On 12/20/23 5:00 AM, WM wrote:
> Le 19/12/2023 à 15:07, Richard Damon a écrit :
>> On 12/19/23 5:16 AM, WM wrote:
>
>>>>>
>>>>>> Right, so NUF(x) is infinite for ANY finite number, and thus for
>>>>>> ANY unit fraction, so it is NEVER 1 at any, so there is no
>>>>>> smallest unit fraction.
>>>>>
>>>>> This is inconsistent with the property, accepted by you: NUF can
>>>>> not increase by more than 1 without being constant for a finite
>>>>> period. NUF(0) = 0 implies NUF(x) = 1.
>>>>
>>>> Yes, at an infintesimally small value of x.
>>>
>>> No unit fraction is infinitesimally small. But perhaps you call dark
>>> values infinitesimally small.
>>
>> But they are UNBOUNDEDLY small, which means there is no smallest.
>
> Each dunit fraction is finite and larger than zero. There is a smallest.
> An increase of NUF(x) before every x > 0 is impossible. There is a first
> increase after 0, but not to more than 1, because after every unit
> fraction there is a distance d_n > 0.
>
> Regards, WM
>

No, there is no smallest finite.

If you disagree, name it.

When you try to name it, cal it x.

x/2 is smaller and finite, so you lied.

Thus, your logic is wrong, and you are proved to be a liar.

On 12/20/23 5:36 AM, WM wrote:
> Le 20/12/2023 à 03:07, Richard Damon a écrit :
>
>> No, they get unboundedly small, but every one is finite, so it isn't
>> "infinitely" small.
>
> Correct.
>>
>> NUF(x) = 1 only at an INFINITELY small value, which is smaller than
>> all UNBOUNDLY small values.
>
> Smaller than all eps > 0 that can be chosen. So it is.

Which isn't a finite number. Your logic doesn't even work for finite
sets. Which of the letters {A, B, C, ... Z} comes before ALL the
elements of that set (not A, as A isn't before A).

There can not be a "finite number" that is smaller than ALL finite
numbers. That is like the set that contains all sets that doesn't
contain itself. *BOOM*

>
> But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo. That
> means the increase from 0 to ℵo cannot be seen and discerned. The axiom
> of choice is invalid here.
>
> Regards, WM

Right, and the points it increases are NOT "finite" numbers, and thus
NOT the "unit fractions" that you talk about.

It increases over a different set of numbers, which are NOT the unit
fractions, or any finite number, they are the infinitesimals, which you
have explicitly said are not your "dark numbers" (and can't be, because
they are all individually namable in a system that can talk about them
at all).

On 12/20/23 4:55 AM, WM wrote:
> Le 19/12/2023 à 15:07, Richard Damon a écrit :
>> On 12/19/23 5:25 AM, WM wrote:
>>> Le 18/12/2023 à 17:27, Richard Damon a écrit :
>>>
>>>> Nope, because every unit fraction is distinguishable individually,
>>>> and there are even rational values between them.
>>>
>>> Then not ℵ₀ unit fractions would remain below every eps which can be
>>> used to distinguish them.
>
>> Why?
>>
>> Remember, you are talking about an UNBOUNDED set
>
> Wrong. The set of unit fractions is bounded by 0 and 1.

0 is not a "Unit Fraction" so not a bound but a limit.

>
>> so it is inexhaustible.
>
> NUF(x) exists for every point.

But is infinite, and thus NOT a finite value for all finite x > 0, so if
you are working in the relm of finite numbers, like you claim, isn't
actually defined for ANY finite x > 0, so your system went *BOOM* in a lie.

>
> Regards, WM
>

On 12/20/23 5:07 AM, WM wrote:
> Le 19/12/2023 à 15:14, Richard Damon a écrit :
>
>> NUF actually does those finite steps at infinitesimals,
>
> Whatever you call them, these steps cannot be investigated as
> individuals. They are dark.
>> your NUF fails to be defined for x > 0.
>
> The function is defined but its values are dark for small x.

No, the function isn't defined for any finite x.

If you disagree, NAME an actual finite value and give its finite value
for that finite value. (name, not just a description like "the smallest"
which doesn't actually exist)

It is only "defined" for 0, and x that are not finite, but
infintesimally small, and thus out of your domain of regard.

The "values" of the function aren't dark, at best you could claim the
domain is dark, but then dark values can't be part of the domain of a
funtion, as inputs to a function must be used "individually", which your
"dark" numbers can not be.

So, you system has good *BOOM* in a puff of illogic.

