tisdag 21 september 2021 kl. 13:13:05 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:18:09 UTC+2:
> > torsdag 16 september 2021 kl. 14:54:57 UTC+2 skrev 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?
> > >
> > Non-sequitor so no, I do not accept it.
> What? All endsegments exhaust all indices? Or all endsegments have aleph_0 elements?
>
> Regards, WM
You are still going on 1 sequence fails, so another must. That is a pure non-sequitor.

On 9/21/2021 6:57 AM, WM wrote:
> Greg Cunt schb
> am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
>> On Sunday, September 19, 2021 at 4:40:47 PM UTC+2,
>> WM wrote:
>>> Greg Cunt schrieb
>>> am Samstag, 18. September 2021 um 20:20:00 UTC+2:

>>>> Let n e IN.
>>>
>>> n does not belong to that set.

n is in N if and only if
n can be counted to in principle..

>> To the set of all natural numbers, IN?
>
> No.
> Every natural number has a unique prime decomposition.
> n has none.

We know some things about n. We don't know other things.
We know n has a unique prime factorization.
We don't know what its UPF is.

We know that,
if P(n),
then a _first_ n1 exists,
P(n1) and, for each k < n1, ~P(k)

We know that,
if non-trivial n doesn't have a UPF,
then a _first_ non-trivial n1 exists such that
n1 doesn't have a UPF, and
for each non-trivial k < n1, k has a UPF.

We know that non-trivial n1 has a non-trivial factor, n1.
So, we know n1 has a _first_ non-trivial factor, j.

We know that j does not have any non-trivial factor i < j
If j did, i would also be a non-trivial factor of n1,
and j would not be first.
So, we know j is a prime factor of n1.

For some k, j*k = n1.
k < n1, so k has a UPF
j has a UPF, which is j.
n1 = j*k has a UPF.
But n1 is the first non-trivial without a UPF.
Contradiction.

Therefore,
we know that non-trivial n has a UPF.

We don't know what the UPF is that n has,
but we don't need to know what the UPF is
in order to know n has it. See above.

In some ways, knowing n has a UPF _without_ knowing
what n is is better.

What is the UPF of floor(pi*(10^(10^(10^(10^(10))))))?
I don't know, and I'm not sure that it's physically
possible to calculate.

Does floor(pi*(10^(10^(10^(10^(10)))))) have a UPF?
Yes. That's trivial to show. I just did, in a paragraph.

----
Also, having a UPF is not part of the _definition_ of an
element of N.

The _definition_ is that n can be counted to in principle.
More facts follow from the definition.
N is well-ordered.
(i*j)*k = i*(j*k)
for non-trivial i,j,k, i*j = k -> i < k and j < k
And from those facts, more facts,
like that n has a UPF.

On Tuesday, September 21, 2021 at 6:50:48 PM UTC+2, Serg io wrote:
> On 9/21/2021 10:13 AM, Greg Cunt wrote:
> > On Tuesday, September 21, 2021 at 12:57:12 PM UTC+2, WM wrote:
> >> Greg Cunt schrieb am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
> >>> On Sunday, September 19, 2021 at 4:40:47 PM UTC+2, WM wrote:
> >>>> Greg Cunt schrieb am Samstag, 18. September 2021 um 20:20:00 UTC+2:
> >
> > Let n e IN.
> >
> >> Every natural number has a unique prime decomposition. n has none.
> >
> > Huh?! I mean... errr?
> >
> > Since every natural number has a unique prime decomposition and n e IN (i.e. n is a natural number), n has a unique prime decomposition.
> >
> > What's wrong with you, Mückenheim?
> >
> >>> So we have
> >>>
> >>> n e IN
> >>>
> >> that is only an abbreviation for: A natural number is element of IN.
> >
> > No, it says/expresses: "n is a natural number" (well, actually it says/expresses "n is an element in IN").
> >
> >>>>> So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?
> >>>>>
> >>> No answer?
> >>>
> >> Of course, since it is fact.
> >
> > What is a fact?
> >
> > That n isn't a natural number even though n e IN?
> >
> > Are you SERIOUS?
> >
> >>>> They are placeholders.
> >>>>
> >>> So placeholders are elements in IN?
> >>>
> >> No. They are place holders for elements of IN.
> >
> > No, n and n+1 aren't "place holders for elements of IN", BUT _elements in IN_.
> >
> > That's what the expressions
> >
> > n e IN
> >
> > and
> >
> > n+1 e IN
> >
> > "say" or "express", namely:
> >
> > n is an element in IN
> >
> > and
> >
> > n+1 is an element in IN .
> >
> >>> It is well-known that matheologians claim counterfactual nonsense
> >
> > In this case you clearly qualify for a matheologian.
> >
> >> They are easily recognizable as fools.
> >
> > Indeed! :-)
> >
> > Again: 'Let n e IN.'
> >
> >>> might be the beginning of a proof.
> >>>
> >> Yes, but the proof shall be valid for all natural numbers like 1, 2, 3, ..., not only for n.
> >
> > Holy shit! :-(
> >
> whoa, that last one is a keeper!
>
> Let n e IN
> n = 1 + 2
> therefore n = 3

