On 23.11.2023 06:38, Daniel Pehoushek wrote:
> On Wednesday, November 22, 2023 at 8:38:27 PM UTC-5, Fritz Feldhase wrote:
>> On Tuesday, November 21, 2023 at 8:19:25 PM UTC+1, WM wrote:
>>
>>> My game tells us
>>
> why sci dot logic is dead.

The reason is the refusal to argue step by step.
Can you demonstrate a difference between Cantor's enumeration of the
rational numbers or the algebraic numbers and a supertask?

Can you demonstrate where my "game like billiards" fails? Which step is
wrong?

(1) My game tells us that the number of positions of A(n)
is constant for every n.

(2) Further it tells us that B(n) is a part of
A(n).

(3) it tells that all A(n) contain all fractions at not
indexed positions.

Which step is mistaken?

Regards, WM

Jim Burns schrieb am Montag, 27. November 2023 um 02:01:33 UTC+1:
> On 11/26/2023 3:20 PM, Transfinity wrote:
> > Jim Burns schrieb am Samstag,
> > 25. November 2023 um 15:52:43 UTC+1:

> > X contains only those natnumbers which,
> > when added to X,
> > leave |ℕ \ X| = ℵo.
>
> X contains only those natnumbers which,
> already being in X,
> cannot be added to X

X is a set function. It can be increased. But never |ℕ \ X| < ℵo will be
accomplished by adding individually definable numbers. Therefore
contains dark numbers: |ℕ \ ℕ| = 0.

Regards, WM

On 11/27/2023 6:52 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 27. November 2023 um 02:01:33 UTC+1:
Message-ID: <877498d2-ee9a-4deb-a806-14e152c11154@att.net>
Newsgroups: sci.math
Subject: Re: The Principle of Mathematical Induction versus Infinity
Date: Sun, 26 Nov 2023 20:01:24 -0500
>> On 11/26/2023 3:20 PM, Transfinity wrote:

>>> X contains only those natnumbers which,
>>> when added to X,
>>> leave |ℕ \ X| = ℵo.
>>
>> X contains only those natnumbers which,
>> already being in X,
>> cannot be added to X
>
> X is a set function.

X is a collection.

You, upthread in sci.math
>>> Collect
>>> all natnumbers n with the above property
>>> into a collection X

Collecting those is a supertask, not a task.
Describing collecting those is a task.

Describe the collection X of all n such that
|ℕ \ {1, 2, 3, ..., n}| = ℵo

∀S: ∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ S ⟹
∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ X ⊆ S

Describing complete.

> It can be increased.

A set increased from X
is not X

X described upthread in sci.math
cannot be increased and also be X

If you want to describe
infinitely-many different sets
which result from
adding all the visible numbers one by one
to the empty set,
you can describe them.

For each set so described,
the visible successor to its last element
is not-in that set.
That set is not X, which all are in.

You can describe them.
Each set so described is not X

> But never |ℕ \ X| < ℵo will be accomplished
> by adding individually definable numbers.

We who are not Chuck Norris can't do supertasks.
That's a supertask.

> Therefore
> contains dark numbers: |ℕ \ ℕ| = 0.

For each visibleᴶᴮ number n
ordered ⟨0,…,n⟩ exists such that
0‖n exists first‖last in ⟨0,…,n⟩ and,
for each split Fᣔ<ᣔH of ⟨0,…,n⟩
i‖i⁺¹ exists last‖first in F‖H

i⁺¹ is non-0 non-doppelgänger non-final

Ridiculously-large visibleᴶᴮ numbers exist.
Do any darkᵂᴹ and visibleᴶᴮ numbers exist?

Each ⟨0,…,n⟩ is not X because
visibleᴶᴮ n⁺¹ is not in ⟨0,…,n⟩

If darkᴶᴮ d ¬∃⟨0,…,d⟩ is in X
X\{d} also contains all visibleᴶᴮ numbers
Why isn't X\{d} the set of all visibleᴶᴮ ?

