On 12/14/2023 2:31 PM, WM wrote:
> On 14.12.2023 20:00, Jim Burns wrote:
>> On 12/14/2023 5:16 AM, WM wrote:

>>> nothing that
>>> could disprove the fact that
>>> by exchanging X and O in
>>>
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ...
>>>
>>> the matrix will never be covered with X.
>>> Simple geometry.
>>
>> "Geometry", he says.
>> Geometry is thronging with
>> sets you deny exist.
>
> Sets exist and
> their elements remain existing when
> redistributing them.

Sets exist and,
for some sets, some proper subsets
match their supersets.

In those superset-matching-instances,
redistributing the proper subset
to its whole superset
will cover some of the elements.

Performing those redistributions would be
a supertask of redistribution.
We cannot perform
a supertask of redistribution.

However,
it's enough for us to perform the _task_ of
making descriptive and not.first.false
claims.

Because we cannot perform a supertask,
there are only finitely.many claims we've made,
and so, among only not-first-false claims,
there cannot be any false claims.

Irony:
It is because we are finite that we know that
our claims are correct about infinity.

>> Sets which are the same "size" as
>> proper subsets is the explanation.
>> It is Cantor's explanation.
>
> It is wrong.
> Consider the unit fractions.
> For every x > 0 that you choose
> you know that
> infinitely many unit fractions are smaller.

Our choosing and our not-choosing
does not change our description of reals,
rationals, integers, naturals, or
unit.fractions to a different description.

Our choosing and our not-choosing
does not change the visibly.not.first.false
to anything less visibly.not.first.false.

Perhaps you're depending on an unstated
| since we aren't infinite beings,
| we can't choose infinitely-many things.

I agree that we finite beings can't choose
infinitely.many things.
But that's not how we know these facts.

> What can you do to distinguish them
> in order to enumerate them? Nothing!
> No chosen x will separate them.

I agree that we finite beings can't distinguish
infinitely.many things.
But that's not how we know these facts.

I agree that we finite beings can't enumerate
infinitely.many things.
But that's not how we know these facts.

I agree that we finite beings can't separate
infinitely.many things.
But that's not how we know these facts.

Chris M. Thomasson submitted this idea :
> On 12/7/2023 4:20 AM, FromTheRafters wrote:
>> on 12/7/2023, WM supposed :
>>> William schrieb am Mittwoch, 6. Dezember 2023 um 18:43:56 UTC+1:
>>>  > On Wednesday, December 6, 2023 at 12:08:01 PM UTC-4, WM wrote:
>>>  >
>>>  > > > Your game of billiards does not define the putative matrix A
>>>  > > Why should that be relevant?
>>>  > You repeatedly claim that B is a proper submatrix of the putative A.
>>>
>>> Irrelevant claim. Only this is relevant: My game of billiards defines and
>>> places every natural number that is defined by
>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> Are you saying that the pairing formula 'defines' the natural numbers?
>
> And vise versa. It's the full circle mapping... From natural number, to
> unique pair and back again; nothing is lost! So they basically, both define
> themselves in and of themselves? Is that fair enough in a sense?

Not to my way of thinking. It would be circular to say that positive
rationals are 'defined' as ordered pairs of naturals and then to define
naturals in terms of those positive rationals.

OTOH it is still an equivalence relation which is invertable, a
bijection between sets.

On Thursday, December 14, 2023 at 3:24:58 PM UTC-4, WM wrote:
> On 14.12.2023 17:34, William wrote:
> > On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
> >> On 14.12.2023 16:45, William wrote:
> >>> A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.
> >> You are wrong. The algorithm is given in the OP.
> > Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,
> What defines
A. I.e. given (n,m) an element of |Nx|N how can you determine the value A[m,n]. Your game of billiards does not give a method.

