On 1/15/24 3:18 AM, WM wrote:
> Le 14/01/2024 à 19:42, Richard Damon a écrit :
>> On 1/14/24 7:35 AM, WM wrote:
>
>>>> That seems to be your problem, you are assuming you can do what
>>>> isn't allowed,
>>>
>>> It is not only disallowed (a revolutionary spirit would violate this
>>> prohibition), but it is impossible. Why? Because it is impossible. I
>>> call that dark. But the notion is irrelevant. The fact is what counts
>>> (and hinders the counting).
>
>> So, your logic is based on the concept that since we can't divide an
>> infinite set into two finite sets, there must be elements that we can
>> not name in the set?
>
> There *are* elements we cannot name, even almost all, namely ℵ, whereas
> we can name only finitely many.
>
> Regards, WM
>

But ℵ isn't a number in the domain of regard. We were talking about the
Natural Numbers, and ℵ isn't a Natural Number.

Note, we CAN name any of a countable infinite number of them, as we have
a countable infinite number of names to use.

We can only name a finitely many at a time, but a countable infinite
number have names, and that number matches the ℵ0 of them that are.

On Tuesday 9 January 2024 at 20:16:06 UTC+1, mitchr...@gmail.com wrote:
> On Tuesday, January 9, 2024 at 9:23:57 AM UTC-8, WM wrote:
> > Le 09/01/2024 à 03:31, Richard Damon a écrit :
> > > On 1/8/24 7:29 AM, WM wrote:
> > >> Le 08/01/2024 à 13:21, Richard Damon a écrit :
> > >>> On 1/8/24 5:56 AM, WM wrote:
> > >>
> > >>>> I understand that every real x > 0 that you can name has ℵ smaller
> > >>>> positive reals ℵ of which you cannot name.
> > >>>
> > >>> why do you think I can not name any of them?
> > >>
> > >> You can name many of them, but every attempt fails to reduce the
> > >> remainder below ℵ. It can be reduced however to the empty set collectively.
> > >
> > > Why do you expect otherwise?
> > I don't. I only note that fact.
> > >
> > > We have an infinite set.
> > >
> > > Individually naming is a finite operation, so can only name a finite
> > > number of them at a time.
> > But if you could name every number, then you could name one with lesser
> > successors.
> > >
> > > Thus there will alway be an infinite number that we haven't produced the
> > > name.
> > >
> > > That doesn't mean the rest can't be named.
> > ℵ will always be without name. But they can be removed such that none
> > remains, collectively.
> > >
> > > "naming" is the opposite of your "dark", you can only do it
> > > individually, but not to an infinite group.
> > So it is. And not for the numbers belonging to the last ℵ.
> > >
> > > It doesn't mean that any particular ones left can't be named.
> > It does. The last ℵ numbers will remain.
> > >
> > > It seems your problem is you can't keep the right domain in focus.
> > >
> > > Your "dark" numbers have some of the properties of the transfinite
> > > numbers,
> > Yes, they are also dark because they have no finite initial segment of
> > natural numbers.
> >
> > Regards, WM
> Infinitesimals of themselves are dark quantity by being below the finite.
> Zero is no quantity to see...
There are no infinitesimals in real numbers, retard.

On 1/16/2024 10:20 PM, zelos...@gmail.com wrote:
> On Tuesday 9 January 2024 at 20:16:06 UTC+1, mitchr...@gmail.com wrote:
>> On Tuesday, January 9, 2024 at 9:23:57 AM UTC-8, WM wrote:
>>> Le 09/01/2024 à 03:31, Richard Damon a écrit :
>>>> On 1/8/24 7:29 AM, WM wrote:
>>>>> Le 08/01/2024 à 13:21, Richard Damon a écrit :
>>>>>> On 1/8/24 5:56 AM, WM wrote:
>>>>>
>>>>>>> I understand that every real x > 0 that you can name has ℵ smaller
>>>>>>> positive reals ℵ of which you cannot name.
>>>>>>
>>>>>> why do you think I can not name any of them?
>>>>>
>>>>> You can name many of them, but every attempt fails to reduce the
>>>>> remainder below ℵ. It can be reduced however to the empty set collectively.
>>>>
>>>> Why do you expect otherwise?
>>> I don't. I only note that fact.
>>>>
>>>> We have an infinite set.
>>>>
>>>> Individually naming is a finite operation, so can only name a finite
>>>> number of them at a time.
>>> But if you could name every number, then you could name one with lesser
>>> successors.
>>>>
>>>> Thus there will alway be an infinite number that we haven't produced the
>>>> name.
>>>>
>>>> That doesn't mean the rest can't be named.
>>> ℵ will always be without name. But they can be removed such that none
>>> remains, collectively.
>>>>
>>>> "naming" is the opposite of your "dark", you can only do it
>>>> individually, but not to an infinite group.
>>> So it is. And not for the numbers belonging to the last ℵ.
>>>>
>>>> It doesn't mean that any particular ones left can't be named.
>>> It does. The last ℵ numbers will remain.
>>>>
>>>> It seems your problem is you can't keep the right domain in focus.
>>>>
>>>> Your "dark" numbers have some of the properties of the transfinite
>>>> numbers,
>>> Yes, they are also dark because they have no finite initial segment of
>>> natural numbers.
>>>
>>> Regards, WM
>> Infinitesimals of themselves are dark quantity by being below the finite.
>> Zero is no quantity to see...
> There are no infinitesimals in real numbers, retard.

