On 5/2/2021 9:59 AM, WM wrote:
> William schrieb am Sonntag, 2. Mai 2021 um 16:04:18 UTC+2:
>> On Sunday, May 2, 2021 at 7:51:06 AM UTC-3, WM wrote:
>>> William schrieb am Sonntag, 2. Mai 2021 um 00:42:02 UTC+2:
>>>> ... while there are an infinite number of
>>>> elements that remain, every element is finite, and no finite element remains for every n.
>>
>>> If none remains for every n and no new elements are added, then there must be an n where each one has gone and hence all have gone.
>> No this does not follow and is not true.
>
> It is.
>
>> n is always finite.
>
> That shows that the sequence is potentially infinite

potentially infinite = finite

> only. This is the characteristic property: Being always finite but having no mximum. Actual infinity on the contrary is a number larger than all finite numbers.
>
>> A finite n cannot exhaust an infinite set.
>
> So it is. Every definable step n belongs to the first percent of all steps (if "all" exist).
>
>> There is no n at which all have gone.
>
> That is correct. Therefore your claim that none remains is wrong. At every step n, there remain ℵo elements which will never have gone because they remain at every step.
>
>> At every n there are an infinite number of elements that remain, and an infinite number or steps still to be done.
>
> Very good! But the steps to be done, infinitely many though, are far less than the elements which remain.

not so, you are comparing infinities, a classic mistake of noobs,
remember oo - k = oo for finite k

> Otherwise the set would become exhausted.

no, as this implies a last n

> This is the difference between potential infinity (never ℵo) and actual infinity (ℵo).

rather, finite vs infinite

>
> Regards, WM
>

On Sunday, May 2, 2021 at 7:04:18 AM UTC-7, William wrote:
> On Sunday, May 2, 2021 at 7:51:06 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 2. Mai 2021 um 00:42:02 UTC+2:
> > > ... while there are an infinite number of
> > > elements that remain, every element is finite, and no finite element remains for every n.
>
> > If none remains for every n and no new elements are added, then there must be an n where each one has gone and hence all have gone.
> No this does not follow and is not true. n is always finite. A finite n cannot exhaust an infinite set.
> There is no n at which all have gone. At every n there are an infinite number of elements that remain, and an infinite number or steps still to be done.

This is all merely a rehash of the (centuries-old) fallacy that an infinite set must have an infinite element.

One notable improvement that the 'dark numbers' fad makes is to use two different terms to denote the distinct (but related) concepts of an infinite set and in infinite (natural) number -- the latter receives the appellation 'dark', thus enabling some clearing-up of the confusion for anyone interested enough to notice it.

Unfortunately, the proponents kerfuffle do not likewise distinguish between the two concepts of 'finite': finite sets and finite numbers. Thus they remain insistent on the impossibility of an infinite set of natural numbers each of which is finite.

On 5/2/2021 7:03 AM, WM wrote:
> Sergio schrieb
> am Sonntag, 2. Mai 2021 um 02:02:14 UTC+2:
>> On 5/1/2021 9:16 AM, WM wrote:

>>> Try to find the limit of an infinite digit sequence
>>> without finite definition.
>>
>> like 0.999... ?
>
> That is not an infinite digit sequence but a finite formula
> (8 symbols) defining an infinite digit sequence, i.e.,
> it enabling you to write every digit you wish.

| Bhikkhus, the teaching is merely a vehicle to describe the
| truth. Don’t mistake it for the truth itself. A finger
| pointing at the moon is not the moon. The finger is needed
| to know where to look for the moon, but if you mistake the
| finger for the moon itself, you will never know the real moon.
| -- Thich Nhat Hanh

The 8 symbols are a finger pointing at an infinite digit
sequence. Don't mistake it for an infinite digit sequence.