On 15.12.2023 09:25, William wrote:
> On Thursday, December 14, 2023 at 3:24:58 PM UTC-4, WM wrote:
>> On 14.12.2023 17:34, William wrote:
>>> On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
>>>> On 14.12.2023 16:45, William wrote:
>>>>> A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.
>>>> You are wrong. The algorithm is given in the OP.
>>> Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,
>> What defines
> A. I.e. given (n,m) an element of |Nx|N how can you determine the value A[m,n]. Your game of billiards does not give a method.

Dark elements cannot be determined as individuals.
The same problem is inherent in enumerating the algebraic numbers.
The same problem is inherent in enumerating the fractions.
For every eps > 0 there are infinitely many fractions between 0 and eps.
How should they be indexed?
All those tasks cannot be executed.

Regards, WM

On 07.12.2023 13:20, FromTheRafters wrote:
> on 12/7/2023, WM supposed :
>> My game of billiards defines
>> and places every natural number that is defined by
>> k = (m + n - 1)(m + n - 2)/2 + m.
>
> Are you saying that the pairing formula 'defines' the natural numbers?

The index k of the position (m, n) is defined by the choice of m and n.
For dark positions (that cannot b chosen) no index can be defined.
Therefore the formula does ot define the whole matrix B.

Regards, WM

On 15.12.2023 02:00, Richard Damon wrote:
> On 12/14/23 10:32 AM, WM wrote:

>> It involves the realization of all natural numbers
>> k = (m + n - 1)(m + n - 2)/2 + m
>> at matrix places
>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
>> 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>>
>> This requires a limit or a last step. Your belief alone is not
>> sufficient.

> So, what IS the "last Step" that is required?
>
> You don't seem to understand that Natural Numbers are Unbounded, and
> thus there isn't a "last" one.

Therefore the complete indexing can only happen in the limit or never.
In fact it will never happen.

Simple example: For all x > 0 that can be chosen there exist infinitely
many smaller fractions. They cannot be distinguished. They cannot be
indexed.

Regards, WM

On 15.12.2023 05:49, Jim Burns wrote:
> On 12/14/2023 2:31 PM, WM wrote:

> Because we cannot perform a supertask,
> there are only finitely.many claims we've made,

How do you enumerate the algebraic numbers?

> Irony:
> It is because we are finite that we know that
> our claims are correct about infinity.

No, because your are stupid, you believe that.
>
>>> Sets which are the same "size" as
>>> proper subsets is the explanation.
>>> It is Cantor's explanation.
>>
>> It is wrong.
>> Consider the unit fractions.
>> For every x > 0 that you choose
>> you know that
>> infinitely many unit fractions are smaller.
>
> Our choosing and our not-choosing
> does not change our description of reals,
> rationals, integers, naturals, or
> unit.fractions to a different description.

But it separates those that can be chosen as individuals from the dark
numbers.
>
> Our choosing and our not-choosing
> does not change the visibly.not.first.false
> to anything less visibly.not.first.false.
>
> Perhaps you're depending on an unstated
> | since we aren't infinite beings,
> | we can't choose infinitely-many things.

But Cantor claims that. Dedekind, the last pupil of Gauss, it is a
shame!, claims it too.
>
> I agree that we finite beings can't choose
> infinitely.many things.
> But that's not how we know these facts.

But that is required to enumerate the fractions smaller than every eps
or the algebraic numbers.
>
>> What can you do to distinguish them
>> in order to enumerate them? Nothing!
>> No chosen x will separate them.
>
> I agree that we finite beings can't distinguish
> infinitely.many things.

That is not the problem! Simply try to distinguish any unit fraction
which has not infinitely many smaller ones. Such unit fractions must
exist, because they occupy separate points in a linear order. Never
infinitely many sit at the same point.

> I agree that we finite beings can't enumerate
> infinitely.many things.
> But that's not how we know these facts.

But that is how Dedekind tried to enumerate the algebraic numbers. You
know Dedekind's approach? If not you can learn it here:
https://www.hs-augsburg.de/~mueckenh/HI/HI11.PPT, p.19.

