On 9/11/2021 4:27 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 11. September 2021 um 20:28:41 UTC+2:
>> On 9/11/2021 9:05 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 11. September 2021 um 00:51:17 UTC+2:

>>>> if all end segments are infinite,
>>>> then there is no last end segment.
>>>
>>> Then there is an infinite set of natural numbers
>>> remaining in all endsegments forever.
>>
>> No, there are no forever-numbers.
>
> All endsegments are infinite.
> What do they contain?

Infinitely-many not-forever-numbers.

( Note that this is by design.
( Forever-numbers can't be counted with.
( Finitely-many numbers risk halting the counting-process
( in error, with things-being-counted left uncounted.

Different end segments contain different numbers.
Because they're different end segments.

Their intersection is either the last end segment
or the empty set. Suppose no last end segment exists.
What does their intersection contain?

( The intersection of closed upwards sets is
( a closed upwards set.
( ( A closed upwards set is an end segment or the empty set.
( ( An intersection of end segments is, _if it is an end segment_
( the last end segment.
( ( Therefore,
( an intersection of end segments is the last end segment
( or the empty set.

WM submitted this idea :
> Jim Burns schrieb am Samstag, 11. September 2021 um 20:28:41 UTC+2:
>> On 9/11/2021 9:05 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 11. September 2021 um 00:51:17 UTC+2:
>>>> if all end segments are infinite,
>>>> then there is no last end segment.
>>>
>>> Then there is an infinite set of natural numbers
>>> remaining in all endsegments forever.
>> No, there are no forever-numbers.
>
> All endsegments are infinite. What do they contain?

Natural numbers, naturally.

On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
> Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
> > On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
> > > William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
> > > > On Friday, September 10, 2021 at 5:35:51 PM UTC-3, WM wrote:
> > > > > William schrieb am Freitag, 10. September 2021 um 20:19:16 UTC+2:
> > >
> > > > > [The endless sequence] consist of only finite initial subsequences.
> > > >
> > > > An endless sequence consists of an infinite number of finite sequences. An infinite sequence has properties that no finite sequence has.
> > > >
> > > > You are fond of the argument:
> > > Every infinite endsegment has an infinite intersection with all its predecessors. If the infinite sequence has only infinite endsegments, then this cannot change and does not change.
> > And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
> And there is no finite n leading to an infinite intersection. But there are onkly finite n. Hece tere is no empty intersection.
Haha. I don't know if you think this funny or are serious. Either way, it is not an argument. For the record, there are infinitely many natural numbers. each of them finite.

On Saturday, September 11, 2021 at 10:22:50 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:

> > For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).

Of course.

> And there is no finite n leading to an [empty] intersection.

Exactly.

But the intersection of (the set of) ALL end segments -i.e. INFINITELY MANY- is EMPTY.

> > The intersection over *INFINITE* (do you see the difference?)

A sane person might "see" (comprehend) it.

Of course, *you*, Mückenheim, are a

> fool

and you should not be

> admit[ed] to mathematics.

On 9/11/2021 3:25 PM, WM wrote:
> Jim Burns schrieb am Samstag, 11. September 2021 um 18:02:11 UTC+2:
>
>> Write "the intersection is an end segment" as "Assumption"
>> and "there is a last end segment" as "Conclusion".
>
> The intersection cannot be less than the smallest endsegment.

Wrong. all endsegments are the same size.

> If all endsegments are infinite, then the smallest endsegment is infinite too.

AND they are all the same size, aleph_null

>
> Regards, WM
>

On 9/11/2021 8:20 AM, WM wrote:
> William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
>> On Friday, September 10, 2021 at 5:35:51 PM UTC-3, WM wrote:
>>> William schrieb am Freitag, 10. September 2021 um 20:19:16 UTC+2:
>
>>> [The endless sequence] consist of only finite initial subsequences.
>>
>> An endless sequence consists of an infinite number of finite sequences. An infinite sequence has properties that no finite sequence has.
>>
>> You are fond of the argument:
>
> Every infinite endsegment has an infinite intersection with all its predecessors.

ok

> If the infinite sequence has only infinite endsegments,

you have switched to vague mode, please clarify what you mean...

>then this cannot change and does not change.

meaningless.

>
> If you claim that a sequence of only infinite endsegments has an empty intersection,

The interesection of All endsegments is empty.

> then this is nonsense and not useful to continue this discussion. It is enough for me to have forced you matheologians to make nonsense claims.
>
> Regards, WM
>

you do nothing, except complain at the keyboard. show some Math, be a Professor! show some Proofs... nope.