Let Xᴶᴮ be the set of
all and only visibleᴶᴮ numbers.
For each n in Xᴶᴮ
|Xᴶᴮ\⟨0,…,n⟩| = |Xᴶᴮ|

For ℕ = Xᴶᴮ
and ℵ₀ the "size" of any 1×1 1.ended
Xᴶᴮ is the set of all and only
numbers n such that |ℕ\⟨0,…,n⟩| = ℵ₀
No darkᴶᴮ numbers are in ℕ

On 11/27/2023 6:52 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 27. November 2023 um 02:01:33 UTC+1:
Message-ID: <877498d2-ee9a-4deb-a806-14e152c11154@att.net>
Newsgroups: sci.math
Subject: Re: The Principle of Mathematical Induction versus Infinity
Date: Sun, 26 Nov 2023 20:01:24 -0500
>> On 11/26/2023 3:20 PM, Transfinity wrote:

>>> X contains only those natnumbers which,
>>> when added to X,
>>> leave |ℕ \ X| = ℵo.
>>
>> X contains only those natnumbers which,
>> already being in X,
>> cannot be added to X
>
> X is a set function.

X is a collection.

You, upthread in sci.math
>>> Collect
>>> all natnumbers n with the above property
>>> into a collection X

Collecting those is a supertask, not a task.
Describing collecting those is a task.

Describe the collection X of all n such that
|ℕ \ {1, 2, 3, ..., n}| = ℵo

∀S: ∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ S ⟹
∀n:|ℕ\⟨1,…,n⟩|=ℵ₀ ⇒ n ∈ X ⊆ S

Describing complete.

> It can be increased.

A set increased from X
is not X

X described upthread in sci.math
cannot be increased and also be X

If you want to describe
infinitely-many different sets
which result from
adding all the visible numbers one by one
to the empty set,
you can describe them.

For each set so described,
the visible successor to its last element
is not-in that set.
That set is not X, which all are in.

You can describe them.
Each set so described is not X

> But never |ℕ \ X| < ℵo will be accomplished
> by adding individually definable numbers.

We who are not Chuck Norris can't do supertasks.
That's a supertask.

> Therefore
> contains dark numbers: |ℕ \ ℕ| = 0.

For each visibleᴶᴮ number n
ordered ⟨0,…,n⟩ exists such that
0‖n exists first‖last in ⟨0,…,n⟩ and,
for each split Fᣔ<ᣔH of ⟨0,…,n⟩
i‖i⁺¹ exists last‖first in F‖H

i⁺¹ is non-0 non-doppelgänger non-final

Ridiculously-large visibleᴶᴮ numbers exist.
Do any darkᵂᴹ and visibleᴶᴮ numbers exist?

Each ⟨0,…,n⟩ is not X because
visibleᴶᴮ n⁺¹ is not in ⟨0,…,n⟩

If darkᴶᴮ d ¬∃⟨0,…,d⟩ is in X
X\{d} also contains all visibleᴶᴮ numbers
Why isn't X\{d} the set of all visibleᴶᴮ ?

Let Xᴶᴮ be the set of
all and only visibleᴶᴮ numbers.
For each n in Xᴶᴮ
|Xᴶᴮ\⟨0,…,n⟩| = |Xᴶᴮ|

For ℕ = Xᴶᴮ
and ℵ₀ the "size" of any 1×1 1.ended
Xᴶᴮ is the set of all and only
numbers n such that |ℕ\⟨0,…,n⟩| = ℵ₀
No darkᴶᴮ numbers are in ℕ

Fritz Feldhase schrieb am Montag, 27. November 2023 um 19:31:37 UTC+1:
> On Monday, November 27, 2023 at 12:53:01 PM UTC+1, WM wrote:
> >
> > X is a set function.
> Nope. You defined X as a "collection" (set, class). [Hint: In ZF(C)
everything is a _set_.]

A collection like the definable natnumbers is a set function.
>
> > It can be increased.
>
> Nope. Sets ("collections") do neither grow nor shrink.

X is a set function.

Regards, WM

Jim Burns schrieb am Montag, 27. November 2023 um 19:05:05 UTC+1:
> On 11/27/2023 6:52 AM, WM wrote:
> > Jim Burns schrieb am Montag,

> >>> X contains only those natnumbers which,
> >>> when added to X,
> >>> leave |ℕ \ X| = ℵo.
> >>
> >> X contains only those natnumbers which,
> >> already being in X,
> >> cannot be added to X
> >
> > X is a set function.
> X is a collection.

A collection is a set function.
>
> You, upthread in sci.math
> >>> Collect
> >>> all natnumbers n with the above property
> >>> into a collection X
>
> Collecting those is a supertask, not a task.

