WM considers a process where at every step n the interval (1/(n+1),1/n] is removed from the interval (0,1] . It is immediately obvious that for any element of the complete ordered field, x, there is an m(x) such that x is an element of (1/(m(x) +1),1/m(x)]. WM terms this removing "individually". He notes that for ever step n, the set S_n-(0,1]\ (1/n,1] contains an infinite number of unit fractions. However, if (0,1] contains only elements of the complete ordered field there can be no element in S_n for every n, and thus the intersection will be empty. WM considers this to be a contradiction. His "solution" is to have the set of real numbers, i.e. members of the complete ordered field, (01] to contain other elements. He calls these elements "dark real numbers" [despite the name they are not real numbers, as they do not have the properties of an element of the complete ordered field]. Note the above process cannot remove "dark real numbers", so the intersection of the S_n will not be empty.

WM also considers a method of removing all the elements of (0,1] in a single step. He terms this removing "collectively". Clearly, "collective: removal can remove dark elements as well, so collective removal is more powerful that individual removal.

My criticism is twofold. Firstly, even if this "contradiction" were real, the cure, adding elements that are not real number to a set of real numbers,
is worse than the disease. Secondly, besides WM's imagined contradiction, there is not one iota of evidence for the "dark real numbers"

-- William Hughes

William schrieb am Freitag, 14. Mai 2021 um 21:15:28 UTC+2:

Thank you for considering this topic seriously!

> WM considers a process where at every step n the interval (1/(n+1),1/n] is removed from the interval (0,1] . It is immediately obvious that for any element of the complete ordered field, x, there is an m(x) such that x is an element of (1/(m(x) +1),1/m(x)].

Yes, every real number of (0, 1] that can be addressed or chosen belongs to one of the intervals.

> WM terms this removing "individually". He notes that for ever step n, the set S_n-(0,1]\ (1/n,1] contains an infinite number of unit fractions. However, if (0,1] contains only elements of the complete ordered field there can be no element in S_n for every n, and thus the intersection will be empty..

If (0, 1] contains only real numbers which can be addressed or chosen, then the intersection will be empty. On the other hand it is impossible to chose an n such that |S_n| < ℵo. Hitherto this has been accepted without much ado as natural: CardLim need not be LimCard. Why this is the case has, to my knwoledge, never been asked, let alone been answered,

> WM considers this to be a contradiction. His "solution" is to have the set of real numbers, i.e. members of the complete ordered field, (01] to contain other elements. He calls these elements "dark real numbers" [despite the name they are not real numbers, as they do not have the properties of an element of the complete ordered field]. Note the above process cannot remove "dark real numbers", so the intersection of the S_n will not be empty.

No-one can remove these numbers. One can only accept that in this case CardLim =/= LimCard.
>
> WM also considers a method of removing all the elements of (0,1] in a single step. He terms this removing "collectively". Clearly, "collective: removal can remove dark elements as well, so collective removal is more powerful that individual removal.
>
> My criticism is twofold. Firstly, even if this "contradiction" were real, the cure, adding elements that are not real number to a set of real numbers,
> is worse than the disease.

It is everyone's decision whether to accept dark numbers or to waste them. I am not sure myself what is the correct choice. But the result would be gaps on the real line. Between the smallest 1/n and 0 there would not exist ℵo unit fractions, ℵo of which are undefinable, but simply nothing, a gap.

How would you remedy this problem?

> Secondly, besides WM's imagined contradiction, there is not one ioa of evidence for the "dark real numbers"

But there is evience that every addressed unit fraction differes from 0 and therefore leaves space which could be occupied by ℵo unit fractions.

Regards, WM

On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:

> ... the result would be gaps on the real line. Between the smallest 1/n

There is no smallest 1/n. There is no smallest definable 1/n

--
William Hughes

On Friday, 14 May 2021 at 16:59:34 UTC-4, William wrote:
> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
>
> > ... the result would be gaps on the real line. Between the smallest 1/n

There is no such thing as a "real number line". There is a "magnitude line", but it cannot be reified in any way whatsoever.

https://drive.google.com/file/d/0B-mOEooW03iLMHVYcE8xcmRZRnc

>
> There is no smallest 1/n. There is no smallest definable 1/n

I have told WM this several times. He keeps repeating the same old drivel.

