On Sunday, 16 May 2021 at 16:16:32 UTC-4, WM wrote:
> 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.

I know. I am speaking from a logical point of view, not any perspective of set theory which I have long ago dismissed as utter garbage having zero worth.

>
> Regards, WM

On Sunday, 16 May 2021 at 17:29:35 UTC-4, William wrote:
> 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.

What is an "infinite number" and how do you "doing an infinite number of steps"?

What a moron you are.

>
> --
> William Hughes

fredag 14 maj 2021 kl. 22:46:51 UTC+2 skrev WM:
> 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
there is no smallest 1/n, how many times do we have to tell you this?

On Monday, 17 May 2021 at 00:49:37 UTC-4, zelos...@gmail.com wrote:
> fredag 14 maj 2021 kl. 22:46:51 UTC+2 skrev WM:
> > 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
> there is no smallest 1/n, how many times do we have to tell you this?

Apparently there is in "set theory" since you can collectively remove all the numbers. But I suppose just wave hands again eh? Too funny. LMAO.

William schrieb am Sonntag, 16. Mai 2021 um 23:21:28 UTC+2:
> 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,
> --
If |N is potentially infinite, then it is a potentially infinite sequence of FISON - not fixed and therefore it has no fixed largest element.
If |N is actually infinite, then all FISONs can be subtracted without getting the empty set. Therefore |N is not the union of FISONs.

Regards, WM

William schrieb am Sonntag, 16. Mai 2021 um 23:29:35 UTC+2:
> 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.

But collectively you can remove all, such that none remains. This is not possible individually, even when doing infinitely many steps. There always ℵo elements remain. That is the dark subset.

Regards, WM

zelos...@gmail.com schrieb am Montag, 17. Mai 2021 um 06:49:37 UTC+2:
> fredag 14 maj 2021 kl. 22:46:51 UTC+2 skrev WM:

> there is no smallest 1/n, how many times do we have to tell you this?

There are unit fractions which can be removed collectively such that none remains. This however is impossible individually. There always remains an infinite set of ℵo unit fractions.

Regards, WM

On Monday, May 17, 2021 at 12:09:29 PM UTC-3, WM wrote:
> This is not possible individually, even when doing infinitely many steps. There always ℵo elements remain.

Nope, this is a result you do not like. After a finite number of steps (last step exists) the set remaining has cardinality ℵo. After an infinite number or steps (no last step) the remaining set is empty.

--
William Hughes

On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
> William schrieb am Sonntag, 16. Mai 2021 um 23:21:28 UTC+2:
> > 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,
> > --
> If |N is potentially infinite, then it is a potentially infinite sequence of FISON - not fixed and therefore it has no fixed largest element.

So N is not a FISON

> If |N is actually infinite, then all FISONs can be subtracted without getting the empty set. Therefore |N is not the union of FISONs.

So N is not a FISON

In both cases the third line is wrong.

--
William Hughes

On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
> If |N is actually infinite, then all FISONs can be subtracted without getting the empty set.

Every element of |N that is connected to the origin by a FISON can be subtracted (Duh). Elements of the set of natural numbers that are not natural numbers are of no interest whatsoever.

==
William Hughes

On 5/16/2021 3:02 PM, WM wrote:
> 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.

not true. Recursion requires no time, 0 time.

What is 0 times an infinite set ? oo * 0 = k

oo = k/0 and 0 = k/oo both true.

> Processing individually the complete set such that non element remains requires to process a last element.

only with sausages, not numbers.

> Only collectively all elements can be processed without processing a last one.

not so.

> That's the advantage of dark elements.

no such thing.

>>
>> 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.

that is what is done in recursion.

> But this state requires to have processed a last element.

no, only you do. recursion can go the infinite distance, instantly.

>>>
>>>>> 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.

that is not a definition of "collective completeness"...

> If none remains, then a last one has been named.

your "last one" tire is stuck in a rut called "none remain".

