On 12/9/2023 6:11 AM, WM wrote:
> On 08.12.2023 21:37, Jim Burns wrote:
>> On 12/8/2023 10:21 AM, WM wrote:
>>> On 08.12.2023 05:09, Jim Burns wrote:
>>>> On 12/7/2023 10:47 AM, WM wrote:

>>>>> That makes potentiel infinity
>>>>> a difficult concept.
>>>>
>>>> Your potential.infinityᵂᴹ wanders
>>>> between being about numbers I can take
>>>> and about numbers which exist.
>>>
>>> With n also n+1 is visible.
>>
>> That matters. Quite a lot.
>>
>> We can say _once_ and
>> know _infinitely.many ways_ that
>> n+1 in ⟨0,…,n,n+1⟩ is not.first.darkᵂᴹ
>>
>> We can follow that by saying once and
>> knowing infinitely.many ways that
>> n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
>
> Yes.
> Nevertheless
> ∀n ∈ ℕ_vis:
> |ℕ \ {1, 2, 3, ..., n}| = ℵo .

Yes.

|ℕ\{1,2,3,...,n}| = ℵ₀
means
ℕ\{1,2,3,...,n} is 1.by.1;1.end.able

We agree that
∀n ∈ ℕ_vis:
ℕ\{1,2,3,...,n} is 1.by.1;1.end.able

Order < (arithmetic) 1.by.1;1.ends
ℕ\{1,2,3,...,n}

Moreover,
order < (arithmetic) 1.by.1;1.ends
ℕ

Not all orders of ℕ 1.by.1;1.end it
but
one 1.by.1;1.end.ing order < is enough
ℕ is 1.by.1;1.end.able
abbreviated as |ℕ| = ℵ₀

Order < (arithmetic)
_does not_ 1.by.1;1.end integers ℤ
'<' 1.by.1;0.ends ℤ = ⟨…,-2,-1,0,1,2,…⟩

However, a different order
'<₂' 1.by.1;1.ends ℤ = ⟨0,1,-1,2,-2,…⟩₂

z₁ <₂ z₂ :⇔ n(z₁) < n(z₂)
n(z) := |2⋅z|-⟦z>0⟧
⟦true⟧ := 1
⟦false⟧ := 0

one 1.by.1;1.end.ing order <₂ is enough
ℤ is 1.by.1;1.end.able
abbreviated |ℤ| = ℵ₀

|ℕ| = ℵ₀ = |ℤ| though ℕ ≠⊂ ℤ
ℕ is a same "sized" proper subset of ℤ

Same "sized" proper subsets distinguish
1.by.1;2.end.able (finite) sets from
not.1.by.1;2.end (infinite) sets.

>> We can say once and
>> know infinitely.many ways that
>> unit fraction ⅟n is not-in the first ℵ₀
>>
>> The first ℵ₀ is empty.
>> And empty is not ℵ₀-many.
>> Contradiction.
>
> If your logic says that
> there is no smallest unit fraction,
> then it is in contradiction with
> mathematics which even a Pisa-pupil
https://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment
> should master:
> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .

We can say once
1/n - 1/(n+1) = d_n > 0
and know it is true in infinitely.many ways
for each number in ℕ

⅟n - ⅟n⁺¹ > 0
⅟n⁺¹ < ⅟n
⅟n is not the first unit fraction

We can say once
⅟n is not the first unit fraction
and know it is true in infinitely.many ways
for each unit fraction in ⅟ℕ

> Note the universal quantifier which
> does not admit ***any*** exception.

We can say once
⅟n is not an exception to being not.first in ⅟ℕ
and know it is true in infinitely-many ways
for each unit fraction in ⅟ℕ

On 09.12.2023 17:32, Richard Damon wrote:
> On 12/9/23 11:11 AM, WM wrote:

>>  From Wikipedia, the free encyclopedia
>> "In mathematics, a partial order or total order < on a set X is said
>> to be dense if, for all x and y in X for which x < y, there is a z in
>> X such that x < z < y." Note for all. You could have looked up that
>> yourself.
>
> Yes, and the extension of that definition, for Dense at a point,
> restricts that x or y is that point. In this case, x, which is 0.

That is not an extension but a wrong application.

>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>>>> with a universal quantifier. No unit fraction can violate this
>>>> statement.
>>>
>>> So?, that isn't saying that there isn't a first point.

You are right.
>>
>> No. There is a first point.
>
> Then what is it?

It is dark.

> The distance is finite, but unboundedly small, thus no first element,
> which would become a bound.

That doesn't make it disappear. Mathematics needs precision, not the
belief that unboundedly small is same as vanishing.

Regards, WM

On 09.12.2023 20:56, Jim Burns wrote:
> On 12/9/2023 6:11 AM, WM wrote:
>> On 08.12.2023 21:37, Jim Burns wrote:

>>> We can follow that by saying once and
>>> knowing infinitely.many ways that
>>> n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
>>
>> Yes.
>> Nevertheless
>> ∀n ∈ ℕ_vis:
>> |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> Yes.
>
> |ℕ\{1,2,3,...,n}| = ℵ₀
> means

that almost all natnumbers are dark.

> |ℕ| = ℵ₀ = |ℤ|   though  ℕ ≠⊂ ℤ
> ℕ is a same "sized" proper subset of ℤ
>
> Same "sized" proper subsets

"Same size" requires a bijection which is only possible for visible
numbers. That is the reason why all infinite sets appear to have "same
size". Of course it is clear that the sizes of ℕ and ℤ differ by a factor 2.