>
> Regards, WM
>
>
>

WM expressed precisely :
> Le 19/12/2023 à 15:07, Richard Damon a écrit :
>> On 12/19/23 5:25 AM, WM wrote:
>>> Le 18/12/2023 à 17:27, Richard Damon a écrit :
>>>
>>>> Nope, because every unit fraction is distinguishable individually, and
>>>> there are even rational values between them.
>>>
>>> Then not ℵ₀ unit fractions would remain below every eps which can be used
>>> to distinguish them.
>
>> Why?
>>
>> Remember, you are talking about an UNBOUNDED set
>
> Wrong. The set of unit fractions is bounded by 0 and 1.

It used to be (0,1] didn't it?

Le 20/12/2023 à 13:42, Richard Damon a écrit :
> On 12/20/23 5:00 AM, WM wrote:
>
>> Each unit fraction is finite and larger than zero. There is a smallest.
>> An increase of NUF(x) before every x > 0 is impossible. There is a first
>> increase after 0, but not to more than 1, because after every unit
>> fraction there is a distance d_n > 0.
>>
> No, there is no smallest finite.
>
> If you disagree, name it.

It is dark. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 holds for all
natnumbers.

You disagree with mathematics and with my logic:

Every positive point has ℵo unit fractions at its left-hand side.
==> There is no positive point with less than ℵo unit fractions
at its left-hand side.
==> All positive points with no exception have ℵo unit fractions at
their left-hand
side.
==> The interval (0, 1] has aleph_0 unit fractions at its left-hand side.
==> Contradiction.

> Probably you support the "logic" of matheologians like Feldhase.

There is no positive point with less than ℵo unit fractions at its
left-hand side.
But that does not imply that ℵo unit fractions lie left of the interval.

But that is nonsense - if all points can be selected as the axiom of
choice proposes.

Regards, WM

Le 20/12/2023 à 13:54, Richard Damon a écrit :
> On 12/20/23 5:36 AM, WM wrote:

>> Smaller than all eps > 0 that can be chosen. So it is.
>
> Which isn't a finite number.

There are many finite numbers smaller than every esp > 0 that can be
chosen.

> There can not be a "finite number" that is smaller than ALL finite
> numbers

Choose an eps > 0 with less than almost all unit fractions in (0, eps].
Fail.

>> But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo. That
>> means the increase from 0 to ℵo cannot be seen and discerned. The axiom
>> of choice is invalid here.
>
> Right, and the points it increases are NOT "finite" numbers, and thus
> NOT the "unit fractions" that you talk about.

How should the number of unit fractions increase before all unit
fractions?
Finest logic!

Regards, WM

On 12/20/23 11:14 AM, WM wrote:
> Le 20/12/2023 à 13:42, Richard Damon a écrit :
>> On 12/20/23 5:00 AM, WM wrote:
>>
>>> Each unit fraction is finite and larger than zero. There is a
>>> smallest. An increase of NUF(x) before every x > 0 is impossible.
>>> There is a first increase after 0, but not to more than 1, because
>>> after every unit fraction there is a distance d_n > 0.
>>>
>> No, there is no smallest finite.
>>
>> If you disagree, name it.
>
> It is dark. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 holds for all natnumbers.
>
> You disagree with mathematics and with my logic:

No, it is NON-EXISTANT.

>
> Every positive point has ℵo unit fractions at its left-hand side.
> ==> There is no positive point with less than ℵo unit fractions
> at its left-hand side.
> ==> All positive points with no exception have ℵo unit fractions at
> their left-hand
> side.
> ==> The interval (0, 1] has aleph_0 unit fractions at its left-hand side.
> ==> Contradiction.

What contradiction?

Every finite natural number has ℵo Natural Numbers above it, so the just
isn't a highest.

Evry finite unit fraction has ℵo finite unit fractions smaller than it,
so no unit fraciton is the smallest.

No contradictions.

The contradiction is in assuming that there needs to be a smallest
"defined" natural number so all the ones less than it are "dark"

ALL Natural Numbers are definable, so the set of "dark" natural numbers
is empty.

>
>> Probably you support the "logic" of matheologians like Feldhase.
>
> There is no positive point with less than ℵo unit fractions at its
> left-hand side.
> But that does not imply that ℵo unit fractions lie left of the interval.

Why not?
>
> But that is nonsense - if all points can be selected as the axiom of
> choice proposes.

Why?

What in the "Axiom of Choice" says we can choose the "End" of an
Unbounded set?

In fact, you could argue that the principle of chioce implies your dark
numbers don't exist, as if it is non-empty, it means you can always
choose an individual in it. After all, you claim your "dark numbers" are
elements of the Natural Numbers, and as such, indexable (since Natural
Numbers are their own index) so it applies to that set.

>
> Regards, WM
>
>

You are just trying to use "Bounded" logic on an unbounded set.

That logic can't HAVE the "Natural Numbers" in it, so is unsuitable for
the job.

