[ Sorry, I unintentionally sent this to FF's email.
[ I did not intend to make this a private conversation.

[ This is my third try to post this to sci.math.
[ (@FF sorry^2)
[ Some days I need a keeper.

On 9/15/2021 10:58 AM, Greg Cunt wrote:
> On Wednesday, September 15, 2021 at 4:32:37 PM UTC+2,
> Jim Burns wrote:

>> We don't know.
>
> Oh, but we *do* know! It's n (!).
>
> Really!
>
> I mean, it's not n-k for all k e {m e {1, 2, 3, ...} : m <= n}
> and it is not n+k for all k e IN = {1, 2, 3, ...}, it's just n.

Natural language is not the best tool for this discussion.

We have some facts about n.

If all we know about n is that n is a natural number,
we _don't_ know anything about n that we don't also know
about any other natural number.
Then, there are a lot of things we don't know about n.
However, what's left, things we know about n anyway,
is not nothing. What's left are some facts about n.

And, starting from some facts, we derive further facts with
truth-preserving inferences.

This is all a bland, nodding-off description of mathematics.
What puts WM's guts in an uproar is the claim that we can
still do this even if we don't have as one of our facts
"n is one of some individuals for which there is a last".

On Wednesday, September 15, 2021 at 3:07:34 PM UTC+2, WM wrote:

> The set of endsgments {E(n) | n ∈ ℕ} <bla bla bla

"As long as an endsegment is infinite, it cannot belong to an infinite set of endsegments." [WM]

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).

On Thursday, September 16, 2021 at 1:11:33 AM UTC+2, Jim Burns wrote:

> Then, there are a lot of things we don't know about n.

Indeed!

> [...] What's left are some facts about n.

Right. For example, we don't even know if n is even or if it is odd.

On the other hand, we know for certain that it is EITHER even OR odd. :-P
> What puts WM's guts in an uproar is the claim that we can
> still do this even if we don't have as one of our facts
> "n is one of some individuals for which there is a last".

:-)

Right.

I still like Russell's claim:

"Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true."

[... "true" in an "absolute" sense.]

On Wednesday, 15 September 2021 at 18:53:49 UTC-3, FromTheRafters wrote:
> WM was thinking very hard :
> > FromTheRafters schrieb am Mittwoch, 15. September 2021 um 13:33:50 UTC+2:
> >> WM presented the following explanation :
> >
> >>> As long as an endsegment is infinite, it cannot belong to an infinite set
> >>> of endsegments.
> >> With all due respect, stop lying.
> >
> > The set of endsgments {E(n) | n ∈ ℕ} absorbs all natural numbers. None are
> > remaining for subsequent contents of the endsegments. But an infinite
> > endsegment can only use the first number of its contents as index enumerating
> > its position. Between its position and ω there remain infinitely many
> > natnumbers enumerating finite endsegments. If you are not convinced, you
> > should confess that you despise mathematics, namely its basic definition of
> > endsegments:
> >
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> You misunderstand. I said *STOP* lying.
You think he knows the difference? He can't figure out quantifiers, either....

onsdag 15 september 2021 kl. 13:22:36 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 15. September 2021 um 08:36:39 UTC+2:
> > tisdag 14 september 2021 kl. 21:54:23 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 13. September 2021 um 14:46:09 UTC+2:
> > > > måndag 13 september 2021 kl. 14:36:20 UTC+2 skrev WM:
> > > > > Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:39:21 UTC+2:
> > > > > > On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
> > > > > > [...]
> > > > > > > Here is the proof
> > > > > > > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> > > > > > > aleph_0 natnumbers before the equals sign and after it is not possible.
> > > > There is nothing in mathematics that says infinitely many endsegments cannot exist.
> > > Infinitely many endsegments can exist. They can be put in bijection with |N by E(n) being mapped on n. But then all natural numbers are exhausted (note that a bijection is surjective). Then not all endsegments can contain infinitely many natural numbers.
> > >
> > empty assertion, they all have infinite cardinality.
> A sequence of ordinal oemga cannot have aleph_0 elements which are followed by aleph_0 elements. But almost all contents of all endsegments follows after the indices of endsegments.
>
> As long as an endsegment is infinite, it cannot belong to an infinite set of endsegments.
>
> Regards, WM
Again, empty assertions. You just make assertions that have no proof behind them. All endsegments have the same cardinality and the set of all endsegments exists.

