On Tuesday, April 19, 2022 at 12:37:23 PM UTC-3, WM wrote:
> William schrieb am Montag, 18. April 2022 um 18:40:41 UTC+2:
> > On Monday, April 18, 2022 at 12:35:14 PM UTC-3, WM wrote:
> > > William schrieb am Sonntag, 17. April 2022 um 20:28:12 UTC+2:
> > > > But induction works for every element of N_p.
> > > It shows: Every element of N_p belongs to a finite set and has infinitely many sucessors.
> > So what? Each of the infinitely many successors is an element of N_p and, as induction works for every element of N_p, we know that all infinitely many successors are not "dark".
> That is potential infinity.

Piffle. A "potentially infinite set" is nonsense. We apply induction exactly once, and show something about each of the infinitely many elements of N_p.

--
William Hughes

On 4/19/2022 10:29 AM, WM wrote:
> horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
>> On Monday, 18 April 2022 at 12:33:01 UTC-3, WM wrote:
>>> horand....@gmail.com schrieb am Sonntag, 17. April 2022 um 20:05:40 UTC+2:
>>>
>>>>>>>> 1/1, oo, oo, oo, ...
>>>>>>>> 1/2, oo, oo, oo, ...
>>>>>>>> 1/3, oo, oo, oo, ...
>>>>>>>> 2/2, oo, oo, oo, ...
>>>>>>>> ...
>>>>>> is the limit of your sequence of matrices.
>>>>> Nonsense. There is no limit in enumerating the rationals
>>>> In *ANY* infinite sequential process such as the one *YOU* introduced, *because* it cannot be finished in stepwise fashion (you'll perhaps remember that there are infinitely many steps), you have to think about any limiting behaviour the process might exhibit.
>>> No, the point is that it goes on and on with no obstacle: "und es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor, p. 239]
>> That is exactly my point, too,
>
> It is not filling the matrix places by oo.

Zug. (that would make it a DIFFERENT MATRIX)

>
>> The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
>
> which is not reached.

Zug Zug. (what does *LIMIT* mean ?)

>
>> To *EVERY* stepwise infinite process one must study the limiting behavior.
>
> I did. It is this: Exchanging X and O will never change the numbers of X's and O's. That is the "limit-behaviour".

Quack. Swapparoo limiting behavior detected!

>
> Regards, WM

On 4/19/2022 10:43 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:39:46 UTC+2:
>> torsdag 14 april 2022 kl. 14:05:20 UTC+2 skrev WM:
>
>>> Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
>
>>> The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.
>>
>> False, in cardinal arithmetic |ℕ| - 1 = |ℕ|
>
> That is why cardinal arithmetic is false.
>
> Regards, WM

Zug.

On 4/19/2022 10:41 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:38:32 UTC+2:
>> torsdag 14 april 2022 kl. 13:30:41 UTC+2 skrev WM:
>>> William schrieb am Mittwoch, 13. April 2022 um 23:41:11 UTC+2:
>>>> On Wednesday, April 13, 2022 at 5:22:05 PM UTC-3, WM wrote:
>>>>> William schrieb am Mittwoch, 13. April 2022 um 21:07:30 UTC+2:
>>>>
>>>>> You can only notice the absence of natnumbers in the intersection
>>>> And thus, be definition of the intersection, the intersection in empty.
>>> But that is contrary to mathematics.
>>
>> Nope, it is according to mathematics.
>
> Non-empty endsegments have a nonempty intersection with all their precedessors. All definable endsegments are predecessors of non-empty endsegments. This fact cannot be countered by quantifier magic.
>
> Regards, WM

use MATH equations to state such. Can you do that?

tisdag 19 april 2022 kl. 17:41:19 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:38:32 UTC+2:
> > torsdag 14 april 2022 kl. 13:30:41 UTC+2 skrev WM:
> > > William schrieb am Mittwoch, 13. April 2022 um 23:41:11 UTC+2:
> > > > On Wednesday, April 13, 2022 at 5:22:05 PM UTC-3, WM wrote:
> > > > > William schrieb am Mittwoch, 13. April 2022 um 21:07:30 UTC+2:
> > > >
> > > > >You can only notice the absence of natnumbers in the intersection
> > > > And thus, be definition of the intersection, the intersection in empty.
> > > But that is contrary to mathematics.
> >
> > Nope, it is according to mathematics.
> Non-empty endsegments have a nonempty intersection with all their precedessors.