Gotta love 10^(-n), gets pretty small and they are all real numbers, n
is an integer.

[0] = 10^(0) = 1
[1] = 10^(-1) = .1
[2] = 10^(-2) = .01
[3] = 10^(-3) = .001
....

Limit zero. On and on, never equals zero, but gets closer and closer to
it, without any bounds... Take n to infinity, 0, -1, -2, -3, ... :^)

Sum them and get, 1.(1) ;^)

On 1/16/2024 10:28 PM, Chris M. Thomasson wrote:
> On 1/16/2024 10:20 PM, zelos...@gmail.com wrote:
>> On Tuesday 9 January 2024 at 20:16:06 UTC+1, mitchr...@gmail.com wrote:
>>> On Tuesday, January 9, 2024 at 9:23:57 AM UTC-8, WM wrote:
>>>> Le 09/01/2024 à 03:31, Richard Damon a écrit :
>>>>> On 1/8/24 7:29 AM, WM wrote:
>>>>>> Le 08/01/2024 à 13:21, Richard Damon a écrit :
>>>>>>> On 1/8/24 5:56 AM, WM wrote:
>>>>>>
>>>>>>>> I understand that every real x > 0 that you can name has ℵ smaller
>>>>>>>> positive reals ℵ of which you cannot name.
>>>>>>>
>>>>>>> why do you think I can not name any of them?
>>>>>>
>>>>>> You can name many of them, but every attempt fails to reduce the
>>>>>> remainder below ℵ. It can be reduced however to the empty set
>>>>>> collectively.
>>>>>
>>>>> Why do you expect otherwise?
>>>> I don't. I only note that fact.
>>>>>
>>>>> We have an infinite set.
>>>>>
>>>>> Individually naming is a finite operation, so can only name a finite
>>>>> number of them at a time.
>>>> But if you could name every number, then you could name one with lesser
>>>> successors.
>>>>>
>>>>> Thus there will alway be an infinite number that we haven't
>>>>> produced the
>>>>> name.
>>>>>
>>>>> That doesn't mean the rest can't be named.
>>>> ℵ will always be without name. But they can be removed such that none
>>>> remains, collectively.
>>>>>
>>>>> "naming" is the opposite of your "dark", you can only do it
>>>>> individually, but not to an infinite group.
>>>> So it is. And not for the numbers belonging to the last ℵ.
>>>>>
>>>>> It doesn't mean that any particular ones left can't be named.
>>>> It does. The last ℵ numbers will remain.
>>>>>
>>>>> It seems your problem is you can't keep the right domain in focus.
>>>>>
>>>>> Your "dark" numbers have some of the properties of the transfinite
>>>>> numbers,
>>>> Yes, they are also dark because they have no finite initial segment of
>>>> natural numbers.
>>>>
>>>> Regards, WM
>>> Infinitesimals of themselves are dark quantity by being below the
>>> finite.
>>> Zero is no quantity to see...
>> There are no infinitesimals in real numbers, retard.
>
> Gotta love 10^(-n), gets pretty small and they are all real numbers, n
> is an integer.
>
> [0] = 10^(0) = 1
> [1] = 10^(-1) = .1
> [2] = 10^(-2) = .01
> [3] = 10^(-3) = .001
> ...
>
> Limit zero. On and on, never equals zero, but gets closer and closer to
> it, without any bounds... Take n to infinity, 0, -1, -2, -3, ... :^)
>
> Sum them and get, 1.(1) ;^)
>

10^(0)+10^(-1)+10^(-3)+10^(-5)+10^(-7)+10^(-9)

;^)

