On 9/16/2021 8:05 AM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 16. September 2021 um 15:02:17 UTC+2:
>> WM formulated on Thursday :
>>> FromTheRafters schrieb am Mittwoch, 15. September 2021 um 23:53:49 UTC+2:
>>>
>>>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>>>> You misunderstand. I said *STOP* lying.
>>>
>>> Start thinking! Infinitely man infinite endsegments require that there are
>>> aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ... and aleph_0
>>> numbers beyond. Do you accept this?
>> Absolutely not!
>
> Then you should not accept that all endsegments with infinite contents of aleph_0 natnumbers can be an infinite set with aleph_0 indices consumed.
>
> Regards, WM
>

lies.

On 9/16/2021 7:48 AM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 01:34:48 UTC+2:
>
>> Consider a listing of all (i.e. infinitely many) endsegments (top-down):
>>
>> {1, 2, 3, 4, 5,. 6, ...}
>> {2, 3, 4, 5, 6, 7, ...}
>> {3, 4, 5, 6, 7, 8, ...}
>> :
>>
>> All endsegments are infinite and they all are elements in the infinite set of all endesegments (since there are infinitely many endsegments).
>
> Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ... and aleph_0 numbers beyond. Do you believe this?
>
> Regards, WM
>

do you believe there are aleph_0 numbers between 0.0001 and 0.0002 ?

How about aleph_0 numbers between .00000000001 and .000000000002 ?

On 9/15/2021 11:08 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 15. September 2021 um 16:32:37 UTC+2:
>> On 9/15/2021 7:18 AM, WM wrote:
>>> Greg Cunt schrieb
>>> am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:
>>
>>>> Hint: After stating this definition we are able to prove
>>>> There is an ordinal x such that x = n ,
>>>> or simpler
>>>> n is an ordinal.
>>>
>>> What is its place in the sequence of ordinals?
>> We don't know. At least, we apparently haven't been told yet.
>
> But you have been told, long ago, the position of 3 and 17 and even 10^10.
>>
>> However, we already know other things about n.
>
> Yes, it should be the placeholder for a natural number which satisfies the following conditions:
>>
>> There is an order-relation '<' between n and other individuals
>> such that, for any collection B of these (ordinal) individuals,
>> B contains a first element, unless B is empty.
>> ('<' is a well-order.)
>>
>> We know facts that can be derived from this fact about n
>> It's interesting that there are a lot of these derived facts,
>> but, interesting or not, we know them.
>
> Of course.
>>
>> Not-knowing which individual n is does not remove from our
>> brains anything we know about n.
>
> But it should remove the silly idea that it is a natural number. On the other hand this simple example explains how fanatical matheologians can defend the obviously wrong results of set theory. To mention only the most striking ones which every sober mind can identify as nonsense:
> - All fractions can be enumerated.
> - There are uncountably many real numbers although only countably many can be distinguished.
> - There are more paths in the Binary Tree than nodes.
> - The bankruptcy of McDuck.
> - aleph_0 rational points separate uncountably many irrational points such that never two are existing without rational between them.
>
> It is painful to see so much fool's crap blocking your minds. I will really enjoy my lectures next semester. It will be relieving to address an audience of unbiased thinkers.
>
> Regards, WM
>

and your class is not required for any real Math Degree, just a fillin elective for those majoring in non science.

On 9/16/2021 8:02 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 15. September 2021 um 21:28:32 UTC+2:
>> On 9/15/2021 12:08 PM, WM wrote:
>
>>>>> What is its place in the sequence of ordinals?
>>>>
>>>> We don't know. At least, we apparently haven't been told yet.
>>>
>>> But you have been told, long ago,
>>> the position of 3 and 17 and even 10^10.
>> The difference between
>> "n is 3", "n is 17", "n is 10^10"
>> and
>> "n is one of these well-ordered things"
>> is that we know
>> "n is one of these well-ordered things".
>
> No is a place holder. In place of n one of these things can be inserted.
>
> Regards, WM
>

no wonder you never progressed in Math.