No, there are only less than ℵo.

> > It can be increased.
>
> A set increased from X
> is not X

X is a function. The arguments has been dropped.

> > But never |ℕ \ X| < ℵo will be accomplished
> > by adding individually definable numbers.
> We who are not Chuck Norris can't do supertasks.

Collect as many as you can. Never |ℕ \ X| < ℵo will be accomplished. But
|ℕ \ ℕ| = ℵo ==> ℕ contains dark numbers.

Regards, WM

Regards, WM

On 11/27/2023 2:36 PM, WM wrote:
> Jim Burns schrieb am Montag,
> 27. November 2023 um 19:05:05 UTC+1:
Newsgroups: sci.logic
Subject: Re: The Principle of Mathematical Induction versus Infinity
Date: Mon, 27 Nov 2023 13:05:02 -0500

>> You, upthread in sci.math
>>>>> Collect
>>>>> all natnumbers n with the above property
>>>>> into a collection X
>>
>> Collecting those is a supertask, not a task.
>
> No, there are only less than ℵo.

Numbers n such that |ℕ\⟨1,…,n⟩|=ℵ₀
are 1×1 1.ended

Each 1×1 1.ended is the same "size" ℵ₀

>>> But never |ℕ \ X| < ℵo will be accomplished
>>> by adding individually definable numbers.
>>
>> We who are not Chuck Norris can't do supertasks.
>
> Collect as many as you can.

Fewer than 10^(10^(10^(10^(10))))

Infinitely fewer than we can reason about.

> Never |ℕ \ X| < ℵo will be accomplished.
> But |ℕ \ ℕ| = ℵo ==>
> ℕ contains dark numbers.

The visibleᴶᴮ numbers are 1×1 1.ended

Each 1×1 1.ended is the same "size" ℵ₀

Each split of 1×1 1.ended is 3.ended.
One part is finite 1×1 2.ended
The other part is ℵ₀-many 1×1 1.ended,
the same "size" as its 1×1 1.ended superset.

On 27.11.2023 22:11, Jim Burns wrote:
> On 11/27/2023 2:36 PM, WM wrote:
> Each 1×1 1.ended is the same "size" ℵ₀

ℵ₀ is not a size but a shorthand for actual infinity. |ℕ| = ℵ₀ as well
as |ℚ| = 2|ℕ|^2 + 1 = ℵ₀ .

>> Collect as many as you can.
>
> Fewer than 10^(10^(10^(10^(10))))
>
> Infinitely fewer than we can reason about.

If you can give a number like above, then all smaller numbers are
automatically defined too. All last numbers of a FISON that you can
reason about and all their predecessors belong to X. You cannot reason
about numbers as individuals which are in the difference |ℕ \ X|, can you?

Try it!

Regards, WM

Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:

> Define X

If you can identify a natnumber, then this number and all smaller
numbers are automatically elements of X. (All last numbers of FISONs
that you can reason about and all their predecessors belong to X.) You
cannot reason about natnumbers as individuals which are in the
difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it
belongs to X. Nevertheless the difference remains actually infinite: ℵo
natnumbers. Therefore they are dark.

Regards, WM

On 27.11.2023 19:05, Jim Burns wrote:
> On 11/27/2023 6:52 AM, WM wrote:

>> It can be increased.
>
> A set increased from X
> is not X

If you can identify a natnumber, then this number and all smaller
numbers are automatically elements of X. (All last numbers of FISONs
that you can reason about and all their predecessors belong to X.) You
cannot reason about natnumbers as individuals which are in the
difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it
belongs to X. Nevertheless the difference remains actually infinite: ℵo
natnumbers. Therefore they are dark.

Regards, WM

On 28.11.2023 15:22, Fritz Feldhase wrote:
> On Tuesday, November 28, 2023 at 10:41:32 AM UTC+1, WM wrote:
>> Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
>>>
>>> Define X
>
> Ich hatte geschrieben:

That is irrelevant. Relevant is the definition of X.
>
> | 1. Define the notion /set function/

A set function is a sequence of sets S(n) depending on an argument n.
Example: The sequence of FISONs (F(n)) with F(n) = {1, 2, 3, ..., n}.