On 1/16/2024 10:30 PM, Chris M. Thomasson wrote:
> On 1/16/2024 10:28 PM, Chris M. Thomasson wrote:
>> On 1/16/2024 10:20 PM, zelos...@gmail.com wrote:
>>> On Tuesday 9 January 2024 at 20:16:06 UTC+1, mitchr...@gmail.com wrote:
>>>> On Tuesday, January 9, 2024 at 9:23:57 AM UTC-8, WM wrote:
>>>>> Le 09/01/2024 à 03:31, Richard Damon a écrit :
>>>>>> On 1/8/24 7:29 AM, WM wrote:
>>>>>>> Le 08/01/2024 à 13:21, Richard Damon a écrit :
>>>>>>>> On 1/8/24 5:56 AM, WM wrote:
>>>>>>>
>>>>>>>>> I understand that every real x > 0 that you can name has ℵ smaller
>>>>>>>>> positive reals ℵ of which you cannot name.
>>>>>>>>
>>>>>>>> why do you think I can not name any of them?
>>>>>>>
>>>>>>> You can name many of them, but every attempt fails to reduce the
>>>>>>> remainder below ℵ. It can be reduced however to the empty set
>>>>>>> collectively.
>>>>>>
>>>>>> Why do you expect otherwise?
>>>>> I don't. I only note that fact.
>>>>>>
>>>>>> We have an infinite set.
>>>>>>
>>>>>> Individually naming is a finite operation, so can only name a finite
>>>>>> number of them at a time.
>>>>> But if you could name every number, then you could name one with
>>>>> lesser
>>>>> successors.
>>>>>>
>>>>>> Thus there will alway be an infinite number that we haven't
>>>>>> produced the
>>>>>> name.
>>>>>>
>>>>>> That doesn't mean the rest can't be named.
>>>>> ℵ will always be without name. But they can be removed such that none
>>>>> remains, collectively.
>>>>>>
>>>>>> "naming" is the opposite of your "dark", you can only do it
>>>>>> individually, but not to an infinite group.
>>>>> So it is. And not for the numbers belonging to the last ℵ.
>>>>>>
>>>>>> It doesn't mean that any particular ones left can't be named.
>>>>> It does. The last ℵ numbers will remain.
>>>>>>
>>>>>> It seems your problem is you can't keep the right domain in focus.
>>>>>>
>>>>>> Your "dark" numbers have some of the properties of the transfinite
>>>>>> numbers,
>>>>> Yes, they are also dark because they have no finite initial segment of
>>>>> natural numbers.
>>>>>
>>>>> Regards, WM
>>>> Infinitesimals of themselves are dark quantity by being below the
>>>> finite.
>>>> Zero is no quantity to see...
>>> There are no infinitesimals in real numbers, retard.
>>
>> Gotta love 10^(-n), gets pretty small and they are all real numbers, n
>> is an integer.
>>
>> [0] = 10^(0) = 1
>> [1] = 10^(-1) = .1
>> [2] = 10^(-2) = .01
>> [3] = 10^(-3) = .001
>> ...
>>
>> Limit zero. On and on, never equals zero, but gets closer and closer
>> to it, without any bounds... Take n to infinity, 0, -1, -2, -3, ... :^)
>>
>> Sum them and get, 1.(1) ;^)
>>
>
> 10^(0)+10^(-1)+10^(-3)+10^(-5)+10^(-7)+10^(-9)
>
> ;^)
>

Another interesting summation:

10^(-2)+10^(-3)+10^(-5)+10^(-7)+10^(-11)+10^(-13)

;^D primes anyone?

Le 15/01/2024 à 13:52, Richard Damon a écrit :
> On 1/15/24 3:18 AM, WM wrote:

>> There *are* elements we cannot name, even almost all, namely ℵ, whereas
>> we can name only finitely many.

> But ℵ isn't a number in the domain of regard.

But the ℵ elements are in the domain.

> Note, we CAN name any of a countable infinite number of them, as we have
> a countable infinite number of names to use.

Why do always infinitely many unit fractions remain between the smallest
named one and zero?

> We can only name a finitely many at a time, but a countable infinite
> number have names, and that number matches the ℵ0 of them that are.

But you cannot use a name of a unit fraction having less than ℵ smaller
ones.

Regards, WM

On 1/17/24 6:36 AM, WM wrote:
> Le 15/01/2024 à 13:52, Richard Damon a écrit :
>> On 1/15/24 3:18 AM, WM wrote:
>
>>> There *are* elements we cannot name, even almost all, namely ℵ,
>>> whereas we can name only finitely many.
>
>> But ℵ isn't a number in the domain of regard.
>
> But the ℵ elements are in the domain.