Regards, WM

On Friday, December 15, 2023 at 6:10:03 AM UTC-4, WM wrote:
> On 15.12.2023 09:25, William wrote:
> > On Thursday, December 14, 2023 at 3:24:58 PM UTC-4, WM wrote:
> >> On 14.12.2023 17:34, William wrote:
> >>> On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
> >>>> On 14.12.2023 16:45, William wrote:
> >>>>> A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.
> >>>> You are wrong. The algorithm is given in the OP.
> >>> Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,
> >> What defines
> > A. I.e. given (n,m) an element of |Nx|N how can you determine the value A[m,n]. Your game of billiards does not give a method.
> Dark elements

Are not elements of the set of natural numbers. Your game of billiards does not provide an algorithm that defines A.

On 15.12.2023 11:57, William wrote:
> On Friday, December 15, 2023 at 6:10:03 AM UTC-4, WM wrote:

>> Dark elements
>
> Are not elements of the set of natural numbers.

Then there are not ℵo natural numbers.
All visible elements belong to an increasing but always finite set.
All visible unit fractions belong to [eps, 1]. If you accept ℵo unit
fractions, then they are not individually choosable. They are dark.

> Your game of billiards does not provide an algorithm that defines A.

Dedekinds enumeration of the algebraic numbers doesn't either.

Regards, WM

On 12/15/23 5:22 AM, WM wrote:
> On 15.12.2023 02:00, Richard Damon wrote:
>> On 12/14/23 10:32 AM, WM wrote:
>
>>> It involves the realization of all natural numbers
>>> k = (m + n - 1)(m + n - 2)/2 + m
>>> at matrix places
>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
>>> 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>>>
>>> This requires a limit or a last step. Your belief alone is not
>>> sufficient.
>
>> So, what IS the "last Step" that is required?
>>
>> You don't seem to understand that Natural Numbers are Unbounded, and
>> thus there isn't a "last" one.
>
> Therefore the complete indexing can only happen in the limit or never.
> In fact it will never happen.

No, it happens "at the limit" since to set is countably infinite.

Your mind just doens't seem to understand the meaning of Unbounded.

>
> Simple example: For all x > 0 that can be chosen there exist infinitely
> many smaller fractions. They cannot be distinguished. They cannot be
> indexed.
>

If x is your unit fractions or the rationals, then they CAN be indexed,
just not from lowest to highest.

Your problem is you think just because one indexing method doesn't work
means that no indexing method works.

> Regards, WM
>

On 12/15/23 5:17 AM, WM wrote:
> On 07.12.2023 13:20, FromTheRafters wrote:
>> on 12/7/2023, WM supposed :
>>> My game of billiards defines and places every natural number that is
>>> defined by
>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> Are you saying that the pairing formula 'defines' the natural numbers?
>
> The index k of the position (m, n) is defined by the choice of m and n.
> For dark positions (that cannot b chosen) no index can be defined.
> Therefore the formula does ot define the whole matrix B.
>
> Regards, WM

The problem is that there are no positions that can not be chosen.

That is your false assumption.

Every position has some definite m and n, thus every position can be
definitely indexed by a k, so no position is "dark"

You are just proving that you brain is to small.

On 12/15/2023 5:35 AM, WM wrote:
> On 15.12.2023 05:49, Jim Burns wrote:
>> On 12/14/2023 2:31 PM, WM wrote:

>>> What can you do to distinguish them
>>> in order to enumerate them? Nothing!
>>> No chosen x will separate them.
>>
>> I agree that we finite beings can't distinguish
>> infinitely.many things.
>
> That is not the problem!
> Simply try to distinguish
> any unit fraction which has
> not infinitely many smaller ones.
> Such unit fractions must exist,
> because
> they occupy separate points in a linear order.

No.