You are missing the foundations in Math, and you resent Math.

On 9/15/2021 10:37 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 15:46:10 UTC+2:
>> On Wednesday, 15 September 2021 at 09:57:18 UTC-3, WM wrote:
>>
>>> Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition
>>>
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (*)
>>>
>>> would be broken.
>>
>> you have no idea what your equation involving the set difference in (*) actually tells you.
>
> It tells me and every person able to read math text, that no empty set can result unless finite endsegments are passed by.
>
> Regards, WM
>

*wrong*.

the equation *only shows the between adjacent endsegments*, that is all.

There is no stepping, not passed by, no broken, ...

if you want that, write another line or two of math.

FromTheRafters schrieb am Donnerstag, 16. September 2021 um 15:17:35 UTC+2:
> After serious thinking WM wrote :

> > Then you should not accept that all endsegments with infinite contents of
> > aleph_0 natnumbers can be an infinite set with aleph_0 indices consumed.
> Wrong again, I said absolutely not because there is no such
> requirement.

If there are aleph_0 endsegments, then their indices exhaust |N. Do you accept that?
What is their contents?

Regards, WM

On 9/16/2021 2:53 PM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 16. September 2021 um 15:17:35 UTC+2:
>> After serious thinking WM wrote :
>
>>> Then you should not accept that all endsegments with infinite contents of
>>> aleph_0 natnumbers can be an infinite set with aleph_0 indices consumed.
>> Wrong again, I said absolutely not because there is no such
>> requirement.
>
> If there are aleph_0 endsegments, then their indices exhaust |N. Do you accept that?

no. Why do you think that ?

> What is their contents?

what kind of question is that ?

>
> Regards, WM
>

WM explained on 9/16/2021 :
> FromTheRafters schrieb am Donnerstag, 16. September 2021 um 15:17:35 UTC+2:
>> After serious thinking WM wrote :
>
>>> Then you should not accept that all endsegments with infinite contents of
>>> aleph_0 natnumbers can be an infinite set with aleph_0 indices consumed.
>> Wrong again, I said absolutely not because there is no such
>> requirement.
>
> If there are aleph_0 endsegments, then their indices exhaust |N. Do you
> accept that?

Exhaust? no, but the infinite cardinality of the individual endsegments
(a sequence taken as a set) is the same cardinality as the cardinality
of the entire infinite set of endsegments.

No problem.

> What is their contents?

Sets.

Serg io wrote :
> On 9/16/2021 7:48 AM, WM wrote:
>> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 01:34:48 UTC+2:
>>
>>> Consider a listing of all (i.e. infinitely many) endsegments (top-down):
>>>
>>> {1, 2, 3, 4, 5,. 6, ...}
>>> {2, 3, 4, 5, 6, 7, ...}
>>> {3, 4, 5, 6, 7, 8, ...}
>>>>
>>>
>>> All endsegments are infinite and they all are elements in the infinite set
>>> of all endesegments (since there are infinitely many endsegments).
>>
>> Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2,
>> ... and aleph_0 numbers beyond. Do you believe this?
>>
>> Regards, WM
>>
>
> do you believe there are aleph_0 numbers between 0.0001 and 0.0002 ?
>
>
> How about aleph_0 numbers between .00000000001 and .000000000002 ?

At least.

On Thursday, September 16, 2021 at 2:48:34 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 01:34:48 UTC+2:
> >
> > Consider a listing of all (i.e. infinitely many) endsegments (top-down):
> >
> > {1, 2, 3, 4, 5,. 6, ...}
> > {2, 3, 4, 5, 6, 7, ...}
> > {3, 4, 5, 6, 7, 8, ...}
> > :
> >
> > All endsegments are infinite and they all are elements in the infinite set of all endesegments (since there are infinitely many endsegments).

> Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ...

Nope. There are exactly n numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ...

> and aleph_0 numbers beyond.

Right.

> Do you believe this?

Sure.

Hint: An e IN: card({1, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 is a theorem in ZF(C).

On 9/16/2021 9:02 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 15. September 2021 um 21:28:32 UTC+2:
>> On 9/15/2021 12:08 PM, WM wrote:

>>>>> What is its place in the sequence of ordinals?
>>>>
>>>> We don't know. At least, we apparently haven't been
>>>> told yet.
>>>
>>> But you have been told, long ago,
>>> the position of 3 and 17 and even 10^10.
>>
>> The difference between
>> "n is 3", "n is 17", "n is 10^10"
>> and
>> "n is one of these well-ordered things"
>> is that we know
>> "n is one of these well-ordered things".
>
> No is a place holder.
> In place of n one of these things can be inserted.

It is not the case that _anything_ can be inserted.
No flying rainbow sparkle ponies allowed, for example.

Suppose that we want to discuss individuals that can be
counted to, in principle.

We could say
"n can be counted to, in principle."

I assume that we have, at that point, excluded flying
rainbow sparkle ponies. Because flying rainbow sparkle
ponies cannot be counted to, not even in principle.

If it were flying rainbow sparkle ponies we intend to
discuss, the description of n would be different.

The description of what we are discussing is the essence,
the keystone of this whole style of argumentation.

----
( If you (WM) want to add another layer of verbiage,
( that's irritating, but nothing worse than that.
( What do you want to say instead?
( "One of those individuals we are discussing can be
( inserted for n."
( ( I don't see any advantage to the extra verbiage.
( But, once we've formalized everything, all we see
( is "n". I don't see any serious disadvantage, either.

----
Note that that being one of those individuals is not
the sort of thing that can be "checked".
Essentially, it's true because _we say it's true_
(This is a rare and valuable situation.)

Assuming we _know_ that we say it's true, we _know_ that
n is (n can be replaced by)(n refers to)(...)
a natural number. A point in the line. A flying rainbow
sparkle pony. Whatever it is we are discussing.

torsdag 16 september 2021 kl. 12:36:02 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:19:37 UTC+2:
> > onsdag 15 september 2021 kl. 17:37:24 UTC+2 skrev WM:
> > > Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 15:46:10 UTC+2:
> > > > On Wednesday, 15 September 2021 at 09:57:18 UTC-3, WM wrote:
> > > >
> > > > > Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition
> > > > >
> > > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (*)
> > > > >
> > > > > would be broken.
> > > >
> > > > you have no idea what your equation involving the set difference in (*) actually tells you.
> > > It tells me and every person able to read math text, that no empty set can result unless finite endsegments are passed by.
> > >
> > It means nothing of the sort!
> You don't like mathematics? It is this: ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. Can you see the universal quantifer?
>
> Regards, WM

I can read it better than you, know mathematics better than you and understand mathematics better than you.

The fact that the intersection of all endsegments is empty does not imply finite endsegments.

On Friday, September 17, 2021 at 7:18:31 AM UTC+2, zelos...@gmail.com wrote:

> The fact that the intersection of all endsegments is empty does not imply finite endsegments.

WM's approach: "I have said it thrice: What I tell you three times is true."

On 9/17/2021 1:03 AM, Greg Cunt wrote:
> On Friday, September 17, 2021 at 7:18:31 AM UTC+2, zelos...@gmail.com wrote:
>
>> The fact that the intersection of all endsegments is empty does not imply finite endsegments.
>
> WM's approach: "I have said it thrice: What I tell you three times is true."
>

WM has screw lose somewhere as endsegments are defined as infinite, so there are no finite endsegments.

So WM can also state "Alice has found her glasses does not imply finite endsegments."

Jim Burns schrieb am Freitag, 17. September 2021 um 02:00:49 UTC+2:

> We could say
> "n can be counted to, in principle."