>> If your logic says that
>> there is no smallest unit fraction,
>> then it is in contradiction with
>> mathematics which even a Pisa-pupil
> https://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment
>> should master:
>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
>
> We can say once
> 1/n - 1/(n+1) = d_n > 0
> and know it is true in infinitely.many ways
> for each number in ℕ

At least for every existing natnumber. If all numbers exist, then it is
true for all natnumbers.
>
> ⅟n - ⅟n⁺¹ > 0
> ⅟n⁺¹ < ⅟n
> ⅟n is not the first unit fraction
>
> We can say once
> ⅟n is not the first unit fraction

Yes, every unit fraction that you can identify.

> and know it is true in infinitely.many ways
> for each unit fraction in ⅟ℕ

It is true for a potential infinity of unit fractions.
>
>> Note the universal quantifier which
>> does not admit ***any*** exception.
>
> We can say once
> ⅟n is not an exception to being not.first in ⅟ℕ
> and know it is true in infinitely-many ways
> for each unit fraction in ⅟ℕ

A sequence of distinct points which starts after zero cannot start with
two or more not-distinct points at zero or afterwards.

Regards, WM

On 12/10/23 3:53 AM, WM wrote:
> On 09.12.2023 17:32, Richard Damon wrote:
>> On 12/9/23 11:11 AM, WM wrote:
>
>>>  From Wikipedia, the free encyclopedia
>>> "In mathematics, a partial order or total order < on a set X is said
>>> to be dense if, for all x and y in X for which x < y, there is a z in
>>> X such that x < z < y." Note for all. You could have looked up that
>>> yourself.
>>
>> Yes, and the extension of that definition, for Dense at a point,
>> restricts that x or y is that point. In this case, x, which is 0.
>
> That is not an extension but a wrong application.

So, you don't understand how logic works.

By your own arguement "Dark" numbers don't exist, as they are just
trying to use an "extention" of logic, arguing that you can't remove an
infinite set with finite operations.

>
>>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>>>>> with a universal quantifier. No unit fraction can violate this
>>>>> statement.
>>>>
>>>> So?, that isn't saying that there isn't a first point.
>
> You are right
>>>
>>> No. There is a first point.
>>
>> Then what is it?
>
> It is dark.

Nope, you just agreed that there wasn't one, so it can't be dark.

The proof shows that there can not be a "first" Natural Number, nor can
there be a first "describable" natural number, as we can always finitely
describe a smaller one.

>
>> The distance is finite, but unboundedly small, thus no first element,
>> which would become a bound.
>
> That doesn't make it disappear. Mathematics needs precision, not the
> belief that unboundedly small is same as vanishing.

Why? What is the difference between the terms?

You just don't seem to understand unbounded things.

>
> Regards, WM

On 12/10/2023 5:13 AM, WM wrote:
> On 09.12.2023 20:56, Jim Burns wrote:
>> On 12/9/2023 6:11 AM, WM wrote:
>>> On 08.12.2023 21:37, Jim Burns wrote:

>>>> We can follow that by saying once and
>>>> knowing infinitely.many ways that
>>>> n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
>>>
>>> Yes.
>>> Nevertheless
>>> ∀n ∈ ℕ_vis:
>>> |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>>
>> Yes.
>>
>> |ℕ\{1,2,3,...,n}| = ℵ₀
>> means
>
> that almost all natnumbers are dark.

That's not what ℵ₀ means.
ℵ₀.many are 1.by.1;1.end.able.

However,
what you (WM) mean is
∀n ∈ ℕ_vis:
almost all in ℕᵂᴹ\{1,2,3,...,n} are darkᵂᴹ

Elsewhere, you:
| With n also n+1 is visible.
| Message-ID: <439026df-be45-4503-a713-26a8303590ab@tha.de>
Date: Fri, 8 Dec 2023 16:21:35 +0100

ℕᵖⁱˢᵃ
(such as PISA students would use)
is the least upper bound of FISONs
∀B: ∀⟨0,…,n⟩ ⊆ B ⟹ ∀⟨0,…,n⟩ ⊆ ℕᵖⁱˢᵃ ⊆ ⊆ B

Because with n also n+1 is visibleᵂᴹ
no number in ℕᵖⁱˢᵃ is first.darkᵂᴹ

Because no number in ℕᵖⁱˢᵃ is first.darkᵂᴹ
no number in ℕᵖⁱˢᵃ is darkᵂᴹ

Also,
no visibleᵂᴹ number is not-in ℕᵖⁱˢᵃ

ℕᵖⁱˢᵃ = ℕ_vis

∀n ∈ ℕ_vis:
none in ℕᵖⁱˢᵃ\{1,2,3,...,n} are darkᵂᴹ

In contrast,
∀n ∈ ℕ_vis:
almost all in ℕᵂᴹ\{1,2,3,...,n} are darkᵂᴹ

It's safe to say ℕᵂᴹ ≠ ℕᵖⁱˢᵃ

>>> If your logic says that
>>> there is no smallest unit fraction,
>>> then it is in contradiction with
>>> mathematics which even a Pisa-pupil
>> https://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment
>>> should master:
>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .

Imagine how shocked I am to learn that
it is _you_ Wolfgang Mückenheim who is contrdicting
the mathematics a PISA student should master.