FredJeffries schrieb am Sonntag, 2. Mai 2021 um 18:52:27 UTC+2:
> On Sunday, May 2, 2021 at 7:04:18 AM UTC-7, William wrote:
> > On Sunday, May 2, 2021 at 7:51:06 AM UTC-3, WM wrote:
> > > William schrieb am Sonntag, 2. Mai 2021 um 00:42:02 UTC+2:
> > > > ... while there are an infinite number of
> > > > elements that remain, every element is finite, and no finite element remains for every n.
> >
> > > If none remains for every n and no new elements are added, then there must be an n where each one has gone and hence all have gone.
> > No this does not follow and is not true. n is always finite. A finite n cannot exhaust an infinite set.
> > There is no n at which all have gone. At every n there are an infinite number of elements that remain, and an infinite number or steps still to be done.
> This is all merely a rehash of the (centuries-old) fallacy that an infinite set must have an infinite element.

Infinite means without end. This does not at all imply an infinite element. The centuries old fallacy of an infinite element? Who has uttered it before Cantor had the idea of completed infinity? Only that has lead him to postulate "whole numbers larger than every finite number".
>
> One notable improvement that the 'dark numbers' fad makes is to use two different terms to denote the distinct (but related) concepts of an infinite set and in infinite (natural) number -- the latter receives the appellation 'dark', thus enabling some clearing-up of the confusion for anyone interested enough to notice it.

Dark natural numbers are not infinite numbers but natural numbers and therefore they are finite .

Regards, WM

Jim Burns schrieb am Sonntag, 2. Mai 2021 um 19:43:07 UTC+2:
> On 5/2/2021 7:03 AM, WM wrote:
> > Sergio schrieb
> > am Sonntag, 2. Mai 2021 um 02:02:14 UTC+2:
> >> On 5/1/2021 9:16 AM, WM wrote:
>
> >>> Try to find the limit of an infinite digit sequence
> >>> without finite definition.
> >>
> >> like 0.999... ?
> >
> > That is not an infinite digit sequence but a finite formula
> > (8 symbols) defining an infinite digit sequence, i.e.,
> > it is enabling you to write every digit you wish.
> | Bhikkhus, the teaching is merely a vehicle to describe the
> | truth. Don’t mistake it for the truth itself. A finger
> | pointing at the moon is not the moon. The finger is needed
> | to know where to look for the moon, but if you mistake the
> | finger for the moon itself, you will never know the real moon.
> |
> -- Thich Nhat Hanh
>
> The 8 symbols are a finger pointing at an infinite digit
> sequence. Don't mistake it for an infinite digit sequence.

Very good!!! I'd wish that not 99 % of all mathematicians were unable to understand this important difference.

Regards, WM

William schrieb am Sonntag, 2. Mai 2021 um 17:26:42 UTC+2:
> On Sunday, May 2, 2021 at 11:59:08 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 2. Mai 2021 um 16:04:18 UTC+2:
>
> >> > There is no n at which all have gone.
>
> > That is correct. Therefore your claim that none remains is wrong.
> I do no claim that none remain. I claim that no element remains forever (but an element may remain a long time),

"no element remains forever" means that every element goes. Since no new elements come in, this is wrong, because an infinite set of cardinality ℵo remains forever - from the first step. Its elements do not go at any n.

Regards, WM

Sergio schrieb am Sonntag, 2. Mai 2021 um 17:32:45 UTC+2:

> > Otherwise the set would become exhausted.
> no, as this implies a last n

Try to learn at least a bit of set theory if you wish to defend it.

Hausdorff: Before that (axiom of choice) it was usual to argue as follows: From the set A to be well-ordered take by arbitrary choice an element and denote it as a0, then from the set A \ {a0} an element a1, then an element from the set A \ {a0, a1} and so on. If the set {a0, a1, a2, ...} is not yet the complete set A, we can choose from A \ {a0, a1, a2, ...} an element aω, then an element aω+1, and so on. This procedure must come to an end, because beyond the set W of ordinal numbers which are mapped on elements of A, there are greater numbers; these obviously cannot be mapped on elements of A.
>
That is what Jerabek, Hamkins, and Cohen claim. Hamkins has explicitly endorsed the above method. And Jerabek has the effrontery to claim that this does not violate the Peano axioms, because here not natural numbers but ordinal numbers are used.