Yes, but the value ℵ isn't

>
>> Note, we CAN name any of a countable infinite number of them, as we
>> have a countable infinite number of names to use.
>
> Why do always infinitely many unit fractions remain between the smallest
> named one and zero?

Becuase the set is unbounded.

>
>> We can only name a finitely many at a time, but a countable infinite
>> number have names, and that number matches the ℵ0 of them that are.
>
> But you cannot use a name of a unit fraction having less than ℵ smaller
> ones.
>
> Regards, WM
>
>

because such a number doesn't exist. You don't need to name the
non-existant.

Le 17/01/2024 à 13:56, Richard Damon a écrit :
> On 1/17/24 6:36 AM, WM wrote:
>> Le 15/01/2024 à 13:52, Richard Damon a écrit :
>>> On 1/15/24 3:18 AM, WM wrote:
>>
>>>> There *are* elements we cannot name, even almost all, namely ℵ,
>>>> whereas we can name only finitely many.
>>
>>> But ℵ isn't a number in the domain of regard.
>>
>> But the ℵ elements are in the domain.
>
> Yes, but the value ℵ isn't
>
>>
>>> Note, we CAN name any of a countable infinite number of them, as we
>>> have a countable infinite number of names to use.
>>
>> Why do always infinitely many unit fractions remain between the smallest
>> named one and zero?
>
> Because the set is unbounded.

Wrong. The set is bounded by its smallest upper bound 0.
>
>>
>>> We can only name a finitely many at a time, but a countable infinite
>>> number have names, and that number matches the ℵ0 of them that are.
>>
>> But you cannot use a name of a unit fraction having less than ℵ smaller
>> ones.

> because such a number doesn't exist. You don't need to name the
> non-existant.

NUF(0) = 0
NUF(x>0) = ℵo

This means an increase. But that increase cannot happen between 0 and (0,
1] because
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
So you are wrong, if mathematics is right.

Regards, WM

Le 17/01/2024 à 15:08, WM a écrit :
> The set is bounded by its smallest upper bound 0.

Should read: Greatest lower bound.
Regards, WM

On 1/17/2024 6:36 AM, WM wrote:
> Le 15/01/2024 à 13:52, Richard Damon a écrit :

>> Note, we CAN name any of
>> a countable infinite number of them, as
>> we have a countable infinite number of
>> names to use.
>
> Why do always
> infinitely many unit fractions remain
> between the smallest named one and zero?

Some cardinalities can change by 1.
They are the cardinalities of
flocks of sheep and of pockets of pebbles.
They are finite.

The cardinality of
cardinalities which can change by 1
is larger.than any cardinality which can change by 1,
is not any cardinality which can change by 1.
That cardinality cannot change by 1.

|⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...

On 1/17/24 9:08 AM, WM wrote:
> Le 17/01/2024 à 13:56, Richard Damon a écrit :
>> On 1/17/24 6:36 AM, WM wrote:
>>> Le 15/01/2024 à 13:52, Richard Damon a écrit :
>>>> On 1/15/24 3:18 AM, WM wrote:
>>>
>>>>> There *are* elements we cannot name, even almost all, namely ℵ,
>>>>> whereas we can name only finitely many.
>>>
>>>> But ℵ isn't a number in the domain of regard.
>>>
>>> But the ℵ elements are in the domain.
>>
>> Yes, but the value ℵ isn't
>>
>>>
>>>> Note, we CAN name any of a countable infinite number of them, as we
>>>> have a countable infinite number of names to use.
>>>
>>> Why do always infinitely many unit fractions remain between the
>>> smallest named one and zero?
>>
>> Because the set is unbounded.
>
> Wrong. The set is bounded by its smallest upper bound 0.

But 0 isn't member of the set, so the set itself is unbounded.

>>
>>>
>>>> We can only name a finitely many at a time, but a countable infinite
>>>> number have names, and that number matches the ℵ0 of them that are.
>>>
>>> But you cannot use a name of a unit fraction having less than ℵ
>>> smaller ones.
>
>> because such a number doesn't exist. You don't need to name the
>> non-existant.
>
> NUF(0) = 0
> NUF(x>0) = ℵo
>
> This means an increase. But that increase cannot happen between 0 and
> (0, 1] because
> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
> So you are wrong, if mathematics is right.
>
> Regards, WM
>
>
>

Which only shows it can't increase to a finite number at any finite
number as that number will have an unbounded number of unit fractions
below it.