For each unit.fraction ⅟n
the same ⅟n...

is in unit.fractions toward ⅟1
which can't match proper subsets
which behave "normally" like
herds of sheep, bags of pebbles, etc,
(each unit fraction is visibleᵂᴹ)
and
is in unit fractions toward.not-reaching 0
which are not.like.sheep
which can match proper subsets,
thus can disappear Bob.
(unit fractions "must" contain darkᵂᴹ)

Each unit fraction
among the must.include.darkᵂᴹ unit.fractions
is not a darkᵂᴹ unit.fraction.

So, no.

> Never infinitely many sit at the same point.

They aren't sitting at the same point.
They are 1.by.1;1.end.ed

WM has brought this to us :
> On 15.12.2023 05:49, Jim Burns wrote:
>> On 12/14/2023 2:31 PM, WM wrote:
>
>> Because we cannot perform a supertask,
>> there are only finitely.many claims we've made,
>
> How do you enumerate the algebraic numbers?

All at once.

Le 15/12/2023 à 18:27, FromTheRafters a écrit :
> WM has brought this to us :
>> On 15.12.2023 05:49, Jim Burns wrote:
>>> On 12/14/2023 2:31 PM, WM wrote:
>>
>>> Because we cannot perform a supertask,
>>> there are only finitely.many claims we've made,
>>
>> How do you enumerate the algebraic numbers?
>
> All at once.

Not true, impossible. All at once is an oximoronic statement.

Le 15/12/2023 à 18:27, FromTheRafters a écrit :
> WM has brought this to us :
>> On 15.12.2023 05:49, Jim Burns wrote:
>>> On 12/14/2023 2:31 PM, WM wrote:
>>
>>> Because we cannot perform a supertask,
>>> there are only finitely.many claims we've made,
>>
>> How do you enumerate the algebraic numbers?
>
> All at once.

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Le 15/12/2023 à 21:52, Jim Reburns a écrit :
> Le 15/12/2023 à 18:27, FromTheRafters a écrit :
>> WM has brought this to us :
>>> On 15.12.2023 05:49, Jim Burns wrote:
>>>> On 12/14/2023 2:31 PM, WM wrote:
>>>
>>>> Because we cannot perform a supertask,
>>>> there are only finitely.many claims we've made,
>>>
>>> How do you enumerate the algebraic numbers?
>>
>> All at once.
>
> Not true, impossible. All at once is an oximoronic statement.

Enumerating is a stepwise process. For algebraic numbers the polynomials
of height n have to be searched one after the other.

Regards, WM

Le 15/12/2023 à 18:58, Jim Burns a écrit :
> On 12/15/2023 5:35 AM, WM wrote:
>> On 15.12.2023 05:49, Jim Burns wrote:
>>> On 12/14/2023 2:31 PM, WM wrote:
>
>>>> What can you do to distinguish them
>>>> in order to enumerate them? Nothing!
>>>> No chosen x will separate them.
>>>
>>> I agree that we finite beings can't distinguish
>>> infinitely.many things.
>>
>> That is not the problem!
>> Simply try to distinguish
>> any unit fraction which has
>> not infinitely many smaller ones.
>> Such unit fractions must exist,
>> because
>> they occupy separate points in a linear order.
>
> No.

Yes. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n.
∀x ∈ (0, 1]: NUF(x) = ℵo is wrong.

Regards, WM

Le 15/12/2023 à 13:53, Richard Damon a écrit :
> On 12/15/23 5:22 AM, WM wrote:
>> Therefore the complete indexing can only happen in the limit or never.
>> In fact it will never happen.
>
> No, it happens "at the limit" since to set is countably infinite.

Nothing happens at the limit i.e., after all finite terms.
>
>> Simple example: For all x > 0 that can be chosen there exist infinitely
>> many smaller fractions. They cannot be distinguished. They cannot be
>> indexed.
>>
>
> If x is your unit fractions or the rationals, then they CAN be indexed,
> just not from lowest to highest.