On Saturday, September 11, 2021 at 1:07:10 PM UTC-7, Ross A. Finlayson wrote:
> On Saturday, September 11, 2021 at 12:50:20 PM UTC-7, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 9:34:11 PM UTC+2, Serg io wrote:
> >
> > > what about midsegment ?
> > > and backsegment ?
> > > or startsegment ?
> > > or unsegment?
> > Please don't forget about the
> >
> > antsegment
> >
> > !
> I read you as "anti-segment".
>
> This is writing the same numbers as 0+ 0, 1, 2, 3, ..., oo- 0, 1, 2, 3, ....
>
> I.e. it would be incomplete to deny the numbers their symmetry.
>
>
> I.e., 0 + 0, 1, 2, 3, ..., 3, 2, 1, 0 -oo.
>
> With that being the compass closing, there are all sorts of
> usual simplest arguments what make guarantees.

Everybody of course is familiar with Georg Cantor about uncountability,
then there's also some informal constructions like "counting backward"
from infinity, that like the "domain principle", another usual notion that
the world exists, this is just for saying that some of the arguments that
WM doesn't carry, Cantor let drop for not carrying: but they're not in
themselves "wrong", only that he let them go for his higher result,
the mw proof the nested intervals proof and antidiagonal proof,
and the powerset result, to cement his legacy in uncountability.

Then, after modern methods like that a family of functions model a
function in the limit, like delta or here or course sweep, allowed entry
of those definitions of functions into the space of well-defined functions,
what I find relevant of course is that Cantor thusly proves that the line is
drawn, being ran(sweep), which of course is only amenable as that in
terms of the mw proof, the nested intervals, and antidiagonal argument,
that it falls through as "Counterexample", to uncountability of a continuous
domain, as what follows, from that sweep is not a Cartesian function.

Modern approaches assign length for measure 1.0 as via definition,
and for various concerns in topology after bundles and sheaves the
ultrafilter, of the countable saturation of what results a continuous domain
or the satisfaction under definition of continuity, usually functions, that
reflects why many efforts, for example as we're talking about Gruenbaum,
find justification in Cantor's line-drawing of ran(sweep), in a mathematics that
of course (or, it happens that it does) presents transfinite cardinals, that it
also works out for the well-ordering of the reals, or, any subset of them dense
in the reals, that the proto-typical implementation of a definition of the
passage of time: makes for a better, fuller mathematics, including as being
naturally in the geometric setting, that modern mathematics needs,
to overcome the inductive impasse, what results that measure 1.0 after
the measurability of a point-set, implements countable additivity of atoms.

On 9/11/2021 4:53 PM, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 10:26:15 AM UTC+2, Ross A. Finlayson wrote:
>
>> Please pick a new name if you're not using your real one,
>
> You may call me Great Cunt, if you like.
>

Finlayson means Ang Mga Ito in Filipino,

which means Izy ireo in Malagasy,

or Lazyo in Samoan or

Lazy Oreo in Kaliforianian

(googletranslate)

On 9/11/2021 7:37 PM, Gus Gassmann wrote:
> On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
>> Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
>>> On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
>>>> William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
>>>>> On Friday, September 10, 2021 at 5:35:51 PM UTC-3, WM wrote:
>>>>>> William schrieb am Freitag, 10. September 2021 um 20:19:16 UTC+2:
>>>>
>>>>>> [The endless sequence] consist of only finite initial subsequences.
>>>>>
>>>>> An endless sequence consists of an infinite number of finite sequences. An infinite sequence has properties that no finite sequence has.
>>>>>
>>>>> You are fond of the argument:
>>>> Every infinite endsegment has an infinite intersection with all its predecessors. If the infinite sequence has only infinite endsegments, then this cannot change and does not change.
>>> And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
>> And there is no finite n leading to an infinite intersection. But there are onkly finite n. Hece tere is no empty intersection.
>
> Haha. I don't know if you think this funny or are serious. Either way, it is not an argument. For the record, there are infinitely many natural numbers. each of them finite.
>
>

I picked up a number at the bar, I thought she was infinite, but she said she was finite, and looking for an endsegment.

