Jim Burns schrieb am Sonntag, 12. September 2021 um 21:40:00 UTC+2:
> On 9/12/2021 12:28 PM, 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.
> No end segment is the last end segment.
> No number is the last number.

But if there are all infinitely many endsegments in bijection with |N, then no numbers are left for their contents.
>
> For each 0,...,k, there is k+1 not in it.

But if the set of endsegments E(n) is in bijection with |N, then nothig is left for their contents.
>
> Also, and without contradicting the previous:
> Each end segment is finitely-preceded.
> Each number is finitely-preceded.

But all natnumbers are not succeeded by further natnumbers.

Regards, WM

WM presented the following explanation :
> Jim Burns schrieb am Sonntag, 12. September 2021 um 21:40:00 UTC+2:
>> On 9/12/2021 12:28 PM, 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.
>> No end segment is the last end segment.
>> No number is the last number.
>
> But if there are all infinitely many endsegments in bijection with |N, then
> no numbers are left for their contents.

Fantasy.

William schrieb am Montag, 13. September 2021 um 14:40:36 UTC+2:
> On Monday, September 13, 2021 at 9:32:54 AM UTC-3, WM wrote:
>
> > But if there are infinitely many starting elements, then there are no successors. You cannot count beyond all infinitely many starting elements.
> There are infinitely many starting elements but each starting element is finite.

If all are used, then none remains.

> There is no starting element that does not have a successor, no starting element that you cannot "count beyond".

That is irrelevant. ℕ has no finite successor, let alone infinitely many. But all elements n of ℕ are used as starting elements for the set of endsegments {E(n) | n ∈ ℕ}.

Regards, WM

On 9/13/2021 10:26 AM, WM wrote:
> FromTheRafters schrieb am Montag, 13. September 2021 um 03:54:55 UTC+2:
>> WM submitted this idea :
>>> FromTheRafters schrieb am Sonntag, 12. September 2021 um 20:36:13 UTC+2:
>>>> 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.
>>>
>>> Your claim is wrong if there are infinitely many endsegments:
>> Your usual horseshit.
>>> 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.
>> The LHS is a finite collection of infinite objects and the RHS is an
>> infinite collection of finite objects.
>
> The RHS is E(n). The natural numbers are used in E(1), E(2), and so on. Only what remains after the "and so on" can be the contents of the next endsegment. But if there is no last endsegment, then nothing remains for the contents.
>
> Regards, WM
>

Facts;

There is no last endsegment.

The intersection of all endsegments is empty.

WM, you do not think in terms of math, or equations. You typically relate things to counting rocks.

On 9/13/2021 10:29 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 12. September 2021 um 21:40:00 UTC+2:
>> On 9/12/2021 12:28 PM, 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.
>> No end segment is the last end segment.
>> No number is the last number.
>
> But if there are all infinitely many endsegments in bijection with |N, then no numbers are left for their contents.

then you did it wrong. Fail.

>>
>> For each 0,...,k, there is k+1 not in it.
>
> But if the set of endsegments E(n) is in bijection with |N, then nothig is left for their contents.

Wrong.

>>
>> Also, and without contradicting the previous:
>> Each end segment is finitely-preceded.
>> Each number is finitely-preceded.
>
> But all natnumbers are not succeeded by further natnumbers.

wrong. Please list such natural numbers.

>
> Regards, WM
>

3 Wrongs in one post, by Always Wrong WM (AWWM).

FromTheRafters schrieb am Montag, 13. September 2021 um 17:36:22 UTC+2:
> WM presented the following explanation :

> > But if there are all infinitely many endsegments in bijection with |N, then
> > no numbers are left for their contents.
> Fantasy.

The complete set of all endsegments is fantasy?
Are tzere other complete infinite sets? Or are they all fantasy?

Regards, WM

On 9/13/2021 2:58 PM, WM wrote:
> FromTheRafters schrieb am Montag, 13. September 2021 um 17:36:22 UTC+2:
>> WM presented the following explanation :
>
>>> But if there are all infinitely many endsegments in bijection with |N, then
>>> no numbers are left for their contents.
>> Fantasy.
>
> The complete set of all endsegments is fantasy?

dude, this is very simple stuff.

Here is your set of all endsegments

E(n) where n is a natural number

> Are tzere other complete infinite sets? Or are they all fantasy?