Yes n is or is representig the typical natural number. That means n is not a concrete natural number, but every natural number can be put at its place.. Let n ∈ ℕ, then n has a uniqe prime decomposition. You cannot know what this prime decomposition is. But you know it in principle from every natural number.

Regards, WM

Greg Cunt schrieb am Donnerstag, 16. September 2021 um 18:04:33 UTC+2:

> No, n is a natural number, not a "place holder".

What is its prime decomposition? How many factors has it?

Regards, WM

FromTheRafters schrieb am Donnerstag, 16. September 2021 um 22:44:42 UTC+2:
> WM explained on 9/16/2021 :
> > FromTheRafters schrieb am Donnerstag, 16. September 2021 um 15:17:35 UTC+2:
> >> After serious thinking WM wrote :
> >
> >>> Then you should not accept that all endsegments with infinite contents of
> >>> aleph_0 natnumbers can be an infinite set with aleph_0 indices consumed.
> >> Wrong again, I said absolutely not because there is no such
> >> requirement.
> >
> > If there are aleph_0 endsegments, then their indices exhaust |N. Do you
> > accept that?
> Exhaust? no,

Then you refuse set theory. There the sequence E(1), E(2), E(3), ... exhausts all natural numbers. None is remaining outside of the sequence.

> but the infinite cardinality of the individual endsegments
> (a sequence taken as a set) is the same cardinality as the cardinality
> of the entire infinite set of endsegments.

Here we have not only same cardinality but precisely the same numbers.
>
> No problem.
>
> > What is their contents?
>
> Sets.

What elements do these sets have?

Regards, WM

Greg Cunt schrieb am Donnerstag, 16. September 2021 um 23:02:49 UTC+2:
> On Thursday, September 16, 2021 at 2:48:34 PM UTC+2, WM wrote:

> > > All endsegments are infinite and they all are elements in the infinite set of all endesegments (since there are infinitely many endsegments).
>
> > Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ...
> Nope. There are exactly n numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ...

In case of all endsegments? What is the last endsegment? Note that n represents a finite number.
>
> > and aleph_0 numbers beyond.
>
> Right.
>
> > Do you believe this?
>
> Sure.
>
> Hint: An e IN: card({1, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 is a theorem in ZF(C).

And card of all endsegments is finite with last element E(n)? What is its last element? Or is the set of endsegments infinite?

Regards, WM

zelos...@gmail.com schrieb am Freitag, 17. September 2021 um 07:18:31 UTC+2:
> torsdag 16 september 2021 kl. 12:36:02 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:19:37 UTC+2:
> > > onsdag 15 september 2021 kl. 17:37:24 UTC+2 skrev WM:
> > > > Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 15:46:10 UTC+2:
> > > > > On Wednesday, 15 September 2021 at 09:57:18 UTC-3, WM wrote:
> > > > >
> > > > > > Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition
> > > > > >
> > > > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (*)
> > > > > >
> > > > > > would be broken.
> > > > >
> > > > > you have no idea what your equation involving the set difference in (*) actually tells you.
> > > > It tells me and every person able to read math text, that no empty set can result unless finite endsegments are passed by.
> > > >
> > > It means nothing of the sort!
> > You don't like mathematics? It is this: ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. Can you see the universal quantifer?
> >
> I can read it better than you

But you don't accept it?

> The fact that the intersection of all endsegments is empty does not imply finite endsegments.

If a finite endsegment E(m) was the minimal endsegment, then the intersection of all would be E(m).

But if no finite endsegment is among them, then the intersection is less than every finite set? Only real fools can accept that.

All infinite endsegments have an infinite intersection with each other. This is proved by inclusion monotony! As long as an endsegment contains elements, it contains them with all preceding endsegments.

