On 6/27/2021 3:44 PM, Gus Gassmann wrote:
> On Sunday, 27 June 2021 at 13:56:46 UTC-3, William wrote:
>> On Sunday, June 27, 2021 at 1:09:15 PM UTC-3, Christa M wrote:
>>> William schrieb am Samstag, 26. Juni 2021 um 19:55:03 UTC+2:
>>>> On Saturday, June 26, 2021 at 2:43:25 PM UTC-3, WM wrote:
>>>>> William schrieb am Samstag, 26. Juni 2021 um 17:24:51 UTC+2:
>>>>>> On Saturday, June 26, 2021 at 11:59:46 AM UTC-3, WM wrote:
>>>>>
>>>>>>>> All intersections occur after a finite number of removals. Only the limit comes after an infinite number of removals.
>>>>>>> So there are many steps between the finite numbers of removeal (by definable endsegments) and the limit. I think you've got it.
>>>>>>>
>>>>>>> What changes the infinite intersection to the empty set? There are not infinitely many definable endsegments
>>>>>> Yes, but the set of endsegments with least value in |N_F has cardinality aleph_0.
>>>>> |N_F ?
>>>> The set of all natural numbers that are the largest element of some FISON (note you may not be able to write down the number or its associated FISON).
>>>>>> It is these endsegments which changes the infinite intersection to the empty set.
>>> Note that all definable endsegments with no exception are infinite. So they have not deleted all natnumbers. But the limit is empty. What causes this sudden leap?
>> Informally speaking, the infinite number of endsegments with smallest element in |N_F, one requires an infinite number or endsegments to "remove an infinite number of natural numbers one at a time". But this is not a very good description. An infinite intersection is not a the result of a stepwise process, it is a limit. All the sets in the sequence are infinite. The limit is empty. This is not a sudden leap.
>
> WM seems to think he should get to the end of the rainbow by taking one step at a time. Me, I wait until sundown, at the latest.
>

WM does not know, or think, in terms of math.

Most can solve problems using math on paper using algebra or calculus.
Many can glance at an equation(s) and see what the next steps are to get
to an answer, or see the answer directly.

WM does not have this skill.

He has only posted 1 or 2 equations in the last year.
He has yet to provide any formal math proof.

That is why we get a few sentences at a time of partial descriptions of
an idea verbally from WM.

No Math.

No Math terms to describe dark numbers either.

anyhow WM is stuck on a simple thing, take the set of natural numbers,
stop at k, from 0 to k is finite, and from k to oo is infinite.

He masks it in Achillies Frames, turtles, and eggsegments, and other
nonsense, because *he is unable to state it in math terms*.

WM does not know math.

On Monday, June 28, 2021 at 4:50:47 AM UTC+2, Sergio wrote:

> anyhow WM is stuck on a simple thing, take the set of natural numbers,
> stop at k, from 0 to k is finite, and from k to oo is infinite.

It's getting worse with him. The other day he has written in de.sci.mathematik:

"Nicht alle natürlichen Zahlen [besitzen] unendlich viele Nachfolger." (WM)
[ "Not every natural number has an infinite number of successors." (WM) ]

No joke. His claim came after the following "dialog":

FF: "Der Beweis bezieht sich auf ALLE natürlichen Zahlen."
[ FF: The proof concerns ALL natural numbers. ]

WM: "Falsch. Der Beweis bezieht sich nur auf natürliche Zahlen, die unendlich viele Nachfolger besitzen."
[ WM: "Wrong. The proof only concerns natural numbers that have an infinite number of successors." ]

Well...

Greg Cunt wrote on 6/28/2021 :
> On Monday, June 28, 2021 at 4:50:47 AM UTC+2, Sergio wrote:
>
>> anyhow WM is stuck on a simple thing, take the set of natural numbers,
>> stop at k, from 0 to k is finite, and from k to oo is infinite.
>
> It's getting worse with him. The other day he has written in
> de.sci.mathematik:
>
> "Nicht alle natürlichen Zahlen [besitzen] unendlich viele Nachfolger." (WM)
> [ "Not every natural number has an infinite number of successors." (WM) ]
>
> No joke. His claim came after the following "dialog":
>
> FF: "Der Beweis bezieht sich auf ALLE natürlichen Zahlen."
> [ FF: The proof concerns ALL natural numbers. ]
>
> WM: "Falsch. Der Beweis bezieht sich nur auf natürliche Zahlen, die unendlich
> viele Nachfolger besitzen." [ WM: "Wrong. The proof only concerns natural
> numbers that have an infinite number of successors." ]
>
> Well...

I've been telling him all along that *each* natural number has *one*
successor as the axioms indicate. Speaking of *all* natural numbers or
*all* successors requires acceptance of induction which WM seems to
have trouble with.

Apparently now, when it is convenient for him to do so, he expresses
the opposite view. He's a fraud.