>
> --
> William Hughes

William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
>
> > ... the result would be gaps on the real line. Between the smallest 1/n
>
> There is no smallest 1/n. There is no smallest definable 1/n
>
William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
>
> > ... the result would be gaps on the real line. Between the smallest 1/n
>
> There is no smallest 1/n. There is no smallest definable 1/n
>
"The sequence of numbers which grows beyond any stage already reached by passing to the next number, is a manifold of possibilities open towards infinity; it remains forever in the status of creation, but is not a closed realm of things existing in themselves." [H. Weyl: "Levels of infinity: Selected writings on mathematics and philosophy", Peter Pesic (ed.), Dover Publications (2012) p. 141]

If we nevertheless co nsider it a closed realm, then we need dark numbers.

There is no largest n, but every n that "there is" such that we can address it, has infinitely many successors.

Regards, WM

Eram semper recta schrieb am Samstag, 15. Mai 2021 um 00:17:13 UTC+2:
> On Friday, 14 May 2021 at 16:59:34 UTC-4, William wrote:

> > There is no smallest 1/n. There is no smallest definable 1/n
> I have told WM this several times. He keeps repeating the same old drivel..
>
There is no largest n, but every n that is the last number of a FISON (1, 2, 3, ..., n) belongs to a FISON. No FISON has ℵo elements. If there are ℵo elements, then there are more elements than every FISON has and that means more than *all* FISONs have. It is simply nonsense to believe that infintely many FISONs could have more elements than every FISON, in particular because there cannot exist ℵo FISONs. In order to distinguish ℵo FISONs, at least one FISONs would need ℵo elements and therefore would not be a FISON.

Regards, WM

William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
>
> > ... the result would be gaps on the real line. Between the smallest 1/n
>
> There is no smallest 1/n. There is no smallest definable 1/n
>

The ultimate argument:

Collectively you can remove all unit fractions.
How could you individually do that without removing a last one?
Individual completeness requires a last one.
Collective completeness is possible without a last one only by dark numbers because they do not show a comprehensible order.

Regards, WM

On Saturday, 15 May 2021 at 07:34:02 UTC-4, WM wrote:
> William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
> > On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
> >
> > > ... the result would be gaps on the real line. Between the smallest 1/n
> >
> > There is no smallest 1/n. There is no smallest definable 1/n
> >
> The ultimate argument:
>
> Collectively you can remove all unit fractions.

Unremarkable. You could remove ALL fractions collectively.

> How could you individually do that without removing a last one?

You can't do it individually. Period. To talk about removing a "last one" means accepting that an infinite task is possible which is nonsense.
> Individual completeness requires a last one.
> Collective completeness is possible without a last one only by dark numbers because they do not show a comprehensible order.

Nonsense. No such thing as dark numbers. Just because there is no last 1/n does not mean it is "dark". It only tells you that there is no greatest n which is a boring fact.

>
> Regards, WM

Eram semper recta schrieb am Samstag, 15. Mai 2021 um 13:39:35 UTC+2:
> On Saturday, 15 May 2021 at 07:34:02 UTC-4, WM wrote:

> > Collectively you can remove all unit fractions.
> You could remove ALL fractions collectively.
> > How could you individually do that without removing a last one?
> You can't do it individually.

Therefore we can do collectively what we cannot do individually.

> No such thing as dark numbers.

Numbers that can be removed collectively but not individually.

> Just because there is no last 1/n does not mean it is "dark". It only tells you that there is no greatest n which is a boring fact.

It would be boring fact also for me, if we could not collectively remove so many that none remains.

Regards, WM

On 5/15/2021 6:33 AM, WM wrote:
> William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
>> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
>>
>>> ... the result would be gaps on the real line. Between the smallest 1/n
>>
>> There is no smallest 1/n. There is no smallest definable 1/n
>>
>
> The ultimate argument:
>
> Collectively you can remove all unit fractions.
> How could you individually do that without removing a last one?

there is no last one,

your refusal to accept recursion leads to your dilemma.

> Individual completeness requires a last one.

there is no last one.

> Collective completeness is possible without a last one only by dark numbers

no. No fictitious doobies hiding behind the curtain, try recursion
instead, it will help your understanding of Math.