> | 2. Define X

X is the sequence and union of FISONs that can be defined as
individuals. You cannot reason about natnumbers as individuals which are
in the difference |ℕ \ X| = ℵo. As soon as you identify a natnumber n
there, its FISON belongs to X. Nevertheless the difference remains
actually infinite: ℵo natnumbers. Therefore they are dark.

> | 3. Show that X is a set function (based on the definitions 1 and 2)
>
Obvious. See above.

Regards, WM

On 11/28/2023 4:33 AM, WM wrote:
> On 27.11.2023 22:11, Jim Burns wrote:

>> Each 1×1 1.ended is the same "size" ℵ₀
>
> ℵ₀ is not a size
> but a shorthand for actual infinity.

ℵ₀ is the smallest infinite cardinality

Cardinality is one of
several different kinds of "size".
The different kinds of "size" have
different properties.

If
one-to-one f: A → B exists
∀x′≠x: f(x′)≠f(x)
then
|A| ≤ |B|
and the _cardinality_ of B is
at least as "large" as
the cardinality of A

For a finite set, a finite order exists.
For an infinite set, no finite order exists.

<′ is a finite order of set A ==
<′ is 1×1 2.ended

<′ is 2.ended ==
first‖last exist in A

<′ is 1×1 ==
for each split Fᣔ<ᣔH of A
last‖first exist in F‖H

<″ is an infinite order of set B ==
<″ is not 1×1 or not 2.ended

For each set C
either
each order of C is finite
and C is finite
or
each order of C is infinite
and C is infinite

> ℵ₀ is not a size
> but a shorthand for actual infinity.
> |ℕ| = ℵ₀ as well
> as |ℚ| = 2|ℕ|^2 + 1 = ℵ₀ .

ℵ₀ is the smallest infinite cardinality

ℕ is infinite.
The standard order 0<1<2<... of ℕ is
an infinite order, not a finite order,
since it is 1×1 1.ended
No order of ℕ is finite, meaning 1x1 2.ended

|ℚ⁺| ≤ |ℕ⁺|
because
one-to-one k: ℚ⁺ → ℕ⁺ exists
k(m/n) = m+(m+n-1)(m+n-2)/2
for m/n in lowest terms

∀m′/n′≠m/n: k(m′/n′)≠k(m/n)

|ℕ⁺| ≤ |ℚ⁺|
because
one-to-one id: ℕ⁺ → ℚ⁺ exists
id(k) = k

|ℚ⁺| = |ℕ⁺|
because
https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem

|ℚ⁺| = |ℕ⁺|
because
arithmetic says,
for m′/n′ ≠ m/n
m′+(m′+n′-1)(m′+n′-2)/2 ≠
m+(m+n-1)(m+n-2)/2

>>> Collect as many as you can.
>>
>> Fewer than 10^(10^(10^(10^(10))))
>>
>> Infinitely fewer than we can reason about.
>
> If you can give a number like above,
> then all smaller numbers are
> automatically defined too.

If
all smaller numbers are automatically defined
because we've described defining them,
and not because we've actually defined them,
then
all larger numbers are automatically defined
because we've described defining them,
even though we haven't actually defined them

> All last numbers of a FISON that
> you can reason about and
> all their predecessors

"All FISON-ends" automatically includes
"all predecessors of all FISON-ends".
Each predecessor ends a FISON, a different one.

> all their predecessors belong to X.
> You cannot reason about numbers as individuals
> which are in the difference |ℕ \ X|,
> can you?

You tell me ∀⟨0,…,n⟩ ⊆ X

Define
ℕ(X) = ⋂{S ⊆ X| ∀⟨0,…,n⟩ ⊆ S}

ℕ(X) contains all FISON-ends and only FISON-ends.
∀A: ∀⟨0,…,n⟩ ⊆ A ⟹ ∀⟨0,…,n⟩ ⊆ ℕ(X) ⊆ A

Consider Y such that ∀⟨0,…,n⟩ ⊆ Y
and Y might contain darkᵂᴹ numbers
(We don't know Y does since they're darkᵂᴹ)

Define
ℕ(Y) = ⋂{S ⊆ Y| ∀⟨0,…,n⟩ ⊆ S}

ℕ(Y) contains all FISON-ends and only FISON-ends.
∀A: ∀⟨0,…,n⟩ ⊆ A ⟹ ∀⟨0,…,n⟩ ⊆ ℕ(Y) ⊆ A

ℕ(Y) ⊆ ℕ(X) ⊆ ℕ(Y) = ℕ(X)

ℕ(X) = ℕ(Y) is unique.
Whichever darkᵂᴹ numbers exist or not-exist,
ℕ(X) is the same.

