/1
| such that,
| for each split,
| some 1/i is last-before
| and 1/i++ is first-after.
|
| ∀j: ∃i:i⁺⁺=j ⟹
| j≠0 ∧ ¬∃h≠i:h⁺⁺=j ∧ ∃k:j⁺⁺=k

I know it's true about them
because
things for which it's not true are
things about which we're not talking.

> How would you describe and
> remove an individual from a collective
> if you cannot reduce it?

I wouldn't.

I would better describe each one and
augment the description with
visibly not-first-possibly-false claims.

>> What you describe,
>> a positive lower bound of FUFs,
>> does not exist in this line.
>>
>> In _this line_
>> no all-continuous function jumps.
>
> Here is the limit of my concessions.

Aah.
You live in a universe in which
some all-continuous functions jump.
Send us a postcard!

> If any idea violates
> ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
> then this idea has to be banned.

We know that
the ones we are talking about do not violate it,
because
we can augment a description of
one of the ones we are talking about
with visibly not-first-possibly-false claims
which include
the claim that it doesn't violate your law.

> That means
> NUF(x) cannot increase from NUF(0) = 0 to NUF(x)
> 0 without passing the value 1.

No,
that doesn't mean that.

We know that,
for each unit fraction,
those after it matches all unit fractions,

because
we know that
there is _no first_ unit fraction such that
those after it _do not_ match all unit fractions.


We know that
infinity is not
a reallyreallyreallyreallyreallyreally large
natural number.

[Ross Finlayson]

Yeah, but isn't "infinity", according to numbers, very large?

[Ross Finlayson]

I figure you agree that when
it's written "point at infinity" it means infinity,
it's written "fixed point" it means infinity, and,
when it's written "infinity" it means "no less than infinity".

For the usual number-sense, of a usual word-sense,
that touch, smell, taste, hearing, and sight are inputs
to matters of making sense, or knowledge, in recognition,
the number-sense of infinity in the usual word-sense of infinity,
has it, in numbers, outside them, large.

Some people hear words, others spell them.

[WM]

Jim Burns schrieb am Dienstag, 14. März 2023 um 00:44:43 UTC+1:
> On 3/13/2023 5:58 PM, WM wrote:

> > How would you describe and
> > remove an individual from a collective
> > if you cannot reduce it?
> I wouldn't.

You can't. The collection consists of dark numbers.

>
> I would better describe each one

Impossible. What you can describe you can subtract.l

> > Here is the limit of my concessions.
> Aah.
> You live in a universe in which
> some all-continuous functions jump.

Between two adjacent unit fractions, there is a finite distance, i.e., many points. With increasing x the function NUF(x) grows. And it cannot grow to NUF(x) > 1 without passing the value NUF(x) = 1, simply because there are points between two adjacent unit fractions.

> > If any idea violates
> > ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
> > then this idea has to be banned.
> We know that
> the ones we are talking about do not violate it,

You can only talk about visible numbers. But all the points with NUF(x) < ℵo are dark.

> > That means
> > NUF(x) cannot increase from NUF(0) = 0 to NUF(x)
> > 0 without passing the value 1.
> No,
> that doesn't mean that.

Between two adjacent unit fractions, there is a finite distance, i.e., many points.

> there is _no first_ unit fraction such that
> those after it _do not_ match all unit fractions.

∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0 proves the existence of a first unit fraction because there cannot be more than one unit fraction be the first.

Regards, WM

[Jim Burns]

On 3/14/2023 11:38 AM, Ross Finlayson wrote:
> On Monday, March 13, 2023
> at 10:09:55 PM UTC-7, Ross Finlayson wrote:
>> On Monday, March 13, 2023
>> at 4:44:43 PM UTC-7, Jim Burns wrote:

>>> We know that
>>> infinity is not
>>> a reallyreallyreallyreallyreallyreally large
>>> natural number.
>>
>> Yeah, but isn't "infinity", according to numbers,
>> very large?

Infinity is what's left over, once we've described
finity.

There are a zillion different things to be described
other than finity. Finity gets good press, though,
because so much of _what we care about_ is finite.
For example, us.

Here's what I think you (RF) mean:
|
| There are the large finites, and then ω
| There are the really large finites, and then ω
| There are the reallyreally large finites,
| and then ω
| There are the reallyreallyreally large finites,
| and then ω
| There are the reallyreallyreallyreally large
| finites, and then ω
| There are the reallyreallyreallyreallyreally large
| finites, and then ω
| There are the reallyreallyreallyreallyreallyreally
| large finites, and then ω
|
| Is ω different from
| a reallyreallyreallyreallyreallyreallyreally
| large finite?

Infinity is what's left over, once we've described
finity.

Describe finity: the finite ordinals.
| It is a finite ordinal iff,
| for each split between it and 0
| some i is last-before the split
| and some j is first-after the split.

Notation:
If j is first-after i, write j = i+1

We've formalized the notion of being able
to get there from here. For each split of
ordinals "on the way" to a finite ordinal,
there is a step across that split:
ordinals before and after next to each other.
Finity described.

The description is true of each finite ordinal.
That's not something finite-us can "check",
even in principle. Nonetheless, we know it
because we know what we mean by "finite ordinal".

Compare to:
We don't (can't, needn't) check that each triangle
has three corners.

>> Yeah, but isn't "infinity", according to numbers,
>> very large?

We can extend our discussion beyond finite ordinals
to _not-necessarily-finite_ ordinals by loosening
our restriction on (AKA our definition of) what we are
talking about.

| It is a not-necessarily-finite ordinal iff,
| for each split between it and 0
| ⬚⬚⬚ ⬚ ⬚ ⬚⬚⬚⬚⬚⬚ ⬚⬚⬚ ⬚⬚⬚
| ⬚⬚⬚ some j is first-after the split.

Clearly, finite ordinals also satisfy the definition
of not-necessarily-finite ordinals.

In addition, for each _finite_ ordinal,
for each split between it and 0
some i is last-before the split.

All the last-befores ==
we can get there from here.

Infinity is what's left over, once we've described
finity.

If there is some λ left over,
after describing the finites,
then there is a split between the finites,
(we can get there from here)
and the others
(we can't, not even in principle)

There is a first-after of that split.
We name the first-after ω

The difference between ω and the large,
really large, reallyreally large,
reallyreallyreally large,
reallyreallyreallyreally large,
reallyreallyreallyreallyreally large,
reallyreallyreallyreallyreallyreally large, and
reallyreallyreallyreallyreallyreallyreally large
numbers is that you _can_ get there from here (0)
for all of those, but not for ω

This is the explanation for the various
results that contradict what we would get for

a reallyreallyreallyreallyreallyreally large
natural number.

Infinity is not

a reallyreallyreallyreallyreallyreally large
natural number.

It is something else, with different properties.

[Jim Burns]

On 3/14/2023 12:53 PM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 14. März 2023 um 00:44:43 UTC+1:
>> On 3/13/2023 5:58 PM, WM wrote:

>>> How would you describe and
>>> remove an individual from a collective
>>> if you cannot reduce it?
>>
>> I wouldn't.
>
> You can't.

I don't need to.

Describe each one.
Augment the description with
visibly not-first-possibly-false claims.
This results in more claims
not-possibly-false-about each one.

> The collection consists of dark numbers.

Describe each one.
Augment the description with
visibly not-first-possibly-false claims.
This results in more claims
not-possibly-false-about each one.

Anyway, you (WM) are an unreliable witness
on what a dark number is.

Here, we have a collection which is all ordered,
you're saying is mostly dark numbers, despite
your (sometimes) saying dark numbers having no order.

>> I would better describe each one
>
> Impossible.
> What you can describe you can subtract.l

A green frog is green and a frog.
A red car is red and a car.
A flying rainbow sparkle pony flies,
is rainbow-colored, sparkles and is a pony.

It is no great trick to describe things
which I cannot subtract.

What I describe are things for which
the description is true.
You (WM) tell your student that
your dark numbers are counter-examples to that.
That is an indictment of your superiors
at Hochschule Augsburg.

>>> Here is the limit of my concessions.
>>
>> Aah.
>> You live in a universe in which
>> some all-continuous functions jump.
>
> Between two adjacent unit fractions,
> there is a finite distance, i.e., many points.

There are as many points between any two points
as there are between any other two points.
Call it "classical geometry".

> With increasing x the function NUF(x) grows.
> And it cannot grow to NUF(x) > 1 without
> passing the value NUF(x) = 1,
> simply because there are points between
> two adjacent unit fractions.

No.
You (WM) have spawned some mutant version of
the Intermediate Value Theorem IVT

However,
because the IVT is _correct_ for that line,
and no all-continuous functions jump,

your (WM's) mutant-IVT is _incorrect_


There is a lower bound for unit fractions.

There is a not-a-lower-bound.

There is a point g between lower-bounds and
not-lower-bounds,
otherwise,
there are all-continuous functions which jump.
But IVT is correct: no all-continuous jumps.

The point-between g can't be positive.
If 0 < g < each FUF 1/n
then 0+g < g+g < 2(each FUF 1/n)
and
g > 0 is not between bounds and not-bounds.

Therefore,
points x exist for which NUF(x) = 0
points x exist for which NUF(x) = ℵ₀
no points exist x for which NUF(x) = 1

Because there are no continuous jumps
and IVT is correct,
your (WM's) mutant-IVT is incorrect.

[WM]

Jim Burns schrieb am Dienstag, 14. März 2023 um 22:17:16 UTC+1:
> On 3/14/2023 12:53 PM, WM wrote:

> > Between two adjacent unit fractions,
> > there is a finite distance, i.e., many points.
> There are as many points between any two points
> as there are between any other two points.

That is nonsense (between 1 and 2 there are only half as many points as between 1 and 3) but irrelevant. Therefore it need not be corrected.

> > With increasing x the function NUF(x) grows.
> > And it cannot grow to NUF(x) > 1 without
> > passing the value NUF(x) = 1,
> > simply because there are points between
> > two adjacent unit fractions.
> No.
> You (WM) have spawned some mutant version of
> the Intermediate Value Theorem IVT

No, it is the simple evaluation of
∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
Between any two unit fractions there are points.

>
> Therefore,
> points x exist for which NUF(x) = 0
> points x exist for which NUF(x) = ℵ₀

Therefore ℵ₀ unit fractions cannot be distinguished.

> no points exist x for which NUF(x) = 1

So you don't accept ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

>
> Because there are no continuous jumps
> and IVT is correct,
> your (WM's) mutant-IVT is incorrect.

∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0 is correct.

You cannot find a mistake.

Regards, WM

[Jim Burns]

On 3/14/2023 6:41 PM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 14. März 2023 um 22:17:16 UTC+1:
>> On 3/14/2023 12:53 PM, WM wrote:

>>> Between two adjacent unit fractions,
>>> there is a finite distance, i.e., many points.
>>
>> There are as many points between any two points
>> as there are between any other two points.
>
> That is nonsense
> (between 1 and 2 there are only half as many points
> as between 1 and 3) but irrelevant.
> Therefore it need not be corrected.