On 10.12.2023 14:07, Richard Damon wrote:
> On 12/10/23 3:53 AM, WM wrote:
>> On 09.12.2023 17:32, Richard Damon wrote:
>>> On 12/9/23 11:11 AM, WM wrote:
>>
>>>>  From Wikipedia, the free encyclopedia
>>>> "In mathematics, a partial order or total order < on a set X is said
>>>> to be dense if, for all x and y in X for which x < y, there is a z
>>>> in X such that x < z < y." Note for all. You could have looked up
>>>> that yourself.
>>>
>>> Yes, and the extension of that definition, for Dense at a point,
>>> restricts that x or y is that point. In this case, x, which is 0.
>>
>> That is not an extension but a wrong application.
>
> So, you don't understand how logic works.

Try to study the meaning of mathematical definitions. Also others here
have pointed out that you are wrong.
>>>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>>>>>> with a universal quantifier. No unit fraction can violate this
>>>>>> statement.
>>>>>
>>>>> So?, that isn't saying that there isn't a first point.
>>
>> You are right
>>>>
>>>> No. There is a first point.
>>>
>>> Then what is it?
>>
>> It is dark.
>
> Nope, you just agreed that there wasn't one, so it can't be dark.

You said: "that isn't saying that there isn't a first point". That means
that there can be a first point. Of course I agreed.

>>> The distance is finite, but unboundedly small, thus no first element,
>>> which would become a bound.
>>
>> That doesn't make it disappear. Mathematics needs precision, not the
>> belief that unboundedly small is same as vanishing.
>
> Why? What is the difference between the terms?

The difference is that between finite and not finite. All natural
numbers, with no exception, are finite.

Regards, WM
>

On 11.12.2023 05:50, Jim Burns wrote:
> On 12/10/2023 5:13 AM, WM wrote:

>>> |ℕ\{1,2,3,...,n}| = ℵ₀
>>> means
>>
>> that almost all natnumbers are dark.
>
> That's not what ℵ₀ means.

It is precisely what ℵ₀ means: After every natural number that you can
name, there are almost all natural numbers. That means that you cannot
reduce this amount of not used numbers. They cannot be used. They are
dark, most of them.

> ℵ₀.many are 1.by.1;1.end.able.
>
> However,
> what you (WM) mean is
> ∀n ∈ ℕ_vis:
> almost all in ℕᵂᴹ\{1,2,3,...,n} are darkᵂᴹ

Obviously. Try to reduce the set of not used numbers. Fail.
>
> Elsewhere, you:
> | With n also n+1 is visible.

And having ℵ₀ successors.
> |
>
>>>> If your logic says that
>>>> there is no smallest unit fraction,
>>>> then it is in contradiction with
>>>> mathematics which even a Pisa-pupil
>>> https://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment
>>>> should master:
>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
>
> Imagine how shocked I am to learn that
> it is _you_ Wolfgang Mückenheim who is contradicting
> the mathematics a PISA student should master.

It is this formula. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Most matheologians fail to interpret it correctly.

No quantifier magic will help you in this case:
If ℵ₀ points are left-hand side of every positive point, then ℵ₀ points
are left-hand side of all positive points. This proves that ℵ₀ points
are left-hand side of the interval (0, 1].

Regards, WM

On 12/11/23 5:48 AM, WM wrote:
> On 10.12.2023 14:07, Richard Damon wrote:
>> On 12/10/23 3:53 AM, WM wrote:
>>> On 09.12.2023 17:32, Richard Damon wrote:
>>>> On 12/9/23 11:11 AM, WM wrote:
>>>
>>>>>  From Wikipedia, the free encyclopedia
>>>>> "In mathematics, a partial order or total order < on a set X is
>>>>> said to be dense if, for all x and y in X for which x < y, there is
>>>>> a z in X such that x < z < y." Note for all. You could have looked
>>>>> up that yourself.
>>>>
>>>> Yes, and the extension of that definition, for Dense at a point,
>>>> restricts that x or y is that point. In this case, x, which is 0.
>>>
>>> That is not an extension but a wrong application.
>>
>> So, you don't understand how logic works.
>
> Try to study the meaning of mathematical definitions. Also others here
> have pointed out that you are wrong.

And which definitions are you trying to use?

>>>>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>>>>>>> with a universal quantifier. No unit fraction can violate this
>>>>>>> statement.
>>>>>>
>>>>>> So?, that isn't saying that there isn't a first point.
>>>
>>> You are right
>>>>>
>>>>> No. There is a first point.
>>>>
>>>> Then what is it?
>>>
>>> It is dark.
>>
>> Nope, you just agreed that there wasn't one, so it can't be dark.
>
> You said: "that isn't saying that there isn't a first point". That means
> that there can be a first point. Of course I agreed.

Not denying a falsehood doesn't make a sentence true, but can make it
irrelevant.

The is nothing in the statement ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 that
even implies that there is or has to be a "first point", so it is just
invlid logic to try to make it say that.

>
>>>> The distance is finite, but unboundedly small, thus no first
>>>> element, which would become a bound.
>>>
>>> That doesn't make it disappear. Mathematics needs precision, not the
>>> belief that unboundedly small is same as vanishing.
>>
>> Why? What is the difference between the terms?
>
> The difference is that between finite and not finite. All natural
> numbers, with no exception, are finite.

Yes, but unbounded, which is the bridge to infinite.