On 12/20/23 11:18 AM, WM wrote:
> Le 20/12/2023 à 13:54, Richard Damon a écrit :
>> On 12/20/23 5:36 AM, WM wrote:
>
>>> Smaller than all eps > 0 that can be chosen. So it is.
>>
>> Which isn't a finite number.
>
> There are many finite numbers smaller than every esp > 0 that can be
> chosen.

SO you agree that there is no "smallest" eps that can be used.

>> There can not be a "finite number" that is smaller than ALL finite
>> numbers
>
> Choose an eps > 0 with less than almost all unit fractions in (0, eps].
> Fail.

Why should you be able to?

Choose a number greater than 1/2 of infinity? That is a nonsense question,

>
>>> But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo.
>>> That means the increase from 0 to ℵo cannot be seen and discerned.
>>> The axiom of choice is invalid here.
>>
>> Right, and the points it increases are NOT "finite" numbers, and thus
>> NOT the "unit fractions" that you talk about.
>
> How should the number of unit fractions increase before all unit fractions?
> Finest logic!

Simply because you are referencing an non-existance, there is no
"before" all unit fractions as a finite value that isn't 0

Make logic based on the existance of the non-existent and you can make
anything you want seem to make sense, but the logic becomes worthless.

>
> Regards, WM
>

Le 20/12/2023 à 17:38, Richard Damon a écrit :
> On 12/20/23 11:14 AM, WM wrote:

>> Every positive point has ℵo unit fractions at its left-hand side.
>> ==> There is no positive point with less than ℵo unit fractions
>> at its left-hand side.
>> ==> All positive points with no exception have ℵo unit fractions at
>> their left-hand
>> side.
>> ==> The interval (0, 1] has aleph_0 unit fractions at its left-hand side.
>> ==> Contradiction.
>
> What contradiction?

There is no unit fraction left-hand of the interval.

>> But that does not imply that ℵo unit fractions lie left of the interval.
>
> Why not?

Because they all are positive.
>>
>> But that is nonsense - if all points can be selected as the axiom of
>> choice proposes.
>
> Why?
>
> What in the "Axiom of Choice" says we can choose the "End" of an
> Unbounded set?

It says that we can choose every element. But here almost all remain below
every choice.

Regards, WM

Le 20/12/2023 à 17:42, Richard Damon a écrit :
> On 12/20/23 11:18 AM, WM wrote:
>> Le 20/12/2023 à 13:54, Richard Damon a écrit :
>>> On 12/20/23 5:36 AM, WM wrote:
>>
>>>> Smaller than all eps > 0 that can be chosen. So it is.
>>>
>>> Which isn't a finite number.
>>
>> There are many finite numbers smaller than every esp > 0 that can be
>> chosen.
>
> SO you agree that there is no "smallest" eps that can be used.

Of course.
>
>>> There can not be a "finite number" that is smaller than ALL finite
>>> numbers
>>
>> Choose an eps > 0 with less than almost all unit fractions in (0, eps].
>> Fail.
>
> Why should you be able to?
>
> Choose a number greater than 1/2 of infinity? That is a nonsense question,

because it can't be done. But if all were there and accessible, it could
be done.

Regards, WM

On 12/20/2023 5:12 AM, WM wrote:
> Le 19/12/2023 à 19:15, Jim Burns a écrit :
>> On 12/19/2023 5:42 AM, WM wrote:
>>> Le 18/12/2023 à 17:48, Jim Burns a écrit :
>>>> On 12/18/2023 6:36 AM, WM wrote:

>>>>> I consider Bob leaving the matrix
>>>>> as impossible.
>>>
>>> Exchanging between two matrix-position
>>> will never decrease
>>> the exchanged elements.
>>
>> I think it can be a good and useful thing
>> to say
>> | 2¹ᐟ² is impossible
>> in a discussion about rationals.
>>
>> However, that type of impossibility
>> has no power outside that discussion.
>> 2¹ᐟ² is possible in other discussions.
>>
>> There are matrices ⟨1,…,i⟩×⟨1,…,j⟩ for which
>> Bob leaving ⟨1,…,i⟩×⟨1,…,j⟩ is impossible.
>
> Only those matrices exist,
> because otherwise logic is violated.

2¹ᐟ² _among the rationals_ violates logic.
meaning: 2¹ᐟ² is irrational.

However,
if we're discussing the real numbers, then,
no,
2¹ᐟ² _among the reals_ doesn't violate logic.

Set A ⊇ each ⟨1,…,i⟩×⟨1,…,j⟩
_among sets without internal de.Bob.ification_
violates logic.
meaning: A is infinite.

