Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 00:47:51 UTC+2:
> 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 actually infinite 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.
> For every n in |N, E(n) is the set {n, n+1, n+2, n+3, ...}.

But you claim that there are all E(n). And you claim that all are infinite. Contradiction.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 06:58:40 UTC+2:
> 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.
> >
> 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

And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible. Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite. In the limit n --> ω we get:

Lim n = ω
Lim E(n) = E(ω) = { }
Lim |E(n)| = 0

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 06:57:54 UTC+2:
> 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.
> >

> >I can deduce that there is no infinite.
> You can deduce nothing of the sort from finite cases.

If no endsegment exists outside of (*), then all are covered by (*).

> >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.

From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on. If all endsegments have infinite contents, then infinite contents is in all of them.

Regards, WM

Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 00:49:30 UTC+2:
> On Monday, 4 October 2021 at 18:23:38 UTC-3, WM wrote:

> > 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

I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo.

Do you know any endsegment that is not covered by (*)? This proves that all endsegments (which can be known) have infinite intersection.

Look: Every infinite endsegment has an infinite intersection with itself and with all its predecessors. But you claim that there are no other endsegments. What else could an infinite set of infinite endsegments contain in addition? Nothing!

Regards, WM

Greg Cunt schrieb am Dienstag, 5. Oktober 2021 um 05:06:45 UTC+2:
> 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"

I can place n at 1, at 2, at 3, at any place I like. This is not possible with a natural number like 7 for instance, and it is not possible with a hitherto unknown number.

Regards, WM

Jim Burns schrieb am Dienstag, 5. Oktober 2021 um 00:42:23 UTC+2:

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

Then there is no set of natural numbers but only a potentially infinite collection, namely the ends of FISONs. These are the only alternatives.

Regards, WM

On Tuesday, 5 October 2021 at 04:57:10 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 00:47:51 UTC+2:
> > 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 actually infinite 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.
> > For every n in |N, E(n) is the set {n, n+1, n+2, n+3, ...}.
> But you claim that there are all E(n). And you claim that all are infinite. Contradiction.

Bullshit. You don't even bother to argue anymore. I repeat: You have no clue about infinity, never have, and never will. But I won't engage in a tit-for-tat on words. Since you clearly have no argument, EOD.

On Tuesday, 5 October 2021 at 05:03:10 UTC-3, WM wrote:
[...]
> Lim n = ω
> Lim E(n) = E(ω) = { }
> Lim |E(n)| = 0

You clearly have *NO* clue about infinity, and now you don't even understand limits any longer. Three lines, three times garbage.

1. Lim n = ω. In order to prove this, you have to first define the topology or metric in which the convergence is to take place. Only *THEN* does that equation make any sense.

2. Lim E(n) = E(ω). Aside from the fact that you have no justification for even *WRITING* E(ω) (and you *CERTAINLY* can't just mechanically substitute ω into any formula defined for natural numbers), you *CAN* define what you mean by the symbol "E(ω)", and to define E(ω) to be the empty set, you at least have consistency. It is even correct in Halmos's set limit that Lim E(n) ={}. But you can't *JUSTIFY* that limit by your definition. Ergo, your line is backwards.

3. Lim |E(n)| = 0. This, of course, is bullshit. For every n, |E(n)| = aleph_0, and hence Lim |E(n)| is *NOT* 0. What is 0 is |Lim E(n)|, which is *NOT* the same thing. As explained hundreds of times, card() is not a continuous function.

4. Three lines, written in isolation do not make a proof.

On Tuesday, 5 October 2021 at 05:12:00 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 00:49:30 UTC+2:
> > On Monday, 4 October 2021 at 18:23:38 UTC-3, WM wrote:
>
> > > 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
> I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo.

Nonsense. You think inclusion monotony for a finite number of sets and inclusion monotony for an in finite number of sets behave in the same way. You accept this without proof, without reflection and therefore without even the whiff of understanding. Your choice of words ("taboo") seems to suggest that you actually, deep down, have an inkling that you might be wrong. Well, let me tell you, you *ARE* wrong.

> Do you know any endsegment that is not covered by (*)? This proves that all endsegments (which can be known) have infinite intersection.

And here you go, slipping this innocent "can be known" into your sentence. You do this so casually -- and knowingly! -- to misdirect. That doesn't make it less wrong, because as soon as you do this, you are no longer arguing about ZFC but Muckemythics. I am not interested in that.

> Look: Every infinite endsegment has an infinite intersection with itself and with all its predecessors. But you claim that there are no other endsegments.

*YOU* claim that, nobody else. Of course there are infinitely many end segments that are not in these finite intersections that you are talking about. And, yes,

intersect {E(k): k < n} =/= intersect {E(k): k <= n}.

But every one of these intersections is over a finite set. To get intersect {E(k): k in |N} you have to use *LIMITS* (about which of course you have no clue).