On Saturday, September 11, 2021 at 9:04:09 PM UTC-7, Serg io wrote:
> On 9/11/2021 4:53 PM, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 10:26:15 AM UTC+2, Ross A. Finlayson wrote:
> >
> >> Please pick a new name if you're not using your real one,
> >
> > You may call me Great Cunt, if you like.
> >
> Finlayson means Ang Mga Ito in Filipino,
>
> which means Izy ireo in Malagasy,
>
> or Lazyo in Samoan or
>
> Lazy Oreo in Kaliforianian
>
> (googletranslate)

Who's buried in the lone highlander's grave?
Was William Wallace really a mack?

You can call me Ross DesCartes Lincoln Finlayson.

It means "son of Finlay Mor,
who was struck by a cannonball,
at the Battle of Pinkie in 1500's".

Sorry, I don't go by any pseudo-nyms.

There's Finn Ecces, though....

Don't forget the anagram generator,
as long as the jabberwock is filling your idea-sheet.

Privat, kak dela.

I don't use words like C*** so I get a lot more that way.

This isn't a moderated forum, but decorum is a thing.

Kline quotes Weyl:

The question of the foundations and the ultimate meaning
of mathematics remains open; we do not know in what direction
it will find its final solution or even whether a final objective answer
can be expected at all. "Mathematizing" may well be a creative
activity of man, like language or music, of primary originality,
whose _historical_ decisions defy complete objective rationalization.
[emph. added]

I find Kline a bit much the pessimist though, for modern axiomless
foundations in natural deduction, what neatly arrive at theory.

Consider (from "Mathematics: The Loss of Certainty"):

It seems intuitively clear that the larger the number of rectangles
the better the sum of their areas would approximate the desired
curvilinear area. However, if one stopped at 50 or 100 rectnagles
the sum would still not be the area sought. The though that occurred
to the 17-th century mathematicians who fashioned this approach was
to let n become infinite. However, just what infinite meant was not clear.

Fermat believed that precise proofs could be made, most likely resorting
to the method of exhaustion, introduced by Eudoxus (a very limited and
rather complicated geometrical method and used so skillfully by Archimedes).

(end quote)

(quoting)

.... perhaps Bonaventura Cavalieri's work deserves attention because it influenced
many contemporary and later men and because it is typical of the vague thinking
of the times. Cavalieri regarded an area ... as a sum of an infinite number of units
which he called indivisibles and which presumably could be lines. ... [Cavalieri]
"explained" that the area in question is made up of indivisibles, just as a necklace
is made up of beads, a cloth is of threads, and a book is of pages. With this concept
he did manage to compare two areas or two volumes and obtain the correct
relationship between the two.

....

Pascal defended Cavalieri. ... [Pascal] affirmed that the geometry of indivisibles
and classical Greek geometry were in agreement. "What is demonstrated by the
true rules of indivisibles could be demonstrated also with the rigor and the manner
of the ancients." They differed only in terminology. Further, the method of indivisibles
had to be accepted by any mathematician who pretended to rank among geometers.

(end quote)

If you might care to read on there page 134, 135, ..., it helps collect some of the
thinking that's part of historiography of foundations of analysis.

(quoting)

And so the 18th century ended with the logic of the calculus and of the branches
of analysis built on the calculus in a totally confused state. In fact, one could say
that the state of the foundations was worse in 1800 than in 1700. Giants, notably
Euler and Lagrange, had given incorrect logical foundations. Because these men
were authorities, many of their colleagues accepted and repeated uncritically what
they proposed and even built more analysis on their foundations. Other lesser lights
were not quite satisfied with what the master s had advanced, but they were confident
that a totally clear foundation could be secured merely by clarification or minor emendation.
Of course they were being led down false trails.

( -- Morris Kline)

On Saturday, 11 September 2021 at 22:40:54 UTC-3, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 10:22:50 PM UTC+2, WM wrote:
> > Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
>
> > > For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
> Of course.
>
> > And there is no finite n leading to an [empty] intersection.
>
> Exactly.
>
> But the intersection of (the set of) ALL end segments -i.e. INFINITELY MANY- is EMPTY.
> > > The intersection over *INFINITE* (do you see the difference?)

As a first step I would be happy for him to acknowledge that FINITE and INFINITE 1. *look* different and 2. do not *mean* the same. I am not even sure that he gets either of those.

William schrieb am Samstag, 11. September 2021 um 23:25:41 UTC+2:
> On Saturday, September 11, 2021 at 5:22:50 PM UTC-3, WM wrote:
>
> > And there is no finite n leading to a [finite] intersection. But there are onkly finite n.
>
> So what? We are talking about a property of the sequence. There are only finite n. But the sequence is made from
> and infinite number of finite n.