If it were only finite sets we were discussing,
that would be fine.
However,
we are having a different discussion here,
a discussion in part about set A
Set A is where you've placed your darknessᵂᴹ

> By definition only
> exchanging X and O without loss
> is allowed.

Perhaps you will define 2¹ᐟ² to be rational.
If you do that,
the difficulties you encounter will not be
because you're a secret genius.

----
𝕍×𝕍 is the least upper bound of
all the ⟨1,…,i⟩×⟨1,…,j⟩

Each ⟨1,…,i⟩×⟨1,…,j⟩ subset 𝕍×𝕍

For any set S such that
each ⟨1,…,i⟩×⟨1,…,j⟩ subset S
𝕍×𝕍 subset S

For example,
each ⟨1,…,i⟩×⟨1,…,j⟩ subset A
𝕍×𝕍 subset A

For each ⟨i,j⟩ in 𝕍×𝕍
i ∈ ⟨1,…,i⟩, j ∈ ⟨1,…,j⟩
⟨i,j⟩ is visibleᵂᴹ
Otherwise, 𝕍×𝕍 isn't least.

𝕍×𝕍 is without darkᵂᴹ

For each ⟨i,j⟩ in 𝕍×𝕍
⟨i,j⟩⇄⟨k,1⟩
k = i+(i+j-1)(i+j-2)/2
is in the swaps,
and the swaps are only those swaps.

(1)
⟨k,1⟩ is in 𝕍×𝕍
⟨i,j⟩⇄⟨k,1⟩ is internal to 𝕍×𝕍

(2)
⟨i,j⟩⇄⟨k,1⟩ is first.swap for ⟨k,1⟩
Before ⟨i,j⟩⇄⟨k,1⟩
⟨k,1⟩ holds Xₖ which ⟨k,1⟩ initially held.

(3)
⟨i,j⟩⇄⟨k,1⟩ is last.swap for ⟨i,j⟩
After ⟨i,j⟩⇄⟨k,1⟩
⟨i,j⟩ holds Xₖ which ⟨k,1⟩ initially held.

(1) (2) (3) follow from
not.violating arithmetic.
I can show you my work, or
you should be able to
work it out for yourself.

For each ⟨i,j⟩ in 𝕍×𝕍
after all swaps
⟨i,j⟩ holds Xₖ which ⟨k,1⟩ initially held.

After all swaps,
no initial Oᵢⱼ is in 𝕍×𝕍
𝕍×𝕍 is internally de.Bob.ified.

Bob isn't in darknessᵂᴹ in 𝕍×𝕍
because 𝕍×𝕍 is without darkᵂᴹ

Bob (initial Oᵢⱼ) isn't in A\(𝕍×𝕍)
because each swap has both ends in 𝕍×𝕍

The explanation for internal de.Bob.ification
isn't darknessᵂᴹ

The explanation for internal de.Bob.ification
is that
only some sets are like ⟨1,…,i⟩×⟨1,…,j⟩

Your set A, containing each ⟨1,…,i⟩×⟨1,…,j⟩
is among the sets not like ⟨1,…,i⟩×⟨1,…,j⟩

Set A has internal de.Bob.ification
because, whatever else A contains,
A contains at least 𝕍×𝕍

Each initial Oᵢⱼ is in 𝕍×𝕍
No swap is to A\(𝕍×𝕍)
After all swaps
no initial Oᵢⱼ is in A\(𝕍×𝕍)

Each initial Oᵢⱼ is in 𝕍×𝕍
Each ⟨i,j⟩ in 𝕍×𝕍 has a swap
after which Xₖ is in ⟨i,j⟩
After all swaps
no initial Oᵢⱼ is in 𝕍×𝕍

After all swaps
no initial Oᵢⱼ is in 𝕍×𝕍 or A\(𝕍×𝕍)
A is internally de.Bob.ified.

On 12/20/2023 1:55 AM, WM wrote:
> Le 19/12/2023 à 15:07, Richard Damon a écrit :
>> On 12/19/23 5:25 AM, WM wrote:
>>> Le 18/12/2023 à 17:27, Richard Damon a écrit :
>>>
>>>> Nope, because every unit fraction is distinguishable individually,
>>>> and there are even rational values between them.
>>>
>>> Then not ℵ₀ unit fractions would remain below every eps which can be
>>> used to distinguish them.
>
>> Why?
>>
>> Remember, you are talking about an UNBOUNDED set
>
> Wrong. The set of unit fractions is bounded by 0 and 1.