>
> Regards, WM
>

Sergio schrieb am Samstag, 15. Mai 2021 um 19:29:17 UTC+2:
> On 5/15/2021 6:33 AM, WM wrote:
> > William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
> >> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
> >>
> >>> ... the result would be gaps on the real line. Between the smallest 1/n
> >>
> >> There is no smallest 1/n. There is no smallest definable 1/n
> >>
> >
> > The ultimate argument:
> >
> > Collectively you can remove all unit fractions.
> > How could you individually do that without removing a last one?
> there is no last one,

But there remain no number when removing collectively.
This is impossible when removing individually,

> > Individual completeness requires a last one.
> there is no last one.

Therefore individual completeness can be excluded.
This would not be astonishing unless collective completeness was possible.

Regards, WM

On Saturday, May 15, 2021 at 10:17:01 AM UTC-7, WM wrote:
> Eram semper recta schrieb am Samstag, 15. Mai 2021 um 13:39:35 UTC+2:
> > On Saturday, 15 May 2021 at 07:34:02 UTC-4, WM wrote:
>
> > > Collectively you can remove all unit fractions.
> > You could remove ALL fractions collectively.
> > > How could you individually do that without removing a last one?
> > You can't do it individually.
> Therefore we can do collectively what we cannot do individually.

That is correct. We cannot determine the limit of a[n infinite] sequence by examining one term or two terms or finitely many terms. We need to examine the sequence as a thing-in-itself. In your own words, we need to examine 'the formula'.

We can remove one term from a sequence without changing the limit. We can remove two terms from the sequence without changing the limit. We can remove any finite collection of terms without changing the limit.

The only way to change the limit is to change 'the formula'.

And so the Axiom of Infinity is NECESSARY in set theory to study the transfinite.

And so Cantor's Second Principle of Ordinal Number Generation is NECESSARY in order to study transfinite ordinals.

On 5/15/2021 12:36 PM, WM wrote:
> Sergio schrieb am Samstag, 15. Mai 2021 um 19:29:17 UTC+2:
>> On 5/15/2021 6:33 AM, WM wrote:
>>> William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
>>>> On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
>>>>
>>>>> ... the result would be gaps on the real line. Between the smallest 1/n
>>>>
>>>> There is no smallest 1/n. There is no smallest definable 1/n
>>>>
>>>
>>> The ultimate argument:
>>>
>>> Collectively you can remove all unit fractions.
>>> How could you individually do that without removing a last one?
>> there is no last one,
>
> But there remain no number when removing collectively.
> This is impossible when removing individually,

not at all, in math we use recursion and limits to handle the infinite.

Also I think you assign an amount of time to "remove" or "process" a
number which leads to infinite time required, but that is not the case
at all.

>
>>> Individual completeness requires a last one.
>> there is no last one.
>
> Therefore individual completeness can be excluded.
> This would not be astonishing unless collective completeness was possible.
>
> Regards, WM
>

how are you defining individual completeness and then collective
completeness ?

On Saturday, May 15, 2021 at 6:31:37 AM UTC-3, WM wrote:

> There is no largest n

And hence there is no smallest 1/n
--
William Hughes

On Saturday, May 15, 2021 at 6:45:44 AM UTC-3, WM wrote:

> There is no largest n, but every n that is the last number of a FISON (1, 2, 3, ..., n) belongs to a FISON. No FISON has ℵo elements. If there are ℵo elements, then there are more elements than every FISON has and that means more than *all* FISONs have

The union of all FISONs is not a FISON. Of course what we get now is WMs famous syllogism:

Every FISON has a fixed largest element
N does not have a fixed largest element
N is a FISON

With logic like this, even the gods can only quote Schiller.

--
William Hughes

On Saturday, May 15, 2021 at 8:34:02 AM UTC-3, WM wrote:
> William schrieb am Freitag, 14. Mai 2021 um 22:59:34 UTC+2:
> > On Friday, May 14, 2021 at 5:46:51 PM UTC-3, WM wrote:
> >
> > > ... the result would be gaps on the real line. Between the smallest 1/n
> >
> > There is no smallest 1/n. There is no smallest definable 1/n
> >
> The ultimate argument:
>
> Collectively you can remove all unit fractions.
> How could you individually do that without removing a last one? D

Don't stop (you cannot stop, there is no last unit fraction). Removing items individually from an infinite set requires that you are able to do an infinite number of steps at each step you remove one item. Removing items collectively from an infinite set takes one step but requires you to be able to remove an infinite number of items per step.