Impossible for definable n:
Representation of the natural numbers: o, oo, ooo,...
Representation of the number of terms: o, oo, ooo, ...
Both sequences will never deviate. Therefore "infinite" here can only mean always finite but without a last term, i.e., potential infinity. What is your counter argument?

> No finite number of finite n can lead to a finite intersection. But an infinite number of finite n can.

As you can see above, it cannot exist (unless having dark elements).

Regards, WM

Greg Cunt schrieb am Samstag, 11. September 2021 um 23:40:49 UTC+2:
> On Saturday, September 11, 2021 at 10:28:05 PM UTC+2, WM wrote:
>
> > All endsegments are infinite. What do they contain?
> Natural numbers?

which are also contained in all predecessor endsegments?
>
Regards, WM

Greg Cunt schrieb am Samstag, 11. September 2021 um 23:56:49 UTC+2:
> On Saturday, September 11, 2021 at 10:25:50 PM UTC+2, WM wrote:
>
> > The intersection cannot be less than the smallest endsegment. If all endsegments are infinite, then the smallest endsegment is infinite too.

> Hint: There IS NO /smallest endsegment/.

Then there are two or more.

Regards, WM

Jim Burns schrieb am Sonntag, 12. September 2021 um 00:42:52 UTC+2:
> On 9/11/2021 4:27 PM, WM wrote:

> > All endsegments are infinite.
> > What do they contain?
> Infinitely-many not-forever-numbers.

If endsegments are forever, then numbers must be forever.
>
> ( Note that this is by design.
> ( Forever-numbers can't be counted with.

Then endsegments can't contain them.
> ( an intersection of end segments is the last end segment
> ( or the empty set.

Forever non-empty endsegments contain more than nothing. This is contained in all endsegments and hence in their intersection.

Regards, WM

FromTheRafters schrieb am Sonntag, 12. September 2021 um 01:15:47 UTC+2:

> > All endsegments are infinite. What do they contain?
> Natural numbers, naturally.

Natural numbers are remaining in endsegments as long as endsegments are infinite. Forever.
Alas, when we denote the endsegments E(n) by their natural numbers n, then only finitely many are available because infinitely many must be spared for the contents of the endsegments. Note that two consecutive infinite sets in order omega are impossible.

Regards, WM

Gus Gassmann schrieb am Sonntag, 12. September 2021 um 02:37:38 UTC+2:
> On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
> > > On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
> > > > William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:

> > > And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
> > And there is no finite n leading to an infinite intersection. But there are only finite n. Hence there is no empty intersection.
> For the record, there are infinitely many natural numbers. each of them finite.

But infinitely many must be spared for the contents of the endsegments. Therefore only finitely many n can be used to index the endsegments E(n). Note that in order omega two consecutive infinite sets of card aleph_0 are not possible. Therefore aleph_0 endsegments with aleph_0 elements each can only be assumed by peopel who cannot count 1 + 1.

Regards, WM

Greg Cunt schrieb am Sonntag, 12. September 2021 um 03:40:54 UTC+2:

> But the intersection of (the set of) ALL end segments -i.e. INFINITELY MANY- is EMPTY.

There is *one* sequence of natural numbers used to index the endsegments and to fill them. As long as aleph_0 numbers are required to fill the endsegments only finitely many can be used to index them.

Here is an example for beginners:

E(1), E(2), E(3), E(4), E(5) = (5, 6, 7, 8, 9, ...)

and here for more advanced readers

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.

> A sane person might "see" (comprehend) it.

Regards, WM

On 9/12/2021 11:24 AM, WM wrote:
> Greg Cunt schrieb am Samstag, 11. September 2021 um 23:40:49 UTC+2:
>> On Saturday, September 11, 2021 at 10:28:05 PM UTC+2, WM wrote:
>>
>>> All endsegments are infinite. What do they contain?
>> Natural numbers?
>
> which are also contained in all predecessor endsegments?
>>
> Regards, WM
>

You got lost again, this is simple stuff, pay attention!

All endsegments are infinite.

All endsegments are composed of Natural numbers.

now what ?

On 9/12/2021 11:28 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 12. September 2021 um 00:42:52 UTC+2:
>> On 9/11/2021 4:27 PM, WM wrote:
>
>>> All endsegments are infinite.
>>> What do they contain?
>> Infinitely-many not-forever-numbers.
>
> If endsegments are forever, then numbers must be forever.