There is an infinity:

1/1 = 1
1/2 = .5
1/3 = .(3) or 0.3333... whatever...
1/4 = .25
1/5 = .2
1/6 = .1(6)

Notice how they get closer to zero, yet never equal zero at any
iteration? It is unbounded. Afaict, you just have a problem wrt
infinity... ;^)

>
>> so it is inexhaustible.
>
> NUF(x) exists for every point.
>
> Regards, WM
>

On 12/19/2023 6:50 PM, Richard Damon wrote:
> On 12/19/23 9:27 PM, Chris M. Thomasson wrote:
>> On 12/19/2023 6:07 PM, Richard Damon wrote:
>>> On 12/19/23 7:58 PM, Chris M. Thomasson wrote:
>>>> On 12/19/2023 2:16 AM, WM wrote:
>>>>> Le 18/12/2023 à 17:21, Richard Damon a écrit :
>>>>>> On 12/18/23 10:56 AM, WM wrote:
>>>>>>> Le 18/12/2023 à 15:21, Richard Damon a écrit :
>>>>>>>> On 12/18/23 6:19 AM, WM wrote:
>>>>>>>
>>>>>>>> Right, so NUF(x) is infinite for ANY finite number, and thus for
>>>>>>>> ANY unit fraction, so it is NEVER 1 at any, so there is no
>>>>>>>> smallest unit fraction.
>>>>>>>
>>>>>>> This is inconsistent with the property, accepted by you: NUF can
>>>>>>> not increase by more than 1 without being constant for a finite
>>>>>>> period. NUF(0) = 0 implies NUF(x) = 1.
>>>>>>
>>>>>> Yes, at an infintesimally small value of x.
>>>>>
>>>>> No unit fraction is infinitesimally small. But perhaps you call
>>>>> dark values infinitesimally small.
>>>>
>>>> Huh? 1/1, 1/2, 1/3, ...
>>>>
>>>> They do get infinitely small, yet never equal zero...
>>>>
>>>
>>> No, they get unboundedly small, but every one is finite, so it isn't
>>> "infinitely" small. The key is unboundedly small means there is no
>>> smallest item, but is just ever getting smaller, but stays a finitely
>>> describable value.
>>>
>>> NUF(x) = 1 only at an INFINITELY small value, which is smaller than
>>> all UNBOUNDLY small values. The "Infinitely small values" are a
>>> different type of number, NOT part of the standard definition of unit
>>> fractions or the unbounded numeber systems like rational or reals.
>>>
>>> These are similar classes of numbers as that which deals with the
>>> mathematics of Omega and its friends. In other words, the
>>> Transfinites, which WM has admitted are out of this domain of
>>> discussion.
>>>
>>> In one sense, his "dark" numbers could be these transfinites, except
>>> he claims that his "dark numbers" are specifically part of the
>>> Natural Numbers / Unit Fractions (which the Transfinite are not).
>>> Also, the Transfinite numbers are "nameable" and "usable
>>> individually" (just not with sets limited to the finite numbers).
>>>
>>
>> Humm... How about, humm, their ability to go to infinity is unbounded?
>> Is that okay? They do get smaller...
>
> Yes, that could be an ok wording (in my opinion). The key is that the
> unit fractions, like the Natural Numbers, never REACH the
> infintesimal/infinite limit, but approach it without a limiting bound.

Agreed! Thanks Richard. :^)

On 12/20/23 12:31 PM, WM wrote:
> Le 20/12/2023 à 17:38, Richard Damon a écrit :
>> On 12/20/23 11:14 AM, WM wrote:
>
>>> Every positive point has ℵo unit fractions at its left-hand side.
>>> ==> There is no positive point with less than ℵo unit fractions
>>> at its left-hand side.
>>> ==> All positive points with no exception have ℵo unit fractions at
>>> their left-hand
>>> side.
>>> ==> The interval (0, 1] has aleph_0 unit fractions at its left-hand
>>> side.
>>> ==> Contradiction.
>>
>> What contradiction?
>
> There is no unit fraction left-hand of the interval.

Bounded thinking,

>
>>> But that does not imply that ℵo unit fractions lie left of the interval.
>>
>> Why not?
>
> Because they all are positive.

Bounded Thinking.

>>>
>>> But that is nonsense - if all points can be selected as the axiom of
>>> choice proposes.
>>
>> Why?
>>
>> What in the "Axiom of Choice" says we can choose the "End" of an
>> Unbounded set?
>
> It says that we can choose every element. But here almost all remain
> below every choice.

So?

That is just part of the strangeness of Unbounded sets. EVERY element is
choosable, but every element when chosen still has an unlimited number
of elements past it.

Your thoughts of "Darkness" are just your inability to understand
unboundness.

No elements are not chooseable, but since we are finite, e can never get
to the end.

That doesn't make an "end" happen, after which things become "dark", and
there is nothing different about any of those "dark" items, they are
just finite number bigger than you first though of.