Right. And since n was an arbitrary natural number, we get that for all x e IN: x = 3.

Well done, Serg io! (Only slightly worse than Mückenheim.)

On 9/21/2021 3:08 PM, Greg Cunt wrote:
> On Tuesday, September 21, 2021 at 6:50:48 PM UTC+2, Serg io wrote:
>> On 9/21/2021 10:13 AM, Greg Cunt wrote:
>>> On Tuesday, September 21, 2021 at 12:57:12 PM UTC+2, WM wrote:
>>>> Greg Cunt schrieb am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
>>>>> On Sunday, September 19, 2021 at 4:40:47 PM UTC+2, WM wrote:
>>>>>> Greg Cunt schrieb am Samstag, 18. September 2021 um 20:20:00 UTC+2:
>>>
>>> Let n e IN.
>>>
>>>> Every natural number has a unique prime decomposition. n has none.
>>>
>>> Huh?! I mean... errr?
>>>
>>> Since every natural number has a unique prime decomposition and n e IN (i.e. n is a natural number), n has a unique prime decomposition.
>>>
>>> What's wrong with you, Mückenheim?
>>>
>>>>> So we have
>>>>>
>>>>> n e IN
>>>>>
>>>> that is only an abbreviation for: A natural number is element of IN.
>>>
>>> No, it says/expresses: "n is a natural number" (well, actually it says/expresses "n is an element in IN").
>>>
>>>>>>> So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?
>>>>>>>
>>>>> No answer?
>>>>>
>>>> Of course, since it is fact.
>>>
>>> What is a fact?
>>>
>>> That n isn't a natural number even though n e IN?
>>>
>>> Are you SERIOUS?
>>>
>>>>>> They are placeholders.
>>>>>>
>>>>> So placeholders are elements in IN?
>>>>>
>>>> No. They are place holders for elements of IN.
>>>
>>> No, n and n+1 aren't "place holders for elements of IN", BUT _elements in IN_.
>>>
>>> That's what the expressions
>>>
>>> n e IN
>>>
>>> and
>>>
>>> n+1 e IN
>>>
>>> "say" or "express", namely:
>>>
>>> n is an element in IN
>>>
>>> and
>>>
>>> n+1 is an element in IN .
>>>
>>>>> It is well-known that matheologians claim counterfactual nonsense
>>>
>>> In this case you clearly qualify for a matheologian.
>>>
>>>> They are easily recognizable as fools.
>>>
>>> Indeed! :-)
>>>
>>> Again: 'Let n e IN.'
>>>
>>>>> might be the beginning of a proof.
>>>>>
>>>> Yes, but the proof shall be valid for all natural numbers like 1, 2, 3, ..., not only for n.
>>>
>>> Holy shit! :-(
>>>
>> whoa, that last one is a keeper!
>>
>> Let n e IN
>> n = 1 + 2
>> therefore n = 3
>
> Right. And since n was an arbitrary natural number, we get that for all x e IN: x = 3.
>
> Well done, Serg io! (Only slightly worse than Mückenheim.)
>

Thank you! Quite an improvement! and now only slightly (within ϵ ?) within range of the Professor !

zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 14:02:41 UTC+2:
> tisdag 21 september 2021 kl. 13:03:30 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:11:54 UTC+2:
> > > fredag 17 september 2021 kl. 22:10:38 UTC+2 skrev WM:
> >
> > > > 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.
> > > >
> > > Nope, the intersection of endsegments is empty. only FINITE intersections of endsegments has infintie cardinality.
> > >
> > Every finite intersection. All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.