Every Natural Number is finite, but as a set, they are infinite.

>
>  Regards, WM
>>
>

On 12/11/2023 5:58 AM, WM wrote:
> On 11.12.2023 05:50, Jim Burns wrote:

>> [...]
>
> No quantifier magic will help you in this case:
> If ℵ₀ points are left-hand side of
> every positive point,
> then ℵ₀ points are left-hand side of
> all positive points.

.... "quantifier magic" in arithmetic which
PISA students must master.

Consider a use of "quantifier magic" for finites.

Four friends are going to have dinner together,
Nora, Sarah, Earl, and William.
I'm setting the table.

I take four glasses, set them at Nora's chair,
and go back into the kitchen to check the roast.
N:1234 S E W

While checking the roast, I tell you
| For each glass 1234
| there is a chair NSEW
| such that
| the chair faces the glass

A bit later,
I go back and distribute the glasses.
N:1 S:2 E:3 W:4

I tell you a second time
| For each glass 1234
| there is a chair NSEW
| such that
| the chair faces the glass

The "quantifier magic" for this table
is to not.conclude that,
the glasses could be N:1 S:2 E:3 W:4 and
| There _isn't_ a chair NSEW
| such that
| for each glass 1234
| the chair faces the glass

Where there are infinitely.many,
you also must not.conclude.

We know that
| For each unit.fraction ⅟n
| there is an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that
| S is bounded above by ⅟n: S ᣔ≤ ⅟n

From only that, you must not-conclude that
| There is an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that,
| for each unit.fraction ⅟n
| S is bounded above by ⅟n: S ᣔ≤ ⅟n

However,
you can conclude the negation of that
| There _isn't_ an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that,
| for each unit.fraction ⅟n
| S is bounded above by ⅟n: S ᣔ≤ ⅟n

| Assume otherwise.
| Assume
| there _is_ an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that,
| for each unit.fraction ⅟n
| S is bounded above by ⅟n: S ᣔ≤ ⅟n
| | However,
| S is not bounded by
| the _second.largest_ unit fraction ⅟mₛ⁺¹ in S
| S is not bounded by each unit.fraction
| Contradiction.

Therefore, there isn't.

> This proves that ℵ₀ points
> are left-hand side of the interval (0, 1].

Your claim here is equivalent to,
in the glasses-and-chairs example,
thinking of only N:1234 S E W
and
forgetting about N:1 S:2 E:3 W:4

On 12/11/2023 4:55 PM, Jim Burns wrote:
> On 12/11/2023 5:58 AM, WM wrote:
>> On 11.12.2023 05:50, Jim Burns wrote:

>>> [...]
>>
>> No quantifier magic will help you in this case:
>> If ℵ₀ points are left-hand side of
>> every positive point,
>> then ℵ₀ points are left-hand side of
>> all positive points.
>
> ... "quantifier magic" in arithmetic which
> PISA students must master.
>
> Consider a use of "quantifier magic" for finites.
>
> Four friends are going to have dinner together,
> Nora, Sarah, Earl, and William.
> I'm setting the table.
>
> I take four glasses, set them at Nora's chair,
> and go back into the kitchen to check the roast.
> N:1234 S E W
>
> While checking the roast, I tell you
> | For each glass 1234
> | there is a chair NSEW
> | such that
> | the chair faces the glass
>
> A bit later,
> I go back and distribute the glasses.
> N:1 S:2 E:3 W:4
>
> I tell you a second time
> | For each glass 1234
> | there is a chair NSEW
> | such that
> | the chair faces the glass
>
> The "quantifier magic" for this table
> is to not.conclude that,

Oops.
Better:
the glasses must be N:1234 S E W and
> | There _is_ a chair NSEW
> | such that
> | for each glass 1234
> | the chair faces the glass

I blame the editor I don't have.

>
> Where there are infinitely.many,
> you also must not.conclude.
>
> We know that
> | For each unit.fraction ⅟n
> | there is an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
> | such that
> | S is bounded above by ⅟n: S ᣔ≤ ⅟n
>
> From only that, you must not-conclude that
> | There is an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
> | such that,
> | for each unit.fraction ⅟n
> | S is bounded above by ⅟n: S ᣔ≤ ⅟n
>
>
> However,
> you can conclude the negation of that
> | There _isn't_ an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
> | such that,
> | for each unit.fraction ⅟n
> | S is bounded above by ⅟n: S ᣔ≤ ⅟n
>
>
> | Assume otherwise.
> | Assume
> | there _is_ an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
> | such that,
> | for each unit.fraction ⅟n
> | S is bounded above by ⅟n: S ᣔ≤ ⅟n
> |
> | However,
> | S is not bounded by
> | the _second.largest_ unit fraction ⅟mₛ⁺¹ in S
> | S is not bounded by each unit.fraction
> | Contradiction.
>
> Therefore, there isn't.
>
>> This proves that ℵ₀ points are
>> left-hand side of the interval (0, 1].
>
> Your claim here is equivalent to,
> in the glasses-and-chairs example,
> thinking of only N:1234 S E W
> and
> forgetting about N:1 S:2 E:3 W:4
>
>

Richard Damon schrieb am Montag, 11. Dezember 2023 um 13:19:35 UTC+1:
> On 12/11/23 5:48 AM, WM wrote:

> > Try to study the meaning of mathematical definitions. Also others here
> > have pointed out that you are wrong.
> And which definitions are you trying to use?