>
> Regards, WM
>
>
>

On 12/20/23 4:29 PM, Chris M. Thomasson wrote:
> On 12/19/2023 6:50 PM, Richard Damon wrote:
>> On 12/19/23 9:27 PM, Chris M. Thomasson wrote:
>>> On 12/19/2023 6:07 PM, Richard Damon wrote:
>>>> On 12/19/23 7:58 PM, Chris M. Thomasson wrote:
>>>>> On 12/19/2023 2:16 AM, WM wrote:
>>>>>> Le 18/12/2023 à 17:21, Richard Damon a écrit :
>>>>>>> On 12/18/23 10:56 AM, WM wrote:
>>>>>>>> Le 18/12/2023 à 15:21, Richard Damon a écrit :
>>>>>>>>> On 12/18/23 6:19 AM, WM wrote:
>>>>>>>>
>>>>>>>>> Right, so NUF(x) is infinite for ANY finite number, and thus
>>>>>>>>> for ANY unit fraction, so it is NEVER 1 at any, so there is no
>>>>>>>>> smallest unit fraction.
>>>>>>>>
>>>>>>>> This is inconsistent with the property, accepted by you: NUF can
>>>>>>>> not increase by more than 1 without being constant for a finite
>>>>>>>> period. NUF(0) = 0 implies NUF(x) = 1.
>>>>>>>
>>>>>>> Yes, at an infintesimally small value of x.
>>>>>>
>>>>>> No unit fraction is infinitesimally small. But perhaps you call
>>>>>> dark values infinitesimally small.
>>>>>
>>>>> Huh? 1/1, 1/2, 1/3, ...
>>>>>
>>>>> They do get infinitely small, yet never equal zero...
>>>>>
>>>>
>>>> No, they get unboundedly small, but every one is finite, so it isn't
>>>> "infinitely" small. The key is unboundedly small means there is no
>>>> smallest item, but is just ever getting smaller, but stays a
>>>> finitely describable value.
>>>>
>>>> NUF(x) = 1 only at an INFINITELY small value, which is smaller than
>>>> all UNBOUNDLY small values. The "Infinitely small values" are a
>>>> different type of number, NOT part of the standard definition of
>>>> unit fractions or the unbounded numeber systems like rational or reals.
>>>>
>>>> These are similar classes of numbers as that which deals with the
>>>> mathematics of Omega and its friends. In other words, the
>>>> Transfinites, which WM has admitted are out of this domain of
>>>> discussion.
>>>>
>>>> In one sense, his "dark" numbers could be these transfinites, except
>>>> he claims that his "dark numbers" are specifically part of the
>>>> Natural Numbers / Unit Fractions (which the Transfinite are not).
>>>> Also, the Transfinite numbers are "nameable" and "usable
>>>> individually" (just not with sets limited to the finite numbers).
>>>>
>>>
>>> Humm... How about, humm, their ability to go to infinity is
>>> unbounded? Is that okay? They do get smaller...
>>
>> Yes, that could be an ok wording (in my opinion). The key is that the
>> unit fractions, like the Natural Numbers, never REACH the
>> infintesimal/infinite limit, but approach it without a limiting bound.
>
> Agreed! Thanks Richard. :^)

Which is part of WM's problem, he can't conceive of something that *IS*
finite but arbitrarily small/big, such that there is no "last" in the
sequence.

On 12/20/23 12:34 PM, WM wrote:
> Le 20/12/2023 à 17:42, Richard Damon a écrit :
>> On 12/20/23 11:18 AM, WM wrote:
>>> Le 20/12/2023 à 13:54, Richard Damon a écrit :
>>>> On 12/20/23 5:36 AM, WM wrote:
>>>
>>>>> Smaller than all eps > 0 that can be chosen. So it is.
>>>>
>>>> Which isn't a finite number.
>>>
>>> There are many finite numbers smaller than every esp > 0 that can be
>>> chosen.
>>
>> SO you agree that there is no "smallest" eps that can be used.
>
> Of course.
>>
>>>> There can not be a "finite number" that is smaller than ALL finite
>>>> numbers
>>>
>>> Choose an eps > 0 with less than almost all unit fractions in (0,
>>> eps]. Fail.
>>
>> Why should you be able to?
>>
>> Choose a number greater than 1/2 of infinity? That is a nonsense
>> question,
>
> because it can't be done. But if all were there and accessible, it could
> be done.
>
> Regards, WM
>
>
>

But all ARE "accessible", just unbounded.

The problem is a problem of unbounded numbers, and infinite sets. Since
the size of the set of Natural Numbers (and Unit Fractions) is infinite,
a size of 1/2 that is not definable.

And that is EXACTLY what choose an eps that has more unit fractions
above it then below is asking,

So, you are just admitting that your criteria is incorrect.