ℕ(X) is what we mean by ℕ
and
∀n ∈ ℕ(X): |ℕ(X)\⟨0,…,n⟩| = ℵ₀
|ℕ(X)\ℕ(X)| = 0

> You cannot reason about numbers as individuals
> which are in the difference |ℕ \ X|,
> can you?

ℕ(X)\X = ∅

Logic says, for elements of ∅
anything is true, anything is false.
But none exist, so
most people find better uses of their time.

On 11/28/23 4:41 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
>
> > Define X
>
> If you can identify a natnumber, then this number and all smaller
> numbers are automatically elements of X. (All last numbers of FISONs
> that you can reason about and all their predecessors belong to X.) You
> cannot reason about natnumbers as individuals which are in the
> difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it
> belongs to X. Nevertheless the difference remains actually infinite: ℵo
> natnumbers. Therefore they are dark.
>
> Regards, WM

Except that this logic is flawed. What is the first number in that
difference, it is the highest number that we can reason about/identify + 1.

Since we CAN reason about that, since we just did, then this number
needs to be in X and not be a "dark" number.

Your logic INCORRECTLY presumes that there IS a finite number that is
the highest we can identify or reason about. There is no actual basis
for assuming that such a number must exist, and the assumption is shown
to make the system inconsistant (as we have a "dark" number that is
identified, so it isn't actually dark).

On 29.11.2023 03:04, Richard Damon wrote:
> On 11/28/23 4:41 AM, WM wrote:
>> Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
>>
>>  > Define X
>>
>> If you can identify a natnumber, then this number and all smaller
>> numbers are automatically elements of X. (All last numbers of FISONs
>> that you can reason about and all their predecessors belong to X.) You
>> cannot reason about natnumbers as individuals which are in the
>> difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it
>> belongs to X. Nevertheless the difference remains actually infinite:
>> ℵo natnumbers. Therefore they are dark.

> What is the first number in that
> difference, it is the highest number that we can reason about/identify + 1.

Dark numbers cannot be used as individuals. The visible nunbers are
potentially infinite. There is no last one. So there is no first dark one.
>
> Since we CAN reason about that, since we just did, then this number
> needs to be in X and not be a "dark" number.

No. Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
>
> Your logic INCORRECTLY presumes that there IS a finite number that is
> the highest we can identify or reason about.

No. With every vivible number n also n^n^n is visible. Pot. Inf.

Regards, WM

Daniel Pehoushek schrieb am Mittwoch, 29. November 2023 um 05:09:26 UTC+1:
> On Tuesday, November 28, 2023 at 10:44:26 PM UTC-5, Daniel Pehoushek
> > is there a less than relation to your set?

Not to the dark numbers.

> > that would necessitate a finite structure to perform comparison.
> > without lessthan the set may contain aleph null natnumbers.
> > daniel2383
> your visible set is ordered.
> the larger set is not.

So it is. Otherwise the dark numbers could not reach until omega.

Regards, WM

On 11/29/23 6:50 AM, WM wrote:
> On 29.11.2023 03:04, Richard Damon wrote:
>> On 11/28/23 4:41 AM, WM wrote:
>>> Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
>>>
>>>  > Define X
>>>
>>> If you can identify a natnumber, then this number and all smaller
>>> numbers are automatically elements of X. (All last numbers of FISONs
>>> that you can reason about and all their predecessors belong to X.)
>>> You cannot reason about natnumbers as individuals which are in the
>>> difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there,
>>> it belongs to X. Nevertheless the difference remains actually
>>> infinite: ℵo natnumbers. Therefore they are dark.
>
>> What is the first number in that difference, it is the highest number
>> that we can reason about/identify + 1.
>
> Dark numbers cannot be used as individuals. The visible nunbers are
> potentially infinite. There is no last one. So there is no first dark one.

So they aren't "numbers", and thus not in the set {1, 2, 3, ... } so the
only number in your sequence they could be in your set is ℵo, but we can
actually reason about that, so they aren't that either, so they just
don't exist in your starting set.