Regards, WM

On Sunday, May 2, 2021 at 2:56:15 PM UTC-3, WM wrote:
1m that none remain. I claim that no element remains forever (but an element may remain a long time),
> "no element remains forever" means that every element goes. Since no new elements come in, this is wrong, because an infinite set of cardinality ℵo remains forever

Not quite. "An infinite set of cardinality ℵo remains forever" is wrong as it implies the set does not change.
Correct is : The set of elements Y(n) that remains always has infinite cardinality. However this set does change, elements leave the set. There is no element in Y(n) that does not leave eventually. This does not mean that there is a step n where all elements of Y(n) leave because Y(n) does not have a last element.

--William Hughes

On 5/2/2021 12:47 PM, WM wrote:
> FredJeffries schrieb am Sonntag, 2. Mai 2021 um 18:52:27 UTC+2:
>> On Sunday, May 2, 2021 at 7:04:18 AM UTC-7, William wrote:
>>> On Sunday, May 2, 2021 at 7:51:06 AM UTC-3, WM wrote:
>>>> William schrieb am Sonntag, 2. Mai 2021 um 00:42:02 UTC+2:
>>>>> ... while there are an infinite number of
>>>>> elements that remain, every element is finite, and no finite element remains for every n.
>>>
>>>> If none remains for every n and no new elements are added, then there must be an n where each one has gone and hence all have gone.
>>> No this does not follow and is not true. n is always finite. A finite n cannot exhaust an infinite set.
>>> There is no n at which all have gone. At every n there are an infinite number of elements that remain, and an infinite number or steps still to be done.
>> This is all merely a rehash of the (centuries-old) fallacy that an infinite set must have an infinite element.
>
> Infinite means without end. This does not at all imply an infinite element. The centuries old fallacy of an infinite element? Who has uttered it before Cantor had the idea of completed infinity? Only that has lead him to postulate "whole numbers larger than every finite number".
>>
>> One notable improvement that the 'dark numbers' fad makes is to use two different terms to denote the distinct (but related) concepts of an infinite set and in infinite (natural) number -- the latter receives the appellation 'dark', thus enabling some clearing-up of the confusion for anyone interested enough to notice it.
>
> Dark natural numbers are not infinite numbers but natural numbers and therefore they are finite .
>
> Regards, WM
>
>

how does a "dark number" know it is a natural number, and not a rational
or irrational number ?

William schrieb am Sonntag, 2. Mai 2021 um 20:14:51 UTC+2:

> Correct is : The set of elements Y(n) that remains always has infinite cardinality.
> However this set does change, elements leave the set.

But no elements enter! So the set shrinks but not to a subset of less than aleph_0 elements. If aleph_0 elements remain, then also less elements remain, then afinite number remains, then also one element remains.

> There is no element in Y(n) that does not leave eventually.

To leave is the complement of to remain. If there is no element that does not leave, then every element does leave, then no element remains. That is wrong.

> This does not mean that there is a step n where all elements of Y(n) leave because Y(n) does not have a last element.

We talk about *all* steps in set theory as we talk about *all elements* of an infinite set without last element. Here the result is that for *all steps* aleph_0 elements remain. These elements make up a subset Y which does not have a last element and does not have first element. But it has for all steps aleph_0 elements, aleph_0 of which are the same remaining from the first step during all steps.

Regards, WM

Sergio schrieb am Sonntag, 2. Mai 2021 um 20:51:40 UTC+2:
> On 5/2/2021 12:47 PM, WM wrote:

> > Dark natural numbers are not infinite numbers but natural numbers and therefore they are finite .
> >
> how does a "dark number" know it is a natural number, and not a rational
> or irrational number ?

Numbers don't know anything. We can talk about dark numbers, naturals, rationals, irrationals, complex and so on.

Regards, WM

WM has brought this to us :
> William schrieb am Sonntag, 2. Mai 2021 um 20:14:51 UTC+2:
>
>> Correct is : The set of elements Y(n) that remains always has infinite
>> cardinality. However this set does change, elements leave the set.
>
> But no elements enter! So the set shrinks but not to a subset of less than
> aleph_0 elements.

To avoid shrinkage, wash your sets gently in cold water.