Irrelevant to the infinite intersection

>All definable endsegments

"Definble" is meaningless

>are predecessors of non-empty endsegments. This fact cannot be countered by quantifier magic.
>
> Regards, WM

None of those are relevant to the INFINITE intersection

tisdag 19 april 2022 kl. 17:43:43 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:39:46 UTC+2:
> > torsdag 14 april 2022 kl. 14:05:20 UTC+2 skrev WM:
>
> > > Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
> > > The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.
> >
> > False, in cardinal arithmetic |ℕ| - 1 = |ℕ|
> That is why cardinal arithmetic is false.
>
> Regards, WM

False, that is how cardinal arithmetic works, it is not the same as ordinary arthmetic nor other arithmetics.

On Tuesday, 19 April 2022 at 18:43:43 UTC+3, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:39:46 UTC+2:
> > torsdag 14 april 2022 kl. 14:05:20 UTC+2 skrev WM:
>
> > > Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
> > > The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.
> >
> > False, in cardinal arithmetic |ℕ| - 1 = |ℕ|
> That is why cardinal arithmetic is false.

Correct response. You cannot convince a crank, no matter how overwhelming is your evidence. For starters, your argument is already lost when you allow the crank to talk about the cardinality of N, that is, |ℕ|. The moment you allow the crank to think he can use |ℕ|, there is nothing you can say that will convince him otherwise. You can't even disprove it using his beliefs because it would require that he is able to reason logically. In Malum's case, this is an impossibility. He will always be an idiot and nothing can fix this.

>
> Regards, WM

Crank John Gabriel, aka Eram semper recta, responding to crank Wolfgang
Mückenheim, wrote:
> On Tuesday, 19 April 2022 at 18:43:43 UTC+3, WM wrote:
>> zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:39:46 UTC+2:
>>> torsdag 14 april 2022 kl. 14:05:20 UTC+2 skrev WM:
>>
>>>> Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
>>>> The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.
>>>
>>> False, in cardinal arithmetic |ℕ| - 1 = |ℕ|
>> That is why cardinal arithmetic is false.
>
> Correct response. You cannot convince a crank, no matter how overwhelming is your evidence. For starters, your argument is already lost when you allow the crank to talk about the cardinality of N, that is, |ℕ|. The moment you allow the crank to think he can use |ℕ|, there is nothing you can say that will convince him otherwise. You can't even disprove it using his beliefs because it would require that he is able to reason logically. In Malum's case, this is an impossibility. He will always be an idiot and nothing can fix this.

Oh, come on cranks! |ℕ| - 1 = |ℕ| means that there is a bijection
between {0, 1, 2, 3 ...} and (for instance) {1, 2, 3 ...}. Any 8 years
old child would notice that n->n+1 is such a bijection.

You kooks are pathetically stupid...

William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:

> A "potentially infinite set" is nonsense.

Alas it is the only set that we can apply in mathematics.

> We apply induction exactly once, and show something about each of the infinitely many elements of N_p.

The most important result of induction is this: Every number covered by induction belongs to a finite set which is followed by an infinite set of dark numbers, ℵ₀ of which will remain dark.

Regards, WM

horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
> On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
> [...]
> > > The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
> > which is not reached.
> Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.

What does this mean? The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.

In set theory however, we do not need the sizes of terms but the distances. That is what you confuse.

Regards, WM

On 4/20/2022 7:47 AM, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
>> On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
>>> horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
>> [...]
>>>> The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
>>> which is not reached.
>> Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.
>
> What does this mean? The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.

f(ω) is wrong usage, Use Limit of f(n) as n = oo

>
> In set theory however, we do not need the sizes of terms but the distances. That is what you confuse.

You are confused, he is not.

>
> Regards, WM

On 4/20/2022 7:38 AM, WM wrote:
> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>
>> A "potentially infinite set" is nonsense.
>
> Alas it is the only set that we can apply in mathematics.

no, it is nonsense. And you are not a mathematician.

>
>> We apply induction exactly once, and show something about each of the infinitely many elements of N_p.
>
> The most important result of induction is this: Every number covered by induction belongs to a finite set which is followed by an infinite set of dark numbers, ℵ₀ of which will remain dark.

No, that is bullshit.

>
> Regards, WM

On Wednesday, 20 April 2022 at 09:47:13 UTC-3, WM wrote:
[...]
>The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps.

Whatever f is in this context? And what is f(ω)? Your habit of simply plugging ω into the functional equation is what has gotten you in trouble time and again. You simply can't do it that way.