On Wednesday, December 20, 2023 at 2:16:34 PM UTC-8, Richard Damon wrote:
> On 12/20/23 12:31 PM, WM wrote:
>>
> >> What in the "Axiom of Choice" says we can choose the "End" of an
> >> Unbounded set?
> >
> > It says that we can choose every element.

Priceless!

Le 20/12/2023 à 23:16, Richard Damon a écrit :

> But all ARE "accessible", just unbounded.

Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀
unit fractions, in a non-vanishing part of the real axis, which cannot be
chosen by any x. How would you choose one of them? Don't claim that it
could be done. Show how it can be done!

Regards, WM

On 12/21/23 5:45 AM, WM wrote:
> Le 20/12/2023 à 23:16, Richard Damon a écrit :
>
>> But all ARE "accessible", just unbounded.
>
> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
> fractions, in a non-vanishing part of the real axis, which cannot be
> chosen by any x. How would you choose one of them? Don't claim that it
> could be done. Show how it can be done!
>
> Regards, WM
>
>

What unit fraction can't be chosen?

Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
that n being a natural number greater than 1/x-1

Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ... are
all smaller, and individually definable.

Note, this can be done for ANY x you choose.

That seems to be your problem, your lack of the ability to access values
aren't caused by the properties of the numbers, but of the ability of
your logic to handle unbounded sets.

This cause you to try to define something not actually definable as your
NUF is just not a properly defined function.

To be a function, it must be a mapping of values in the domain to values
in its range.

You claim this range is finite numbers, but they can not be.

There is no finite value of x (>0) where NUF(x) has a finite value.

Thus, the function is just not defined.

If you accepted the range of an extended number system that included
infinitesimals, then perhaps you could define it, but that then breaks
your concepts, as your "dark" numbers end up just being the
infintesimals, but then we have the fact that once you allow them, they
are nameable and usable individually, so aren't "dark".

Perhaps where you get into a problem is when you try to do you operation
for EVERY x at once, so the problem isn't that there are numbers that
can't be used individually, but your logic can't handle using an
unbounded set collectively, which isn't surprising as it makes claims
that are bounded.

Le 21/12/2023 à 16:35, Richard Damon a écrit :
> On 12/21/23 5:45 AM, WM wrote:
>> Le 20/12/2023 à 23:16, Richard Damon a écrit :
>>
>>> But all ARE "accessible", just unbounded.
>>
>> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
>> fractions, in a non-vanishing part of the real axis, which cannot be
>> chosen by any x. How would you choose one of them? Don't claim that it
>> could be done. Show how it can be done!
>>
>
> What unit fraction can't be chosen?

Those which cannot be chosen, almost all.
>
> Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
> that n being a natural number greater than 1/x-1

and almost all unit fractions, namely ℵo being smaller. You cannot
discern them, separate them (although they are separated by uncountably
many points, none of which you can access), choose them.
>
> Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ... are
> all smaller, and individually definable.
>
> Note, this can be done for ANY x you choose.

Of course, but ℵo unit fractions, almost all unit fractions, remain
unchosen.

They exist inside of the interval (0, 1] and have uncountably many points
between each other. Hence, ther is a first one.

Regards, WM

On 12/21/23 2:35 PM, WM wrote:
> Le 21/12/2023 à 16:35, Richard Damon a écrit :
>> On 12/21/23 5:45 AM, WM wrote:
>>> Le 20/12/2023 à 23:16, Richard Damon a écrit :
>>>
>>>> But all ARE "accessible", just unbounded.
>>>
>>> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
>>> fractions, in a non-vanishing part of the real axis, which cannot be
>>> chosen by any x. How would you choose one of them? Don't claim that
>>> it could be done. Show how it can be done!
>>>
>>
>> What unit fraction can't be chosen?
>
> Those which cannot be chosen, almost all.

If you can't show them, you can't say they exist.

I guess you believe Russel's teapot has been prove to exist.

You are needing to assume their existance without actual proof.

>>
>> Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
>> that n being a natural number greater than 1/x-1
>
> and almost all unit fractions, namely ℵo being smaller. You cannot
> discern them, separate them (although they are separated by uncountably
> many points, none of which you can access), choose them.

So, you can't actually show your unicorn exists.

Your "proof" is you can't see them, so they must be "dark", but they
don't actually need to exist.

All your arguement shows is that they are not bounded, but that was
already known. Your logic system just can't handle unbounded sets, and
thus can't actually work with the Natural Numbers, giving you your errors.

>>
>> Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ...
>> are all smaller, and individually definable.
>>
>> Note, this can be done for ANY x you choose.
>
> Of course, but ℵo unit fractions, almost all unit fractions, remain
> unchosen.

Please describe which one can't be?