>
> Regards, WM
>

On 5/17/2021 10:11 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 17. Mai 2021 um 06:49:37 UTC+2:
>> fredag 14 maj 2021 kl. 22:46:51 UTC+2 skrev WM:
>
>> there is no smallest 1/n, how many times do we have to tell you this?
>
> There are unit fractions which can be removed collectively such that none remains. This however is impossible individually. There always remains an infinite set of ℵo unit fractions.
>
> Regards, WM
>

looks like you are stuck.

Try using Recursion, or limits, those advanced techniques will get you
our of your hole

Meritocracy formulated on Monday :
> On 5/16/2021 3:02 PM, WM wrote:
>> 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.
>
> not true. Recursion requires no time, 0 time.
>
> What is 0 times an infinite set ? oo * 0 = k
>
> oo = k/0 and 0 = k/oo both true.
>
>
>
>
>> Processing individually the complete set such that non element remains
>> requires to process a last element.
>
> only with sausages, not numbers.
>
>> Only collectively all elements can be processed without processing a last
>> one.
>
> not so.
>
>> That's the advantage of dark elements.
>
> no such thing.
>
>>>
>>> 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.
>
> that is what is done in recursion.
>
>> But this state requires to have processed a last element.
>
> no, only you do. recursion can go the infinite distance, instantly.
>
>>>>
>>>>>> 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.
>
> that is not a definition of "collective completeness"...
>
>> If none remains, then a last one has been named.
>
> your "last one" tire is stuck in a rut called "none remain".

Now make him add back all those misappropriated naturals he subtracted,
other people might need them.

On 5/17/2021 3:18 PM, FromTheRafters wrote:
> Meritocracy formulated on Monday :
>> On 5/16/2021 3:02 PM, WM wrote:
>>> 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.
>>
>> not true.  Recursion requires no time, 0 time.
>>
>>       What is 0 times an infinite set ?  oo * 0 = k
>>
>> oo = k/0   and   0 = k/oo    both true.
>>
>>
>>
>>
>>> Processing individually the complete set such that non element
>>> remains requires to process a last element.
>>
>> only with sausages, not numbers.
>>
>>> Only collectively all elements can be processed without processing a
>>> last one.
>>
>> not so.
>>
>>> That's the advantage of dark elements.
>>
>> no such thing.
>>
>>>>
>>>> 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.
>>
>> that is what is done in recursion.
>>
>>> But this state requires to have processed a last element.
>>
>> no, only you do. recursion can go the infinite distance, instantly.
>>
>>>>>
>>>>>>> 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.
>>
>> that is not a definition of "collective completeness"...
>>
>>> If none remains, then a last one has been named.
>>
>> your "last one" tire is stuck in a rut called "none remain".
>
> Now make him add back all those misappropriated naturals he subtracted,
> other people might need them.

I am sure there are alien civilizations that are still counting the
naturals all they way up, and they need those numbers!

On 5/16/2021 4:29 PM, William wrote:
> 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.
>

like being in the military, every day is an infinity long.

Meritocracy has brought this to us :
> On 5/17/2021 3:18 PM, FromTheRafters wrote:
>> Meritocracy formulated on Monday :
>>> On 5/16/2021 3:02 PM, WM wrote:
>>>> 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.
>>>
>>> not true.  Recursion requires no time, 0 time.
>>>
>>>       What is 0 times an infinite set ?  oo * 0 = k
>>>
>>> oo = k/0   and   0 = k/oo    both true.
>>>
>>>
>>>
>>>
>>>> Processing individually the complete set such that non element
>>>> remains requires to process a last element.
>>>
>>> only with sausages, not numbers.
>>>
>>>> Only collectively all elements can be processed without processing a
>>>> last one.
>>>
>>> not so.
>>>
>>>> That's the advantage of dark elements.
>>>
>>> no such thing.
>>>
>>>>>
>>>>> 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.
>>>
>>> that is what is done in recursion.
>>>
>>>> But this state requires to have processed a last element.
>>>
>>> no, only you do. recursion can go the infinite distance, instantly.
>>>
>>>>>>
>>>>>>>> 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.
>>>
>>> that is not a definition of "collective completeness"...
>>>
>>>> If none remains, then a last one has been named.
>>>
>>> your "last one" tire is stuck in a rut called "none remain".
>>
>> Now make him add back all those misappropriated naturals he subtracted,
>> other people might need them.
>
> I am sure there are alien civilizations that are still counting the
> naturals all they way up, and they need those numbers!