> nope, infinite intersection yields empty,

But all endsegments contain numbers? What numbers?

Regards, WM

On Wednesday, 22 September 2021 at 12:36:16 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 14:02:41 UTC+2:
> > tisdag 21 september 2021 kl. 13:03:30 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:11:54 UTC+2:
> > > > fredag 17 september 2021 kl. 22:10:38 UTC+2 skrev WM:
> > >
> > > > > 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.
> > > > >
> > > > Nope, the intersection of endsegments is empty. only FINITE intersections of endsegments has infintie cardinality.
> > > >
> > > Every finite intersection. All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.

Oh, that's really subtle: "all finite sets"! What the fuck is that, other than the *INFINITE* union of infinitely many finite sets?
And then you should take the intersection over this infinite set. Of course you are no longer able to do so, due to the state of your progressive dementia.

> > nope, infinite intersection yields empty,
> But all endsegments contain numbers? What numbers?

Subtle, again, or totally incomprehended. Every end segment E(n) contains the aleph_0 natural numbers n, n+1, n+2, ... . No natural number n is contained in the end segment E(n+1) or indeed any of the aleph_0 end segments E(m) with m > n. That's why infinite intersections of end segments are empty. Clearly you have never been able to grasp that. I conclude that you know *SHIT* about anything mathematical.

On Wednesday, September 22, 2021 at 5:36:16 PM UTC+2, WM wrote:

> But all endsegments contain numbers? What numbers?

If E is an endsegments it contains the (infinitly many) natural numbers that are >= min(E) (and nothing else).

Too complicated for you?

Here's an example:

The endsegment {3, 4, 5, ...} contains the (infinitly many) natural numbers 3, 4, 5, ... (and nothing else).

We seen to have disconnected by the breakup of the counterexample topic. See new topic on confusion of Set and Element of Set.

--
William Hughes

Gus Gassmann schrieb am Mittwoch, 22. September 2021 um 18:33:50 UTC+2:
> On Wednesday, 22 September 2021 at 12:36:16 UTC-3, WM wrote:

> > But all endsegments contain numbers? What numbers?
> Every end segment E(n) contains the aleph_0 natural numbers n, n+1, n+2, ... .

The question is for all, not for every. If there are all endsegments, then they are indexed by *all* natural numbers.

> No natural number n is contained in the end segment E(n+1) or indeed any of the aleph_0 end segments E(m) with m > n.

But infinitely many natural numbers are contained in all endsegments which exhaust |N already with their indices.

> That's why infinite intersections of end segments are empty.

Infinitely many infinite endsegments means infinitely many non-empty-intersection endsegments.

Regards, WM

Greg Cunt schrieb am Mittwoch, 22. September 2021 um 18:45:00 UTC+2:
> On Wednesday, September 22, 2021 at 5:36:16 PM UTC+2, WM wrote:
>
> > But all endsegments contain numbers? What numbers?
> If E is an endsegments it contains

something. The question is whether all natural numbers are used as indices such that none remains as contents.

> Here's an example:
>
> The endsegment {3, 4, 5, ...} contains the (infinitly many) natural numbers 3, 4, 5, ... (and nothing else).

Here is another example: The set of endsegments E(n) exhausts all natural numbers n. What remains as the contents?

Regards, WM

On Wednesday, 22 September 2021 at 15:32:40 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 22. September 2021 um 18:33:50 UTC+2:
> > On Wednesday, 22 September 2021 at 12:36:16 UTC-3, WM wrote:
>
> > > But all endsegments contain numbers? What numbers?
> > Every end segment E(n) contains the aleph_0 natural numbers n, n+1, n+2, ... .
> The question is for all, not for every.