If you knew how to read math, this would not be a concern at all.

your logic is fantasy when you say you run out of numbers with an infinite set

>
> Regards, WM
>

We cannot help you with your mis-understanding of infinity,

but be sure => you are not fooling anybody <=

WM pretended :
> FromTheRafters schrieb am Montag, 13. September 2021 um 17:36:22 UTC+2:
>> WM presented the following explanation :
>
>>> But if there are all infinitely many endsegments in bijection with |N, then
>>> no numbers are left for their contents.
>> Fantasy.
>
> The complete set of all endsegments is fantasy?

Yes, because you use "complete" as a synonym for "has a last element".

> Are tzere other complete infinite sets? Or are they all fantasy?

The reals are complete, but not so in your unique lexicon.

On Monday, September 13, 2021 at 10:52:46 PM UTC+2, Serg io wrote:

> dude, this is very simple stuff.
>
> Here is your set of all endsegments[:]

{E(n) : n e IN} ,
with

> E(n)

= {m e IN : m >= n} ,

> where n is a natural number[.]

Indeed!

On 9/13/2021 11:29 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 12. September 2021 um 21:40:00 UTC+2:
>> On 9/12/2021 12:28 PM, 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.
>>
>> No end segment is the last end segment.
>> No number is the last number.
>
> But if there are all infinitely many endsegments in bijection
> with |N, then no numbers are left for their contents.

index 2 contents 2
2 <---> { 2,3,4,5,6,7,... }

>> For each 0,...,k, there is k+1 not in it.
>
> But if the set of endsegments E(n) is in bijection with |N,
> then nothig is left for their contents.
>
>> Also, and without contradicting the previous:
>> Each end segment is finitely-preceded.
>> Each number is finitely-preceded.
>
> But all natnumbers are not succeeded by further natnumbers.

A natural number which cannot be counted to,
even in principle, is not a natural number.

For each {0,...,k}, there is k+1 not in it.

----
By {0,...,k}, I mean
a _crowd_ {0,...,k}

of things to which the successor '+1' can be applied,

which has a transitive and connected order '<',

begins with 0, ends with k, and

in which, for each _split_ B,A of {0,...,k},
there are _adjacent_ i,j such that
i in B, j in A, j = i+1.

By "k can be counted to, in principle", I mean
{0,...,k} exists as described.

----
Lemma 1
For each k with {0,...,k} as described,
for each j in {1,...,k}, j-1 is in {0,...,k-1}

(Proof: Each split has a crossing-pair.)

----
By lemma 1,
if j = k+1 in {1,...,k}, then j-1 = k in {0,...,k-1}.

That is equivalent to
if k not-in {0,...,k-1}, then k+1 not-in {1,...,k}

Since, k+1 not-in {0},
(inductive step)
If k not-in {0,...,k-1}, then k+1 not-in {0,...,k}

Also (base step)
0+1 not-in {0}

By induction,
for each k with {0,...,k} as described,
k+1 not-in {0,...,k}

> But all natnumbers are not succeeded by further natnumbers.

A natural number which cannot be counted to,
even in principle, is not a natural number.

One consequence of being countable-to is that
a natural number is succeeded by further natural numbers.
Without exception.

On 9/13/2021 5:08 PM, FromTheRafters wrote:
> WM pretended :

>> Are tzere other complete infinite sets? Or are they all fantasy?
>
> The reals are complete, but not so in your unique lexicon.

In his "dyslexicon"?

Jim Burns brought next idea :
> On 9/13/2021 5:08 PM, FromTheRafters wrote:
>> WM pretended :
>
>>> Are tzere other complete infinite sets? Or are they all fantasy?
>>
>> The reals are complete, but not so in your unique lexicon.
>
> In his "dyslexicon"?

Indeed! :)

måndag 13 september 2021 kl. 17:29:28 UTC+2 skrev WM:
> Jim Burns schrieb am Sonntag, 12. September 2021 um 21:40:00 UTC+2:
> > On 9/12/2021 12:28 PM, 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.
> > No end segment is the last end segment.
> > No number is the last number.
>
> But if there are all infinitely many endsegments in bijection with |N, then no numbers are left for their contents.
> >
> > For each 0,...,k, there is k+1 not in it.
>
> But if the set of endsegments E(n) is in bijection with |N, then nothig is left for their contents.
> >
> > Also, and without contradicting the previous:
> > Each end segment is finitely-preceded.
> > Each number is finitely-preceded.
>
> But all natnumbers are not succeeded by further natnumbers.
>
> Regards, WM
all natural numbers have successors. No exceptions.