No.
It is both correct and relevant that
there are as many points between any two points

as there are between any other two points.

Correct:

y ∈ (0,1] ⟷ xy ∈ (0,x]

Relevant:

In order for a point y to exist
last-before or first-after a point x,
there would need to be two points which
have no points _seen or unseen_ between them.

However,
as many points between is not
the necessary zero between.

No last-before or first-after points exist.

_Lists_ have last-before and first-after.
A line is not correctly described as
a _list_ of points.

>>> With increasing x the function NUF(x) grows.
>>> And it cannot grow to NUF(x) > 1 without
>>> passing the value NUF(x) = 1,
>>> simply because there are points between
>>> two adjacent unit fractions.
>>
>> No.
>> You (WM) have spawned some mutant version of
>> the Intermediate Value Theorem IVT
>
> No, it is the simple evaluation of
> ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
> Between any two unit fractions there are points.

>>> simply because there are points between
>>> two adjacent unit fractions.

You (WM) are assuming that
two sets with a one-element difference
always have different cardinalities.

No,
they don't always have different cardinalities.

There is a one-to-one map of each FISON
to its first-after,
thus, of all FISONs
onto all not-first FISONs.
The same cardinality, a one-element difference.

⟪0⟫ ⟪0 1⟫ ⟪0 1 2⟫ ⟪0 1 2 3⟫ ...
↕ ↕ ↕ ↕
⟪0 1⟫ ⟪0 1 2⟫ ⟪0 1 2 3⟫ ⟪0 1 2 3 4⟫ ...

Even if you declare some FISONS dark(!),
that is still
not-possibly-false-about _FISONs_
both visible and dark(!).
Your declaration changes nothing.

>> Therefore,
>> points x exist for which NUF(x) = 0
>> points x exist for which NUF(x) = ℵ₀
>
> Therefore ℵ₀ unit fractions cannot be
> distinguished.

No.

If 0 < g < each visible unit fraction VUF
then 0+g < g+g < 2(each VUF)
and g is not the greatest lower bound of VUFs

For x > 0
x is not a lower bound of VUFs, 0 is greatest.
mₓ = ⌈1/x⌉
|⟨ 1/mₓ 1/(mₓ+1) 1/(mₓ+2) ... ⟩| = ℵ₀

| Assume otherwise.
| Assume |⟨ 1/mₓ 1/(mₓ+1) 1/(mₓ+2) ... ⟩| < ℵ₀
|
| ⟨ 1/mₓ 1/(mₓ+1) 1/(mₓ+2) ... ⟩ ends.
| The end of ⟨ 1/mₓ 1/(mₓ+1) 1/(mₓ+2) ... g ⟩
| is a VUF
| g is a VUF
| A VUF is positive.
| g is a positive GLB of VUFs
|
| However,
| no positive GLB of VUFs exists.
| Contradiction.

Therefore,
for x > 0
|⟨ 1/mₓ 1/(mₓ+1) 1/(mₓ+2) ... ⟩| = ℵ₀
NVUF(x) ≠ 1
NUF(x) ≠ 1

>>> With increasing x the function NUF(x) grows.
>>> And it cannot grow to NUF(x) > 1 without
>>> passing the value NUF(x) = 1,

No. It can.

>> no points exist x for which NUF(x) = 1
>
> So you don't accept
> ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

I don't accept your provably-wrong
mutant intermediate-value-theorem.

[Fritz Feldhase]

On Tuesday, March 14, 2023 at 11:41:53 PM UTC+1, WM wrote:

> So you don't accept ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

Indeed. We'd prefer

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
_^_

> ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0 is correct.
>
> You cannot find a mistake.

We can. Should read:

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
_^_

[Fritz Feldhase]

On Tuesday, March 14, 2023 at 11:41:53 PM UTC+1, WM wrote:

> Jim Burns schrieb am Dienstag, 14. März 2023 um 22:17:16 UTC+1:

Def. NUF(x) := card {u e U : u <= x} (x e IR) with U = {1/n e IR : n e IN}.

> > points x exist for which NUF(x) = 0

Actually, for all x e IR, x <= 0: NUF(x) = 0.

Especially, NUF(0) = 0.

> > points x exist for which NUF(x) = ℵ₀

Actually, for all x e IR, x > 0: NUF(x) = ℵ₀.

Especially, NUF(1) = ℵ₀.

> > no points exist x for which NUF(x) = 1

Actually, there is no x e IR such that NUF(x) = 1, since Ax e IR: x <= 0 v x > 0.

(You see: Ax e IR: x < 0 v x = 0 v x > 0.)

Indeed: {NUF(x) : x e IR} = {0, ℵ₀}.

> So you don't <bla>

Yeah whatever, Mückenheim.

[WM]

Jim Burns schrieb am Mittwoch, 15. März 2023 um 15:47:05 UTC+1:
> On 3/14/2023 6:41 PM, WM wrote:

> > So you don't accept

> > ∀n∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

> I don't accept your provably-wrong
> mutant intermediate-value-theorem.

The formula is now correct. The linearity of the real line and the finite distances between *all* unit fractions show that one is the first.

Regards, WM

[WM]

Fritz Feldhase schrieb am Mittwoch, 15. März 2023 um 22:15:49 UTC+1:
> On Tuesday, March 14, 2023 at 11:41:53 PM UTC+1, WM wrote:
>
> > So you don't accept ∀∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
> Indeed. We'd prefer
>
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
> _^_

Thanks, I overlooked that mistake. But now the formula is correct. The linearity of the real line and the finite distances between *all* unit fractions show that one is the first.

Regards, WM

[WM]

Fritz Feldhase schrieb am Mittwoch, 15. März 2023 um 22:35:09 UTC+1:
> On Tuesday, March 14, 2023 at 11:41:53 PM UTC+1, WM wrote:
> > Jim Burns schrieb am Dienstag, 14. März 2023 um 22:17:16 UTC+1:
> Def. NUF(x) := card {u e U : u <= x} (x e IR) with U = {1/n e IR : n e IN}.
> > > points x exist for which NUF(x) = 0
> Actually, for all x e IR, x <= 0: NUF(x) = 0.
>
> Especially, NUF(0) = 0.

Yes.

> > > points x exist for which NUF(x) = ℵ₀
> Actually, for all x e IR, x > 0: NUF(x) = ℵ₀.

No.

>
> Especially, NUF(1) = ℵ₀.

Yes.

> > > no points exist x for which NUF(x) = 1
> Actually, there is no x e IR such that NUF(x) = 1, since Ax e IR: x <= 0 v x > 0.
>

> Indeed: {NUF(x) : x e IR} = {0, ℵ₀}.

That contradicts mathematics according to which an increase is only possible one by one, separated by finite distances.

Regards, WM

[Fritz Feldhase]

On Wednesday, March 15, 2023 at 11:20:47 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 15. März 2023 um 22:35:09 UTC+1:
> > On Tuesday, March 14, 2023 at 11:41:53 PM UTC+1, WM wrote:
> > > Jim Burns schrieb am Dienstag, 14. März 2023 um 22:17:16 UTC+1:
> > Def. NUF(x) := card {u e U : u <= x} (x e IR) with U = {1/n e IR : n e IN}.
> > > > points x exist for which NUF(x) = 0
> > Actually, for all x e IR, x <= 0: NUF(x) = 0.
> >
> > Especially, NUF(0) = 0.
> Yes.
> > > > points x exist for which NUF(x) = ℵ₀
> > Actually, for all x e IR, x > 0: NUF(x) = ℵ₀.
> No.

Yes.

> <nonsense deleted>

[WM]

∀n∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0

The function number of unit fractions between zero and x is a step function which increases from 0 to 1. If it increased without intermediate value 1 from 0 to 2 or more or even ℵo, then there would not be a distance between 2 or more or even ℵo unit fractions where the function would be constant.

Regards, WM

[Fritz Feldhase]

On Thursday, March 16, 2023 at 11:12:31 AM UTC+1, WM wrote:

Def. NUF(x) := card {u e U : u <= x} (x e IR) with U = {1/n e IR : n e IN}.

Hence:

(1) For all x e IR, x <= 0: NUF(x) = 0 ,

and especially NUF(0) = 0.

(2) For all x e IR, x > 0: NUF(x) = ℵ₀ .

In other words: img(NUF) = {NUF(x) : x e IR} = {0, ℵ₀}.

> The function /number of unit fractions between zero and x/ is a step function which [jumps] from 0 to [ℵ₀].

Yes.

> <nonsense deleted>

[Jim Burns]

On 3/15/2023 6:15 PM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 15. März 2023 um 15:47:05 UTC+1:
>> On 3/14/2023 6:41 PM, WM wrote:

>>> So you don't accept
>>> ∀n∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.
>>
>> I don't accept your provably-wrong
>> mutant intermediate-value-theorem.
>
> The formula is now correct.

Thank you for correcting it.
Thanks to FF for his assistance.

I am not interested in holding you to account
for what you _unintentionally_ say.
There's too much to say about what you
_intentionally_ say, to spend time on that.

> The linearity of the real line and
> the finite distances between
> *all* unit fractions
> show that one is the first.

No, it doesn't show that.

Apparently, the argument that you are suggesting
without being willing to baldly state
seems to be:
|
| No last point and non-zero distances
| implies[?]
| infinite total distance.
|
| However,
| non-zero distances and finite total distance.
|
| Therefore,
| last point.

I will try to remember that as an example
the next time I revisit the augmenting with
visibly not-first-possibly-false claims.

Yes,
"last point" is
visibly not-first-possibly-false
in that sequence of claims.
_That step_ is correct.

However,
_that sequence_ is not _all_
not-first-possibly-false, so
we can't use here what we use elsewhere
|
| If
| any claim in the sequence is
| possibly-false
| then
| some claim is
| first possibly-false.
|
| If
| each claim is
| not-first-possibly-false
| then
| each claim in the sequence
| is not-possibly-false

The first claim
|
| No last point and non-zero distances
| implies[?]
| infinite total distance.
|
_seems_ correct, but it isn't.

For any real number x > 0
1/1 ∈ (1/1-x/4, 1/1+x/4)
1/2 ∈ (1/2-x/8, 1/2+x/8)
2/1 ∈ (2/1-x/16, 2/1+x/16)
1/3 ∈ (1/3-x/32, 1/3+x/32)
2/2 ∈ (2/2-x/64, 2/2+x/64)
3/1 ∈ (3/1-x/128, 3/1+x/128)
1/4 ∈ (1/4-x/256, 1/4+x/256)
...

If we summed all the lengths of all of
those intervals, it would be _greater than_
the sum of all fraction-to-fraction distances,
because intervals overlap and because
1/1 2/2 3/3 ... are the same point etc.

The sum of all those lengths is
(x/2+x/4+x/8+x/16+...)

(x/2+x/4+x/8+x/16+...) - x/2 =
(x/4+x/8+x/16+...)