--
William Hughes

On 5/15/2021 4:20 PM, William wrote:
> On Saturday, May 15, 2021 at 6:45:44 AM UTC-3, WM wrote:
>
>> There is no largest n, but every n that is the last number of a FISON (1, 2, 3, ..., n) belongs to a FISON. No FISON has ℵo elements. If there are ℵo elements, then there are more elements than every FISON has and that means more than *all* FISONs have
>
> The union of all FISONs is not a FISON. Of course what we get now is WMs famous syllogism:
>
> Every FISON has a fixed largest element
> N does not have a fixed largest element
> N is a FISON
>
> With logic like this, even the gods can only quote Schiller.
>

same sort of logic;

All men have two legs
Monkeys have two legs
therefore all Men are monkeys.

List of fallacies
https://en.wikipedia.org/wiki/List_of_fallacies

On Saturday, 15 May 2021 at 13:17:01 UTC-4, WM wrote:
> Eram semper recta schrieb am Samstag, 15. Mai 2021 um 13:39:35 UTC+2:
> > On Saturday, 15 May 2021 at 07:34:02 UTC-4, WM wrote:
>
> > > Collectively you can remove all unit fractions.
> > You could remove ALL fractions collectively.
> > > How could you individually do that without removing a last one?
> > You can't do it individually.
> Therefore we can do collectively what we cannot do individually.

No. That too is a delusion. It's the same flawed logic one applies when talking about Zeno's faux paradox.

0_____1

You can't remove ALL the numbers in the above interval because you don't know what they all are. By removing the line, you aren't actually proving anything, because the line is only symbolic of the distance 1. To remove all the numbers, you would first have to know what they are, but since you can never know what they all are, you can't remove them collectively because they are NEVER all there.

> > No such thing as dark numbers.
> Numbers that can be removed collectively but not individually.

Nonsense. You can't remove individually or collectively.

> > Just because there is no last 1/n does not mean it is "dark". It only tells you that there is no greatest n which is a boring fact.
> It would be boring fact also for me, if we could not collectively remove so many that none remains.

You cannot collectively remove the numbers. That is a misguided thought n your brain.

>
> Regards, WM

FredJeffries schrieb am Samstag, 15. Mai 2021 um 20:32:18 UTC+2:
> On Saturday, May 15, 2021 at 10:17:01 AM UTC-7, WM wrote:
> > Eram semper recta schrieb am Samstag, 15. Mai 2021 um 13:39:35 UTC+2:
> > > On Saturday, 15 May 2021 at 07:34:02 UTC-4, WM wrote:
> >
> > > > Collectively you can remove all unit fractions.
> > > You could remove ALL fractions collectively.
> > > > How could you individually do that without removing a last one?
> > > You can't do it individually.
> > Therefore we can do collectively what we cannot do individually.
> That is correct. We cannot determine the limit of a[n infinite] sequence by examining one term or two terms or finitely many terms. We need to examine the sequence as a thing-in-itself. In your own words, we need to examine 'the formula'.

That is the precondition for knowing the terms and for calculating the limit, if a limit exists. But it does not allow for individually processing all terms.

Individual processing of the complete set, such that no element is remaining, requires processing a last term. That is impossible for infinite sets. The problem of completeness without last term cannot be solved individually but only collectively by means of dark elements.
>
> We can remove one term from a sequence without changing the limit. We can remove two terms from the sequence without changing the limit. We can remove any finite collection of terms without changing the limit.

Yes, but the limit is not of interest, if we investigate the completeness problem. Let's consider the problem of intervals: [1/(n+1), 1/n]. The limit is 0, but here we are interested in the fact that the union of all of them is (0, 1] while individually it is impossible to union more than [1/n, 1] for any possible n. The dark intervals comprising (0, 1/n) are required in order to avoid a last interval of the form [1/(n+1), 1/n] and nevertheless processing all intervals with no remainder.

Regards, WM

Sergio schrieb am Samstag, 15. Mai 2021 um 20:36:54 UTC+2:
> On 5/15/2021 12:36 PM, WM wrote:

> > But there remain no number when removing collectively.
> > This is impossible when removing individually,
> not at all, in math we use recursion and limits to handle the infinite.

Recursion does never end and therefore never finished a set. Processing individually the complete set such that non element remains requires to process a last element.
Only collectively all elements can be processed without processing a last one. That's the advantage of dark elements.
>
> Also I think you assign an amount of time to "remove" or "process" a
> number which leads to infinite time required, but that is not the case
> at all.