Regards, WM

On 9/17/2021 2:44 PM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 18:04:33 UTC+2:
>
>> No, n is a natural number, not a "place holder".
>
> What is its prime decomposition? How many factors has it?
>
> Regards, WM
>

Please refer to "Highly Composite Numbers" by S. Ramanujan, Proc London Mathematical Society 2, XIV, 1915, 347-409

On Friday, September 17, 2021 at 9:55:31 PM UTC+2, WM wrote:

> Or is the set of endsegments infinite?

Errrr, right, it is infinite.

Hint there's a bijection between IN and {E(n) : n e IN}.

For example: f: IN ---> {E(n) : n e IN} with f(n) = E(n) for all n e IN is a bijektion.

Hence card({E(n) : n e IN}) = card(IN) = aleph_0. Which means that {E(n) : n e IN} is infinite. qed

On Friday, September 17, 2021 at 9:43:23 PM UTC+2, WM wrote:

Let n e IN.

> n is not a concrete natural number

It's certainly a "concrete" natural number, though we don't KNOW which one.

> but every natural number can be put at its place.

Nope it can't. Hint n+1 =/= n.

> Let n ∈ ℕ, then n has a unique prime decomposition.

Exactly (if n > 1).

> You cannot know what this prime decomposition is.

Right.

> But you know [there is one for each and] every natural number [> 1].

Right.

On Friday, September 17, 2021 at 9:44:43 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 18:04:33 UTC+2:
> >
> > No, n is a natural number, not a "place holder".
> >
> What is its prime decomposition? How many factors has it?

I/we don't *know*.

Though we DO KNOW that n has a unique prime factorization (if n > 1), SINCE n is a natural number. See?!

(Hint: Does a "place holder"have a unique prime factorization ...and can it even be > 1?)

On 9/17/2021 2:50 PM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 16. September 2021 um 22:44:42 UTC+2:
>> WM explained on 9/16/2021 :
>>> FromTheRafters schrieb am Donnerstag, 16. September 2021 um 15:17:35 UTC+2:
>>>> After serious thinking WM wrote :
>>>
>>>>> Then you should not accept that all endsegments with infinite contents of
>>>>> aleph_0 natnumbers can be an infinite set with aleph_0 indices consumed.
>>>> Wrong again, I said absolutely not because there is no such
>>>> requirement.
>>>
>>> If there are aleph_0 endsegments, then their indices exhaust |N. Do you
>>> accept that?
>> Exhaust? no,
>
> Then you refuse set theory. There the sequence E(1), E(2), E(3), ... exhausts all natural numbers.

so you admit the three dots ... leads to infinity!

> None is remaining outside of the sequence.

all natural numbers were used.

>
>> but the infinite cardinality of the individual endsegments
>> (a sequence taken as a set) is the same cardinality as the cardinality
>> of the entire infinite set of endsegments.
>
> Here we have not only same cardinality but precisely the same numbers.
>>
>> No problem.
>>
>>> What is their contents?
>>
>> Sets.
>
> What elements do these sets have?

natural numbers... why do you ask ?

>
> Regards, WM
>

On 9/17/2021 2:55 PM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 23:02:49 UTC+2:
>> On Thursday, September 16, 2021 at 2:48:34 PM UTC+2, WM wrote:
>
>>>> All endsegments are infinite and they all are elements in the infinite set of all endesegments (since there are infinitely many endsegments).
>>
>>> Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ...
>> Nope. There are exactly n numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ...
>
> In case of all endsegments? What is the last endsegment?

there is no LAST endsegment.

> Note that n represents a finite number.

so what ?

>>
>>> and aleph_0 numbers beyond.
>>
>> Right.
>>
>>> Do you believe this?
>>
>> Sure.
>>
>> Hint: An e IN: card({1, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 is a theorem in ZF(C).
>
> And card of all endsegments is finite with last element E(n)? What is its last element? Or is the set of endsegments infinite?

you got confused again.

there is no last endsegment. there is no last element.

>
> Regards, WM
>