Use proper quantifiers if you are still able. As written, your objection is baseless gobbledygook.

Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:40:14 UTC+2:
> On Tuesday, September 21, 2021 at 3:26:47 PM UTC+2, WM wrote:
>
> > 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not in both positions an infinite set can exist.
> Why don't you express this using standard notation?
>
> Say, An e IN: card({1, 2, 3, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 ,
>
> or: An e IN: {1, 2, 3, ..., n} is finite & {n, n+1, n+2, ...} is infinite .

Because this would show for the set of all endsegments card of indices aleph_0 and nothing remaining for the contents. It would be less instructive.

> > Your claim is n infinite set of indices and an infinite set following every index. That is wrong.

> Yeah, there are infinitely many ("finite") numbers in IN and each one is followed by infinitely many natural numbers.

No, the numbers followede by infinitely many can only make up a finite set. An infinite set by an infinite set is impossible.
>
> If the natural numbers in our system (i.e. our set theory) are defined due to von Neumann (or due to Zermelo), we actually have:

a contradiction.
>
> There are infinitely many numbers in IN [all of which are finite sets] and each one is followed by infinitely many natural numbers.

Impossible.

Regards, WM

Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:29:32 UTC+2:

>All endsegments have the same cardinality and the set of all endsegments exists (at least in the context of set theory, say ZFC).

That is the reason why ZFC is bound for the rubbish heap or garbage dump of history.

Regards, WM

On Wednesday, September 22, 2021 at 8:34:46 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Mittwoch, 22. September 2021 um 18:45:00 UTC+2:
> >
> > The question is whether all natural numbers are used as indices

The answer is yes, since the set of natural numbers, IN, is adopted as index set for the SEQUENCE of endsegments: (E_n)_(n e IN). Now THAT was easy!

> such that none remains as contents.

Why do you thing that using the natural numbers as indices PROHIBITS us from using them for other purposes too? What's the matter with you Mückenheim?

Look I'll use the natural numbers 1, 2 as indices and also as terms here:

Let k_1 = 1 and k_2 = 2. Then k_1 + k_2 = 3.

A miracle?

Again, dumbo, if E is an endsement, then min(E) is its index and it contains all natural numbers >= min(E).

This holds for each and every ensegments, i.e. for ALL endsegments.

> > Here's an example:
> >
> > The endsegment {3, 4, 5, ...} contains the (infinitly many) natural numbers 3, 4, 5, ... (and nothing else).

Another example:

Let n e IN. Then the endsegment {n, n+1, n+2, ...} contains the (infinitly many) natural numbers n, n+1, n+2, ... (and nothing else).

Serg io schrieb am Dienstag, 21. September 2021 um 15:34:58 UTC+2:
> On 9/21/2021 5:57 AM, WM wrote:

> 6 - All endsegments are infinite but their intersection is empty.

All endsegments which have an infinite intersection with all endsegments have an empty intersection with each other.

Regards, WM

Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:13:38 UTC+2:
> On Tuesday, September 21, 2021 at 12:57:12 PM UTC+2, WM wrote:

> Let n e IN.
> > Every natural number has a unique prime decomposition. n has none.

> Since every natural number has a unique prime decomposition and n e IN (i.e. n is a natural number), n has a unique prime decomposition.

Specify it. That can be done for every natural number less than 1 billion. Or is n larger?

Regards, WM

Jim Burns schrieb am Dienstag, 21. September 2021 um 19:42:24 UTC+2:
> On 9/21/2021 6:57 AM, WM wrote:

> > Every natural number has a unique prime decomposition.
> > n has none.
> We know some things about n. We don't know other things.

Every natural number less than 1 billion has a known prime decomposition. Is n larger?

> We know n has a unique prime factorization.
> We don't know what its UPF is.

For a natural number you would know what it is. You don't. Therefore n is not a natural number.