On Monday, June 28, 2021 at 2:01:17 PM UTC+2, FromTheRafters wrote:

> I've been telling him all along that *each* natural number has *one*
> successor as the axioms indicate. Speaking of *all* natural numbers or
> *all* successors requires acceptance of induction which WM seems to
> have trouble with.
>
> Apparently now, when it is convenient for him to do so, he expresses
> the opposite view. He's a fraud.

Right. To avoid confusion we sometimes speak of the "immediate successor" (in contrast to the other "successors" of a certain natural number) when "allowing" for more than one "successor" (in the context of a certain mathematica text).

WMs claims above refer to these "successors" of a natural number.

It happens that Greg Cunt formulated :
> On Monday, June 28, 2021 at 2:01:17 PM UTC+2, FromTheRafters wrote:
>
>> I've been telling him all along that *each* natural number has *one*
>> successor as the axioms indicate. Speaking of *all* natural numbers or
>> *all* successors requires acceptance of induction which WM seems to
>> have trouble with.
>>
>> Apparently now, when it is convenient for him to do so, he expresses
>> the opposite view. He's a fraud.
>
> Right. To avoid confusion we sometimes speak of the "immediate successor" (in
> contrast to the other "successors" of a certain natural number) when
> "allowing" for more than one "successor" (in the context of a certain
> mathematica text).
>
> WMs claims above refer to these "successors" of a natural number.

Yes, I know what he meant. However, he does it while denying the
implication that stating *all* means no longer merely step by step
"potentially" infinite sets but *actually* infinite sets.

Then, when it suits him, he flip-flops to the other argument. Why treat
*all* natural numbers as only potentially infinite and in need of dark
numbers at the same time as trying to prove his dark numbers by
accepting *all* endsegments without batting an eyelash?

William schrieb am Sonntag, 27. Juni 2021 um 18:56:46 UTC+2:
> An infinite intersection is not a the result of a stepwise process

Up to every finite index n it can be described as a stepwise process. At which finite index does this change? At none. But there are only finite indices. Therefore your claim is wrong. Only beyond all finite indices the process ceases because mathematics ceases.

>, it is a limit. All the sets in the sequence are infinite. The limit is empty.

All the sets contain aleph_0 natnumbers. Therefore the argument that the limit is empty because every natnumber has gone is wrong. In allo cases aleph_0 elements have not gone.

> This is not a sudden leap.

If there is nothing between the infinite endsegments and the limit, then it is a sudden leap. Moreover not all natnumbers have gone as is prescribed and required by

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Therefore your calculation of the limit would be wrong without dark numbers..

Regards, WM

Greg Cunt schrieb am Sonntag, 27. Juni 2021 um 18:56:51 UTC+2:
> On Sunday, June 27, 2021 at 6:05:09 PM UTC+2, Christa M wrote:
> > Jim Burns schrieb am Samstag, 26. Juni 2021 um 20:12:46 UTC+2:
> > >
> > > Each k in N is not in E(k+1), so it's not in all end segments.
> > >
> > The endsegments E(1), E(2), E(3), ... don't delete all natnumbers because they all are infinite.
> Endsegements don't do anything, they just are.

Wrong. ∀k ∈ ℕ: E(k+1) = E(k) \ {k} means E(k+1) is deleating k.

Regards, WM

Jim Burns schrieb am Sonntag, 27. Juni 2021 um 21:02:45 UTC+2:
> On 6/27/2021 12:05 PM, Christa M wrote:
> > Jim Burns schrieb
> > am Samstag, 26. Juni 2021 um 20:12:46 UTC+2:
> >> Each k in N is not in E(k+1), so it's not in all end segments.
> >
> > The endsegments E(1), E(2), E(3), ... don't delete all natnumbers
> > because they all are infinite.
> There are no numbers that each end segment deletes.

There are aleph_0 numbers which no infinite endsegment deletes.
>
> Each number is deleted by at least one end segment.

Each infinite endsegment keeps aleph_0 natnumbers.

> No number is in the intersection.

If only infinite endsegments would exist, the intersection would be infinite.

> > What are their remaining elements?
> {}.
> > What deletes them finally such that the limit is empty?
> No *FINAL* end segment before {} exists.

But there exist all endsegments, and if all are infinite, then not all natnumbers can have gone by action of endsegments as is prescribed and required by

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Therefore your calculation of the limit would be wrong without dark numbers..

Regards, WM

Gus Gassmann schrieb am Sonntag, 27. Juni 2021 um 22:44:51 UTC+2:

> WM seems to think he should get to the end of the rainbow by taking one step at a time. Me, I wait until sundown, at the latest.

If there is nothing between the infinite endsegments and the limit, then not all natnumbers have gone by endsegments as is prescribed and required by

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

and as it is the basis of your calculation of the limit.

Regards, WM

Each endsegment that is not the last endsegment is followed by an infinite number of endsegments
Between each endsegment that is not the last endsegment and the limit there are an infinite number of steps.
There is no last endsegment

So we have two facts,

Each Endsegment contains an infinite number of elements.
Each Endsegment is an infinite number of steps from the limit.