The problem is not time but the principle. The proof of having processed all elements individually, i.e., one after the other, is the absence of further elements. But this state requires to have processed a last element.
> >
> >>> Individual completeness requires a last one.
> >> there is no last one.
> >
> > Therefore individual completeness can be excluded.
> > This would not be astonishing unless collective completeness was possible.

> how are you defining individual completeness and then collective
> completeness ?

Individual completeness requires to address every element. If none remains, then a last one has been named.

Regards, WM

William schrieb am Samstag, 15. Mai 2021 um 23:20:10 UTC+2:
> On Saturday, May 15, 2021 at 6:45:44 AM UTC-3, WM wrote:
>
> > There is no largest n, but every n that is the last number of a FISON (1, 2, 3, ..., n) belongs to a FISON. No FISON has ℵo elements. If there are ℵo elements, then there are more elements than every FISON has and that means more than *all* FISONs have
> The union of all FISONs is not a FISON.

That is wrong. Every union of FISONs is a FISON.

> Of course what we get now is WMs famous syllogism:
>
> Every FISON has a fixed largest element
> N does not have a fixed largest element
> N is a FISON

Wrong. FISONs are not dark. |N is much more than all FISONs.
Proof: All FISONs can be subtracted from |N without getting the empty set.
Try it. But stay with mathematics. Don't believe that if infinitely many FISONs are applied the rules of logic will fail.

Regards, WM

William schrieb am Samstag, 15. Mai 2021 um 23:29:50 UTC+2:
> On Saturday, May 15, 2021 at 8:34:02 AM UTC-3, WM wrote:

> > Collectively you can remove all unit fractions.
> > How could you individually do that without removing a last one? D
>
> Don't stop (you cannot stop, there is no last unit fraction).

But collectively you can remove all unit fractions.

> Removing items individually from an infinite set requires that you are able to do an infinite number of steps at each step you remove one item.

Not even that would help. You cannot get ready individually, even after whatever step you try, because that would require a last element. Impossible.

> Removing items collectively from an infinite set takes one step but requires you to be able to remove an infinite number of items per step.

According to set theory it is possible to remove the union of all intervals [1/(n+1), 1/n] which is (0, 1]. Individually it is impossible to union more than [1/n, 1] for any accessible n. The dark intervals comprising (0, 1/n) are required in order to avoid a last interval of the form [1/(n+1), 1/n] and nevertheless processing all intervals with no remainder.

Regards, WM

Eram semper recta schrieb am Sonntag, 16. Mai 2021 um 01:41:52 UTC+2:
> On Saturday, 15 May 2021 at 13:17:01 UTC-4, WM wrote:
> > Eram semper recta schrieb am Samstag, 15. Mai 2021 um 13:39:35 UTC+2:
> > > On Saturday, 15 May 2021 at 07:34:02 UTC-4, WM wrote:
> >
> > > > Collectively you can remove all unit fractions.
> > > You could remove ALL fractions collectively.
> > > > How could you individually do that without removing a last one?
> > > You can't do it individually.
> > Therefore we can do collectively what we cannot do individually.
> No. That too is a delusion.

Based on set theory is fact.

> You can't remove ALL the numbers in the above interval because you don't know what they all are.

If we would know them all such that none remains, then we would know a last one.

> By removing the line, you aren't actually proving anything, because the line is only symbolic of the distance 1. To remove all the numbers, you would first have to know what they are, but since you can never know what they all are, you can't remove them collectively because they are NEVER all there.

I assume set theory.

> You cannot collectively remove the numbers.

Please note that I assume set theory.

Regards, WM

On Sunday, May 16, 2021 at 5:07:29 PM UTC-3, WM wrote:
WM: Every FISON has a fixed largest element
WM: N does not have a fixed largest element
WM: N is a FISON

Which line is wrong,
--
William Hughes

On Sunday, May 16, 2021 at 5:12:32 PM UTC-3, WM wrote:
> William schrieb am Samstag, 15. Mai 2021 um 23:29:50 UTC+2:
> > On Saturday, May 15, 2021 at 8:34:02 AM UTC-3, WM wrote:

> > Removing items individually from an infinite set requires that you are able to do an infinite number of steps at each step you remove one item.
> Not even that would help. You cannot get ready individually, even after whatever step you try, because that would require a last element. Impossible.

Nope, you are doing an infinite number of steps, There is no last step.

--
William Hughes