>>
>> Since we CAN reason about that, since we just did, then this number
>> needs to be in X and not be a "dark" number.
>
> No. Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
>>
>> Your logic INCORRECTLY presumes that there IS a finite number that is
>> the highest we can identify or reason about.
>
> No. With every vivible number n also n^n^n is visible. Pot. Inf.
>
> Regards, WM
>
>

So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.

On 29.11.2023 14:11, Daniel Pehoushek wrote:
> On Wednesday, November 29, 2023 at 7:30:19 AM UTC-5, Fritz Feldhase wrote:
>> On Wednesday, November 29, 2023 at 12:56:14 PM UTC+1, WM wrote:
>>> Daniel Pehoushek schrieb am Mittwoch, 29. November 2023 um 05:09:26 UTC+1:
>>>>
>>>> is there a less than relation to your set?
>>>>
>>> Not to the dark numbers.
>> Which implies that your "dark numbers" aren't natural numbers.
>>
>> Hint: Any 'two' natural numbers n, m are /comparable/ (with respect to <=).
>>
>> In other words, for all n, m e IN: n <= m or m <= n.
>>>> your visible set is ordered. the larger set is not.
>>>>
>>> So it is.
>> Then "the larger set" is not IN.

ℕ_vis contains numbers which have ℵo successors. ℕ contains all numbers,
also these successors. The successors cannot be well-ordered because
they follow upon every number that belongs to a well-order. Therefore
there is definitely a difference.

Regards, WM

On 30.11.2023 04:42, Richard Damon wrote:
> On 11/29/23 6:50 AM, WM wrote:

>> Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
>>>
>>> Your logic INCORRECTLY presumes that there IS a finite number that is
>>> the highest we can identify or reason about.
>>
>> No. With every visible number n also n^n^n is visible. Pot. Inf.

>
> So, all you have done is shown that your "dark numbers" aren't
> "numbers", and thus don't exist in the initial set that they are a
> remainder of, and thus are just an empty set.

Wrong.
ℕ_vis contains numbers which have ℵo successors. ℕ contains all numbers,
also these successors. The successors cannot be well-ordered because
they follow upon every number that belongs to a well-order. Therefore
there is definitely a difference.

Regards, WM

On 11/30/23 10:49 AM, WM wrote:
> On 30.11.2023 04:42, Richard Damon wrote:
>> On 11/29/23 6:50 AM, WM wrote:
>
>>> Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
>>>>
>>>> Your logic INCORRECTLY presumes that there IS a finite number that
>>>> is the highest we can identify or reason about.
>>>
>>> No. With every visible number n also n^n^n is visible. Pot. Inf.
>
>>
>> So, all you have done is shown that your "dark numbers" aren't
>> "numbers", and thus don't exist in the initial set that they are a
>> remainder of, and thus are just an empty set.
>
> Wrong.
> ℕ_vis contains numbers which have ℵo successors. ℕ contains all numbers,
> also these successors. The successors cannot be well-ordered because
> they follow upon every number that belongs to a well-order. Therefore
> there is definitely a difference.
>
> Regards, WM
>

Except that ALL numbers in ℕ contain ℵo successors (since for any number
in ℕ there exists a "well-ordered" number above it, and then another,
and so on), so all ℕ can be in ℕ_vis

Your problem is you are trying to reason about infinite sets with rules
that only apply to finite sets.

ANY member of ℕ, call it x, will be found in some set of the numbers
{1, 2, 3, ..., n} for a sufficiently large number n, if just requires n
>= x, and for any member of ℕ such a number exists.

Try to give me a member of ℕ that this doesn't hold for.

It is a basic property of the Natural Numbers.

On 30.11.2023 04:42, Richard Damon wrote:

> So, all you have done is shown that your "dark numbers" aren't
> "numbers", and thus don't exist in the initial set that they are a
> remainder of, and thus are just an empty set.

Here is a proof that you will easier understand:
Between every definable unit fraction 1/n and 0, there are ℵo smaller
unit fractions. You cannot reduce this amount to less than ℵo. You
cannot distinguish ℵo of them. But they must exist in the interval (0,
1]. So they are not an empty set, but existing.

