Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 18:12:24 UTC+2:
> On Friday, October 1, 2021 at 8:00:48 PM UTC+2, WM wrote:
>
> > [Each and] Every natural number is missing [in at least one endsegment] because there is at least one predecessor endsegment not containing it.
>
> Exactly.

Only an empty endsegment can cause an emty intersection.

Regards, WM

Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 18:13:59 UTC+2:
> On Friday, October 1, 2021 at 8:04:37 PM UTC+2, WM wrote:
>
> > Why are the endsegments *not* empty?
> Because they (i.e. all of them) are infinite sets BY DEFINITION (and hence not empty).

Then they (i.e. all of them) have an infinite intersection with all infinite endsegments:

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo.

Hence a mathematician cannot believe that there is a sudden drop from all endsegments having infinite intersection to not more endsegments having empty intersection.

Regards, WM

Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 18:15:57 UTC+2:

> Actually, the intersection of all endsegments (each of which is infinite) is empty. A trivial fact.

In mathematics this would be an inconsistecy, because all infinite endsegments have an infinite intersection with each other.

Regards, WM

Serg io schrieb am Samstag, 2. Oktober 2021 um 18:58:21 UTC+2:

> An Endsegment, E(k), is only shorthand for this E(k) = {k,k+1,k+2,...} a fixed (by k) infinite set of natural numbers that does not change.
>
> There are no empty endsegment.

All existing infinite endsegments have an infinite intersection with each other.

Regards, WM

On Sunday, 3 October 2021 at 09:54:57 UTC-3, WM wrote:
> Serg io schrieb am Samstag, 2. Oktober 2021 um 18:58:21 UTC+2:
>
> > An Endsegment, E(k), is only shorthand for this E(k) = {k,k+1,k+2,...} a fixed (by k) infinite set of natural numbers that does not change.
> >
> > There are no empty endsegment.
> All existing infinite endsegments have an infinite intersection with each other.

You've been very busy this morning, and very successful at demonstrating again your utter incompetence. You have no clue of how infinity works, and why it is an important concept. In particular, you have not even a scintilla of a clue as to how infinite intersections and unions differ from finite ones. Five different answers within the span of ten minutes, and not a single correct thought in any of them. That has to be a record of some kind.

fredag 1 oktober 2021 kl. 20:22:53 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 14:21:00 UTC+2:
>
> > All natural numbers have infinitely many natural numbers after,
> So they are not an actuallyinfinite set. Note that two consecutive infinite sets of card aleph_0 are impossible in |N.
>
> Regards, WM

If each natural number has infinitely many successors that means the set of natural numbers is infinite.

fredag 1 oktober 2021 kl. 19:58:10 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 07:03:30 UTC+2:
> > torsdag 30 september 2021 kl. 20:56:13 UTC+2 skrev WM:
>
> > > Just this is impossible for endsegments. Inclusion monotony!
> > >
> > Nope, that only shows that any FINITE intersection is non-empty, not that the intersection of ALL is emtpy
> Is one endsegment among all that is not the last of a finite intersection?
>
> Regards, WM

Irrelevasnt. You cannot deduce anything from the finite about the infinite.

fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 07:04:41 UTC+2:
> > torsdag 30 september 2021 kl. 20:12:13 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Dienstag, 28. September 2021 um 06:56:41 UTC+2:
> > > > måndag 27 september 2021 kl. 21:34:57 UTC+2 skrev WM:
> > >
> > > > > If you claim that n has a decimal representation and this is doubted, then you have to prove it by giving it.
> > > > >
> > > > Nope, I only have to prove all natural numbers have one and that is done already.
> > > Irrelevant since it is just doubted that n is a natural number. If a honest mathematician is asked to show n's decimal representation, then he will do so, if he can.
> > We do not need to know decimal expansions to know there is one and to do arguments about them and you do not doubt n is a natural number. If I say it is from the set of natural number it is by definition a natural number.
> It is not your competence to define what natural numbers are.
> Please mark the place of n on the following scale:
>
> ...1...10...100...1000...10000...
>
> Regards, WM
why? Again, I don't need to do that, it is somewhere