> But the number of terms between n and ω remains infinite: |ω - n| = ℵo.

Which is why you need a topology or a distance measure to study convergence.. Didn't they teach you *ANY* of that?

On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>
> > A "potentially infinite set" is nonsense.
> Alas it is the only set that we can apply in mathematics.
Piffle. A result you do not like is not a contradiction,

--
William Hughes

On Wednesday, April 20, 2022 at 9:47:13 AM UTC-3, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
> > On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
> > [...]
> > > > The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
> > > which is not reached.
> > Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.
> What does this mean?

The lim sup and the lim inf are equal. Both are definined in terms union and intersection. There is no metric

> The difference between the size

"difference" and "size" assume that the set limit is based on a metric. It is not. The set limit is not covered in Calc 101. No concept of "distance" between the sets of the sequence and the limit is useful. The elements of the sequence do not get "close" to the limit.

--
William Hughes

On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>

> > ...We apply induction exactly once, and show something about each of the infinitely many elements of N_p.
> The most important result of induction is this: Every number covered by induction

I.e any element. a we know by definition that each element of a Peano set is covered by induction.

>belongs to a finite set which is followed by an infinite set

trivial. Note that, by the induction we have already done, we know that, as this infinite set contains only elements of N_p, the infinite set does not contain a "dark number".

--
William Hughes

onsdag 20 april 2022 kl. 11:32:02 UTC+2 skrev Eram semper recta:
> On Tuesday, 19 April 2022 at 18:43:43 UTC+3, WM wrote:
> > zelos...@gmail.com schrieb am Dienstag, 19. April 2022 um 09:39:46 UTC+2:
> > > torsdag 14 april 2022 kl. 14:05:20 UTC+2 skrev WM:
> >
> > > > Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
> > > > The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.
> > >
> > > False, in cardinal arithmetic |ℕ| - 1 = |ℕ|
> > That is why cardinal arithmetic is false.
> Correct response.

It is however incorrect

>You cannot convince a crank,

I will conitnue to try with you and WM

>no matter how overwhelming is your evidence.

You both never admit you are wrong given evidence, you are correct.

>For starters, your argument is already lost when you allow the crank to talk about the cardinality of N, that is, |ℕ|.

Why? It is perfectly valid in mathematics.

>The moment you allow the crank to think he can use |ℕ|, there is nothing you can say that will convince him otherwise.

There is nothing in mathematics that says we can't do it.

>You can't even disprove it using his beliefs because it would require that he is able to reason logically.

You could, you would also win a lot of money :) The fact neither of you do is demonstration of how flawed your ideas are.

>In Malum's case, this is an impossibility. He will always be an idiot and nothing can fix this.

Given I am more educated than you, that would mean you are an even bigger idiot :D

>
> >
> > Regards, WM

onsdag 20 april 2022 kl. 14:38:20 UTC+2 skrev WM:
> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>
> > A "potentially infinite set" is nonsense.
> Alas it is the only set that we can apply in mathematics.

Nope, because you cannot even define what it is.

> > We apply induction exactly once, and show something about each of the infinitely many elements of N_p.
> The most important result of induction is this: Every number covered by induction belongs to a finite set which is followed by an infinite set of dark numbers, ℵ₀ of which will remain dark.
>
> Regards, WM

horand....@gmail.com schrieb am Mittwoch, 20. April 2022 um 15:57:15 UTC+2:
> On Wednesday, 20 April 2022 at 09:47:13 UTC-3, WM wrote:
> [...]
> >The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps.
> Whatever f is in this context?

f is a function like (1/n).

> And what is f(ω)?

f(ω) is its limit like 0.

> Your habit of simply plugging ω into the functional equation is what has gotten you in trouble time and again. You simply can't do it that way.

If ω is the limit of the sequence (n), then this can be done.

> > But the number of terms between n and ω remains infinite: |ω - n| = ℵo.

> Which is why you need a topology or a distance measure to study convergence.

Convergence has no relation to this distance, which is always ℵo. All functions f(n) = k/n or k/n^2 or k/sqrt(n), and many more have the limit 0. No topology is necessary to find that limit.

> Didn't they teach you *ANY* of that?

Topologists teach that they are required. In case of the above this is a lie.

Regards, WM

William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> >
> > > A "potentially infinite set" is nonsense.
> > Alas it is the only set that we can apply in mathematics.
> Piffle. A result you do not like is not a contradiction,

By induction we prove that every definable natural number, i.e., every natural number which is subject to induction, belongs to a FISON but has ℵo successors which cannot be removed whatever natural number you consider. That means they cannot be used as individuals.