(1/2)(x/2+x/4+x/8+x/16+...) =
(x/4+x/8+x/16+...)

(x/2+x/4+x/8+x/16+...) - x/2 =
(1/2)(x/2+x/4+x/8+x/16+...)

(x/2+x/4+x/8+x/16+...) = x

The sum of all distances between
elements of ℚ⁺ is _less than_ any x > 0

Infinity is not

[Ross Finlayson]

Here's some more reading from Maugin. https://www.youtube.com/watch?v=UkZGZ6FRpS0

[Jim Burns]

On 3/16/2023 9:14 PM, Ross Finlayson wrote:

> Here's some more reading from Maugin.
> https://www.youtube.com/watch?v=UkZGZ6FRpS0

"Reading from Maugin, pages nine-eleven. "

Nonlinear Electromechanical Effects and Applications
(Theoretical and Applied Mechanics)
by Gerard A Maugin (Author)
https://www.amazon.com/Nonlinear-Electromechanical-Applications-Theoretical-Mechanics/dp/9971978431

Ross: why make a video of you reading those pages?


There is an incantation made by many technical
writers:
| Tell'em what you'll tell'em.
| Tell'em.
| Tell'em what you told'em.

You're missing the first and the last parts.
What will you tell us?
What did you tell us?

Thanks in advance.

[WM]

Jim Burns schrieb am Donnerstag, 16. März 2023 um 17:45:00 UTC+1:

> | No last point and non-zero distances
> | implies[?]
> | infinite total distance.

The distance between 1/1 and 0 is precisely 1, namely the sum of all distances
SUM (1/n - 1/(n+1)) = SUM1/(n*(n+1)) = 1
1/1 -1/2 + 1/2 - 1/3 +-...

None of the distances is zero.
Therefore after every increase of the function NUF(x) from NUF(0) = 0 to any x > 0 there are some points x' > x where NUF(x') = NUF(x). That is a step function.

> Infinity is not
> a reallyreallyreallyreallyreallyreally large
> natural number.

And it does not allow for violating mathematics:

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

Regards, WM

[Jim Burns]

On 3/17/2023 5:19 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 16. März 2023 um 17:45:00 UTC+1:

>> | No last point and non-zero distances
>> | implies[?]
>> | infinite total distance.

However,
no last point and non-zero distances
*DOES NOT* imply
infinite total distance.

1/2 + 1/4 + 1/8 + 1/16 + ...
= 1

That means:
For each x in (1/2, 1),
there is a last finite sum not-in (x, 1)
and no last finite sum in (x, 1).

Almost all of these finite sums are in (x, 1)

1/2
= 1 - 1/2

1/2 + 1/4
= 1 - 1/4

1/2 + 1/4 + 1/8
= 1 - 1/8

...

> The distance between 1/1 and 0 is
> precisely 1,
> namely the sum of all distances
> SUM (1/n - 1/(n+1)) =
> SUM1/(n*(n+1)) =
> 1
> 1/1 -1/2 + 1/2 - 1/3 +-...

Yes.
That means:
For each x in (1/2, 1),
there is a last finite sum not-in (x, 1)
and no last finite sum in (x, 1).

Almost all of these finite sums are in (x, 1).

1/2
= 1 - 1/2

1/2 + 1/6
= 1 - 1/3

1/2 + 1/6 + 1/12
= 1 - 1/4

1/2 + 1/6 + 1/12 + 1/20
= 1 - 1/5

1/2 + 1/6 + 1/12 + 1/20 + 1/30
= 1 - 1/6

...

> None of the distances is zero.

Yes.
None of the summands is zero.
None of the finite sums is 1

Almost all of the summands are near zero.
Almost all of the finite sums are near 1

> Therefore
> after every increase of the function NUF(x)
> from NUF(0) = 0
> to any x > 0
> there are some points x' > x
> where NUF(x') = NUF(x).
> That is a step function.

That looks like a typo.
You (WM) allegedly taught physics.
You have Google.
How could you not-know
what a step function is?

| Informally speaking, a step function is
| a piecewise constant function having
| only finitely many pieces.
|
-- Wikipedia

Step functions are discontinuous at riser points.

Define f: ℝ ⟶ {0,1}
if x > 0
then f(x) = 1
else f(x) = 0

No point d, visible or dark, exists
between
points x at which f(x) = 1 and
points x at which f(x) = 0

f() is discontinuous at 0

[Jim Burns]

On 3/17/2023 5:19 AM, WM wrote:

> Jim Burns schrieb am Donnerstag,
> 16. März 2023 um 17:45:00 UTC+1:

>> | No last point and non-zero distances
>> | implies[?]
>> | infinite total distance.

However,
no last point and non-zero distances
*DOES NOT* imply

infinite total distance.

> The distance between 1/1 and 0 is
> precisely 1,
> namely the sum of all distances
> SUM (1/n - 1/(n+1)) =
> SUM1/(n*(n+1)) =
> 1
> 1/1 -1/2 + 1/2 - 1/3 +-...

> None of the distances is zero.

> Therefore
> after every increase of the function NUF(x)
> from NUF(0) = 0
> to any x > 0
> there are some points x' > x
> where NUF(x') = NUF(x).
> That is a step function.

Define "exists least FUF" function
LUF: (-1,1) ⟶ {0,1}
such that,
if exists FUF 1/i: 1/i⁺⁺ < x ≤ 1/i
then LUF(x) = 1
else LUF(x) = 0

A finite unit fraction (FUF) 1/n ends
a sequence from 1/1
in which,

for each split,
some 1/i is last-before

and 1/i⁺⁺ is first-after.

If exists FUF 1/n < x < 1/1
then
exists FUF 1/i: 1/i⁺⁺ < x ≤ 1/i
and
LUF(x) = 1

If x ≤ 0
then
x < each FUF 1/i⁺⁺
and
not exists FUF 1/i: 1/i⁺⁺ < x ≤ 1/i
and
LUF(x) = 0

No other points in the line exist.

| Assume x exists such that
| not( exists FUF 1/n < x < 1/1 )
| and not( x ≤ 0 )
|
| LUF(x) = 0
|
| 0 < x < each FUF
| thus
| 0+x < x+x < 2(each FUF)
| and
| LUF(2x) = 0
| and
| y ≤ 2x ⇒ LUF(y) = 0
| and
| LUF() is constant near x
|
| LUF() is constant near x
| if exists FUF 1/n < x < 1/1
| if x ≤ 0
| if not (exists FUF 1/n < x < 1/1) and
| not(x ≤ 0)
|
| LUF() is constant everywhere in (-1,1)
|
| However,
| LUF(0) = 0 and LUF(1/2) = 1
|
| _In this line_
| there are no constant-everywhere functions
| with different values.
| Contradiction.

Therefore,
there are no other points.

LUF() is a step function discontinuous at 0

LUF() is defined
such that,
if exists FUF 1/i: 1/i⁺⁺ < x ≤ 1/i
then LUF(x) = 1
else LUF(x) = 0

If x > 0
then LUF(x) = 1
else LUF(x) = 0

>> Infinity is not
>> a reallyreallyreallyreallyreallyreally large
>> natural number.
>
> And it does not allow for violating mathematics:
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

Are step functions a violation of mathematics?

[WM]

Jim Burns schrieb am Freitag, 17. März 2023 um 19:35:52 UTC+1:

> Almost all of the summands are near zero.

But none is zero.

Therefore the function NUF(x) has a step at every single unit fraction with a horizontal graph. The increasing function pauses for a short while. If the universal quantifier is correct, then this holds for all unit fractions.This imples the existence of a first step, i.e., a first unit fraction.

> | Informally speaking, a step function is
> | a piecewise constant function having
> | only finitely many pieces.

Not necessary. But the first steps are finitely many, of course.

Regards, WM

[Ross Finlayson]

It's really a great book and I must recommend it,
compared to something like
"Differential Geometry: Curves - Surfaces - Manifolds",
Kuehnel, AMS, 2002, AMS MSC 52-01
that I just picked ip, it relies somewhat less on diagrams
than the development of the concept of the properties involved.

Also I'm reading Lam's "Topics in Contemporary Mathematical Physics"
but I've already studied Geometric Algebra, about Hestenes and Pezzaglia
and Baylis and Lounesto and so on, then there's something like "Reference
Manual of Telecommunications Engineering" from Freeman that's rather
doorstop-ish. I already have a degree.

Some people need a book read to them vocally before they can read it.
It's a traditional role of seminars called "reading".

In Maugin what I'm researching is that for the non-linear forms
of various boundary problems he has some latest things, then
about Fritz London and his non-linear forms he has some latest things,
that I'm looking for some footholds of some various results in quasi-invariance
to work out some formal methods for some of these non-linear forms
for the "theory of potentials" and "superclassical", for continuum mechanics.

[Jim Burns]

On 3/17/2023 5:29 PM, WM wrote:
> Jim Burns schrieb am Freitag,
> 17. März 2023 um 19:35:52 UTC+1:

>> Yes.
>> None of the summands is zero.
>> None of the finite sums is 1
>>

>> Almost all of the summands are near zero.

>> Almost all of the finite sums are near 1
>

> But none is zero.

Yes.
None of the summands are zero.
None of the finite sums are 1

Almost all of the summands are near zero.

Almost all of the finite sums are near 1

> Therefore
> the function NUF(x) has a step at
> every single unit fraction
> with a horizontal graph.
> The increasing function pauses for a short while.
> If the universal quantifier is correct,
> then this holds for all unit fractions.

Okay.

> This imples the existence of a first step,
> i.e., a first unit fraction.

No,
it does not imply that.

There is no step from infinite to finite.

Infinity is not
a reallyreallyreallyreallyreallyreally large

(finite) natural number.


Consider a (Dedekind) infinite set S
For example, the set UF(0,x] of
unit fractions in (0,x]

S is infinite.
There is a function f: S ⟶ S
which is one-to-one
| x ≠ y ⟹ f(x) ≠ f(y)
and not onto S
| exists b ∈ S
| not exists x ∈ S, f(x) = b

Starting from f
we can define function g: S ⟶ S\{b}
which is one-to-one
| x ≠ y ⟹ g(x) ≠ g(y)
and _onto_ S\{b}
| not exists c ∈ S\{b}
| not exists x ∈ S, g(x) = c

S is infinite.
S\{b} is one step from S
S\{b} has _just as many_ as S

For example,

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0.