fredag 1 oktober 2021 kl. 19:51:36 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 07:06:23 UTC+2:
> > torsdag 30 september 2021 kl. 20:33:05 UTC+2 skrev WM:
> > > Greg Cunt schrieb am Donnerstag, 30. September 2021 um 20:19:56 UTC+2:
> > > > On Thursday, September 30, 2021 at 8:12:13 PM UTC+2, WM wrote:
> > > >
> > > > Let n be a natural number. [Or: Let n e IN.]
> > > > > it is just doubted that n is a natural number.
> > > > BY WHOM,
> > > Irrelevant.
> > > If n is a natural number, then its decimal representation can be given.
> > > Give it!
> > > > Hint: BY DEFINITION (see above) n is an element in the set of all natural numbers.
> > > No, you misunderstand. "n ∈ ℕ" is simply an abbreviation for the sentence: "Consider any natural number". By the way, you can also use k ∈ ℕ or m ∈ ℕ. Are they in trichotomy? k < m < n? Or m < k, n?
> > > > Does this set contain anything else except natural numbers?
> > > No, therefore it does not contain k, m, n.
> > >
> > are you insane or just mentally retarded?
> Do I claim that n is a number?
>
> Regards, WM

You claim a lot of shit that makes no sense

zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
> fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:
> > If I say it is from the set of natural number it is by definition a natural number.
> > It is not your competence to define what natural numbers are.
> > Please mark the place of n on the following scale:
> >
> > ...1...10...100...1000...10000...
> >
> why?

To prove your claim.

> Again, I don't need to do that, it is somewhere

A number is not somewhere but has a fixed place.

Regards, WM

zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:29:16 UTC+2:
> fredag 1 oktober 2021 kl. 20:22:53 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 14:21:00 UTC+2:
> >
> > > All natural numbers have infinitely many natural numbers after,
> > So they are not an actuallyinfinite set. Note that two consecutive infinite sets of card aleph_0 are impossible in |N.
> >
> If each natural number has infinitely many successors that means the set of natural numbers is infinite.

This set |N is used to index all endegments. Therefore not all endsegments can have infinite contents.

Regards, WM

Gus Gassmann schrieb am Sonntag, 3. Oktober 2021 um 15:58:13 UTC+2:
> On Sunday, 3 October 2021 at 09:54:57 UTC-3, WM wrote:

> > All existing infinite endsegments have an infinite intersection with each other.
> You've been very busy this morning, and very successful at demonstrating again your utter incompetence. You have no clue of how infinity works,

I can disprove the shit you believe.

> In particular, you have not even a scintilla of a clue as to how infinite intersections and unions differ from finite ones.

Do you know any endsegment that is not belonging to one of the following sets?
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
No. Therefore all infinite endsegments have an infinite intersection. Inclusion monotony is always true.

Regards, WM

zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:34:31 UTC+2:
> fredag 1 oktober 2021 kl. 19:58:10 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 07:03:30 UTC+2:
> > > torsdag 30 september 2021 kl. 20:56:13 UTC+2 skrev WM:
> >
> > > > Just this is impossible for endsegments. Inclusion monotony!
> > > >
> > > Nope, that only shows that any FINITE intersection is non-empty, not that the intersection of ALL is emtpy
> > Is one endsegment among all that is not the last of a finite intersection?
> >
> You cannot deduce anything from the finite about the infinite.

I can deduce that there is no infinite. I can deduce from
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
that all existing infinite endsegments have an infinite intersection.
Inclusion monotony is always true.

Regards, WM

On 10/4/2021 5:16 PM, WM wrote:
> zelos...@gmail.com schrieb
> am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
>> fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:

>>> If I say it is from the set of natural number
>>> it is by definition a natural number.
>>> It is not your competence to define what natural numbers are.

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

>>> Please mark the place of n on the following scale:
>>>
>>> ...1...10...100...1000...10000...
>>
>> why?
>
> To prove your claim.
>
>> Again, I don't need to do that, it is somewhere
>
> A number is not somewhere but has a fixed place.

Each natural number has a fixed successor.

0 has a unique successor, fixed.
Each successor has a unique successor, fixed.
Each successor has a unique predecessor, fixed.
0 does not have a predecessor, fixed.

Either
a totally-ordered {0,...,k} exists which
begins at 0, ends at k, and,
for each split, there is a crossing-pair j,j+1,
or
{0,...,k} does not exist.

Whether {0,...,k} exists is fixed.

Define N such that,
for each k, k is in N iff
{0,...,k} exists which is totally-ordered,
begins at 0, ends at k, and,
for each split, there is a crossing-pair j,j+1.

Whether k is in N is fixed.

----
N does not change, which I suppose means
N is not potentially infinite.

N does not have a last element, which I suppose means
N is not finite and N is not actually infinite.

I suppose that is a problem for you. Everyone else
just has to get on with reasoning about k in N.