Regards, WM

William schrieb am Mittwoch, 20. April 2022 um 19:11:02 UTC+2:

> "difference" and "size" assume that the set limit is based on a metric. It is not.

For numbers and sets of numbers it is.

> The set limit is not covered in Calc 101.

Luckily, because it is nonsense. A limit is approached better and better but usually it is not reached.

> No concept of "distance" between the sets of the sequence and the limit is useful.

Of course it is the only useful concept. Scrooge McDuck who earns a billion and spends only one dollar every day will not get bankrupt "in the limit". That is a silly idea. It should be banned from human culture but at least from mathematics.

> The elements of the sequence do not get "close" to the limit.

The elements get close to the limit by size but not by step. The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.

We can use ω as the limit of the sequence (n) because the reciprocals (1/n) have the limit 0. But in both cases there remain ℵo terms in between - forever! They are dark and cannot be passed. The amount of dollars earned by McDuck cannot become ω or ℵo because dark dollars cannot be earned. Therefore the argument that he has to lose every dollar is wrong to.

Regards, WM

William schrieb am Mittwoch, 20. April 2022 um 19:26:48 UTC+2:
> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> >
> > > ...We apply induction exactly once, and show something about each of the infinitely many elements of N_p.
> > The most important result of induction is this: Every number covered by induction
> I.e any element. a we know by definition that each element of a Peano set is covered by induction.
> >belongs to a finite set which is followed by an infinite set
> trivial. Note that, by the induction we have already done, we know that, as this infinite set contains only elements of N_p, the infinite set does not contain a "dark number".

They are the buffer between N_p and the limit ω. Proof by induction for all elements of N_p. You don't like it? Irrelevant.

Regards, WM

On Thursday, 21 April 2022 at 11:14:46 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 20. April 2022 um 15:57:15 UTC+2:
> > On Wednesday, 20 April 2022 at 09:47:13 UTC-3, WM wrote:
> > [...]
> > >The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps.
> > Whatever f is in this context?
> f is a function like (1/n).

"Like (1/n)"? What the hell is that supposed to mean. If f is not "(1/n)" (using your non-sensical notation, even though I have no clue what you might mean by that), how do you know what properties it has?

> > And what is f(ω)?
>
> f(ω) is its limit like 0.
> > Your habit of simply plugging ω into the functional equation is what has gotten you in trouble time and again. You simply can't do it that way.
> If ω is the limit of the sequence (n), then this can be done.

Only if f is continuous, you ignorant idiot.

On 4/21/2022 9:14 AM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 20. April 2022 um 15:57:15 UTC+2:
>> On Wednesday, 20 April 2022 at 09:47:13 UTC-3, WM wrote:
>> [...]
>>> The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps.
>> Whatever f is in this context?
>
> f is a function like (1/n).

you need to specify that before hand.

>
>> And what is f(ω)?
>
> f(ω) is its limit like 0.

No. If so, then what is f(ω-1) ?

>
>> Your habit of simply plugging ω into the functional equation is what has gotten you in trouble time and again. You simply can't do it that way.
>
> If ω is the limit of the sequence (n), then this can be done.

No. You are out of math.

>
>>> But the number of terms between n and ω remains infinite: |ω - n| = ℵo.
>
>> Which is why you need a topology or a distance measure to study convergence.
>
> Convergence has no relation to this distance, which is always ℵo. All functions f(n) = k/n or k/n^2 or k/sqrt(n), and many more have the limit 0. No topology is necessary to find that limit.

again you demonstrate your complete lack of knowledge of math.

>
>> Didn't they teach you *ANY* of that?
>
> Topologists teach that they are required. In case of the above this is a lie.

you are the one that lies.

>
> Regards, WM

On 4/21/2022 9:30 AM, WM wrote:
> William schrieb am Mittwoch, 20. April 2022 um 19:11:02 UTC+2:
>
>> "difference" and "size" assume that the set limit is based on a metric. It is not.
>

<snip oppositional crap>

>
>> The set limit is not covered in Calc 101.
>

<snip oppositional crap>

>
>> No concept of "distance" between the sets of the sequence and the limit is useful.
>

<snip oppositional crap>

>
>> The elements of the sequence do not get "close" to the limit.
>

<snip oppositional crap>

>

<snip oppositional crap>
>

> Regards, WM