On Tuesday, 14 September 2021 at 02:01:11 UTC-3, zelos...@gmail.com wrote:
> måndag 13 september 2021 kl. 17:29:28 UTC+2 skrev WM:
[..]
> > But all natnumbers are not succeeded by further natnumbers.
> all natural numbers have successors. No exceptions.

Note his devious and deceitful use of language: "[A]ll natnumbers" can very well refer to *the set* of all natural numbers, which -- quite correctly -- can be said not to be "succeeded by further natnumbers". This allows him to wiggle out of any attempt at nailing him down to a mistake later. He constantly blurs the distinction between "every natural number" as individuals and "all natural numbers" as a set. (Not that the loose usage of colloquial phrases when talking mathematics among mathematicians who understand the correct meaning doesn't do its part to help him to hide behind.)

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.

Regards, WM

Greg Cunt schrieb am Montag, 13. September 2021 um 16:38:05 UTC+2:
> On Monday, September 13, 2021 at 4:04:40 PM UTC+2, Serg io wrote:
>
> > what else can a natural number n serve ?
> See? You are calling it (n) a natural number, here. :-)

If we call n a natural number, then we mean the natural number inserted at n.
"Let n be a natural number" does not put n in the sequence of ordinals but means that any natural number can be chosen to fill the place occupied by n.

Regards, WM

FromTheRafters schrieb am Montag, 13. September 2021 um 23:09:44 UTC+2:
> WM pretended :
> > FromTheRafters schrieb am Montag, 13. September 2021 um 17:36:22 UTC+2:
> >> WM presented the following explanation :
> >
> >>> But if there are all infinitely many endsegments in bijection with |N, then
> >>> no numbers are left for their contents.
> >> Fantasy.
> >
> > The complete set of all endsegments is fantasy?
> Yes, because you use "complete" as a synonym for "has a last element".

I don't do so here. I use it in Cantor's sense: There is no endsegment remaining when we consider the set being in bijection with |N. And there is no natural number remaining. Note the technical term bijection. Wallflowers are excluded.

Regards, WM

Jim Burns schrieb am Dienstag, 14. September 2021 um 00:14:14 UTC+2:
> On 9/13/2021 11:29 AM, WM wrote:

> > But if there are all infinitely many endsegments in bijection
> > with |N, then no numbers are left for their contents.
> index 2 contents 2
> 2 <---> { 2,3,4,5,6,7,... }

Yes, it would have been better from the psychological point of view to chose E(n) = (n+1, n+2, ...), but it does not matter much.

> > But if the set of endsegments E(n) is in bijection with |N,
> > then nothig is left for their contents.
> >
> >> Also, and without contradicting the previous:
> >> Each end segment is finitely-preceded.
> >> Each number is finitely-preceded.
> >
> > But all natnumbers are not succeeded by further natnumbers.
> A natural number which cannot be counted to,
> even in principle, is not a natural number.

Are there natural numbers beyond the set |N which is completely used in the bijection with the endsegments?

> A natural number which cannot be counted to,
> even in principle, is not a natural number.

But all natural numbers which can be counted to are in the bijection with all endsegments. The contents of all endsegments must contain numbers which cannot be counted to.
>
> One consequence of being countable-to is that
> a natural number is succeeded by further natural numbers.
> Without exception.

All are indexing endsegments. Without exception.
Can all endsegments contain only the natnumbers used for indexing them?

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 14. September 2021 um 07:01:11 UTC+2:
> måndag 13 september 2021 kl. 17:29:28 UTC+2 skrev WM:

> > But all natnumbers are not succeeded by further natnumbers.

> all natural numbers have successors. No exceptions.

All successors are used to index endsegments. No natural number is remaining beyond the set |N. But infinitely many numbers are required as the contents of the endsegments.

Regards, WM

On 9/14/2021 3:19 PM, WM wrote:
> Jim Burns schrieb am Dienstag, 14. September 2021 um 00:14:14 UTC+2:
>> On 9/13/2021 11:29 AM, WM wrote:
>
>>> But if there are all infinitely many endsegments in bijection
>>> with |N, then no numbers are left for their contents.
>> index 2 contents 2
>> 2 <---> { 2,3,4,5,6,7,... }
>
> Yes, it would have been better from the psychological point of view to chose E(n) = (n+1, n+2, ...), but it does not matter much.

there is no "psychological" in math.