To start with, we know that {0,...,k} exists which is
totally-ordered, begins at 0, ends at k, and,
for each split, there is a crossing-pair j,j+1.

On Monday, 4 October 2021 at 18:18:46 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:29:16 UTC+2:
> > fredag 1 oktober 2021 kl. 20:22:53 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 14:21:00 UTC+2:
> > >
> > > > All natural numbers have infinitely many natural numbers after,
> > > So they are not an actuallyinfinite set. Note that two consecutive infinite sets of card aleph_0 are impossible in |N.
> > >
> > If each natural number has infinitely many successors that means the set of natural numbers is infinite.
> This set |N is used to index all endegments. Therefore not all endsegments can have infinite contents.

Bullshit. For every n in |N, E(n) is the set {n, n+1, n+2, n+3, ...}. There always are aleph_0 integers larger than n (or n - 1). Did you not learn *ANYTHING*?

On Monday, 4 October 2021 at 18:23:38 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 3. Oktober 2021 um 15:58:13 UTC+2:
> > On Sunday, 3 October 2021 at 09:54:57 UTC-3, WM wrote:
>
> > > All existing infinite endsegments have an infinite intersection with each other.
> > You've been very busy this morning, and very successful at demonstrating again your utter incompetence. You have no clue of how infinity works,
> I can disprove the shit you believe.

You couldn't disprove *SHIT* if you tried, you washed up, demented, lying son of a bitch.

> > In particular, you have not even a scintilla of a clue as to how infinite intersections and unions differ from finite ones.
> Do you know any endsegment that is not belonging to one of the following sets?
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> No. Therefore all infinite endsegments have an infinite intersection. Inclusion monotony is always true.

You don't know what inclusion monotony is, you fucking moron.

On 10/3/2021 6:37 AM, WM wrote:
> Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 02:32:38 UTC+2:
>> On Friday, October 1, 2021 at 7:51:03 PM UTC+2, WM wrote:
>>
>>> Please mark the place of n on the following scale:
>>>
>>> ...1...10...100...1000...10000...
>> Please mark the place of Schnirelmann's constant on the follwoing scale,
>
>> 1, 2, 3, 4, 5, 6, 7, ...
>>
> If it is a constant number, then it has a place there which can be determined in principle. n has no place there.
>
> Regards, WM
>

Let n be a natural number. Any problems with that statement ?

On 10/4/2021 4:16 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
>> fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:
>>> If I say it is from the set of natural number it is by definition a natural number.
>>> It is not your competence to define what natural numbers are.
>>> Please mark the place of n on the following scale:
>>>
>>> ...1...10...100...1000...10000...
>>>
>> why?
>
> To prove your claim.
>
>> Again, I don't need to do that, it is somewhere
>
> A number is not somewhere but has a fixed place.

n does too, n-1 is in front of it, and n+1 is in back of it.

No Variables for you!!

So WM rejects Sets, Algebra, Calculus, Proofs, Equations, along with variables, and solving equations....

WM is NOT a mathematician at all, and never was.

>
> Regards, WM
>

On Tuesday, October 5, 2021 at 4:46:47 AM UTC+2, Serg io wrote:
> On 10/4/2021 4:16 PM, WM wrote:
> > zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
> > > fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:

Let n be a natural number.

> > > > Please mark the place of n on the following scale:
> > > >
> > > > ...1...10...100...1000...10000...
> > > >
> > > why?
> > >
> > To prove your claim.
> > >
> > > I don't need to do that, it is somewhere.
> > >
> > A number is not somewhere but has a fixed place.
> >
> n does too, n-1 is in front of it [if n > 0 --GC], and n+1 is in back of it.
>
> No Variables for you!!

In this case "n" is not a variable, but a constant (or in the context of ND a "parameter").

On Sunday, October 3, 2021 at 1:37:39 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 02:32:38 UTC+2:
> >
> > Please mark the place of Schnirelmann's constant on the following "scale",
> > 1, 2, 3, 4, 5, 6, 7, ...
> >
> If it is a constant [...], then it has a place there [...]. n has no place there.

Nope. Since n is a natural number it certainly has "a place there" (just like the Schnirelmann's constant). We just don't KNOW which one (same for the Schnirelmann's constant). On the other hand, we may still state the truth: n is the n-th term in the sequence (1, 2, 3, ...). Actually, there are many things we DO know about n (since n is a natural number).