UF(0,1/(n+1)] = UF(0,1/n]\{1/n}
But
UF(0,1/(n+1)] has _just as many_ as UF(0,1/n]

----
S is infinite.
There is a function f: S ⟶ S
which is one-to-one
| x ≠ y ⟹ f(x) ≠ f(y)
and not onto S
| exists b ∈ S
| not exists x ∈ S, f(x) = b

From f
we can define function g: S ⟶ S\{b}
which is one-to-one
| x ≠ y ⟹ g(x) ≠ g(y)
and _onto_ S\{b}
| not exists c ∈ S\{b}
| not exists x ∈ S, g(x) = c

Finᶠ(b) is the set of finite sequences of
iterations of f starting from b
⟨ b ⟩
⟨ b f(b) ⟩
⟨ b f(b) f∘f(b) ⟩
⟨ b f(b) f∘f(b) f∘f∘f(b) ⟩
...
⟨ b f(b) f∘f(b) f∘f∘f(b) ... ○ⁿf(b) ⟩
...

where
○⁰f(b) = b
○ⁱ⁺⁺f(b) = f(○ⁱf(b))

In the sequence
⟨ b f(b) f∘f(b) f∘f∘f(b) ... ○ⁿf(b) ⟩
b is first
○ⁿf(b) is last
and, for each split
some x is last-before
and f(x) is first-after.

⋃Finᶠ(b) is the union of Finᶠ(b)
⋃Finᶠ(b) is the set of
finite iterations of f on b

∀x ∈ ⋃Finᶠ(b): f(x) ∈ ⋃Finᶠ(b)
b ∈ ⋃Finᶠ(b)
¬∃x ∈ ⋃Finᶠ(b): f(x) = b

Define
g: S ⟶ S\{b}
such that
if x ∈ ⋃Finᶠ(b)
then g(x) = f(x)
else g(x) = x

g is one-to-one and onto S\{b}

[Jim Burns]

On 3/18/2023 12:50 AM, Ross Finlayson wrote:
> On Thursday, March 16, 2023
> at 9:43:03 PM UTC-7, Jim Burns wrote:
>> On 3/16/2023 9:14 PM, Ross Finlayson wrote:

>>> Here's some more reading from Maugin.
>>> https://www.youtube.com/watch?v=UkZGZ6FRpS0
>> "Reading from Maugin, pages nine-eleven. "
>>
>> Nonlinear Electromechanical Effects and Applications
>> (Theoretical and Applied Mechanics)
>> by Gerard A Maugin (Author)
>> https://www.amazon.com/Nonlinear-Electromechanical-Applications-Theoretical-Mechanics/dp/9971978431
>>
>> Ross: why make a video of you reading those pages?

> Some people need a book read to them vocally
> before they can read it.
> It's a traditional role of seminars
> called "reading".

My own experience is that, with
highly-compressed texts, as in physics, math, etc,
I do better, at least initially, introducing
myself to the material by reading at my own pace.
Glancing up to remind myself what ⋃Finᶠ(b) means,
and so on.
YMMV

[Ross Finlayson]

It's surely so that "visual readers" and "visual learners"
are different things, and that readers may read a page at a glance,
that reading out loud may be a gentle metered pace.

My profession is in information technology,
often it's "vim a ton of sources and map tab to buffer-next,
and shift-tab to buffer-previous, and page through them".

This is vis-a-vis the design of algorithms and well-defined behavior.

One time I was coding and asked Karl, "Karl, where are the docs",
and he looked at me and said "the source code is the documentation".
By the same token many processes are highly proscribed.

Here, knowing what "set theory" is and "classical mathematics", is,
involves a lot of facility in word-sense and number-sense.

It's of course similar that formal languages for the deterministic,
for definitions and the abstraction of reasoning as a process,
and abstraction of process as a reasoning, there's anaphora and cataphora
and analysis and synthesis and this kind of thing, that we
have a theory and philosophy of science, and it conditions our beliefs,
include that it's the least baggage necessary to well-define deterministic
beliefs scientifically. (Sometimes you can throw ram at it, sometimes
you can throw nodes at it, sometimes you can throw money at it,
sometimes you can throw people at it, over time, here though one can't.
Sometimes the design is ram-bound or process-bound or people-bound
or higher-level-head-bound or money-bound where it's time-bound.)

Please consider yourself lucky in recognition that reading
performance and comprehension is basically exponential
that for many it's flat.

Everyone watches the same movie / everyone reads their own book.

[WM]

Jim Burns schrieb am Samstag, 18. März 2023 um 17:52:58 UTC+1:
> On 3/17/2023 5:29 PM, WM wrote:

> > the function NUF(x) has a step at
> > every single unit fraction
> > with a horizontal graph.
> > The increasing function pauses for a short while.
> > If the universal quantifier is correct,
> > then this holds for all unit fractions.
> Okay.
> > This imples the existence of a first step,
> > i.e., a first unit fraction.
> No,
> it does not imply that.
>
> There is no step from infinite to finite.

But there is a step from nothing to something. And the something is a single unit fraction because after every unit fraction the increasing function pauses for a short while.

> Infinity is not
> a reallyreallyreallyreallyreallyreally large
> (finite) natural number.

And infinity is not self-contradictory:

> > The increasing function pauses for a short while.
> > If the universal quantifier is correct,
> > then this holds for all unit fractions.
> Okay.

See?

>
> Consider a (Dedekind) infinite set S

Nonsense cannot prove other nonsense to be correct.

Regards, WM

[Fritz Feldhase]

On Saturday, March 18, 2023 at 5:52:58 PM UTC+1, Jim Burns wrote:
> On 3/17/2023 5:29 PM, WM wrote:
> >
> > the function NUF(x) has a step at
> > every single unit fraction
> > with a horizontal graph.
> > The increasing function pauses for a short while.

> > [...]
> >
> Okay.

Nope. Not ok, but nonsense.

There's just ONE step, from 0 to aleph_0, at 0.

Hint: NUF(x) = 0 for all x e IR, x <= 0
and: NUF(x) = aleph_0 for all x e IR, x > 0.

> > If the universal quantifier is correct,
> > then this holds for all unit fractions.

What does it even mean that "the universal quantifier is correct"?!

How about: "If it holds for all unit fractions, then this holds for all unit fractions."?

Seems to me that that's what he tried to express here.

[WM]

Fritz Feldhase schrieb am Samstag, 18. März 2023 um 20:37:44 UTC+1:
> On Saturday, March 18, 2023 at 5:52:58 PM UTC+1, Jim Burns wrote:
> > On 3/17/2023 5:29 PM, WM wrote:
> > >
> > > the function NUF(x) has a step at
> > > every single unit fraction
> > > with a horizontal graph.
> > > The increasing function pauses for a short while.
> > > [...]
> > >
> > Okay.
>
> Nope. Not ok, but nonsense.

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
>

> There's just ONE step, from 0 to aleph_0, at 0.

If that were true, then these unit fractions would sit at a single point with no chance to distinguish them. So you advocate dark unit fractions?

> > > If the universal quantifier is correct,
> > > then this holds for all unit fractions.
> What does it even mean that "the universal quantifier is correct"?!

It means that never two or more unit fractions sit at a single point together.

>
> How about: "If it holds for all unit fractions, then this holds for all unit fractions."?

Yes, never two unit fractions enter the function NUF(x) together at the same x.

Regards WM

[Ross Finlayson]

In infinite quantities the cardinal has other business than that, which are counts,
you would be only talking about an ordinal or a constant named "infinity".

Then this is a usual notion of "the limit and the sum" or the mental process
so arrived at what results beyond finite differences is completion and fixed-point.

Or, Zeno's paradoxes long since arrived at, Zeno's resolutions are also
long since arrived at.

Anyways, you are mixing apples and oranges and would as well
frame it as the cart instead.

[Mostowski Collapse]

He is demonstrating how to read with a potato in mouth.

[Mostowski Collapse]

If its not the potato, he is stuttering all the time
with "eh", "em" etc.. and then sometimes he
goes into unmotivated higher pitch or lower pitch.

Its the most horrible reading I have ever heard.

[Mostowski Collapse]

You can try here, at t=652, and count the "eh",
he has really problems reading the text.

https://youtu.be/UkZGZ6FRpS0?t=652

If its not nigher pitch or lower pitch, he
prolongs words, or suddently speeds up,

like instead "is two", he says "is twooooooo".

LMAO!

[Ross Finlayson]

I was describing the usual orthornomal vector basis
and that R^3's is usually called e sub i or e sub 1,
so, generously, read out "e, e, e, e sub 1". Here the configuration
space being discussed is flux through a surface, with R^2 and
a time-like dimension.

It's a good stutter though if such things as the perceived
failings of other people are what activate your brain's pleasure
centers, where that's Shadenfreude and is a usual trait of
psychopaths and the sadistic. Surely it aggrieved me.

Please though, if I ever make an intonation that "it ends
like a question?" that something is terribly wrong, because
that is a grating-nails-the-chalkboard-the-teacher-doesn't-know,
and such "presenters" are "defective".

Also there's one good "uhhh....". Sega Dreamcast: "It's thinking."


Now that you've interjected after the wonderful preceding post, and
please feel certain that I read my own posts with certainty
and much as they are written as they are written to be read,
discuss Brouwer's concept of the continuum and see if it
makes sense or that otherwise perhaps it doesn't fit in your head.

[Jeffrey Rubard]

"Wow, that is some sense of self-satisfaction. Otherwise, it doesn't work as text."

[Jim Burns]

On 3/18/2023 3:37 PM, Fritz Feldhase wrote:
> On Saturday, March 18, 2023
> at 5:52:58 PM UTC+1, Jim Burns wrote:
>> On 3/17/2023 5:29 PM, WM wrote:

>>> the function NUF(x) has a step at
>>> every single unit fraction
>>> with a horizontal graph.
>>> The increasing function pauses for a short while.
>>> [...]
>>
>> Okay.
>
> Nope. Not ok, but nonsense.

Damn.
Point to you.

I think what I _meant_ was
| Instead of arguing about all that
| [which I think I might not have read],
| let's argue about this next bit.

Sorry. My bad. Etc.

> There's just ONE step, from 0 to aleph_0, at 0.
>
> Hint: NUF(x) = 0 for all x e IR, x <= 0
> and: NUF(x) = aleph_0 for all x e IR, x > 0.
>
>>> If the universal quantifier is correct,
>>> then this holds for all unit fractions.
>
> What does it even mean that
> "the universal quantifier is correct"?!
>
> How about:
> "If it holds for all unit fractions,
> then this holds for all unit fractions."?
>
> Seems to me that that's
> what he tried to express here.

Well, it's hard to argue with that.

On the other hand,
I have been pressing claims like
| Each natural number is a natural number.

That's kind of where we're at, here.

I hope you can bear the excitement.

[Ross Finlayson]

On Saturday, March 18, 2023 at 2:03:22 PM UTC-7, Ross Finlayson wrote:

Well, here's reading another 2/3 page or so from Maugin's book, though
mostly it's my interpretations and explanations, then skimming a few sections
and a few dozen pages and the appendix, and a quote from opening Kuehnel,
then pointing to Weyl.

https://www.youtube.com/watch?v=GIMKbXBEQno

[Jim Burns]

On 3/18/2023 3:20 PM, WM wrote:
> Jim Burns schrieb am Samstag,
> 18. März 2023 um 17:52:58 UTC+1:
>> On 3/17/2023 5:29 PM, WM wrote:

>>> the function NUF(x) has a step at
>>> every single unit fraction
>>> with a horizontal graph.
>>> The increasing function pauses for a short while.
>>> If the universal quantifier is correct,
>>> then this holds for all unit fractions.
>>
>> Okay.