Thus NUF(x) with x finite, jumps, perhaps in domain of the
infinitesimals between 0 and the bottom of (0,1]

Le 17/01/2024 à 21:04, Jim Burns a écrit :
> On 1/17/2024 6:36 AM, WM wrote:
>> Le 15/01/2024 à 13:52, Richard Damon a écrit :
>
>>> Note, we CAN name any of
>>> a countable infinite number of them, as
>>> we have a countable infinite number of
>>> names to use.
>>
>> Why do always
>> infinitely many unit fractions remain
>> between the smallest named one and zero?
>
> Some cardinalities can change by 1.
> They are the cardinalities of
> flocks of sheep and of pockets of pebbles.
> They are finite.

All sets can change by single elements.
For instance we can remove single elements from every infinite set.
>
> The cardinality of
> cardinalities which can [not] change by 1
> is larger.than any cardinality which can change by 1,
> is not any cardinality which can change by 1.
> That cardinality cannot change by 1.
>
> |⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...

That statement is correct for nameable unit fractions, but in general it
is in contradiction with mathematics because of
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .

Regards, WM

Le 18/01/2024 à 02:36, Richard Damon a écrit :
> On 1/17/24 9:08 AM, WM wrote:

>>>>> Note, we CAN name any of a countable infinite number of them, as we
>>>>> have a countable infinite number of names to use.
>>>>
>>>> Why do always infinitely many unit fractions remain between the
>>>> smallest named one and zero?
>>>
>>> Because the set is unbounded.
>>
>> Wrong. The set is bounded by its largest lower bound 0
>
> But 0 isn't member of the set, so the set itself is unbounded.

No. The bound exists. All unit fractions must fit into the positive
interval. They have internal distances. Hence there is a linear chain
having a first element. The only alternative, namely infinitely many
between 0 and (0, 1], can be excluded by mathematics.

> Thus NUF(x) with x finite, jumps, perhaps in domain of the
> infinitesimals between 0 and the bottom of (0,1]

That is merely another name for dark real numbers. We cannot address them
as individuals.

Regards, WM

On 1/18/24 3:54 AM, WM wrote:
> Le 18/01/2024 à 02:36, Richard Damon a écrit :
>> On 1/17/24 9:08 AM, WM wrote:
>
>>>>>> Note, we CAN name any of a countable infinite number of them, as
>>>>>> we have a countable infinite number of names to use.
>>>>>
>>>>> Why do always infinitely many unit fractions remain between the
>>>>> smallest named one and zero?
>>>>
>>>> Because the set is unbounded.
>>>
>>> Wrong. The set is bounded by its largest lower bound 0
>>
>> But 0 isn't member of the set, so the set itself is unbounded.
>
> No. The bound exists. All unit fractions must fit into the positive
> interval. They have internal distances. Hence there is a linear chain
> having a first element. The only alternative, namely infinitely many
> between 0 and (0, 1], can be excluded by mathematics.
>

There is not need for a first element, since the chain is unbounded in
length.

Your logic is just flawed in that assumption.

Yes, ZFC says there is a "first", but that is in numbering sequence, not
value sequence, and that first is 1/1, and we number infinitely towards 0.

>> Thus NUF(x) with x finite, jumps, perhaps in domain of the
>> infinitesimals between 0 and the bottom of (0,1]
>
> That is merely another name for dark real numbers. We cannot address
> them as individuals.

But they are not elements of the finite numbers, and thus not Real,
Rational, or Unit Fractions.

Yes, if you admit your "Dark Numbers" are not part of those sets, we
might find them existing, but then we have the transfinite theories that
describe them, and they no longer stay dark.

You are just asserting contradictory claims.

If you want to call that area you do not know about "dark", go ahead,
just realize that others have explored it and mapped it, so it really
isn't unknown, except to you.

>
> Regards, WM
>
>

Le 18/01/2024 à 13:46, Richard Damon a écrit :
> On 1/18/24 3:54 AM, WM wrote:

> There is not need for a first element, since the chain is unbounded in
> length.

No, it is bounded in length by the origin 0.

>> That is merely another name for dark real numbers. We cannot address
>> them as individuals.
>
> But they are not elements of the finite numbers, and thus not Real,
> Rational, or Unit Fractions.

All unit fractions ate unit fractions and are elements of the finite
numbers.

Regards, WM

On 1/18/2024 3:48 AM, WM wrote:
> Le 17/01/2024 à 21:04, Jim Burns a écrit :

>> The cardinality of
>> cardinalities which can [not] change by 1
>> is larger.than
>> any cardinality which can change by 1,
>> is not
>> any cardinality which can change by 1.
>> That cardinality cannot change by 1.
>>
>> |⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...
>
> That statement is correct for
> nameable unit fractions, but in general
> it is in contradiction with
> mathematics because of
> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .

Thank you for the correcting "not".

I will give
cardinal which can change by 1
a shorter name: final ordinal.

"Final ordinal" is analogous to
"initial ordinal"
https://en.wikipedia.org/wiki/Von_Neumann_cardinal_assignment#Initial_ordinal_of_a_cardinal

An initial von Neumann ordinal is
the first ordinal _with that cardinality_

0 = {} is first with cardinality 0
5 = {0,1,2,3,4} is first with cardinality 5
ω = {0,1,2,…} is first with cardinality ℵ₀
0, 5, and ω are initial ordinals.

0 is last with cardinality 0
5 is last with cardinality 5
ω _is not_ last with cardinality ℵ₀

Guests from {0,1,…;ω} = ω+1 fit in
rooms from {[0],[1],[2],…} = ω in this way:
ω:[0] 0:[1] 1:[2] 2:[3] ...

ω+1 = {0,1,…;ω} is also with cardinality ℵ₀
|ω| = |ω+1|

0 and 5 are final ordinals.
ω isn't a final ordinal, it is non.final.
ω is defined as the initial non.final ordinal.

The final ordinals are also known as
the natural numbers.

With that terminology change, my proof is
much less work to write,
much less work to read.

An ordinal after each final ordinal
is not any of the final ordinals.
It is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

>> |⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...
>
> That statement is correct for
> nameable unit fractions, but in general
> it is in contradiction with
> mathematics because of
> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .

Between unit.fraction ⅟j and
unit.fraction ⅟k: 0 < ⅟k < ⅟j
⟨⅟k,⅟k⁻¹,…,⅟j⁺¹,⅟j⟩ has a final ordinal.
|⅟k,⅟k⁻¹,…,⅟j⁺¹,⅟j| ≠ |⅟k⁺¹,⅟k,⅟k⁻¹,…,⅟j⁺¹,⅟j|

Between ⅟j and 0
for each final ordinal k
all the between.unit.fractions
do not fit in ⟨⅟j⁺ᵏ,…,⅟j⁺¹⟩

Between ⅟j and 0
all the between.unit.fractions
do not have any final ordinal.

It is a non.final ordinal.
|⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...

On 1/18/24 12:34 PM, WM wrote:
> Le 18/01/2024 à 13:46, Richard Damon a écrit :
>> On 1/18/24 3:54 AM, WM wrote:
>
>> There is not need for a first element, since the chain is unbounded in
>> length.
>
> No, it is bounded in length by the origin 0.

Which isn't an element of the set, so isn't the "first" of the set, or
even implies that there IS a first element.

>
>>> That is merely another name for dark real numbers. We cannot address
>>> them as individuals.
>>
>> But they are not elements of the finite numbers, and thus not Real,
>> Rational, or Unit Fractions.
>
> All unit fractions ate unit fractions and are elements of the finite
> numbers.

yes, and all unit fractions, being one divided by a Natural Number, are
individually definable by that Natural Number, and thus not "dark"

You are unable to show a set that actually show the need for "dark" numbers.

Your attempts just fail.

>
> Regards, WM

On 1/13/2024 2:23 AM, WM wrote:
> Le 13/01/2024 à 00:46, Richard Damon a écrit :
[...]
> It is caused by your inability to reduce the amounts to less than ℵ.
>

WM can be useful? If you accidentally drink poison, read some of WM's
thoughts and automatically puke your brains out? Might save your life?

Le 19/01/2024 à 02:26, Richard Damon a écrit :
> On 1/18/24 12:34 PM, WM wrote:
>> Le 18/01/2024 à 13:46, Richard Damon a écrit :
>>> On 1/18/24 3:54 AM, WM wrote:
>>
>>> There is not need for a first element, since the chain is unbounded in
>>> length.
>>
>> No, it is bounded in length by the origin 0.
>
> Which isn't an element of the set,

But NUF(x) is well defined.

>> All unit fractions are unit fractions and are elements of the finite
>> numbers.
>
> yes, and all unit fractions, being one divided by a Natural Number, are
> individually definable by that Natural Number, and thus not "dark"

Most unit fractions are dark because most natural numbers are dark.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0

Regards, WM

Le 18/01/2024 à 19:20, Jim Burns a écrit :
> I will give
> cardinal which can change by 1
> a shorter name: final ordinal.

A good point.