Those of text books like W. Mückenheim: "Mathematik für die ersten
Semester", 4th ed., De Gruyter, Berlin (2015) or W. Mückenheim: "Die
Mathematik des Unendlichen", Shaker-Verlag, Aachen 2006.

> The is nothing in the statement ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 that
> even implies that there is or has to be a "first point",

It is. You are only too blind to see it. Between all points there is are
gaps. And there is no point on the negative axis.

> > The difference is that between finite and not finite. All natural
> > numbers, with no exception, are finite.
> Yes, but unbounded, which is the bridge to infinite.

Bridge maybe, but no natnumber enters this bridge.
>
> Every Natural Number is finite, but as a set, they are infinite.

We talk about unit fractions, not about the set.

Regards, WM

>

On 11.12.2023 22:55, Jim Burns wrote:
> On 12/11/2023 5:58 AM, WM wrote:

>> This proves that ℵ₀ points are left-hand side of the interval (0, 1].
>
> Your claim here is equivalent to,
> in the glasses-and-chairs example,
> thinking of only N:1234 S E W
> and
> forgetting about N:1 S:2 E:3 W:4

I need not think about the individuals when proving that
∀x ∈ (0, 1]: NUF(x) = ℵo is wrong.
Even ∀x ∈ (0, 1]: NUF(x) > 0 is wrong because it claims that at least
one x is smaller then every x. This is wrong because the x itself is in
(0, 1].

For every x > 0 there are ℵo smaller points implies that there are ℵo
smaller points for every x > 0 and for the whole positive axis.

Regards, WM

On 12/12/23 6:57 AM, WM wrote:
> Richard Damon schrieb am Montag, 11. Dezember 2023 um 13:19:35 UTC+1:
>> On 12/11/23 5:48 AM, WM wrote:
>
>> > Try to study the meaning of mathematical definitions. Also others
>> here > have pointed out that you are wrong.
>> And which definitions are you trying to use?
>
> Those of text books like W. Mückenheim: "Mathematik für die ersten
> Semester", 4th ed., De Gruyter, Berlin (2015) or W. Mückenheim: "Die
> Mathematik des Unendlichen", Shaker-Verlag, Aachen 2006.

SO use it.

>
>> The is nothing in the statement ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 that
>> even implies that there is or has to be a "first point",
>
> It is. You are only too blind to see it. Between all points there is are
> gaps. And there is no point on the negative axis.

But that does not say there is a lowest value.

By your logic, Achillies can't pass the Tortoise.

>
>> > The difference is that between finite and not finite. All natural >
>> numbers, with no exception, are finite.
>> Yes, but unbounded, which is the bridge to infinite.
>
> Bridge maybe, but no natnumber enters this bridge.

The SET of Natural Numbers does, because the SET is unbounded, which
means that while every Natural Number is finite, the count of the
collection of them is infinite, and the set is Unbounded.

The Size of the Set of Natural Numbers is not a Natural Number.

>>
>> Every Natural Number is finite, but as a set, they are infinite.
>
> We talk about unit fractions, not about the set.

And the Set of Unit Fractions directly corresponds to the Set of Natural
Numbers (being that each element of the Set of Unit Fractions is the
reciprical of a Natural Number)

There is not least Unit Fraction, because that would imply a maximum
Natural Number, which there isn't one.

Thus, for Unit Fractions, there is no least element, as the set is
Unbounded, just as there is not last Natural Number as that set is
Unbounded in the same way.

>
> Regards, WM
>
>>
>

On 12.12.2023 13:21, Richard Damon wrote:
> On 12/12/23 6:57 AM, WM wrote:

> There is not least Unit Fraction, because that would imply a maximum
> Natural Number, which there isn't one.

There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
the same point. This proves a first one.

For natnumbers there is a corresponding last one. They are dark like the
smallest unit fractions.
>
> Thus, for Unit Fractions, there is no least element, as the set is
> Unbounded, just as there is not last Natural Number as that set is
> Unbounded in the same way.

When you step down from omega, you get to a definable natural number n.
Between n and omega however, there are ℵo natnumbers. Why can't you step
down to any of them? Because they are dark.

Regards, WM

On 12/12/23 5:12 PM, WM wrote:
> On 12.12.2023 13:21, Richard Damon wrote:
>> On 12/12/23 6:57 AM, WM wrote:
>
>> There is not least Unit Fraction, because that would imply a maximum
>> Natural Number, which there isn't one.
>
> There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
> the same point. This proves a first one.

Which "NUF(x)" are you trying to use?

NUF(x) being the number of unit fractions below x is 0 for x <=0 and
infinite for x > 0, so has a major discontinuity at 0

NUF(x) being the number of unit fractions above x isn't 0 for x == 0,
but infinite.

>
> For natnumbers there is a corresponding last one. They are dark like the
> smallest unit fractions.

Which is the "Last" Natural Number?

>>
>> Thus, for Unit Fractions, there is no least element, as the set is
>> Unbounded, just as there is not last Natural Number as that set is
>> Unbounded in the same way.
>
> When you step down from omega, you get to a definable natural number n.
> Between n and omega however, there are ℵo natnumbers. Why can't you step
> down to any of them? Because they are dark.

You CAN'T "Step Down" from omega. It is a different class of number than
the natural numbers, and not "next" to any of them.

So, it seems your logic is based on the proverbial Fairy Dust Powered
Unicorns of mythology and not applicable to real logic.