Not okay. My bad.

>>> This imples the existence of a first step,
>>> i.e., a first unit fraction.

No.

If 1/n < x
then 1/n⁺⁺ < x

NUF(x) = 0 or NUF(x) ≥ 2

----
By a similar argument
NUF(x) = 0 or NUF(x) ≥ 3
NUF(x) = 0 or NUF(x) ≥ 4
NUF(x) = 0 or NUF(x) ≥ 5
NUF(x) = 0 or NUF(x) ≥ 6
NUF(x) = 0 or NUF(x) ≥ 7
...

>> No,
>> it does not imply that.
>> There is no step from infinite to finite.
>
> But there is a step from nothing to something.

NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³

No first FUF exists.

>> Infinity is not
>> a reallyreallyreallyreallyreallyreally large
>> (finite) natural number.
>
> And infinity is not self-contradictory:

For each FISON F
not all n in F has its successor n⁺⁺
in the same FISON F

For each FISON F
all n in F has its successor n⁺⁺
in another FISON F₂ ≠ F

Those are different claims.
They are not contradictory.

[WM]

Jim Burns schrieb am Samstag, 18. März 2023 um 23:19:48 UTC+1:
> On 3/18/2023 3:37 PM, Fritz Feldhase wrote:
> > On Saturday, March 18, 2023
> > at 5:52:58 PM UTC+1, Jim Burns wrote:
> >> On 3/17/2023 5:29 PM, WM wrote:
>
> >>> the function NUF(x) has a step at
> >>> every single unit fraction
> >>> with a horizontal graph.
> >>> The increasing function pauses for a short while.
> >>> [...]
> >>
> >> Okay.
> >
> > Nope. Not ok, but nonsense.
> Damn.
> Point to you.

Not in real mathematics.

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0

After 1/(n+1) the function pauses for 1/(n*(n+1)) before it grows by 1 at 1/n.
>
Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 08:00:12 UTC+1:

> If 1/n < x
> then 1/n⁺⁺ < x

No. If 1/2 < x then 1/1 is not necessarily < x.

Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 08:00:12 UTC+1:

> > But there is a step from nothing to something.
> NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³

Can you understand that there are never more than 1 unit fractions in one point?
Can you understand that NUF(x) cannot grow by more than 1 in one point (and the point after it)?

Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 08:00:12 UTC+1:

> On 3/18/2023 3:20 PM, WM wrote:

> > But there is a step from nothing to something.
> NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³

More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line. They are dark.

Regards, WM

[Jim Burns]

On 3/19/2023 4:56 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 19. März 2023 um 08:00:12 UTC+1:

>> If 1/n < x
>> then 1/n⁺⁺ < x
>
> No.

? If 1/2 < x

> then 1/1 is not necessarily < x.

Yes.
If 1/2 < x
then 1/2⁺⁺ = 1/3 < x

[Fritz Feldhase]

On Sunday, March 19, 2023 at 9:53:48 AM UTC+1, WM wrote:
> Jim Burns schrieb am Samstag, 18. März 2023 um 23:19:48 UTC+1:
> > On 3/18/2023 3:37 PM, Fritz Feldhase wrote:
> > > On Saturday, March 18, 2023
> > > at 5:52:58 PM UTC+1, Jim Burns wrote:
> > >> On 3/17/2023 5:29 PM, WM wrote:
> > >>>
> > >>> the function NUF(x) has a step at
> > >>> every single unit fraction
> > >>> with a horizontal graph.
> > >>> The increasing function pauses for a short while.
> > >>> [...]
> > >>
> > >> Okay.
> > >
> > > Nope. Not ok, but nonsense.
> > >
> > Damn.
> > Point to you.

Indeed!

> Not in [muckenmatics].

> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
> After 1/(n+1) the function pauses for 1/(n*(n+1)) before it grows by 1 at 1/n.

Nope.

By definition: NUF(x) = aleph_0 for all x e IR,x > 0.

Since An e IN: 1/(n+1) e IR & 1/(n+1) > 0 as well as An e IN: 1/n e IR & 1/n > 0,
An e IN: NUF(1/(n+1)) = aleph_0 as well as NUF(1/n) = aleph_0.
Hence An e IN: NUF(1/(n+1)) = NUF(1/n). No step, no growth.

[Jim Burns]

If 1/n < x
then
1/(n+1) < x
1/(n+2) < x
...
1/(n+6.02214076×10²³) < x

and are all different points.

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 11:37:48 UTC+1:

> If 1/2 < x
> then 1/2⁺⁺ = 1/3 < x

So you mean the successor of n, not of 1/n.

> NUF(x) = 0 or NUF(x) ≥ 2

No, NUF(x) ≥ 2.

Regards, WM

[Fritz Feldhase]

On Sunday, March 19, 2023 at 9:56:59 AM UTC+1, WM wrote:
> Jim Burns schrieb am Sonntag, 19. März 2023 um 08:00:12 UTC+1:
> >

> > If 1/n < x then 1/n+ < x
> >
> No.

Yes.

Hint: Ak e IN: k+ > k. Hence Ak e IN: 1/k+ < 1/k.

Hence especially for some (certain) n e IN: 1/n+ < 1/n.

If now 1/n < x , then (by transitivity of <) 1/n+ < x. qed

> If 1/2 < x then 1/1 is not necessarily < x.

Right. But, at least in the context of mathematics, 1 is clearly NOT the successor of 2.

Hint: If n = 2, then n+ = 3.

Hence Jim Burn's claim implies:
If 1/2 < x, then 1/3 < x,
which is indeed true for all x e IR, since 1/3 < 1/2.

[WM]

Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 11:47:43 UTC+1:
> On Sunday, March 19, 2023 at 9:53:48 AM UTC+1, WM wrote:

> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
> > After 1/(n+1) the function pauses for 1/(n*(n+1)) before it grows by 1 at 1/n.
> Nope.

1/(n*(n+1)) > 0 means a pause. If that example is too difficult for you, take 1/3. After 1/3 the function pauses to increase until 1/2 is met.

>
> By definition: NUF(x) = aleph_0 for all x e IR,x > 0.

That is not a definition but nonsense. But it does not help your case, because these unit fractions are indiscernible, i.e., dark.

>
> Since An e IN: 1/(n+1) e IR & 1/(n+1) > 0 as well as An e IN: 1/n e IR & 1/n > 0,
> An e IN: NUF(1/(n+1)) = aleph_0 as well as NUF(1/n) = aleph_0.
> Hence An e IN: NUF(1/(n+1)) = NUF(1/n). No step, no growth.

Consider the set function. It grows in steps until 1/1.

Regards, WM

[Fritz Feldhase]

On Sunday, March 19, 2023 at 10:33:25 AM UTC+1, WM wrote:
> Jim Burns schrieb am Sonntag, 19. März 2023 um 08:00:12 UTC+1:
> > >

> > > But there is a step from nothing to something. [WM]

Or rather from 0 to aleph_0 in our case.

> > NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³

Since by definition of NUF: {NUF(x) : x e IR} = {0, aleph_0} and aleph_0 ≥ 6.02214076×10²³.

> Can you understand that there are never more than 1 unit fractions in one point?

I guess he can.

> Can you understand that NUF(x) cannot grow by more than 1 in one point (and the point after it)?

NUF does not "grow" in this way. Actually, it jumps from 0 to aleph_0 at 0, that's all.

Hint: NUF(x) = 0 for all x e IR, x <= 0
and: NUF(x) = aleph_0 for all x e IR, x > 0.

[WM]

ℵo unit fractions with their distances cannot sit between 0 and (0 1]. You violate mathematics here as you did with inclusion-monotonic sequences and JB did with his not conserved Bob. We see matheology is hostile to mathematics.

Three facts discard set theory: The present topic, the endsegments, the X-O-matrix.

Regards, WM

[Jim Burns]

On 3/19/2023 5:33 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 19. März 2023 um 08:00:12 UTC+1:

>>> But there is a step from nothing to something.
>>
>> NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³
>
> Can you understand that there are never
> more than 1 unit fractions in one point?

Yes.

> Can you understand that
> NUF(x) cannot grow by
> more than 1 in one point

No.
If 1/n < x then 1/n⁺⁺ < x

NUF(x) ≠ 1

ℵ₀+1 = ℵ₀

Infinity is not
a reallyreallyreallyreallyreallyreally large
(finite) natural number.

> (and the point after it)?

No two points have none between.

[Fritz Feldhase]

Well, since 6.02214076×10²³ was just an arbitrarily chosen natural number your claim shoould hold for any natural number.

Hence your "argument" implies, for example: More than 2 unit fractions cannot be distinguished by their points on the real line.

Does that sound reasonable?

Hint: 1/1 =/= 1/2, 1/1 =/= 1/3, 1/2 =/= 1/3. Hence card {1/1, 1/2, 1/3} = 3.

__________________________________

Another hint: card {1/n : n <= 6.02214076×10²³ + 1} = 6.02214076×10²³ + 1.

Actually, for all n, m <= 6.02214076×10²³ + 1: n =/= m -> 1/n =/= 1/m.

[WM]

Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 12:18:40 UTC+1:
> On Sunday, March 19, 2023 at 10:44:32 AM UTC+1, WM wrote:
> > Jim Burns schrieb am Sonntag, 19. März 2023 um 08:00:12 UTC+1:
> > >
> > > NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³
> > >
> > More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line.
> Well, since 6.02214076×10²³ was just an arbitrarily chosen natural number your claim shoould hold for any natural number.

It does not. There are many unit fractions which can be discerned by their points.

>
> Hence your "argument" implies, for example: More than 2 unit fractions cannot be distinguished by their points on the real line.
>
> Does that sound reasonable?

ℵo unit fractions cannot be distinguished by their points. Because their points are dark. But they cannot sit between 0 and (0, 1].

Regards, WM

[Fritz Feldhase]

HUH?!

Hint: P implies Q or P.

Moreover: NUF(x) ≥ 2 for all x e IR is wrong, since NUF(0) = 0.

On the other hand, NUF(1) ≥ 2 , since aleph_0 ≥ 2 and NUF(1) = aleph_0.

Actually, {NUF(x) : x e IR} = {0, aleph_0}, hence NUF(x) = 0 or NUF(x) ≥ 2 for all x e IR.

[Fritz Feldhase]

On Sunday, March 19, 2023 at 12:07:08 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 11:47:43 UTC+1:
> >

> > By definition: NUF(x) = aleph_0 for all x e IR, x > 0.

> >
> That is not a definition but

Learn to read, you fucking asshole full of shit!

I didn't state the definition of NUF here.

Mückenheim, Du bist wirklich für jede Form der Mathematik zu dumm und zu doof.