Show us how you do that for the ℵ₀ unit fractions which are smaller
than every chosen x.
Regards, WM

Le 15/12/2023 à 13:55, Richard Damon a écrit :
> On 12/15/23 5:17 AM, WM wrote:

>> The index k of the position (m, n) is defined by the choice of m and n.
>> For dark positions (that cannot b chosen) no index can be defined.
>> Therefore the formula does ot define the whole matrix B.
>>
> The problem is that there are no positions that can not be chosen.

You don't see them. But they are there.
Try to identify those which I cannot identify because they are larger than
every identified natural number:

> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

> That is your false assumption.

No that is mathematics, contrary to the belief of matheologial cranks.

Regards, WM

On 12/15/2023 2:17 AM, WM wrote:
> On 07.12.2023 13:20, FromTheRafters wrote:
>> on 12/7/2023, WM supposed :
>>> My game of billiards defines and places every natural number that is
>>> defined by
>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> Are you saying that the pairing formula 'defines' the natural numbers?
>
> The index k of the position (m, n) is defined by the choice of m and n.

Huh? k is the index and it pairs into (m, n). Now, (m, n) can be brought
right back to k. See? Full circle... Wow.

> For dark positions (that cannot b chosen) no index can be defined.
> Therefore the formula does ot define the whole matrix B.
>
> Regards, WM

On 12/15/2023 4:42 PM, Chris M. Thomasson wrote:
> On 12/15/2023 2:17 AM, WM wrote:
>> On 07.12.2023 13:20, FromTheRafters wrote:
>>> on 12/7/2023, WM supposed :
>>>> My game of billiards defines and places every natural number that is
>>>> defined by
>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>
>>> Are you saying that the pairing formula 'defines' the natural numbers?
>>
>> The index k of the position (m, n) is defined by the choice of m and n.
>
> Huh? k is the index and it pairs into (m, n). Now, (m, n) can be brought
> right back to k. See? Full circle... Wow.

If you define (m, n) then it has an index k. To get all of them, a
simple way is to take k from 0 to infinity, take all of the pairs. See
the order? Heck I even made some music with the notes using two
different instruments. This goes on forever, no limit in sight!

>
>
>> For dark positions (that cannot b chosen) no index can be defined.
>> Therefore the formula does ot define the whole matrix B.
>>
>> Regards, WM
>

On 12/15/23 5:32 PM, WM wrote:
> Le 15/12/2023 à 18:58, Jim Burns a écrit :
>> On 12/15/2023 5:35 AM, WM wrote:
>>> On 15.12.2023 05:49, Jim Burns wrote:
>>>> On 12/14/2023 2:31 PM, WM wrote:
>>
>>>>> What can you do to distinguish them
>>>>> in order to enumerate them? Nothing!
>>>>> No chosen x will separate them.
>>>>
>>>> I agree that we finite beings can't distinguish
>>>> infinitely.many things.
>>>
>>> That is not the problem!
>>> Simply try to distinguish
>>> any unit fraction which has
>>> not infinitely many smaller ones.
>>> Such unit fractions must exist,
>>> because
>>> they occupy separate points in a linear order.
>>
>> No.
>
> Yes. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n.
> ∀x ∈ (0, 1]: NUF(x) = ℵo is wrong.
>
> Regards, WM

Then name a value that contradicts it!

Your concept of the function says it only has "finite values>0" for
input values that can not be used as individual numbers, and thus,
nowhere is your NUF finite and not 0.

You problem is that you logic can't handle what you want to talk about,
so rather than the issue being with what you are trying to talk about,
the actual issue is with your logic.

Your "Dark Numbers" are just frm the "darkness" of your logic not being
up to the task you are using it for.