onsdag 15 september 2021 kl. 15:08:54 UTC+2 skrev WM:
> Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 14:03:00 UTC+2:
> > On Wednesday, 15 September 2021 at 08:18:20 UTC-3, 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?
> > It follows its immediate predecessor and is followed by its successor S(n).
> Here is the sequence: 1, 2, 3, 4, 5, 6, ...
>
> Please point to the position of n.
>
> Regards, WM
before n+1 but after n-1

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.
>
> Regards, WM

If you think that you're an idiot. It means nothing of the sort!

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

On Thursday, 16 September 2021 at 07:36:02 UTC-3, WM wrote:
> 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?

Can you? And do you have a fucking clue what it means? It sure as hell does *NOT* imply that there are finite end segments. *EVERY* end segment has cardinality aleph_0. You do not have slightest idea of how infinity works.

Gus Gassmann schrieb am Donnerstag, 16. September 2021 um 13:41:17 UTC+2:
> On Thursday, 16 September 2021 at 07:36:02 UTC-3, WM wrote:
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. Can you see the universal quantifer?
> Can you? And do you have a fucking clue what it means?

I know it. It means that no endsegment can violate this stepwise process.

> It sure as hell does *NOT* imply that there are finite end segments.

If there is an empty limit, then there are finite endsegments according to the formula.

> *EVERY* end segment has cardinality aleph_0.

If there are aleph_0 endsegments, then there are not aleph_0 numbers available as their contents. 1, 2 3, ..., n | n, n+1, n+2, ... Here not aleph_0 numbers can be before the | while as many are behind.

Regards, WM

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

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?

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:05:53 UTC+2:

> All endsegments have the same cardinality and the set of all endsegments exists.

Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ... and aleph_0 numbers beyond. Can you accept this?

Regards, WM

Greg Cunt schrieb am Donnerstag, 16. September 2021 um 01:42:41 UTC+2:

> I still like Russell's claim:
>
> "Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true."
>
I prefer what he says about such fools: "This is an instance of the amazing power of desire in blinding even very able men to fallacies which would otherwise be obvious at once."

Regards, WM

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!

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

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

After serious thinking 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.

Wrong again, I said absolutely not because there is no such
requirement. Saying it is required is just another of your *LIES*.

On Thursday, 16 September 2021 at 09:42:34 UTC-3, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 16. September 2021 um 13:41:17 UTC+2:
> > On Thursday, 16 September 2021 at 07:36:02 UTC-3, WM wrote:
> > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. Can you see the universal quantifer?
> > Can you? And do you have a fucking clue what it means?
> I know it. It means that no endsegment can violate this stepwise process.
> > It sure as hell does *NOT* imply that there are finite end segments.
> If there is an empty limit, then there are finite endsegments according to the formula.
Erm, no. Your logic is ridiculously false.

> > *EVERY* end segment has cardinality aleph_0.
> If there are aleph_0 endsegments, then there are not aleph_0 numbers available as their contents. 1, 2 3, ..., n | n, n+1, n+2, ... Here not aleph_0 numbers can be before the | while as many are behind.

Your usual bullshit. That was to be expected. At least you can still remember your usual spiel, which should be a bright spot in your otherwise miserable life. Enjoy the time before your last few memory cells finally give out..

On 9/16/2021 5:35 AM, WM wrote:
> 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
>

that is not a universal quantifier at all!

It is a simple equation only relating two endsegments, and each endsegment is fixed infinite set.