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 12:16:31 UTC+1:
> On 3/19/2023 5:33 AM, WM wrote:
> > Jim Burns schrieb am Sonntag,
> > 19. März 2023 um 08:00:12 UTC+1:
>
> >>> But there is a step from nothing to something.
> >>
> >> NUF(x) = 0 or NUF(x) ≥ 6.02214076×10²³
> >
> > Can you understand that there are never
> > more than 1 unit fractions in one point?
> Yes.
> > Can you understand that
> > NUF(x) cannot grow by
> > more than 1 in one point
> No.

Then try to learn thinking.

> If 1/n < x then 1/n⁺⁺ < x

Between 1/n and 1/(n+1) the function has grownby 1.
> NUF(x) ≠ 1

Between
NUF(x) = 0
and
NUF(x) = ℵo
there are at least ℵo points with growth 1.
>
> ℵ₀+1 = ℵ₀

But first SBZ must increase from 0 to soemwhat. That cannot happen in steps of more than 1 because, as you can understand, there are never more than 1 unit fractions in one point.

> > (and the point after it)?
> No two points have none between.

I know. Therefore there is a constant horizontal level of NUF(x) behind every 1/n.

Regards, WM

[Fritz Feldhase]

On Sunday, March 19, 2023 at 12:12:33 PM UTC+1, WM wrote:

> ℵo unit fractions with their distances cannot sit between 0 and (0 1].

Did anyone claim that nonsense - except you?

Hint: NO x e IR "sits between 0 and (0 1]": ~Ex e IR: Ay e (0 1]: 0 < x < y.

[Fritz Feldhase]

On Sunday, March 19, 2023 at 12:22:13 PM UTC+1, WM wrote:

> ℵo unit fractions cannot be distinguished by their points.

YOUR CLAIM WAS: "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line." you lying piece of shit!

[Jim Burns]

NUF(1/2) ≥ 2

NUF(0) = 0

[Fritz Feldhase]

On Sunday, March 19, 2023 at 12:28:53 PM UTC+1, WM wrote:

> Jim Burns schrieb am Sonntag, 19. März 2023 um 12:16:31 UTC+1:
> >
> > If 1/n < x then 1/n⁺⁺ < x
> >
> Between 1/n and 1/(n+1) the function has grown by 1.

Nope.

| Hint: By definition NUF(x) = aleph_0 for all x e IR, x > 0.

|
| Since An e IN: 1/(n+1) e IR & 1/(n+1) > 0 as well as An e IN: 1/n e IR & 1/n > 0,
| An e IN: NUF(1/(n+1)) = aleph_0 as well as NUF(1/n) = aleph_0.
|

| Hence An e IN: NUF(1/(n+1)) = NUF(1/n). No growth.

You are dumb like shit, Mückenheim.

[WM]

Yes. ℵo are more than 6.02214076×10²³.

Regards, WM

[WM]

Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 12:40:41 UTC+1:

> | Hint: By definition NUF(x) = aleph_0 for all x e IR, x > 0.

This is wrong because ℵo unit fractions and their finite internal distances require a length of δ which is much, much larger than 0. Therefore NUF(x) = ℵo can be true only for x ≥ δ > 0. This is simple high-school mathematics.

Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 12:39:25 UTC+1:

> NUF(0) = 0
> NUF(x) = 0 or NUF(x) ≥ 2

ℵo unit fractions and their finite internal distances require a length of δ which is much, much larger than 0. Therefore NUF(x) = ℵo can be true only for x ≥ δ > 0. This is simple high-school mathematics. NUF(x) = ℵo is wrong for x < δ.

Regards, WM

[Jim Burns]

On 3/19/2023 7:28 AM, WM wrote:
> Jim Burns schrieb am Sonntag,

> 19. März 2023 um 12:16:31 UTC+1:

>> NUF(x) ≠ 1
>
> Between
> NUF(x) = 0
> and
> NUF(x) = ℵo
> there are at least ℵo points with growth 1.

if NUF(x) > 0
then
1/n < x for some 1/n
and
1/n⁺⁺ < 1/n < x
and
NUF(x) > 1

NUF(x) = 0 or NUF(x) > 1

NUF(x) ≠ 1

>> ℵ₀+1 = ℵ₀
>
> But first SBZ must increase from 0 to soemwhat.

No.
Your mutant intermediate-value-theorem
says that, but it is wrong.

1/n⁺⁺ < 1/n < x

> That cannot happen in steps of more than 1

1/n⁺⁺ < 1/n < x

> because, as you can understand,
> there are never more than 1 unit fractions
> in one point.

1/n⁺⁺ ≠ 1/n
1/n⁺⁺ < 1/n < x
NUF(x) ≠ 1

>>> (and the point after it)?
>>
>> No two points have none between.
>
> I know.

Then you know that
this line is not correctly described as
a sequence of points, with
last-befores and first-afters for splits.

[Fritz Feldhase]

On Sunday, March 19, 2023 at 1:21:46 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 12:40:41 UTC+1:
> >
> > | Hint: By definition NUF(x) = aleph_0 for all x e IR, x > 0.
> >
> This is wrong because

No, this is not wrong.

Hint: Ax e IR, x > 0: card {u e U: u <= x} = aleph_0 (with: U := {1/n : n e IN}).

Proof: Let x e IR, x > 0. Let n := |' 1/x '|, then 1/n <= x, 1/(n+1) < x, 1/(n+2) < x, ...
In general: Ak e IN u {0}: 1/(n+k) <= x. In other words, Au e U(x): u <= x with
U(x) := {1/n : n e IN & n >= 1/x}. Now Ax e IR, x > 0: card U(x) = aleph_0.
Since Ax e IR, x > 0: U(x) c U, we get: Ax e IR, x > 0: card {u e U: u <= x} = aleph_0. qed

Learn some logic, learn some math, Mückenheim.

[Fritz Feldhase]

On Sunday, March 19, 2023 at 1:16:22 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 12:37:39 UTC+1:
> > On Sunday, March 19, 2023 at 12:22:13 PM UTC+1, WM wrote:
> > >
> > > ℵo unit fractions cannot be distinguished by their points.
> > >
> > YOUR CLAIM WAS: "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line."
> >
> Yes.

Look, you psychotic asshole full of shit: YOUR CLAIM "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line." is either nonsensical or wrong.

Hint: Since 6.02214076×10²³ was just an arbitrarily chosen natural number your claim should hold for any natural number.

Hence your "(non)argument" implies, for example: "More than 2 unit fractions cannot be distinguished by their points on the real line."

[Fritz Feldhase]

On Sunday, March 19, 2023 at 12:22:13 PM UTC+1, WM wrote:

> ℵo unit fractions [...] cannot sit between 0 and (0, 1].

Did anyone--except you--claim that nonsense?

Hint: NO x e IR (and hence NO unit fraction) "sits between 0 and (0 1]": ~Ex e IR: Ay e (0 1]: 0 < x < y.

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 13:30:23 UTC+1:
> On 3/19/2023 7:28 AM, WM wrote:

> > But first SBZ must increase from 0 to soemwhat.
> No.
> Your mutant intermediate-value-theorem
> says that, but it is wrong.

It is not my theorem but the distribution of the unit fractions, which are separated one by one on the real line. Do you refuse to believe in the one by one separation of all unit fractions, separated by gaps of size 1/(n*(n+1))? Do you refuse that the separating gaps are real numbers and can be arguments of SBZ(x)?

> >> No two points have none between.
> >
> > I know.
> Then you know that
> this line is not correctly described as
> a sequence of points, with
> last-befores and first-afters for splits.

I need not describe this line at all. I only need one point of the horizontal line which has the same height as its first point at the step 1/n.

Regards, WM

[WM]

Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 14:37:42 UTC+1:
> On Sunday, March 19, 2023 at 1:21:46 PM UTC+1, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 12:40:41 UTC+1:
> > >
> > > | Hint: By definition NUF(x) = aleph_0 for all x e IR, x > 0.
> > >
> > This is wrong because
> No, this is not wrong.

ℵo unit fractions and their finite internal distances require a length of δ which is much, much larger than 0.

Do you agree?
HInt: Even one distance 1/(n*(n+1)) for any natnumber n is much, much larger than a point, whilst your claim contains not even a point but nothing.

[Jim Burns]

On 3/19/2023 8:23 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 19. März 2023 um 12:39:25 UTC+1:

>> NUF(0) = 0
>> NUF(x) = 0 or NUF(x) ≥ 2
>
> ℵo unit fractions and
> their finite internal distances require
> a length of δ which is much, much larger than 0.

You (WM) have previously munged
the distinction between length and cardinality.
That makes you accidentally correct here.
|(0,δ)| > ℵ₀ >> |(0,0)|

> Therefore
> NUF(x) = ℵo can be true only for x ≥ δ > 0.
> This is simple high-school mathematics.

That would be sad if it were true.

It is apparently Hochschule Augsburg mathematics,
though. Yes, that's sad.

> NUF(x) = ℵo is wrong for x < δ.

Still no.

mₓ = ⌈1/x⌉

x >
1/mₓ >
1/(mₓ+1) >
... >
1/(mₓ+6.02214076×10²³) >
...

[Fritz Feldhase]

On Sunday, March 19, 2023 at 2:43:41 PM UTC+1, Fritz Feldhase wrote:
> On Sunday, March 19, 2023 at 1:16:22 PM UTC+1, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 12:37:39 UTC+1:
> > >
> > > YOUR CLAIM WAS: "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line."
> > >
> > Yes.
> >
> Look, you psychotic asshole full of shit: YOUR CLAIM "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line." is either nonsensical or wrong.
>

> Hint: Since 6.02214076×10²³ was just an arbitrarily chosen natural number your claim should hold for _any_ natural number.

Or do you think that Avogadro's number has some special meaning in math/analysis (comparable to AP's borderline number 1*10^604)?

> Hence your "(non)argument" implies [as well], for example: "More than 2 unit fractions cannot be distinguished by their points on the real line."

[Fritz Feldhase]

On Sunday, March 19, 2023 at 3:37:49 PM UTC+1, WM wrote:

> ℵo unit fractions and their finite internal distances require a length of δ

Except "they" don't.

Hint: For every δ e IR, δ > 0: there are ℵo unit fractions in (0, δ/2).

Actually, for every δ e IR, δ > 0:
For every x in IR, δ > x > 0, there are ℵo unit fractions in (0, x).

> Therefore NUF(x) = ℵo can be true only for x ≥ δ > 0.

Nope. Actually, for every δ e IR, δ > 0:
For every x in IR, δ > x > 0: NUF(x) = ℵo.

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 17:42:36 UTC+1:
> On 3/19/2023 8:23 AM, WM wrote:
> > ℵo unit fractions and
> > their finite internal distances require
> > a length of δ which is much, much larger than 0.

> You (WM) have previously munged
> the distinction between length and cardinality.

ℵo finite distances are a finite length not less than every finite length x > 0.

> That makes you accidentally correct here.
> |(0,δ)| > ℵ₀ >> |(0,0)|

That is wrong. |(0,δ)| is a length, ℵ₀ is a number.