On 12/15/23 5:09 AM, WM wrote:
> On 15.12.2023 09:25, William wrote:
>> On Thursday, December 14, 2023 at 3:24:58 PM UTC-4, WM wrote:
>>> On 14.12.2023 17:34, William wrote:
>>>> On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
>>>>> On 14.12.2023 16:45, William wrote:
>>>>>> A matrix is defined if there is an algorithm which gives the
>>>>>> value, given an element of |Nx|N. There is no algorithm for your
>>>>>> putative A.
>>>>> You are wrong. The algorithm is given in the OP.
>>>> Nope, your billiards game defines A(n) for each n. Your billiards
>>>> game does not define A,
>>> What defines
>> A.   I.e. given (n,m) an element of |Nx|N how can you determine the
>> value A[m,n].  Your game of billiards does not give a method.
>
> Dark elements cannot be determined as individuals.
> The same problem is inherent in enumerating the algebraic numbers.
> The same problem is inherent in enumerating the fractions.
> For every eps > 0 there are infinitely many fractions between 0 and eps.
> How should they be indexed?
> All those tasks cannot be executed.
>
> Regards, WM
>

And thus, "Darkness" isn't a property of the Numbers, but of the method
being use to look at them.

Every Natural Number is describable individually, we just can't describe
them all at once with your methods. Thus, your logic just fails to
handle unbounded sets of items.

On 12/15/23 5:43 PM, WM wrote:
> Le 15/12/2023 à 13:53, Richard Damon a écrit :
>> On 12/15/23 5:22 AM, WM wrote:
>
>>> Therefore the complete indexing can only happen in the limit or never.
>>> In fact it will never happen.
>>
>> No, it happens "at the limit" since to set is countably infinite.
>
> Nothing happens at the limit i.e., after all finite terms.
>>
>>> Simple example: For all x > 0 that can be chosen there exist
>>> infinitely many smaller fractions. They cannot be distinguished. They
>>> cannot be indexed.
>>>
>>
>> If x is your unit fractions or the rationals, then they CAN be
>> indexed, just not from lowest to highest.
>
> Show us how you do that for the ℵ₀ unit fractions which are smaller than
> every chosen x.
>
> Regards, WM
>
>
> Given n such that 1/n+1 < x <= 1/n, (or n == 0 if 1 < x) then the ℵ₀
unit fractions smaller than that x are the set of 1/(n+k) for k = 1, 2,
3, ...

That IS an unbounded enumeration that visits EVERY unit fraction smaller
than x. It enumerates them from highest to lowest, since the unit
fractions form a decreasing series.

On 12/15/23 5:48 PM, WM wrote:
> Le 15/12/2023 à 13:55, Richard Damon a écrit :
>> On 12/15/23 5:17 AM, WM wrote:
>
>>> The index k of the position (m, n) is defined by the choice of m and n.
>>> For dark positions (that cannot b chosen) no index can be defined.
>>> Therefore the formula does ot define the whole matrix B.
>>>
>> The problem is that there are no positions that can not be chosen.
>
> You don't see them. But they are there.
> Try to identify those which I cannot identify because they are larger
> than every identified natural number:
>
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
>> That is your false assumption.
>
> No that is mathematics, contrary to the belief of matheologial cranks.
>
> Regards, WM
>
>

So, you need to misquote to make you point? Shows the "strength" of your
arguement.

I said:
The problem is that there are no positions that can not be chosen.

And that means that there is no element of ℕ that can't become a member
of ℕ_def, and thus, there are no member of ℕ that are always left
behind, and thus no elements left to be "dark".

You say ℕ_def is "potentially" infinite, which means "Dark Numbers" are
"potentially non-existent", and since the largest (and ony comlete)
ℕ_def IS "infinite" your "Dark Numbers" are the empty set.

And ℕ_def that isn't infinite, isn't the complete set of possible
entries for it, as n+1 is definable, but excluded from the set of
"definable" numbers, so that set is incomplete.

All this shows is that you construction logic can't handle the set you
are trying to work with, and thus has failed itself.