> 0 = {} is first with cardinality 0
> 5 = {0,1,2,3,4} is first with cardinality 5
> ω = {0,1,2,…} is first with cardinality ℵ₀
> 0, 5, and ω are initial ordinals.

But ω and most of its predecessors have no FISONs. They are dark numbers.
>
> 0 is last with cardinality 0
> 5 is last with cardinality 5
> ω _is not_ last with cardinality ℵ₀

ℵ₀ of set theory is a self-contradictory non-mathematical and
non-logical notion. Proof: Bob.
>
> ω+1 = {0,1,…;ω} is also with cardinality ℵ₀
> |ω| = |ω+1|

That is true if we simply understand "infinitely many" by ℵ₀. In order
to excorzise the bijective meaning I will henceforth use only ℵ, meaning
"infinitely many". |ℕ| = ℵ, |ℚ| = ℵ, |ℝ| = ℵ,
although |ℚ| = 2|ℕ|^2 + 1 and |ℝ| = 2B^|2ℕ| where B is the base, B
= 2 in binary notation and B = 10 in decimal notation.

Regards, WM

On 1/19/24 3:55 AM, WM wrote:
> Le 19/01/2024 à 02:26, Richard Damon a écrit :
>> On 1/18/24 12:34 PM, WM wrote:
>>> Le 18/01/2024 à 13:46, Richard Damon a écrit :
>>>> On 1/18/24 3:54 AM, WM wrote:
>>>
>>>> There is not need for a first element, since the chain is unbounded
>>>> in length.
>>>
>>> No, it is bounded in length by the origin 0.
>>
>> Which isn't an element of the set,
>
> But NUF(x) is well defined.

No, it isn't.

You words define NUF(x) to be

>
>>> All unit fractions are unit fractions and are elements of the finite
>>> numbers.
>>
>> yes, and all unit fractions, being one divided by a Natural Number,
>> are individually definable by that Natural Number, and thus not "dark"
>
> Most unit fractions are dark because most natural numbers are dark.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ {1, 2, 3, ...}| = 0
>
> Regards, WM
>
>

But that doesn't define a set of just dark number.

Your problem is you don't have a definiton of what "N_def" actually is,
so you can't define what your Dark Numbers are.

You are implicitly assuming that there is a "highest" defined number,
but there isn't, and thus you have a wrong definition for "dark".

Le 19/01/2024 à 16:54, Richard Damon a écrit :
> On 1/19/24 3:55 AM, WM wrote:
>>>>
>>>> No, it is bounded in length by the origin 0.
>>>
>>> Which isn't an element of the set,
>>
>> But NUF(x) is well defined.
>
> No, it isn't.

NUF(x) is the number of unit fractions between 0 and x.
>
>> Most unit fractions are dark because most natural numbers are dark.
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>> |ℕ \ {1, 2, 3, ...}| = 0
>
> But that doesn't define a set of just dark number.

There is no set of only dark numbers. There is an infinite set ℕ, a
finite part of which is visible, the complement is dark.
>
> Your problem is you don't have a definiton of what "N_def" actually is,

Every number that is defined individually is visible.

> You are implicitly assuming that there is a "highest" defined number,

Not a constant number but only temporarily.

Regards, WM

Richard Damon formulated the question :
> On 1/19/24 3:55 AM, WM wrote:
>> Le 19/01/2024 à 02:26, Richard Damon a écrit :
>>> On 1/18/24 12:34 PM, WM wrote:
>>>> Le 18/01/2024 à 13:46, Richard Damon a écrit :
>>>>> On 1/18/24 3:54 AM, WM wrote:
>>>>
>>>>> There is not need for a first element, since the chain is unbounded in
>>>>> length.
>>>>
>>>> No, it is bounded in length by the origin 0.
>>>
>>> Which isn't an element of the set,
>>
>> But NUF(x) is well defined.
>
> No, it isn't.
>
> You words define NUF(x) to be
>
>>
>>>> All unit fractions are unit fractions and are elements of the finite
>>>> numbers.
>>>
>>> yes, and all unit fractions, being one divided by a Natural Number, are
>>> individually definable by that Natural Number, and thus not "dark"
>>
>> Most unit fractions are dark because most natural numbers are dark.
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>> |ℕ \ {1, 2, 3, ...}| = 0
>>
>> Regards, WM
>>
>>
>
> But that doesn't define a set of just dark number.
>
> Your problem is you don't have a definiton of what "N_def" actually is, so
> you can't define what your Dark Numbers are.