Regards, WM

On Wednesday, 22 September 2021 at 15:54:11 UTC-3, WM wrote:
> Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:29:32 UTC+2:
>
> >All endsegments have the same cardinality and the set of all endsegments exists (at least in the context of set theory, say ZFC).
> That is the reason why ZFC is bound for the rubbish heap or garbage dump of history.

Much as it grieves me to say that: The prime candidate for the garbage dump of history is one notorious professor of mathematics at that paragon of education, the Technical University of Augsburg. (His initials are WM...)

On 9/22/2021 3:14 PM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 21. September 2021 um 19:42:24 UTC+2:
>> On 9/21/2021 6:57 AM, WM wrote:

>>> Every natural number has a unique prime decomposition.
>>> n has none.
>>
>> We know some things about n. We don't know other things.
>
> Every natural number less than 1 billion has a
> known prime decomposition. Is n larger?

It's possible that n is larger.
Whether larger or smaller, it has a UPF.

>> We know n has a unique prime factorization.
>> We don't know what its UPF is.
>
> For a natural number you would know what it is.

I know other things about n, but not that.

> You don't.
> Therefore n is not a natural number.

n can be counted to in principle.
Therefore, n is a natural number.

If you change the definition of "natural number" to include
"must have its UPF known", then you are talking about
something else, something other than natural numbers.

No doubt, for your next trick, you will "disprove" the
Pythagorean theorem by changing "triangle" to include
four-cornered figures.

On Wednesday, September 22, 2021 at 8:51:26 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:40:14 UTC+2:

> > Why don't you express this using standard notation?

> Because <bla bla bla>

The true reason for this is that you are a crank and "cranks frequently [...] do not use standard notation and terminology [...]" (Wikipedia)

> It would be less instructive.

Right: "That is, cranks [...] often assert that these innovations obscure rather than clarify the situation." (Wikipedia)

Now...

> > Yeah, there are infinitely many ("finite") numbers in IN and each one is followed by infinitely many natural numbers.
> >
> No

Yes.

Hint (if IN is defined due to von Neumann or Zermelo):

1. card(IN) = aleph_0 (An e IN: card(n) < aleph_0)

2. An e IN: card({m e IN : m > n}) = aleph_0

> the numbers followed by infinitely many [numbers] can only make up a finite set.

Huh?!

Let S := {n e IN : card({m e IN : m > n}) = aleph_0}. Then S is infinite. (Actually, S = IN.)

> > If the natural numbers in our system (i.e. our set theory) are defined due to von Neumann (or due to Zermelo), we actually have:
> >
> a contradiction.

Nonsense. You are deluded, man. (Or just a pathological liar.)
> > There are infinitely many numbers in IN [all of which are finite sets] and each one is followed by infinitely many natural numbers.
> >
> Impossible.

No, (provably) true.

Hint (if IN is defined due to von Neumann or Zermelo):

1. card(IN) = aleph_0 [An e IN: card(n) < aleph_0]

2. An e IN: card({m e IN : m > n}) = aleph_0

On Wednesday, September 22, 2021 at 9:14:35 PM UTC+2, WM wrote:
> Jim Burns schrieb am Dienstag, 21. September 2021 um 19:42:24 UTC+2:
> >
> > We know n has a unique prime factorization.
> > We don't know what its UPF is.
> >
> For a natural number you would know what it is. You don't. Therefore n is not a natural number.

Look, you silly asshole:

There are some coins in your pocket. You put your hand in your pocket and just grab one (not pulling it out).

| For a coin you would know what it is. You don't. Therefore the thing in your hand is not a coin.

Holy shit!

On Wednesday, September 22, 2021 at 9:10:39 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:13:38 UTC+2:
> > On Tuesday, September 21, 2021 at 12:57:12 PM UTC+2, WM wrote:
> >
> > Let n e IN.
> >
> > > Every natural number has a unique prime decomposition. n has none.
> >
> > Since every natural number has a unique prime decomposition and n e IN (i.e. n is a natural number), n has a unique prime decomposition.

DID YOU GET THAT, ASSHOLE?!