>
> Regards, WM
>

On 13.12.2023 00:46, Richard Damon wrote:
> On 12/12/23 5:12 PM, WM wrote:
>> On 12.12.2023 13:21, Richard Damon wrote:
>>> On 12/12/23 6:57 AM, WM wrote:
>>
>>> There is not least Unit Fraction, because that would imply a maximum
>>> Natural Number, which there isn't one.
>>
>> There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
>> the same point. This proves a first one.
>
> Which "NUF(x)" are you trying to use?
>
> NUF(x) being the number of unit fractions below x is 0 for x <=0 and
> infinite for x > 0, so has a major discontinuity at 0

Impossible. Contradicting mathematics: Never two or more unit fractions
sit at the same place.
>
> NUF(x) being the number of unit fractions above x

No, below x. NUF(x) is the number between 0 and x.

>> For natnumbers there is a corresponding last one. They are dark like
>> the smallest unit fractions.
>
> Which is the "Last" Natural Number?

Dark. Impossiblr to discern.
>
>>>
>>> Thus, for Unit Fractions, there is no least element, as the set is
>>> Unbounded, just as there is not last Natural Number as that set is
>>> Unbounded in the same way.
>>
>> When you step down from omega, you get to a definable natural number
>> n. Between n and omega however, there are ℵo natnumbers. Why can't you
>> step down to any of them? Because they are dark.
>
>
> You CAN'T "Step Down" from omega.

It can be done. It is a general rule, forced by the well-foundedness
(see 2.12.8) of the sequence of ordinal numbers, that every strictly
decreasing sequence of ordinal numbers reaches its smallest element
after a finite number of steps. Since limit ordinals have no direct
predecessors, we have to jump down from them to some predecessor.
Ask for help elsewhere if you don't trust me.

Regards, WM

On 12/13/23 7:01 AM, WM wrote:
> On 13.12.2023 00:46, Richard Damon wrote:
>> On 12/12/23 5:12 PM, WM wrote:
>>> On 12.12.2023 13:21, Richard Damon wrote:
>>>> On 12/12/23 6:57 AM, WM wrote:
>>>
>>>> There is not least Unit Fraction, because that would imply a maximum
>>>> Natural Number, which there isn't one.
>>>
>>> There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
>>> the same point. This proves a first one.
>>
>> Which "NUF(x)" are you trying to use?
>>
>> NUF(x) being the number of unit fractions below x is 0 for x <=0 and
>> infinite for x > 0, so has a major discontinuity at 0
>
> Impossible. Contradicting mathematics: Never two or more unit fractions
> sit at the same place.

Right, no two at the same space, but there "density" becomes infinite,
and thus we get the discontinuity,

You just don't understand what you need to do to work with infinite sets.

>>
>> NUF(x) being the number of unit fractions above x
>
> No, below x. NUF(x) is the number between 0 and x.

So, it is INFINITE for all x > 0

If you claim otherwise, choose an x > 0, give the finite value, and list
all the unit fractions.

I can add more, thus your claim is just a lie

>
>>> For natnumbers there is a corresponding last one. They are dark like
>>> the smallest unit fractions.
>>
>> Which is the "Last" Natural Number?
>
> Dark. Impossiblr to discern.

So, what is the last discernable number?

>>
>>>>
>>>> Thus, for Unit Fractions, there is no least element, as the set is
>>>> Unbounded, just as there is not last Natural Number as that set is
>>>> Unbounded in the same way.
>>>
>>> When you step down from omega, you get to a definable natural number
>>> n. Between n and omega however, there are ℵo natnumbers. Why can't
>>> you step down to any of them? Because they are dark.
>>
>>
>> You CAN'T "Step Down" from omega.
>
> It can be done. It is a general rule, forced by the well-foundedness
> (see 2.12.8) of the sequence of ordinal numbers, that every strictly
> decreasing sequence of ordinal numbers reaches its smallest element
> after a finite number of steps. Since limit ordinals have no direct
> predecessors, we have to jump down from them to some predecessor.
> Ask for help elsewhere if you don't trust me.

What is 2.12.8? You are referencing random unfinded material. I don't
have your library memorized.

Your arguemnt is assuming its conclusion. Omega isn't a Natural Number,
so you can't use arguements about Natural Numbers to talk about it.

>
> Regards, WM
>

On 13.12.2023 13:36, Richard Damon wrote:
> On 12/13/23 7:01 AM, WM wrote:
>> On 13.12.2023 00:46, Richard Damon wrote:

>>> NUF(x) being the number of unit fractions below x is 0 for x <=0 and
>>> infinite for x > 0, so has a major discontinuity at 0
>>
>> Impossible. Contradicting mathematics: Never two or more unit
>> fractions sit at the same place.
>
> Right, no two at the same space, but there "density" becomes infinite,
> and thus we get the discontinuity,

No. All distances remain finite.
>
> You just don't understand what you need to do to work with infinite sets.

I understand that you are believing in nonsense,not mathematics.
>
>>>
>>> NUF(x) being the number of unit fractions above x
>>
>> No, below x. NUF(x) is the number between 0 and x.
>
> So, it is INFINITE for all x > 0

Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
>
> So, what is the last discernable number?