On Saturday, October 2, 2021 at 5:20:36 PM UTC+2, Serg io wrote:
> On 9/27/2021 3:20 PM, WM wrote:
> > Jim Burns schrieb am Sonntag, 26. September 2021 um 19:14:45 UTC+2:
> > >
> > > "That" is a pronoun.
> > > That is a natural number, when I refer to a natural number.
> > >
> > Yes, if you refer to a certain number.
> >
> no.

Actually, yes. With "that" we refer to a certain nummer (if so), even if this number is not KNOWN to us. (Though the referent of "that" may differ in different contexts.)

For example I may state:

| Consider an arbitrary number in {1, 2, 3}. That number ...

Clearly "that" refers to a CERTAIN number in {1, 2, 3}, though we don't and can't KNOW which one.

Using symbols we might state:

| Let n e {1, 2, 3}. Then n ...

instead.

Here, in the same way, "n" refers to a CERTAIN number in {1, 2, 3}, though we don't and can't KNOW which one.

Please note the difference: IN THIS CONTEXT "n" is a constant (or "parameter", if you like).

On the other hand, in the statement

| For all n in IN: n >= 0.

"n" is (used as) a variable.

Or even in a statement like

| n e IN --> n >= 0,

or some algebraic "equations", etc.

måndag 4 oktober 2021 kl. 23:16:48 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
> > fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:
> > > If I say it is from the set of natural number it is by definition a natural number.
> > > It is not your competence to define what natural numbers are.
> > > Please mark the place of n on the following scale:
> > >
> > > ...1...10...100...1000...10000...
> > >
> > why?
>
> To prove your claim.
> > Again, I don't need to do that, it is somewhere
> A number is not somewhere but has a fixed place.
>
> Regards, WM

>To prove your claim.

Placing it there in no way proves my claim.

>A number is not somewhere but has a fixed place.

Coming from the man who think sets changes!

Of course it is at a specific fixed place, but my ability to place it does not demonstrate it. It is somewhere, even if I cannot point out where :)

måndag 4 oktober 2021 kl. 23:25:42 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:34:31 UTC+2:
> > fredag 1 oktober 2021 kl. 19:58:10 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 07:03:30 UTC+2:
> > > > torsdag 30 september 2021 kl. 20:56:13 UTC+2 skrev WM:
> > >
> > > > > Just this is impossible for endsegments. Inclusion monotony!
> > > > >
> > > > Nope, that only shows that any FINITE intersection is non-empty, not that the intersection of ALL is emtpy
> > > Is one endsegment among all that is not the last of a finite intersection?
> > >
> > You cannot deduce anything from the finite about the infinite.
> I can deduce that there is no infinite. I can deduce from
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> that all existing infinite endsegments have an infinite intersection.
> Inclusion monotony is always true.
>
> Regards, WM

>I can deduce that there is no infinite.

You can deduce nothing of the sort from finite cases.

>that all existing infinite endsegments have an infinite intersection.

Nope, you can only deduce that FINITE intersections give infinite cardinality. You cannot deduce that an infinite intersection has infinite cardinality.

måndag 4 oktober 2021 kl. 23:18:46 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:29:16 UTC+2:
> > fredag 1 oktober 2021 kl. 20:22:53 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 1. Oktober 2021 um 14:21:00 UTC+2:
> > >
> > > > All natural numbers have infinitely many natural numbers after,
> > > So they are not an actuallyinfinite set. Note that two consecutive infinite sets of card aleph_0 are impossible in |N.
> > >
> > If each natural number has infinitely many successors that means the set of natural numbers is infinite.
> This set |N is used to index all endegments. Therefore not all endsegments can have infinite contents.
>
> Regards, WM

non-sequitor. There is nothign that says they cannot have infinite cardinality.
And they all demonstrably do have infinite cardinality because each one is in bijection with N

zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 06:53:57 UTC+2:
> måndag 4 oktober 2021 kl. 23:16:48 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
> > > fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:
> > > > If I say it is from the set of natural number it is by definition a natural number.
> > > > It is not your competence to define what natural numbers are.
> > > > Please mark the place of n on the following scale:
> > > >
> > > > ...1...10...100...1000...10000...
> > > >
> > > why?
> >
> > To prove your claim.
> > > Again, I don't need to do that, it is somewhere
> > A number is not somewhere but has a fixed place.

> >To prove your claim.
>
> Of course it is at a specific fixed place, but my ability to place it does not demonstrate it. It is somewhere, even if I cannot point out where :)

My ability to place it everywhere, at as many places as I like, however, shows that n is not a number.

Regards, WM