Regards, WM

[WM]

Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 20:04:53 UTC+1:
> On Sunday, March 19, 2023 at 2:43:41 PM UTC+1, Fritz Feldhase wrote:
> > YOUR CLAIM "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line." is either nonsensical or wrong.

It is true because ℵo unit fractions cannot be distinguished, as you say, since every attempt to distinguish them by any x > 0 includes ℵo unit fractions in (0, x).

Regards, WM

[WM]

Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 20:24:19 UTC+1:
> On Sunday, March 19, 2023 at 3:37:49 PM UTC+1, WM wrote:
>
> > ℵo unit fractions and their finite internal distances require a length of δ
> Except "they" don't.
>
> Hint: For every δ e IR, δ > 0: there are ℵo unit fractions in (0, δ/2).

For every definable δ, yes. But for the distrance between the first and the second unit fraction, for instance.

>
> Actually, for every δ e IR, δ > 0:
> For every x in IR, δ > x > 0, there are ℵo unit fractions in (0, x).

Therefore you cannot distinguish them by definable x.

> > Therefore NUF(x) = ℵo can be true only for x ≥ δ > 0.
> Nope. Actually, for every δ e IR, δ > 0:
> For every x in IR, δ > x > 0: NUF(x) = ℵo.

Not for the δ made up by the distances of the first 100 unit fractions. Note that since no unit fractions sit at the same point, there is a first one. That, by the way, are the only alternatives: Some unit fractions enter NUF(x) first at the same point x or only one unit fraction enters NUF(x) first. Any other proposal?

Regards, WM

[Jim Burns]

On 3/19/2023 5:06 PM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 19. März 2023 um 17:42:36 UTC+1:
>> On 3/19/2023 8:23 AM, WM wrote:

>>> ℵo unit fractions and
>>> their finite internal distances require
>>> a length of δ which is much, much larger than 0.
>
>> You (WM) have previously munged
>> the distinction between length and cardinality.
>
> ℵo finite distances are
> a finite length not less than
> every finite length x > 0.

No.

(δ/2+δ/4+δ/8+δ/16+δ/32+...) - δ/2 =
(δ/4+δ/8+δ/16+δ/32+...)

(1/2)(δ/2+δ/4+δ/8+δ/16+δ/32+...) =
(δ/4+δ/8+δ/16+δ/32+...)

(δ/2+δ/4+δ/8+δ/16+δ/32+...) - δ/2 =
(1/2)(δ/2+δ/4+δ/8+δ/16+δ/32+...)

(δ/2+δ/4+δ/8+δ/16+δ/32+...) = δ

>> That makes you accidentally correct here.
>> |(0,δ)| > ℵ₀ >> |(0,0)|
>
> That is wrong.
> |(0,δ)| is a length, ℵ₀ is a number.

|(0,δ)| is the cardinality of
the set of points in R between 0 and δ

ℵ₀ is the smallest infinite cardinal number.

ℵ₀ is not

a reallyreallyreallyreallyreallyreally large
(finite) natural number.

For example,
ℵ₀+1 = ℵ₀

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 22:34:37 UTC+1:
> On 3/19/2023 5:06 PM, WM wrote:

> > ℵo finite distances are
> > a finite length not less than
> > every finite length x > 0.
> No.
>

Yes. To be clear: ℵo finite distances together are more than ℵo points. ℵo points are more than one point. One point is more than fits between 0 and (0, 1]. That means NUF(x) = ℵo does not hold for every x ∈ (0, 1].

Regards, WM

[Jim Burns]

On 3/19/2023 5:47 PM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 19. März 2023 um 22:34:37 UTC+1:
>> On 3/19/2023 5:06 PM, WM wrote:

>>> ℵo finite distances are
>>> a finite length not less than
>>> every finite length x > 0.
>>
>> No.
>
> Yes. To be clear:
> ℵo finite distances together are more than
> ℵo points.
> ℵo points are more than
> one point.
> One point is more than
> fits between 0 and (0, 1].

For each unit fraction left of 1/1,
there is a last-before and a first-after.

For each x in (0,1)
there is a last and first to its right,
and, for each split, last-before and first-after.
To the right of x, finitely-many.

For each x in (0,1)
there is a first to its right
and, for each split, last-before and first-after.
However, there is no last.
To the left of x, infinitely-many.

At 1/1,
there is no last unit fraction between it and 0

As x moves toward 0
a last unit fraction does not appear.
There continues to be infinitely-many before x

Infinity is not

a reallyreallyreallyreallyreallyreally large
(finite) natural number.

> That means NUF(x) = ℵo
> does not hold for every x ∈ (0, 1].

No.
NUF(x) = ℵ₀ continues to hold.
A last unit fraction does not appear.

[Ross Finlayson]

On Saturday, March 18, 2023 at 5:01:15 PM UTC-7, Ross Finlayson wrote:
> On Saturday, March 18, 2023 at 2:03:22 PM UTC-7, Ross Finlayson wrote:
> > On Saturday, March 18, 2023 at 1:37:39 PM UTC-7, Mostowski Collapse wrote:
> > > You can try here, at t=652, and count the "eh",
> > > he has really problems reading the text.
> > >
> > > https://youtu.be/UkZGZ6FRpS0?t=652
> > >
> > > If its not nigher pitch or lower pitch, he
> > > prolongs words, or suddently speeds up,
> > >
> > > like instead "is two", he says "is twooooooo".
> > >
> > > LMAO!
> > > Mostowski Collapse schrieb am Samstag, 18. März 2023 um 21:34:29 UTC+1:
> > > > If its not the potato, he is stuttering all the time
> > > > with "eh", "em" etc.. and then sometimes he
> > > > goes into unmotivated higher pitch or lower pitch.
> > > >
> > > > Its the most horrible reading I have ever heard.
> > > > Mostowski Collapse schrieb am Samstag, 18. März 2023 um 21:31:00 UTC+1:
> > > > > He is demonstrating how to read with a potato in mouth.
> > > > > Jim Burns schrieb am Samstag, 18. März 2023 um 19:28:13 UTC+1:
> > > > > > On 3/18/2023 12:50 AM, Ross Finlayson wrote:
> > > > > > > On Thursday, March 16, 2023
> > > > > > > at 9:43:03 PM UTC-7, Jim Burns wrote:
> > > > > > >> On 3/16/2023 9:14 PM, Ross Finlayson wrote:
> > > > > >
> > > > > > >>> Here's some more reading from Maugin.
> > > > > > >>> https://www.youtube.com/watch?v=UkZGZ6FRpS0
> > > > > > >> "Reading from Maugin, pages nine-eleven. "
> > > > > > >>
> > > > > > >> Nonlinear Electromechanical Effects and Applications
> > > > > > >> (Theoretical and Applied Mechanics)
> > > > > > >> by Gerard A Maugin (Author)
> > > > > > >> https://www.amazon.com/Nonlinear-Electromechanical-Applications-Theoretical-Mechanics/dp/9971978431
> > > > > > >>
> > > > > > >> Ross: why make a video of you reading those pages?
> > > > > > > Some people need a book read to them vocally
> > > > > > > before they can read it.
> > > > > > > It's a traditional role of seminars
> > > > > > > called "reading".
> > > > > > My own experience is that, with
> > > > > > highly-compressed texts, as in physics, math, etc,
> > > > > > I do better, at least initially, introducing
> > > > > > myself to the material by reading at my own pace.
> > > > > > Glancing up to remind myself what ⋃Finᶠ(b) means,
> > > > > > and so on.
> > > > > > YMMV
> > I was describing the usual orthornomal vector basis
> > and that R^3's is usually called e sub i or e sub 1,
> > so, generously, read out "e, e, e, e sub 1". Here the configuration
> > space being discussed is flux through a surface, with R^2 and
> > a time-like dimension.
> >
> > It's a good stutter though if such things as the perceived
> > failings of other people are what activate your brain's pleasure
> > centers, where that's Shadenfreude and is a usual trait of
> > psychopaths and the sadistic. Surely it aggrieved me.
> >
> > Please though, if I ever make an intonation that "it ends
> > like a question?" that something is terribly wrong, because
> > that is a grating-nails-the-chalkboard-the-teacher-doesn't-know,
> > and such "presenters" are "defective".
> >
> > Also there's one good "uhhh....". Sega Dreamcast: "It's thinking."
> >
> >
> > Now that you've interjected after the wonderful preceding post, and
> > please feel certain that I read my own posts with certainty
> > and much as they are written as they are written to be read,
> > discuss Brouwer's concept of the continuum and see if it
> > makes sense or that otherwise perhaps it doesn't fit in your head.
> Well, here's reading another 2/3 page or so from Maugin's book, though
> mostly it's my interpretations and explanations, then skimming a few sections
> and a few dozen pages and the appendix, and a quote from opening Kuehnel,
> then pointing to Weyl.
>
> https://www.youtube.com/watch?v=GIMKbXBEQno


Here's some more reading of Maugin. (Any faults are mine not necessarily his.)

https://www.youtube.com/watch?v=3iB85GtOduc

[Fritz Feldhase]

On Sunday, March 19, 2023 at 10:47:55 PM UTC+1, WM wrote:

> NUF(x) = ℵo does not hold for every x ∈ (0, 1].

Nope. NUF(x) = ℵo for every x ∈ (0, 1].

Hint: Ax e IR, x > 0: card {u e U: u <= x} = aleph_0 (with: U := {1/n : n e IN}).

Proof: Let x e IR, x > 0. Let n := |' 1/x '|, then 1/n <= x, 1/(n+1) < x, 1/(n+2) < x, ...
In general: Ak e IN u {0}: 1/(n+k) <= x. In other words, Au e U(x): u <= x with
U(x) := {1/n : n e IN & n >= 1/x}. Now Ax e IR, x > 0: card U(x) = aleph_0.
Since Ax e IR, x > 0: U(x) c U, we get: Ax e IR, x > 0: card {u e U: u <= x} = aleph_0. qed

[Fritz Feldhase]

On Monday, March 20, 2023 at 7:44:25 AM UTC+1, Fritz Feldhase wrote:
> On Sunday, March 19, 2023 at 10:47:55 PM UTC+1, WM wrote:
> >
> > NUF(x) = ℵo does not hold for every x ∈ (0, 1].
> >
> Nope. NUF(x) = ℵo for every x ∈ (0, 1].
>

> Hint: [...]

Jim's approach is even better (simpler, clearer):

Let x ∈ (0, 1] and mₓ = ⌈1/x⌉. Then 1/mₓ <= x.
Hence for all k e IN u {0}: 1/(mₓ+k) <= x. (Since m > n implies1/m < 1/n.)
Hence there are countably-infinitely many unit fractions which are <= x.
Hence there are countably-infinitely many unit fractions in {u e U : u <= x}.
Actually, card {u e U : u <= x} = ℵo. Hence NUF(x) = ℵo. qed

[Fritz Feldhase]

On Sunday, March 19, 2023 at 10:11:19 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 19. März 2023 um 20:04:53 UTC+1:
> >
> > YOUR CLAIM "More than 6.02214076×10²³ unit fractions cannot be distinguished by their points on the real line." is either nonsensical or wrong.
> >

> It is true because ℵo unit fractions cannot be distinguished [...]