The set is potentially infinite. But all its are instances are finite
and are followed by the infinity of dark numbers.
>
>>>
>>> You CAN'T "Step Down" from omega.
>>
>> It can be done. It is a general rule, forced by the well-foundedness
>> (see 2.12.8) of the sequence of ordinal numbers, that every strictly
>> decreasing sequence of ordinal numbers reaches its smallest element
>> after a finite number of steps. Since limit ordinals have no direct
>> predecessors, we have to jump down from them to some predecessor.
>> Ask for help elsewhere if you don't trust me.
>
> What is 2.12.8?

It is the axiom of foundation in
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

> Your arguemnt is assuming its conclusion. Omega isn't a Natural Number,
> so you can't use arguements about Natural Numbers to talk about it.

For steps down from omega see:
https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228

Regards, WM

On 12/13/23 11:51 AM, WM wrote:
> On 13.12.2023 13:36, Richard Damon wrote:
>> On 12/13/23 7:01 AM, WM wrote:
>>> On 13.12.2023 00:46, Richard Damon wrote:
>
>>>> NUF(x) being the number of unit fractions below x is 0 for x <=0 and
>>>> infinite for x > 0, so has a major discontinuity at 0
>>>
>>> Impossible. Contradicting mathematics: Never two or more unit
>>> fractions sit at the same place.
>>
>> Right, no two at the same space, but there "density" becomes infinite,
>> and thus we get the discontinuity,
>
> No. All distances remain finite.
>>
>> You just don't understand what you need to do to work with infinite sets.
>
> I understand that you are believing in nonsense,not mathematics.
>>
>>>>
>>>> NUF(x) being the number of unit fractions above x
>>>
>>> No, below x. NUF(x) is the number between 0 and x.
>>
>> So, it is INFINITE for all x > 0
>
> Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n

Then your "mathematics" are broken.

Give me a finite value of NUF(x) for some definite x.

If the only values you can try to put in for x to get a finite value,
your NUF is undefined, since functions take their parameters as
individual values, and "dark numbers" can't be used individually.

So, you are just admitting that your logic is based on broken mathematics.

>>
>> So, what is the last discernable number?
>
> The set is potentially infinite. But all its are instances are finite
> and are followed by the infinity of dark numbers.

Sets have FIXED membership, so you are talking nonsense.

You might have a SET of sets of partial ℕ_def based on different n's,
but none of them are actually what you are defining ℕ_def to be, the SET
of all definable Natural Numbers.

>>
>>>>
>>>> You CAN'T "Step Down" from omega.
>>>
>>> It can be done. It is a general rule, forced by the well-foundedness
>>> (see 2.12.8) of the sequence of ordinal numbers, that every strictly
>>> decreasing sequence of ordinal numbers reaches its smallest element
>>> after a finite number of steps. Since limit ordinals have no direct
>>> predecessors, we have to jump down from them to some predecessor.
>>> Ask for help elsewhere if you don't trust me.
>>
>> What is 2.12.8?
>
> It is the axiom of foundation in
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
>
>> Your arguemnt is assuming its conclusion. Omega isn't a Natural
>> Number, so you can't use arguements about Natural Numbers to talk
>> about it.
>
> For steps down from omega see:
> https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
>
> Regards, WM
>
>

So, you are just ASSUMING a falsehood

Great work.

If X isn't a set (like it is a Natural Number) the S intersect X isn't
defined.

Also, we are talking about a MAXIMAL Natural Number, not a minimal.

And since Omega isn't in the set of Natural Numbers, you can't talk
about jumping from Omega to a Natural Number by this logic.

On 12/13/23 11:51 AM, WM wrote:
> On 13.12.2023 13:36, Richard Damon wrote:
>> On 12/13/23 7:01 AM, WM wrote:
>>> On 13.12.2023 00:46, Richard Damon wrote:
>
>>>> NUF(x) being the number of unit fractions below x is 0 for x <=0 and
>>>> infinite for x > 0, so has a major discontinuity at 0
>>>
>>> Impossible. Contradicting mathematics: Never two or more unit
>>> fractions sit at the same place.
>>
>> Right, no two at the same space, but there "density" becomes infinite,
>> and thus we get the discontinuity,
>
> No. All distances remain finite.
>>
>> You just don't understand what you need to do to work with infinite sets.
>
> I understand that you are believing in nonsense,not mathematics.
>>
>>>>
>>>> NUF(x) being the number of unit fractions above x
>>>
>>> No, below x. NUF(x) is the number between 0 and x.
>>
>> So, it is INFINITE for all x > 0
>
> Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n

Which says nothing about NUF(x)

Your

>>
>> So, what is the last discernable number?
>
> The set is potentially infinite. But all its are instances are finite
> and are followed by the infinity of dark numbers.

So, it doesn't exist. A "Last Number" is not a "Set" but a specific number.

>>
>>>>
>>>> You CAN'T "Step Down" from omega.
>>>
>>> It can be done. It is a general rule, forced by the well-foundedness
>>> (see 2.12.8) of the sequence of ordinal numbers, that every strictly
>>> decreasing sequence of ordinal numbers reaches its smallest element
>>> after a finite number of steps. Since limit ordinals have no direct
>>> predecessors, we have to jump down from them to some predecessor.
>>> Ask for help elsewhere if you don't trust me.
>>
>> What is 2.12.8?
>
> It is the axiom of foundation in
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
>
>> Your arguemnt is assuming its conclusion. Omega isn't a Natural
>> Number, so you can't use arguements about Natural Numbers to talk
>> about it.
>
> For steps down from omega see:
> https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
>
> Regards, WM
>
>

But they aren't "Natural Numbers".