Since n is a natural number it has a unique prime decomposition.

> Specify it. That can be done for every natural number less than 1 billion.

So what?

Can it be done for every natural number less than 10^10^10^10^10^10^10^10^80 too?

So this is certainly not a sensible criterium.

Hint: There are many "precisely specified" natural numbers, the prime decomposition of which we do not know (yet). :-)

Please specify the unique prime decomposition for:

2519590847565789349402718324004839857142928212620403202777713783604366202070
7595556264018525880784406918290641249515082189298559149176184502808489120072
8449926873928072877767359714183472702618963750149718246911650776133798590957
0009733045974880842840179742910064245869181719511874612151517265463228221686
9987549182422433637259085141865462043576798423387184774447920739934236584823
8242811981638150106748104516603773060562016196762561338441436038339044149526
3443219011465754445417842402092461651572335077870774981712577246796292638635
6373289912154831438167899885040445364023527381951378636564391212010397122822
120720357

See: https://en.wikipedia.org/wiki/RSA_numbers#RSA-2048

On Wednesday, September 22, 2021 at 9:14:35 PM UTC+2, WM wrote:
> Jim Burns schrieb am Dienstag, 21. September 2021 um 19:42:24 UTC+2:
> >
> > We know n has a unique prime factorization.
> > We don't know what its UPF is.
> >
> For a natural number you would know what it is. You don't. Therefore n is not a natural number.

Oh, really?

Please specify the unique prime decomposition for RSA-2048:

2519590847565789349402718324004839857142928212620403202777713783604366202070
7595556264018525880784406918290641249515082189298559149176184502808489120072
8449926873928072877767359714183472702618963750149718246911650776133798590957
0009733045974880842840179742910064245869181719511874612151517265463228221686
9987549182422433637259085141865462043576798423387184774447920739934236584823
8242811981638150106748104516603773060562016196762561338441436038339044149526
3443219011465754445417842402092461651572335077870774981712577246796292638635
6373289912154831438167899885040445364023527381951378636564391212010397122822
120720357

For a natural number you would know what it is. You don't. Therefore RSA-2048 (the decimal expansion of which is given above) is not a natural number.

See: https://en.wikipedia.org/wiki/RSA_numbers#RSA-2048

onsdag 22 september 2021 kl. 20:54:11 UTC+2 skrev WM:
> Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:29:32 UTC+2:
>
> >All endsegments have the same cardinality and the set of all endsegments exists (at least in the context of set theory, say ZFC).
> That is the reason why ZFC is bound for the rubbish heap or garbage dump of history.
>
> Regards, WM

Why? What about this is making it so? There is nothing odd it is saying.

onsdag 22 september 2021 kl. 20:51:26 UTC+2 skrev WM:
> Greg Cunt schrieb am Dienstag, 21. September 2021 um 17:40:14 UTC+2:
> > On Tuesday, September 21, 2021 at 3:26:47 PM UTC+2, WM wrote:
> >
> > > 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not in both positions an infinite set can exist.
> > Why don't you express this using standard notation?
> >
> > Say, An e IN: card({1, 2, 3, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 ,
> >
> > or: An e IN: {1, 2, 3, ..., n} is finite & {n, n+1, n+2, ...} is infinite .
> Because this would show for the set of all endsegments card of indices aleph_0 and nothing remaining for the contents. It would be less instructive.
> > > Your claim is n infinite set of indices and an infinite set following every index. That is wrong.
> > Yeah, there are infinitely many ("finite") numbers in IN and each one is followed by infinitely many natural numbers.
> No, the numbers followede by infinitely many can only make up a finite set. An infinite set by an infinite set is impossible.
> >
> > If the natural numbers in our system (i.e. our set theory) are defined due to von Neumann (or due to Zermelo), we actually have:
> a contradiction.
> >
> > There are infinitely many numbers in IN [all of which are finite sets] and each one is followed by infinitely many natural numbers.
> Impossible.
>
> Regards, WM

Very possible because that is the definition of natural numbers. There is always a successor to each and every one of them. Therefore each one is followed by infinitely many natural numbers