If he creates a set by taking a subset of the naturals and filtering
out all of the FISONs, he has nothing. He needs at least one FISON
(say, maybe zero?) for his new set to again be the smallest inductive
set. Without the induction he has no infinite endsegments to keep him
awake at night and no predecessor function connection to zero (and no
zero to connect to anyway).

He's got nothing.

William schrieb am Montag, 17. Mai 2021 um 18:23:04 UTC+2:
> On Monday, May 17, 2021 at 12:09:29 PM UTC-3, WM wrote:
> > This is not possible individually, even when doing infinitely many steps. There always ℵo elements remain.
> Nope, this is a result you do not like.

No, I can prove it. No number is definable such that less than ℵo numbers remain.

> After a finite number of steps (last step exists) the set remaining has cardinality ℵo. After an infinite number or steps (no last step) the remaining set is empty.

"After an infinite number of steps" implies that all steps have been done. Otherwise there is no "after". Individual "after" implies a last step has been done. Contradiction.

Not every natnumber can be defined. Proof: With n also all smaller natunumbers are defined. But in order to prove zthe definability of every natnumber, you must define a natnumber that leaves nothing remaining. Contradiction..

Regards, WM

William schrieb am Montag, 17. Mai 2021 um 18:58:19 UTC+2:
> On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
> > If |N is actually infinite, then all FISONs can be subtracted without getting the empty set.
> Every element of |N that is connected to the origin by a FISON can be subtracted

without exhausting |N.

> Elements of the set of natural numbers that are not natural numbers are of no interest whatsoever.

But they are required to get cardinality ℵo. Every last element n of a FISON has ℵo successors because n + ℵo = ℵo. n + less than ℵo would yield less than ℵo. Contradiction
if the set of all natnumbers has cardinality ℵo.

Regards, WM

William schrieb am Montag, 17. Mai 2021 um 18:28:59 UTC+2:
> On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
> > William schrieb am Sonntag, 16. Mai 2021 um 23:21:28 UTC+2:
> > > 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,
> > > --
> > If |N is potentially infinite, then it is a potentially infinite sequence of FISON - not fixed and therefore it has no fixed largest element.
> So N is not a FISON

It is a FISON in growth.

> > If |N is actually infinite, then all FISONs can be subtracted without getting the empty set. Therefore |N is not the union of FISONs.
> So N is not a FISON

That is true.
>
> In both cases the third line is wrong.

If a fixed FISON is assumed, yes.
>
Regards, WM

On 5/18/2021 6:37 AM, WM wrote:
> William schrieb am Montag, 17. Mai 2021 um 18:23:04 UTC+2:
>> On Monday, May 17, 2021 at 12:09:29 PM UTC-3, WM wrote:
>>> This is not possible individually, even when doing infinitely many steps. There always ℵo elements remain.
>> Nope, this is a result you do not like.
>
> No, I can prove it. No number is definable such that less than ℵo numbers remain.
>
>> After a finite number of steps (last step exists) the set remaining has cardinality ℵo. After an infinite number or steps (no last step) the remaining set is empty.
>
> "After an infinite number of steps" implies that all steps have been done. Otherwise there is no "after". Individual "after" implies a last step has been done. Contradiction.

no. not at all.

you are flailing about because you do not understand convergance and limits

Isaac Newton describes the process very well and invented calculus using
it in the 1600's

>
> Not every natnumber can be defined. Proof: With n also all smaller natunumbers are defined. But in order to prove zthe definability of every natnumber, you must define a natnumber that leaves nothing remaining. Contradiction.