Regards, WM

On 12/1/2023 7:30 AM, WM wrote:
> On 30.11.2023 04:42, Richard Damon wrote:

>> So, all you have done is shown that
>> your "dark numbers" aren't "numbers",
>> and thus
>> don't exist in the initial set that
>> they are a remainder of,
>> and thus
>> are just an empty set.
>
> Here is a proof that
> you will easier understand:
>
> Between
> every definable unit fraction 1/n and 0,

....where
the definable unit fractions are 1×1 1.ended

> there are ℵo smaller unit fractions.

Each 1×1 1.ended is ℵ₀-many.

The 1×1 1.ended definable unit fractions
between ⅟n and 0 are ℵ₀-many

> You cannot reduce this amount to less than ℵo.
> You cannot distinguish ℵo of them.

Performing a supertask is impossible,
if you aren't Chuck Norris.

> But they must exist in the interval (0,1].
> So they are not an empty set, but existing.

Describing performing a supertask and
augmenting finitely the description with
only not-first-false claims
is possible.

We know that
the augmenting claims are true of
what's described.

On 12/1/23 7:30 AM, WM wrote:
> On 30.11.2023 04:42, Richard Damon wrote:
>
>> So, all you have done is shown that your "dark numbers" aren't
>> "numbers", and thus don't exist in the initial set that they are a
>> remainder of, and thus are just an empty set.
>
> Here is a proof that you will easier understand:
> Between every definable unit fraction 1/n and 0, there are ℵo smaller
> unit fractions. You cannot reduce this amount to less than ℵo. You
> cannot distinguish ℵo of them. But they must exist in the interval (0,
> 1]. So they are not an empty set, but existing.
>
> Regards, WM

Why do you say you can not "distinguish" them, they are merely the unit
fractions 1/x for all Natural Numbers x greater than n (all ℵo of them)

There are ℵo Natural numbers, and removing n of them, still leaves you
with ℵo of them (that's how mathematics of trans-finite numbers work).

There is no need for some "dark" numbers to explain this, they are just
the orinary natural numbers.

What "Unit Fraction" exists in that interval that isn't one of those 1/x's?

Yes, no matter how high you choose for n, you still have ℵo numbers
abort it, but that is just the nature of infinite sets, which I guess is
above your ability to understand.

On 02.12.2023 01:44, Fritz Feldhase wrote:
> On Friday, December 1, 2023 at 1:30:44 PM UTC+1, WM wrote:
>
>> Between every [...] unit fraction 1/n and 0, there are ℵo smaller unit fractions.
>
> Indeed!
>
>> You cannot reduce this amount to less than ℵo.

>
> It's just a fakt. Period.

Yes, dark numbers are just a fact.
>
>> You cannot distinguish ℵo of them.
>
> ??? Actually, the ARE already "distinguished".

They are certainly distinguished. But it is impossible to apply almost
all as individuals. In every case ℵo will remain not distinguishable by you.

> No matter if *I* can "distinguish ℵo of them" or not.

The matter is dark numbers.
>
>> But they must exist in the interval (0, 1].
>
> Right. They do. :-)
>
> Hint: An e IN: 0 < 1/n <= 1.
>
>> So they are not an empty set, but existing.
>
> Sure.
>
> {1/n : n e IN} =/= { }.
>
> Actually,
>
> card {1/n : n e IN} = ℵo.
For all definable 1/n we get:
∀x ∈ (1/n, 1] are larger than ℵo unit fractions.
==> ℵo unit fractions lie at the left-hand side of (1/n, 1].

For all x > 0 we get:
∀x ∈ (0, 1] are larger than ℵo unit fractions.
=/=> ℵo unit fractions lie at the left-hand side of (0, 1].

Why this difference?

Gruß, WM

On 02.12.2023 00:41, Richard Damon wrote:
> On 12/1/23 7:30 AM, WM wrote:
>> On 30.11.2023 04:42, Richard Damon wrote:
>>
>>> So, all you have done is shown that your "dark numbers" aren't
>>> "numbers", and thus don't exist in the initial set that they are a
>>> remainder of, and thus are just an empty set.
>>
>> Here is a proof that you will easier understand:
>> Between every definable unit fraction 1/n and 0, there are ℵo smaller
>> unit fractions. You cannot reduce this amount to less than ℵo. You
>> cannot distinguish ℵo of them. But they must exist in the interval (0,
>> 1]. So they are not an empty set, but existing.
>
> Why do you say you can not "distinguish" them,

Because it is fact. You cannot reduce this amount to less than ℵo.