It is a particularly sticky wicket for him because a set is a
collection of well defined objects with well defined meaning that you
can always tell whether or not a particular object is a member of said
set or not.

A set of undefinable or undefined objects is an absurdity, so he will
say it is not a set but a potentially infinite collection. This
nullifies his idea that such a collection is a subset of naturals or
any other actual set.

> You are implicitly assuming that there is a "highest" defined number, but
> there isn't, and thus you have a wrong definition for "dark".

What is the most interesting number? reframed.

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

On 1/19/24 12:07 PM, WM wrote:
> Le 19/01/2024 à 16:54, Richard Damon a écrit :
>> On 1/19/24 3:55 AM, WM wrote:
>>>>>
>>>>> No, it is bounded in length by the origin 0.
>>>>
>>>> Which isn't an element of the set,
>>>
>>> But NUF(x) is well defined.
>>
>> No, it isn't.
>
> NUF(x) is the number of unit fractions between 0 and x.

Which is infinite for all x > 0.

And thus NUF(x) doesn't have a finite value answer for any x > 0, and if
you are saying you are working in the domain of finite values, NUF(x)
isn't actually defined for any x > 0, since its value isn't defined to
something in the domain of regard.

Nothing in that definition allows it to have values other than 0 or
infinity, so claiming it has other values is just an error.

To presume it has the value 1 somewhere, presumes that there exists a
smallest unit fraction, which is a false assumption.

>>
>>> Most unit fractions are dark because most natural numbers are dark.
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>> |ℕ \ {1, 2, 3, ...}| = 0
>>
>> But that doesn't define a set of just dark number.
>
> There is no set of only dark numbers. There is an infinite set ℕ, a
> finite part of which is visible, the complement is dark.

But why is it only a finite part that is visible?

>>
>> Your problem is you don't have a definiton of what "N_def" actually is,
>
> Every number that is defined individually is visible.

So, all Natural Numbers are visible, as we can find the individual
definition of any of them.

>
>> You are implicitly assuming that there is a "highest" defined number,
>
> Not a constant number but only temporarily.

So not actually existing.

All you are doing is showing that there exist number bigger than HAVE
BEEN named, not bigger than CAN be named.

This becomes a problem of what we have seen, not of what is, so not a
property of the numbers themselves, but of the observer.

>
> Regards, WM

Le 19/01/2024 à 19:09, Richard Damon a écrit :
> On 1/19/24 12:07 PM, WM wrote:
>> Le 19/01/2024 à 16:54, Richard Damon a écrit :
>>> On 1/19/24 3:55 AM, WM wrote:
>>>>>>
>>>>>> No, it is bounded in length by the origin 0.
>>>>>
>>>>> Which isn't an element of the set,
>>>>
>>>> But NUF(x) is well defined.
>>>
>>> No, it isn't.
>>
>> NUF(x) is the number of unit fractions between 0 and x.
>
> Which is infinite for all x > 0.

Not when mathematics is applied.
>
> And thus NUF(x) doesn't have a finite value answer for any x > 0, and if
> you are saying you are working in the domain of finite values, NUF(x)
> isn't actually defined for any x > 0, since its value isn't defined to
> something in the domain of regard.
>
> Nothing in that definition allows it to have values other than 0 or
> infinity, so claiming it has other values is just an error.
>
> To presume it has the value 1 somewhere, presumes that there exists a
> smallest unit fraction, which is a false assumption.

Not when mathematics is applied.
>
>>>
>>>> Most unit fractions are dark because most natural numbers are dark.
>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>> |ℕ \ {1, 2, 3, ...}| = 0
>>>
>>> But that doesn't define a set of just dark number.
>>
>> There is no set of only dark numbers. There is an infinite set ℕ, a
>> finite part of which is visible, the complement is dark.
>
> But why is it only a finite part that is visible?

Try to make more visible. Fail. Then you know it.
>
>>>
>>> Your problem is you don't have a definiton of what "N_def" actually is,
>>
>> Every number that is defined individually is visible.
>
> So, all Natural Numbers are visible, as we can find the individual
> definition of any of them.

Try it.
>>
>>> You are implicitly assuming that there is a "highest" defined number,
>>
>> Not a constant number but only temporarily.
>
> So not actually existing.

The number is actually existing like the biggest known prime number.
>
> All you are doing is showing that there exist number bigger than HAVE
> BEEN named, not bigger than CAN be named.

Try to name all. Fail
>
> This becomes a problem of what we have seen, not of what is, so not a
> property of the numbers themselves, but of the observer.

Exactly. Nevertheless no observer can see all numbers.

Regards, WM