You may ask: "How do we get to the limit if every Endsegment is an infinite number of steps from the limit?" The question has no answer because it is meaningless. The limit is not something reached be a stepwise process (despite some informal descriptions that illustrate the limit by using something that is like a stepwise process). You have a sequence with no last element, and an operator that is called limit. You apply the operator to the sequence and you get a value (in this case a set). If it offends you to have this value called a limit call it something else, e.g. nightingale. It is a consequence of the definition of nightingale, that the nightingale of a series with every element an infinite set, can be the empty set.

> ... not all natnumbers have gone as is prescribed and required by
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

They are gone, (of course the is no "step" at which they are gone) one at each element of |N_F (which of course as a Peano set has no last element).

--
William Hughes

On 6/28/2021 2:37 PM, WM wrote:
> Greg Cunt schrieb am Sonntag, 27. Juni 2021 um 18:56:51 UTC+2:
>> On Sunday, June 27, 2021 at 6:05:09 PM UTC+2, Christa M wrote:
>>> Jim Burns schrieb am Samstag, 26. Juni 2021 um 20:12:46 UTC+2:
>>>>
>>>> Each k in N is not in E(k+1), so it's not in all end segments.
>>>>
>>> The endsegments E(1), E(2), E(3), ... don't delete all natnumbers because they all are infinite.
>> Endsegements don't do anything, they just are.
>
> Wrong. ∀k ∈ ℕ: E(k+1) = E(k) \ {k} means E(k+1) is deleating k.

wrong E(k+1) never had k in it at all.

>
> Regards, WM
>

On 6/28/2021 2:45 PM, WM wrote:
> Jim Burns schrieb am Sonntag, 27. Juni 2021 um 21:02:45 UTC+2:
>> On 6/27/2021 12:05 PM, Christa M wrote:
>>> Jim Burns schrieb
>>> am Samstag, 26. Juni 2021 um 20:12:46 UTC+2:
>>>> Each k in N is not in E(k+1), so it's not in all end segments.
>>>
>>> The endsegments E(1), E(2), E(3), ... don't delete all natnumbers
>>> because they all are infinite.
>> There are no numbers that each end segment deletes.
>
> There are aleph_0 numbers which no infinite endsegment deletes.
>>
>> Each number is deleted by at least one end segment.
>
> Each infinite endsegment keeps aleph_0 natnumbers.
>
>> No number is in the intersection.
>
> If only infinite endsegments would exist, the intersection would be infinite.
>
>>> What are their remaining elements?
>> {}.
>>> What deletes them finally such that the limit is empty?
>> No *FINAL* end segment before {} exists.
>
> But there exist all endsegments, and if all are infinite, then not all natnumbers can have gone by action of endsegments as is prescribed and required by
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>
> Therefore your calculation of the limit would be wrong without dark numbers.
>
> Regards, WM
>

no.

∀k ∈ ℕ: E(k+1) = E(k) \ {k} is trivial, and why do you stop at k ?

On Monday, 28 June 2021 at 16:48:35 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 27. Juni 2021 um 22:44:51 UTC+2:
>
>
> > WM seems to think he should get to the end of the rainbow by taking one step at a time. Me, I wait until sundown, at the latest.
> If there is nothing between the infinite endsegments and the limit, then not all natnumbers have gone by endsegments as is prescribed and required by
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> and as it is the basis of your calculation of the limit.

I am not calculating a limit, you imbecile. There is no limit in the definition of intersect { (n in N) E(n) }. Surely even you can verify that the 28 characters immediately to the left of the previous period do not contain the string "lim".

Gus Gassmann schrieb am Montag, 28. Juni 2021 um 23:29:04 UTC+2:
> On Monday, 28 June 2021 at 16:48:35 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Sonntag, 27. Juni 2021 um 22:44:51 UTC+2:
> >
> >
> > > WM seems to think he should get to the end of the rainbow by taking one step at a time. Me, I wait until sundown, at the latest.
> > If there is nothing between the infinite endsegments and the limit, then not all natnumbers have gone by endsegments as is prescribed and required by
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > and as it is the basis of your calculation of the limit.
> I am not calculating a limit,

Whatever you calculate. As long as all endsegments are infinite, infinitely many natnumbers must be inside each of them. What else could make them infinite? Therefore not all have gone.

Regards, WM

Sergio schrieb am Montag, 28. Juni 2021 um 23:04:30 UTC+2:
> On 6/28/2021 2:45 PM, WM wrote:

> > But there exist all endsegments, and if all are infinite, then not all natnumbers can have gone by action of endsegments as is prescribed and required by
> >
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> >
> > Therefore your calculation of the limit would be wrong without dark numbers.
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} is trivial, and why do you stop at k ?

Because this covers all natural numbers. Is there a natural number that is not among ∀k ∈ ℕ?

Regards, WM