Yes, we can have some "Transfinite" numbers BEYOND the Natural Numbers,
but they are not members of the Natural Numbers and do not obey all the
properties of them.

So they are not "Dark", but just "Transfinite" and in a very real sense
definable.

There is just a discontinuity in the number systems between the Finite
Numbers, and the "Infinite" numbers, and the "Countable Infinites" sit
on that discontinuity. Each member is finite, but they form an infinite set.

On 15.12.2023 02:55, Richard Damon wrote:
> On 12/13/23 11:51 AM, WM wrote:

>>>> No, below x. NUF(x) is the number between 0 and x.
>>>
>>> So, it is INFINITE for all x > 0
>>
>> Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
>
> Which says nothing about NUF(x)

It says that NUF cannot increase by more than 1 without being constant
afterwards o.ver a finite distance
>

>>> Your arguemnt is assuming its conclusion. Omega isn't a Natural
>>> Number, so you can't use arguements about Natural Numbers to talk
>>> about it.
>>
>> For steps down from omega see:
>> https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
>>
>
> But they aren't "Natural Numbers".

Every step down from omega ends at a visible natural number.
>
> Yes, we can have some "Transfinite" numbers BEYOND the Natural Numbers,
> but they are not members of the Natural Numbers and do not obey all the
> properties of them.

Every number below omega is a natural number.

Regards, WM

On 15.12.2023 02:54, Richard Damon wrote:
> On 12/13/23 11:51 AM, WM wrote:

>>> Your arguemnt is assuming its conclusion. Omega isn't a Natural
>>> Number, so you can't use arguements about Natural Numbers to talk
>>> about it.
>>
>> For steps down from omega see:
>> https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
>
> So, you are just ASSUMING a falsehood

No, I have shown your sentence wrong:
>
> And since Omega isn't in the set of Natural Numbers, you can't talk
> about jumping from Omega to a Natural Number by this logic.

You can't understand logic. EOD.

Regards, WM

On 12/15/23 12:41 PM, WM wrote:
> On 15.12.2023 02:54, Richard Damon wrote:
>> On 12/13/23 11:51 AM, WM wrote:
>
>>>> Your arguemnt is assuming its conclusion. Omega isn't a Natural
>>>> Number, so you can't use arguements about Natural Numbers to talk
>>>> about it.
>>>
>>> For steps down from omega see:
>>> https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
>>
>> So, you are just ASSUMING a falsehood
>
> No, I have shown your sentence wrong:

No, you THINK you have because you don't understand the basics of
infinities.

>>
>> And since Omega isn't in the set of Natural Numbers, you can't talk
>> about jumping from Omega to a Natural Number by this logic.
>
> You can't understand logic. EOD.

So, you think that a step down followed by a step back up can lead you
to a very different place on a straight line?

If a step down from omega was a Natural Number n, then the step back up
from n should be omega again, but it is the Natual Number n+1.

Note, if you accept mathematics that makes omega+1 not the same as
omega, then you are in transfinite mathematics, and omega-1 is still a
transfinte number, not a Natural Number.

Otherwise x-1+1 = x is no longer true.

You need to try to CONSISTANTLY define your mathematics you are using.

Either omega-1 is another transfinite number, or it just doesn't exist
(just like in the domain of Natural Numbers 0-1 doesn't exist, you need
to extend you system to the integers)

A Transfinite system that allows omega + Natural, doesn't define the
value of omega-1 just like the Natural Numbers don't have a predecessor
to 0 to express 0 - 1.

>
> Regards, WM
>

On 12/15/23 7:21 AM, WM wrote:
> On 15.12.2023 02:55, Richard Damon wrote:
>> On 12/13/23 11:51 AM, WM wrote:
>
>>>>> No, below x. NUF(x) is the number between 0 and x.
>>>>
>>>> So, it is INFINITE for all x > 0
>>>
>>> Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
>>
>> Which says nothing about NUF(x)
>
> It says that NUF cannot increase by more than 1 without being constant
> afterwards o.ver a finite distance

Yes, NUF can not increase by more than 1 without bein constant for a
finite period, but that finite period is unboundedly small, and the in
any finite interval at the be

>>
>
>>>> Your arguemnt is assuming its conclusion. Omega isn't a Natural
>>>> Number, so you can't use arguements about Natural Numbers to talk
>>>> about it.
>>>
>>> For steps down from omega see:
>>> https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
>>>
>>
>> But they aren't "Natural Numbers".
>
> Every step down from omega ends at a visible natural number.

But then the step up needs to get back to omega, but it doesn't

>>
>> Yes, we can have some "Transfinite" numbers BEYOND the Natural
>> Numbers, but they are not members of the Natural Numbers and do not
>> obey all the properties of them.
>
> Every number below omega is a natural number.
>
> Regards, WM
>

Le 16/12/2023 à 02:05, Richard Damon a écrit :
> On 12/15/23 7:21 AM, WM wrote:

> Yes, NUF can not increase by more than 1 without being constant for a
> finite period,

Fine that you accept mathematics, after all.

> but that finite period is unboundedly small,

It is finite, greater than zero, so Cantor would say it has uncountably
many points. I say: It has at least one point.
>> Every step down from omega ends at a visible natural number.
>
> But then the step up needs to get back to omega, but it doesn't

You are mistaken. Ask a set theorist whom you trust.

Regards, WM