On Sunday, May 2, 2021 at 5:32:45 PM UTC-3, WM wrote:
> William schrieb am Sonntag, 2. Mai 2021 um 20:14:51 UTC+2:

> > This does not mean that there is a step n where all elements of Y(n) leave because Y(n) does not have a last element.

> We talk about *all* steps in set theory as we talk about *all elements* of an infinite set without last element. Here the result is that for *all steps* aleph_0 elements remain.

Indeed, for every n there is a set Y(n) of elements that remain. Y(n) contains aleph_0 elements. Y(n)
has a first element but no last element

> These elements make up a subset Y which does not have a last element and does not have first element

Y consists of natural numbers. The natural numbers are well-ordered, so the only set of natural numbers which does
not have a first element is the empty set.

>. But [Y] has for all steps aleph_0 elements,

No, we can not show this. We know that Y(n) has aleph_0 elements not that the subset Y has aleph_0 elements. But we also know that Y(n) has aleph_0 elements that do not remain for every n (i.e. are not elements of Y). So there is no need for the subset Y to have aleph_0 elements to insure that Y(n) has aleph_0 elements.

zelos...@gmail.com schrieb am Montag, 3. Mai 2021 um 06:59:38 UTC+2:
> >Or Zermelo: Every set can be well-ordered. Of course that is impossible.
>
> Nope, it is very possible and is provably so. Of course it is equivalent toa xiom of choice so if you reject it this one does not follow but in ZFC, all sets can be well ordered

This shows that ZFC is nonsense. Not all real numbers can be chosen because they have no names. Ordering without choosing is impossible in rational thinking. The axiom of choise is as silly as the axiom that 3 even prime numbers exist. Of course this could be "proven" with such an axiom. But to state this axiom would be the same nonsense as the axiom of choice for uncountable sets. Of course the axiom is true since there are no uncountable sets.

Regards, WM

William schrieb am Montag, 3. Mai 2021 um 02:34:20 UTC+2:
> On Sunday, May 2, 2021 at 5:32:45 PM UTC-3, WM wrote:
> > William schrieb am Sonntag, 2. Mai 2021 um 20:14:51 UTC+2:
>
>
> > > This does not mean that there is a step n where all elements of Y(n) leave because Y(n) does not have a last element.
>
> > We talk about *all* steps in set theory as we talk about *all elements* of an infinite set without last element. Here the result is that for *all steps* aleph_0 elements remain.
> Indeed, for every n there is a set Y(n) of elements that remain. Y(n) contains aleph_0 elements. Y(n)
> has a first element but no last element
> > These elements make up a subset Y which does not have a last element and does not have first element
> Y consists of natural numbers. The natural numbers are well-ordered, so the only set of natural numbers which does
> not have a first element is the empty set.

The dark numbers are not well-ordered.
>
> >. But [Y] has for all steps aleph_0 elements,
>
> No, we can not show this. We know that Y(n) has aleph_0 elements not that the subset Y has aleph_0 elements. But we also know that Y(n) has aleph_0 elements that do not remain for every n (i.e. are not elements of Y).

Wrong. Never will aleph_0 elements have left. (Note that aleph_0 is a number larger than *all* finite numbers.)

> So there is no need for the subset Y to have aleph_0 elements to insure that Y(n) has aleph_0 elements.

The other way round: For all definable n, Y(n) has aleph_0 elements. This shows that the set Y, the intersection of all Y(n) has aleph_0 elements.

Regards, WM

On Monday, May 3, 2021 at 7:05:28 AM UTC-3, WM wrote:
> William schrieb am Montag, 3. Mai 2021 um 02:34:20 UTC+2:
> > On Sunday, May 2, 2021 at 5:32:45 PM UTC-3, WM wrote:

R(k_0) is a set of natural numbers. Let Y be the subset of R(k_0) with the property that if x is an element of Y then x is an element of R(k_n) for every n.

> The dark numbers are not well-ordered

but the natural numbers are, and Y, a subset of a set of natural numbers, is a set of natural numbers. Y does not have a first element. The only set of natural numbers with no first element is the empty set.

--
William Hughes