> they are merely the unit
> fractions 1/x for all Natural Numbers x greater than n (all ℵo of them).

I do not deny that they exist. But I denay that they can be used as
individuals. They can only be used collectively. I call them dark.
>
> There are ℵo Natural numbers, and removing n of them, still leaves you
> with ℵo of them (that's how mathematics of trans-finite numbers work).

You cannot reduce this amount to less than ℵo:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
But you can use them collectively:
|ℕ \ {1, 2, 3, ...}| = 0.
They are dark.
>
> There is no need for some "dark" numbers to explain this, they are just
> the orinary natural numbers.

But it is fact that they cannot be used as individuals. Therefore all of
Cantor's "bijections" are wrong. He assumes that all natnumbers can be
used as individuals.
>
> What "Unit Fraction" exists in that interval that isn't one of those 1/x's?
>
> Yes, no matter how high you choose for n, you still have ℵo numbers
> abort it, but that is just the nature of infinite sets, which I guess is
> above your ability to understand.

If you are better, then try to explain this:

For all definable eps > 0 we get:
∀x ∈ (eps, 1] are larger than ℵo unit fractions.
==> ℵo unit fractions lie at the left-hand side of (eps, 1].

For all x > 0 we get:
∀x ∈ (0, 1] are larger than ℵo unit fractions.
=/=> ℵo unit fractions lie at the left-hand side of (0, 1].

Why this difference?

Gruß, WM

On 12/2/23 5:43 AM, WM wrote:
> On 02.12.2023 00:41, Richard Damon wrote:
>> On 12/1/23 7:30 AM, WM wrote:
>>> On 30.11.2023 04:42, Richard Damon wrote:
>>>
>>>> So, all you have done is shown that your "dark numbers" aren't
>>>> "numbers", and thus don't exist in the initial set that they are a
>>>> remainder of, and thus are just an empty set.
>>>
>>> Here is a proof that you will easier understand:
>>> Between every definable unit fraction 1/n and 0, there are ℵo smaller
>>> unit fractions. You cannot reduce this amount to less than ℵo. You
>>> cannot distinguish ℵo of them. But they must exist in the interval
>>> (0, 1]. So they are not an empty set, but existing.
>>
>> Why do you say you can not "distinguish" them,
>
> Because it is fact. You cannot reduce this amount to less than ℵo.

And why you need to?

>
>
>> they are merely the unit fractions 1/x for all Natural Numbers x
>> greater than n (all ℵo of them).
>
> I do not deny that they exist. But I denay that they can be used as
> individuals. They can only be used collectively. I call them dark.

Which number x greater than your first chosen n can't be used as an
individual?

>>
>> There are ℵo Natural numbers, and removing n of them, still leaves you
>> with ℵo of them (that's how mathematics of trans-finite numbers work).
>
> You cannot reduce this amount to less than ℵo:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> But you can use them collectively:
> |ℕ \ {1, 2, 3, ...}| = 0.
> They are dark.

Again, why can't you use any of those individually.

>>
>> There is no need for some "dark" numbers to explain this, they are
>> just the orinary natural numbers.
>
> But it is fact that they cannot be used as individuals. Therefore all of
> Cantor's "bijections" are wrong. He assumes that all natnumbers can be
> used as individuals.

And which ones can't be?

>>
>> What "Unit Fraction" exists in that interval that isn't one of those
>> 1/x's?
>>
>> Yes, no matter how high you choose for n, you still have ℵo numbers
>> abort it, but that is just the nature of infinite sets, which I guess
>> is above your ability to understand.
>
> If you are better, then try to explain this:
>
> For all definable eps > 0 we get:
> ∀x ∈ (eps, 1] are larger than ℵo unit fractions.
> ==> ℵo unit fractions lie at the left-hand side of (eps, 1].
>
> For all x > 0 we get:
> ∀x ∈ (0, 1] are larger than ℵo unit fractions.
> =/=> ℵo unit fractions lie at the left-hand side of (0, 1].
>
> Why this difference?
>
> Gruß, WM
>
>

This just comes down to the fact that there isn't a "last" natural
number or a "first" unit fraction.

You just seem to think becuase we can't "name" the highest natural
number (because there isn't one) at some point they become "dark".

It seems your mind just can't handle infinity. Which is actually a
fairly common problem.