no. your proof is spoof

All natural number can be "defined" using your "defined" with the raps,
taps, blinking lights, rudy-toot-toots,

Proof: Pick any natural number, and start tooting away

>
> Regards, WM
>

On 5/18/2021 6:40 AM, WM wrote:
> William schrieb am Montag, 17. Mai 2021 um 18:58:19 UTC+2:
>> On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
>>> If |N is actually infinite, then all FISONs can be subtracted without getting the empty set.
>> Every element of |N that is connected to the origin by a FISON can be subtracted
>
> without exhausting |N.
>
>> Elements of the set of natural numbers that are not natural numbers are of no interest whatsoever.
>
> But they are required to get cardinality ℵo.

you are stating that numbers that are not natural numbers, are in the
set of natural number ? bullshit.

> Every last element n of a FISON has ℵo successors because n + ℵo = ℵo. n + less than ℵo would yield less than ℵo. Contradiction

no contradiction. There is no last element of a FISON. Fail.

> if the set of all natnumbers has cardinality ℵo.
>
> Regards, WM
>

It happens that Meritocracy formulated :
> On 5/18/2021 6:40 AM, WM wrote:
>> William schrieb am Montag, 17. Mai 2021 um 18:58:19 UTC+2:
>>> On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
>>>> If |N is actually infinite, then all FISONs can be subtracted without
>>>> getting the empty set.
>>> Every element of |N that is connected to the origin by a FISON can be
>>> subtracted
>>
>> without exhausting |N.
>>
>>> Elements of the set of natural numbers that are not natural numbers are of
>>> no interest whatsoever.
>>
>> But they are required to get cardinality ℵo.
>
> you are stating that numbers that are not natural numbers, are in the
> set of natural number ? bullshit.

Exactly, pure WM matheology.

>> Every last element n of a FISON has ℵo successors because n + ℵo = ℵo. n +
>> less than ℵo would yield less than ℵo. Contradiction
>
> no contradiction. There is no last element of a FISON. Fail.

Actually, there is. The FISONS are all finite and have both a greatest
and a least element. WM fails to *create* (not deplete) a set of
natural numbers because he needs at least one FISON for induction to
have a base case in his newly *CREATED* set.

Sets don't shrink and grow despite his insistence that his rubber ones
do.

On Tuesday, May 18, 2021 at 8:43:45 AM UTC-3, WM wrote:
> William schrieb am Montag, 17. Mai 2021 um 18:28:59 UTC+2:
> > On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
> > > William schrieb am Sonntag, 16. Mai 2021 um 23:21:28 UTC+2:
> > > > 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,
> > > > --
> > > If |N is potentially infinite, then it is a potentially infinite sequence of FISON - not fixed and therefore it has no fixed largest element.
> > So N is not a FISON
> It is a FISON in growth.

WM: Every FISON has a fixed largest element
Maybe, since it is not actually a FISON, we should call |N a "dark FISON".

--
William Hughes

On Tuesday, May 18, 2021 at 8:40:19 AM UTC-3, WM wrote:
> William schrieb am Montag, 17. Mai 2021 um 18:58:19 UTC+2:
> > On Monday, May 17, 2021 at 12:07:43 PM UTC-3, WM wrote:
> > > If |N is actually infinite, then all FISONs can be subtracted without getting the empty set.
> > Every element of |N that is connected to the origin by a FISON can be subtracted
> without exhausting |N.
> > Elements of the set of natural numbers that are not natural numbers are of no interest whatsoever.
> But they are required to get cardinality ℵo.

Elements that are not natural numbers are not part of the set of natural numbers and thus have no influence on the cardinality of the set of natural numbers.

--
William Hughes

Meritocracy schrieb am Dienstag, 18. Mai 2021 um 14:18:21 UTC+2:

> you are flailing about because you do not understand convergance and limits

Limits mean collective processing.

Regards, WM