Remember, ALL Natural Numbers ARE describable, as all Natural Numbers
are finite, and thus have a finite name, being built by a finite
sequence of steps.

The fact that there is no upper bound to the length of that finite
number, doesn't make it non-finite, and thus unnamable by a finite name.

>
> They exist inside of the interval (0, 1] and have uncountably many
> points between each other. Hence, ther is a first one.

But if there was, then there is also a smalller one (since x/2 will also
be a unit fraction, and smaller than it), and thus your logic goes *BOOM*

Your logic is just shown to be flawed.

>
> Regards, WM

On 12/21/2023 11:35 AM, WM wrote:
> Le 21/12/2023 à 16:35, Richard Damon a écrit :
>> On 12/21/23 5:45 AM, WM wrote:
>>> Le 20/12/2023 à 23:16, Richard Damon a écrit :
>>>
>>>> But all ARE "accessible", just unbounded.
>>>
>>> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
>>> fractions, in a non-vanishing part of the real axis, which cannot be
>>> chosen by any x. How would you choose one of them? Don't claim that
>>> it could be done. Show how it can be done!
>>>
>>
>> What unit fraction can't be chosen?
>
> Those which cannot be chosen, almost all.

Then, name one that cannot be chosen? Humm...

>>
>> Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
>> that n being a natural number greater than 1/x-1
>
> and almost all unit fractions, namely ℵo being smaller. You cannot
> discern them, separate them (although they are separated by uncountably
> many points, none of which you can access), choose them.
>>
>> Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ...
>> are all smaller, and individually definable.
>>
>> Note, this can be done for ANY x you choose.
>
> Of course, but ℵo unit fractions, almost all unit fractions, remain
> unchosen.
>
> They exist inside of the interval (0, 1] and have uncountably many
> points between each other. Hence, ther is a first one.
>
> Regards, WM

On 12/20/2023 8:18 AM, WM wrote:
> Le 20/12/2023 à 13:54, Richard Damon a écrit :
>> On 12/20/23 5:36 AM, WM wrote:
>
>>> Smaller than all eps > 0 that can be chosen. So it is.
>>
>> Which isn't a finite number.
>
> There are many finite numbers smaller than every esp > 0 that can be
> chosen.
>> There can not be a "finite number" that is smaller than ALL finite
>> numbers
>
> Choose an eps > 0 with less than almost all unit fractions in (0, eps].
> Fail.
>
>>> But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo.
>>> That means the increase from 0 to ℵo cannot be seen and discerned.
>>> The axiom of choice is invalid here.
>>
>> Right, and the points it increases are NOT "finite" numbers, and thus
>> NOT the "unit fractions" that you talk about.
>
> How should the number of unit fractions increase before all unit fractions?
> Finest logic!

You must think akin to this crap, being a bit sarcastic here:

Ahhh, nobody will be ever be able to think of this number... Therefore
it does not exist. Oh wait, lets call it dark... ;^)

WM says, that number will never be thought of. Its dark.

On 12/21/2023 7:35 AM, Richard Damon wrote:
> On 12/21/23 5:45 AM, WM wrote:
>> Le 20/12/2023 à 23:16, Richard Damon a écrit :
>>
>>> But all ARE "accessible", just unbounded.
>>
>> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
>> fractions, in a non-vanishing part of the real axis, which cannot be
>> chosen by any x. How would you choose one of them? Don't claim that it
>> could be done. Show how it can be done!
>>
>> Regards, WM
>>
>>
>
> What unit fraction can't be chosen?
>
> Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
> that n being a natural number greater than 1/x-1
>
> Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ... are
> all smaller, and individually definable.
>
> Note, this can be done for ANY x you choose.
>
> That seems to be your problem, your lack of the ability to access values
> aren't caused by the properties of the numbers, but of the ability of
> your logic to handle unbounded sets.
^^^^^^^^^^^^^^^^^^^^^

BIG TIME! ...

;^)

>
> This cause you to try to define something not actually definable as your
> NUF is just not a properly defined function.
>
> To be a function, it must be a mapping of values in the domain to values
> in its range.
>
> You claim this range is finite numbers, but they can not be.
>
> There is no finite value of x (>0) where NUF(x) has a finite value.
>
> Thus, the function is just not defined.
>
> If you accepted the range of an extended number system that included
> infinitesimals, then perhaps you could define it, but that then breaks
> your concepts, as your "dark" numbers end up just being the
> infintesimals, but then we have the fact that once you allow them, they
> are nameable and usable individually, so aren't "dark".
>
> Perhaps where you get into a problem is when you try to do you operation
> for EVERY x at once, so the problem isn't that there are numbers that
> can't be used individually, but your logic can't handle using an
> unbounded set collectively, which isn't surprising as it makes claims
> that are bounded.