You are dumb like shit, Mückenheim.

If something is true for ℵo things of some "type", then it needn't be true for a finite number of things of the same "type".

For example: The sum of ℵo natural numbers is not a natural number; but the sum of 6.02214076×10²³ + 1 natural numbers is certainly a natural number.

Hint: 6.02214076×10²³ (as well as 6.02214076×10²³ + 1) unit fractions are FINITELY many unit fraction, while ℵo unit fractions are infinitely many unit fractions.

___________________________________

An additional consideration:

6.02214076×10²³ + 1 "unit fractions" (actually, rather their names) could _in principle_ be written down. We can think of a list that starts with

(1) I/I
(2) I/II
(3) I/III
:

and ends (after finitely many lines between start and end) with

:
(6.02214076×10²³) I/I..I
(6.02214076×10²³ + 1) I/I..II

where the numbers of "I"'s after the "/" in the term refered to by "I/I..I" (in the text above) is 6.02214076×10²³ and the numbers os "I"'s after the "/" in the term refered to by "I/I..II" (in the text above) is 6.02214076×10²³ + 1.

Clearly any term of the form I/I..I (consisting of an "I" followed by an "/" and a _finite_ number of "I"'s) "corresponds" (names, denotes) a certain unit fraction.

Moreover, if two terms of this form differ, then the unit fractions denoted by these terms differ.

You see "I/I" denotes the unit fraction 1/1, "I/II" denotes the unit fraction 1/2, "I/III" denotes the unit fraction 1/3, etc.

Since the terms in the list mentioned above clearly "can be distinguished" (by checking the number of "I"'s after the "/"-symbol), the 6.02214076×10²³ + 1 unit fractions they denote "can be distinguished".

Now the Dedekind-Cantor-Axiom "implies" that 6.02214076×10²³ + 1 (and hence MORE that 6.02214076×10²³) unit fractions "can be distinguished" by their points on the real line". (You see: different unit fractions corresponds with different points on the line.) Actually, a "very powerful microscope" should suffice. Hint: The smallest distance between theses unit fractions is 1/(6.02214076×10²³ + 1). A magnification of 6.02214076×10²³ + 1 would allow to observe a "visible length" of 1 between the (6.02214076×10²³ + 1)-th unit fraction and the (6.02214076×10²³)-th unit fraction. (Where "the n-th unit fraction" is the unit fraction referred to by the n-th term in the list mentioned above.)

[Fritz Feldhase]

On Monday, March 20, 2023 at 8:40:22 AM UTC+1, Fritz Feldhase wrote:

> Now the Dedekind-Cantor-Axiom "implies" that 6.02214076×10²³ + 1 (and hence MORE that 6.02214076×10²³) unit fractions "can be distinguished" by their points on the real line". (You see: different unit fractions corresponds with different points on the line.) Actually, a "very powerful microscope" should suffice. Hint: The smallest distance between these unit fractions is 1/(6.02214076×10²³ + 1). A magnification of 6.02214076×10²³ + 1 would allow to observe a "visible length" of 1 between the (6.02214076×10²³ + 1)-th unit fraction and the (6.02214076×10²³)-th unit fraction. (Where "the n-th unit fraction" is the unit fraction referred to by the n-th term in the list mentioned above.)

Remember that the Planck length is around 10^(−35) m.

https://www.forbes.com/sites/startswithabang/2019/06/26/what-is-the-smallest-possible-distance-in-the-universe/?sh=7063b12148a1

[Fritz Feldhase]

On Monday, March 20, 2023 at 9:00:50 AM UTC+1, Fritz Feldhase wrote:
> On Monday, March 20, 2023 at 8:40:22 AM UTC+1, Fritz Feldhase wrote:
>
> > Now the Dedekind-Cantor-Axiom "implies" that 6.02214076×10²³ + 1 (and hence MORE that 6.02214076×10²³) unit fractions "can be distinguished" by their points on the real line". (You see: different unit fractions corresponds with different points on the line.) Actually, a "very powerful microscope" should suffice. Hint: The smallest distance between these unit fractions is 1/(6.02214076×10²³ + 1).

Of course, that's wrong. Sorry. It's 1/((6.02214076×10²³) x (6.02214076×10²³ + 1)).

[WM]

Jim Burns schrieb am Sonntag, 19. März 2023 um 23:35:10 UTC+1:
> On 3/19/2023 5:47 PM, WM wrote:

> For each unit fraction left of 1/1,
> there is a last-before and a first-after.
>

Yes, but that is irrelevant.

> > That means NUF(x) = ℵo
> > does not hold for every x ∈ (0, 1].
> No.
> NUF(x) = ℵ₀ continues to hold.

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
That implies: At least 10^100^1000^10000^100000 different finite distances 1/(n*(n+1)) lie in (0, 1] for all points x of which NUF(x) = ℵo is wrong.

That is mathematics. If you prefer matheology, tht does not change mathematics.

Regards, WM

[WM]

Fritz Feldhase schrieb am Montag, 20. März 2023 um 07:44:25 UTC+1:
> On Sunday, March 19, 2023 at 10:47:55 PM UTC+1, WM wrote:
>
> > NUF(x) = ℵo does not hold for every x ∈ (0, 1].
> Nope. NUF(x) = ℵo for every x ∈ (0, 1].
> Hint: Ax e IR, x > 0: card {u e U: u <= x} = aleph_0 (with: U := {1/n : n e IN}).
>

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
That implies: At least 10^100^1000^10000^100000 different finite distances 1/(n*(n+1)) lie in (0, 1] for all points x of which NUF(x) = ℵo is wrong.

That is mathematics. If you prefer matheology, that does not change mathematics.

Regards, WM

[Fritz Feldhase]

On Monday, March 20, 2023 at 10:36:45 AM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Montag, 20. März 2023 um 07:44:25 UTC+1:
> > On Sunday, March 19, 2023 at 10:47:55 PM UTC+1, WM wrote:
> > >
> > > NUF(x) = ℵo does not hold for every x ∈ (0, 1].
> > >
> > Nope. NUF(x) = ℵo for every x ∈ (0, 1].
> >
> > Hint:

Let x ∈ (0, 1] and mₓ = ⌈1/x⌉. Then 1/mₓ <= x.
Hence for all k e IN u {0}: 1/(mₓ+k) <= x. (Since m > n implies1/m < 1/n.)
Hence there are countably-infinitely many unit fractions which are <= x.
Hence there are countably-infinitely many unit fractions in {u e U : u <= x}.
Actually, card {u e U : u <= x} = ℵo. Hence NUF(x) = ℵo. qed

> for all points x of which NUF(x) = ℵo is wrong.

There is no point x e IR, x > 0, for which NUF(x) = ℵo is wrong.

So true: "[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

[Fritz Feldhase]

On Monday, March 20, 2023 at 10:35:42 AM UTC+1, WM wrote:

> for all points x of which NUF(x) = ℵo is wrong.

There is no point x e IR, x > 0, for which NUF(x) = ℵo is wrong.

So true: "[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

> That is mathematics. If you prefer matheology, that does not change mathematics.

Shut up, you silly crank.

[Jim Burns]

On 3/20/2023 5:35 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 19. März 2023 um 23:35:10 UTC+1:

>> For each unit fraction left of 1/1,
>> there is a last-before and a first-after.
>
> Yes,

Then the unit fractions are one-by-one 1×1
For each split, there is a last-before and
a first-after.

('1×1' doesn't mean injective 'one-to-one'.)

Since the unit fractions are 1×1
the question of finiteness vs infiniteness
takes an especially simple form.
How many ends are there?

Each FISON is 1×1 and 2-ended.
Each FISON is finite.

ℕ and ℤ are 1×1 but not 2-ended.
ℕ and ℤ are not finite.

The set of naturals from 10^(10^(10^(10)))
to 10^(10^(10^(10^(10)))) is 1×1 and 2-ended.
The set of naturals from 10^(10^(10^(10)))
to 10^(10^(10^(10^(10)))) are finite.

The set of naturals from 10^(10^(10^(10)))
to 10^(10^(10^(10^(10)))) has Bob-conservation.
We cannot know this by "checking" it but
we know by its being 1×1 and 2-ended.

The set of ordinals from 0 to ω is 2-ended
but it's not 1×1
The set of ordinals from 0 to ω is not finite.

The set of ordinals from 0 to ω
does not have Bob-conservation.
ω is not

a reallyreallyreallyreallyreallyreally large
(finite) natural number.

Note that
the ordinals from 0 to ω are
very nearly 1×1 but not quite 1×1
One split, between ω and the finites,
only has ω first-after, no last-before.

"Very nearly" won't do.
The ordinals from 0 to ω are not finite.

>> For each unit fraction left of 1/1,
>> there is a last-before and a first-after.
>
> Yes,
> but that is irrelevant.

The set of unit fractions is 1×1
The question of finiteness vs infiniteness,
of Bob-conservations vs Bob-non-conservation,
becomes: how many ends are there?

The set of all unit fractions is 1-ended.

For each unit fraction 1/n
the set from 1/1 to 1/n is 2-ended.
the set from 1/1 to 1/n is finite.

The set of unit fractions after 1/n
is 1-ended.
The set of unit fractions after 1/n
is not finite.

Thus,
for each unit fraction,
almost all unit fractions are after it

because they're 1×1 and 1-ended.

>>> That means NUF(x) = ℵo
>>> does not hold for every x ∈ (0, 1].
>>
>> No.
>> NUF(x) = ℵ₀ continues to hold.
>
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0
>
> That implies:
> At least 10^100^1000^10000^100000 different
> finite distances 1/(n*(n+1))
> lie in (0, 1]
> for all points x of which NUF(x) = ℵo is wrong.

NUF(1/1) = ℵ₀
There is no second end to {1/n | 1/n e (0,1]}

As x moves toward 0

no second end appears in {1/n | 1/n e (0,x]}
NUF(x) = ℵ₀

[WM]

Jim Burns schrieb am Montag, 20. März 2023 um 15:35:09 UTC+1:
> On 3/20/2023 5:35 AM, WM wrote:

> Thus,
> for each unit fraction,
> almost all unit fractions are after it

This cannot be true for a unit fraction in the last 10 % which exist if 100 % exist.

> As x moves toward 0
> no second end appears in {1/n | 1/n e (0,x]}

The end appears at zero. And if before zero the unit fractions do not disappear one by one, then they are not different unit fractions. Then

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n*(n+1)) > 0

would be invalid.

> NUF(x) = ℵ₀

For all x which have a far enough distance from 0. Not for thosewhich are closer. Again: Different unit fractions disappear at different positions, never together.

Regards, WM

[Ross Finlayson]

Yeah, Planck length is about ten orders magnitude smaller the atom.