>
>>> But if the set of endsegments E(n) is in bijection with |N,
>>> then nothig is left for their contents.
>>>
>>>> Also, and without contradicting the previous:
>>>> Each end segment is finitely-preceded.
>>>> Each number is finitely-preceded.
>>>
>>> But all natnumbers are not succeeded by further natnumbers.
>> A natural number which cannot be counted to,
>> even in principle, is not a natural number.
>
> Are there natural numbers beyond the set |N which is completely used in the bijection with the endsegments?

what is your answer, professor ?

>
>> A natural number which cannot be counted to,
>> even in principle, is not a natural number.
>
> But all natural numbers which can be counted to are in the bijection with all endsegments. The contents of all endsegments must contain numbers which cannot be counted to.

that is totally warped. Sorry to see you have so much difficulty with this, but you should have taken a math class, Algebra.

>>
>> One consequence of being countable-to is that
>> a natural number is succeeded by further natural numbers.
>> Without exception.
>
> All are indexing endsegments. Without exception.
> Can all endsegments contain only the natnumbers used for indexing them?

endsegments are fixed sets of numbers

all endsegments already have their numbers.

you dont put numbers into, or out of, an endsegment.

>
> Regards, WM
>

On 9/14/2021 3:00 PM, WM wrote:
> Greg Cunt schrieb am Montag, 13. September 2021 um 16:38:05 UTC+2:
>> On Monday, September 13, 2021 at 4:04:40 PM UTC+2, Serg io wrote:
>>
>>> what else can a natural number n serve ?
>> See? You are calling it (n) a natural number, here. :-)
>
> If we call n a natural number, then we mean the natural number inserted at n.
> "Let n be a natural number" does not put n in the sequence of ordinals but means that any natural number can be chosen to fill the place occupied by n.
>
> Regards, WM
>

what "place occupied" by n ??

Leave n alone, and let n grow up the way he wants too!

On 9/14/2021 3:23 PM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 14. September 2021 um 07:01:11 UTC+2:
>> måndag 13 september 2021 kl. 17:29:28 UTC+2 skrev WM:
>
>>> But all natnumbers are not succeeded by further natnumbers.
>
>> all natural numbers have successors. No exceptions.
>
> All successors are used to index endsegments. No natural number is remaining beyond the set |N. But infinitely many numbers are required as the contents of the endsegments.
>
> Regards, WM
>

in one mapping, yes. but that does not apply to other mappings, that is foolish thinking.

On Tuesday, September 14, 2021 at 10:00:23 PM UTC+2, WM wrote:

> If we call /n/ a natural number, then we mean

that n is a natural number, naturally.

> "Let n be a natural number" does not put n in the sequence of ordinals

*lol* That's EXACTLY what it does, Mucke.

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.

Hope this helps.

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.
>
> Regards, WM

empty assertion, they all have infinite cardinality.

On Tuesday, 14 September 2021 at 17:19:44 UTC-3, WM wrote:
> Jim Burns schrieb am Dienstag, 14. September 2021 um 00:14:14 UTC+2:
> > On 9/13/2021 11:29 AM, WM wrote:
>
> > > But if there are all infinitely many endsegments in bijection
> > > with |N, then no numbers are left for their contents.
> > index 2 contents 2
> > 2 <---> { 2,3,4,5,6,7,... }
> Yes, it would have been better from the psychological point of view to chose E(n) = (n+1, n+2, ...), but it does not matter much.
Interesting that you should say that. Yes, it would have been better, but not just from a psychological standpoint, but to illustrate the fatal flaw in your thinking: With the definition E'(n) = {n+1, n+2, n+3, ...} you have immediately that FISON(n) and E'(n) have an empty intersection, and their union is |N. In other words, for every n you can partition |N into two sets: FISON(n) and |N \ FISON(n) = E'(n). And then you have a process with interesting limiting properties:

What is lim{n -> oo} FISON(n)? The beauty, of course, is that lim{n -> oo} [ |N \ FISON(n) ] *IS* |N \ [ lim{n -> oo} FISON(n) ].

Note: No dark numbers. Put that in your pipe and smoke it!