On Thursday, September 16, 2021 at 3:02:56 PM UTC+2, WM wrote:
> Jim Burns schrieb am Mittwoch, 15. September 2021 um 21:28:32 UTC+2:

Let n e IN.

> > 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, [n] is a place holder. [bla bla]

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

On 9/16/2021 7:42 AM, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 16. September 2021 um 13:41:17 UTC+2:
>> On Thursday, 16 September 2021 at 07:36:02 UTC-3, WM wrote:
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. Can you see the universal quantifer?
>> Can you? And do you have a fucking clue what it means?
>
> I know it. It means that no endsegment can violate this stepwise process.

liar your equation is not a step wise process at all, fool.

>
>> It sure as hell does *NOT* imply that there are finite end segments.
>
> If there is an empty limit, then there are finite endsegments according to the formula.

liar. provide a proof. I proved intersection of all endsegments is empty

>
>> *EVERY* end segment has cardinality aleph_0.
>
> If there are aleph_0 endsegments, then there are not aleph_0 numbers available as their contents.

liar.

> 1, 2 3, ..., n | n, n+1, n+2, ... Here not aleph_0 numbers can be before the | while as many are behind.

liar.

>
> Regards, WM
>

4 lies in one post

On 9/15/2021 9:19 AM, Greg Cunt wrote:
> On Wednesday, September 15, 2021 at 4:14:54 PM UTC+2, Gus Gassmann wrote:
>> On Wednesday, 15 September 2021 at 10:51:21 UTC-3, Greg Cunt wrote:
>>> On Wednesday, September 15, 2021 at 2:57:18 PM UTC+2, WM wrote:
>>>>
>>>> Between every infinite endsegment and {ω} there <bla>
>>>>
>>> Huh?! How is "betweenness" defined in this case. In respect to which order relation?
>>>
>>> After stating a proper definition, please prove your claim.
>>>
>>> Hint: We may use "c" to order the endsegements. Then
>>>
>>> ... c E_3 c E_2 c E_1.
>>>
>>> But {w} isn't an endsegment; moreover for no endsegment E: E c {ω} or {ω} c E.
>>>
>>> Please elaborate.
>>>
>> What for?
>
> Just for fun!
>
>> You know that all he can spout is nonsense,
>
> Right.
>
>> and you know what nonsense he is going to spout.
>
> Not really, that's the "funny" part. He's constantly cumming up with new nonsense.
>

new ANTS !

On 9/15/2021 6:22 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 15. September 2021 um 08:36:39 UTC+2:
>> tisdag 14 september 2021 kl. 21:54:23 UTC+2 skrev WM:
>>> zelos...@gmail.com schrieb am Montag, 13. September 2021 um 14:46:09 UTC+2:
>>>> måndag 13 september 2021 kl. 14:36:20 UTC+2 skrev WM:
>>>>> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:39:21 UTC+2:
>>>>>> On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
>>>>>> [...]
>>>>>>> Here is the proof
>>>>>>> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
>>>>>>> aleph_0 natnumbers before the equals sign and after it is not possible.
>>>> There is nothing in mathematics that says infinitely many endsegments cannot exist.
>>> Infinitely many endsegments can exist. They can be put in bijection with |N by E(n) being mapped on n. But then all natural numbers are exhausted (note that a bijection is surjective). Then not all endsegments can contain infinitely many natural numbers.
>>>
>> empty assertion, they all have infinite cardinality.
>
> A sequence of ordinal oemga cannot have aleph_0 elements which are followed by aleph_0 elements. But almost all contents of all endsegments follows after the indices of endsegments.

liar.

>
> As long as an endsegment is infinite, it cannot belong to an infinite set of endsegments.

liar.

>
> Regards, WM
>

On 9/15/2021 8:07 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 15. September 2021 um 13:33:50 UTC+2:
>> WM presented the following explanation :
>
>>> As long as an endsegment is infinite, it cannot belong to an infinite set of
>>> endsegments.
>> With all due respect, stop lying.
>

<snip crap>

your done.