Forever Ants
Infinitely-many not-forever-Ants

Please define what you mean by "forever" ? it usually refers to time

>>
>> ( Note that this is by design.
>> ( Forever-numbers can't be counted with.
>
> Then endsegments can't contain them.

wrong. Endsegments are composed of natural numbers.

>
>> ( an intersection of end segments is the last end segment
>> ( or the empty set.
>
> Forever non-empty endsegments contain more than nothing. This is contained in all endsegments and hence in their intersection.

meaningless drival.

>
> Regards, WM
>

On 9/12/2021 11:33 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 12. September 2021 um 01:15:47 UTC+2:
>
>>> All endsegments are infinite. What do they contain?
>> Natural numbers, naturally.
>
> Natural numbers are remaining in endsegments as long as endsegments are infinite. Forever.

Wrong. all endsegment are static, they do not change, they area always infinite, there is no "remaining in" or not.

what do you mean by Forever ??

> Alas, when we denote the endsegments E(n) by their natural numbers n,

OK

>then only finitely many are available

Wrong

>because infinitely many must be spared for the contents of the endsegments.

Wrong

>Note that two consecutive infinite sets in order omega are impossible.

Wrong.

>
> Regards, WM
>

4 wrongs out of five, a slight improvement, but still in the "Always Wrong" department.

Memorize this
Aleph0 = k * Aleph0
Aleph0 = K + Aleph0

On 9/12/2021 11:25 AM, WM wrote:
> Greg Cunt schrieb am Samstag, 11. September 2021 um 23:56:49 UTC+2:
>> On Saturday, September 11, 2021 at 10:25:50 PM UTC+2, WM wrote:
>>
>>> The intersection cannot be less than the smallest endsegment. If all endsegments are infinite, then the smallest endsegment is infinite too.
>
>> Hint: There IS NO /smallest endsegment/.
>
> Then there are two or more.

Wrong, all endsegments are the same size, Aleph0

>
> Regards, WM
>

get with the program, you are stuck not knowing infinity

WM wrote on 9/12/2021 :
> FromTheRafters schrieb am Sonntag, 12. September 2021 um 01:15:47 UTC+2:
>
>>> All endsegments are infinite. What do they contain?
>> Natural numbers, naturally.
>
> Natural numbers are remaining in endsegments as long as endsegments are
> infinite.

Which they are, so no need to mention it as if it were a choice.

On 9/12/2021 11:42 AM, Transfinity wrote:
> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 02:37:38 UTC+2:
>> On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
>>>> On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
>>>>> William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
>
>>>> And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
>>> And there is no finite n leading to an infinite intersection. But there are only finite n. Hence there is no empty intersection.
>> For the record, there are infinitely many natural numbers. each of them finite.
>
> But infinitely many must be spared for the contents of the endsegments.

Wrong. no need to spare, we got lota numbers in infinity

> Therefore only finitely many n can be used to index the endsegments E(n).

Wrong. 100% wrong, endsegments are infinite sets, and infinity is required to index

>Note that in order omega two consecutive infinite sets of card aleph_0 are not possible.

Wrong. aleph_0 = k * aleph_0

> Therefore aleph_0 endsegments with aleph_0 elements each can only be assumed by peopel who cannot count 1 + 1.

Wrong. there are aleph_0 endsegments each endsegment has aleph_0 elements, deal with it.

>
> Regards, WM
>

take some time and educate yourself, it is embarrassing showing for you not to know basics of infinity, and these posts stay around forever.

On 9/12/2021 11:58 AM, Transfinity wrote:
> Greg Cunt schrieb am Sonntag, 12. September 2021 um 03:40:54 UTC+2:
>
>> But the intersection of (the set of) ALL end segments -i.e. INFINITELY MANY- is EMPTY.
>
> There is *one* sequence of natural numbers used to index the endsegments and to fill them. As long as aleph_0 numbers are required to fill the endsegments only finitely many can be used to index them.

Wrong.

>
> Here is an example for beginners:
>
> E(1), E(2), E(3), E(4), E(5) = (5, 6, 7, 8, 9, ...)

wrong, that is unclear and misleading, instead show;

endsegment E(5) is defined as E(5) = (5, 6, 7, 8, 9, ...)

you cannot even write an equation !!!

>
> and here for more advanced readers
>
> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)

see above.

>
> aleph_0 natnumbers before the equals sign and after it is not possible.

unclear + misleading

>
>> A sane person might "see" (comprehend) it.
>
> Regards, WM
>

4 wrongs in one post