On Monday, 4 October 2021 at 23:43:34 UTC-3, Serg io wrote:

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

Yes, you made it, not he. (It does make a difference to a psychotic fool like WM, you know.)

Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 12:44:16 UTC+2:
> On Monday, 4 October 2021 at 23:43:34 UTC-3, Serg io wrote:
>
> > Let n be a natural number. Any problems with that statement ?
> Yes, you made it, not he.

Of course I often use this phrase. But I know what it means: Any natural number can be inserted at position of n.

Regards, WM

Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 12:31:08 UTC+2:
> On Tuesday, 5 October 2021 at 05:03:10 UTC-3, WM wrote:
> [...]
> > Lim n = ω
> > Lim E(n) = E(ω) = { }
> > Lim |E(n)| = 0
> You clearly have *NO* clue about infinity, and now you don't even understand limits any longer. Three lines, three times garbage.
>
> 1. Lim n = ω. In order to prove this, you have to first define the topology or metric in which the convergence is to take place.

No. Only if you use some basis where Lim n = ω is wrong you have to define and justify it. Same as with addition. 1 + 1 = 2. Every other meaning of + or basis of numbers has to be indicated.

> 2. Lim E(n) = E(ω). Aside from the fact that you have no justification for even *WRITING* E(ω)

The justification is given under point 1.
>
> 3. Lim |E(n)| = 0. This, of course, is bullshit. For every n, |E(n)| = aleph_0, and hence Lim |E(n)| is *NOT* 0.

If E(ω) is empty, then the stepwise reduction ∀k ∈ ℕ: E(k+1) = E(k) \ {k} forces 3. You see in mathematics there is no other alternative.

> 4. Three lines, written in isolation do not make a proof.

Of course the person reading them must be able to fill the gaps.

Regards, WM

On Tuesday, 5 October 2021 at 08:38:46 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 12:31:08 UTC+2:
> > On Tuesday, 5 October 2021 at 05:03:10 UTC-3, WM wrote:
> > [...]
> > > Lim n = ω
> > > Lim E(n) = E(ω) = { }
> > > Lim |E(n)| = 0
> > You clearly have *NO* clue about infinity, and now you don't even understand limits any longer. Three lines, three times garbage.
> >
> > 1. Lim n = ω. In order to prove this, you have to first define the topology or metric in which the convergence is to take place.
> No. Only if you use some basis where Lim n = ω is wrong you have to define and justify it. Same as with addition. 1 + 1 = 2. Every other meaning of + or basis of numbers has to be indicated.
> > 2. Lim E(n) = E(ω). Aside from the fact that you have no justification for even *WRITING* E(ω)
> The justification is given under point 1.

It is *NOT*. 1. deals with a limit of numbers, 2. deals with a limit of *SETS*. These are not the same, although they *CAN* (sometimes!) give consistent results.

> > 3. Lim |E(n)| = 0. This, of course, is bullshit. For every n, |E(n)| = aleph_0, and hence Lim |E(n)| is *NOT* 0.
> If E(ω) is empty, then the stepwise reduction ∀k ∈ ℕ: E(k+1) = E(k) \ {k} forces 3. You see in mathematics there is no other alternative.

Bullshit! All I see is that you have *ZERO* comprehension of anything! In particular, you cannot comprehend that Lim |E(n)| is not the same as |Lim E(n)|. The former is a limit of natural numbers, the latter is a limit of sets, and here is a situation, where the two give different results. *ZERO* comprehension on your part, as I said.

> > 4. Three lines, written in isolation do not make a proof.
> Of course the person reading them must be able to fill the gaps.

The person writing things down must be able to comprehend what they write. Since you clearly are not, these three lines, taken together, are gibberish, and they would be gibberish even if all of them were correct. (Have you *EVER* produced a correct mathematical proof of *ANYTHING* without copying from someone?)

On 10/4/2021 10:23 PM, Greg Cunt wrote:
> 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.
>

a subtle difference there;

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

n is constant or parameter (member of the set)

| n e IN: n >= 0

"n" is variable

tisdag 5 oktober 2021 kl. 10:06:49 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 06:57:54 UTC+2:
> > 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.
> > >
> > >I can deduce that there is no infinite.
> > You can deduce nothing of the sort from finite cases.
> If no endsegment exists outside of (*), then all are covered by (*).
> > >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.
> From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on. If all endsegments have infinite contents, then infinite contents is in all of them.
>
> Regards, WM

>If no endsegment exists outside of (*), then all are covered by (*).

It is true for all endsegments but that does not mean what you want.

>From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on. If all endsegments have infinite contents, then infinite contents is in all of them.

Correct but that doesn't mean that the infinite intesrection of all is not empty. It IS empty.

>I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo

No one is surpressing anything. It just doesn't fucking mean what you want it to!

tisdag 5 oktober 2021 kl. 10:03:10 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 06:58:40 UTC+2:
> > 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.
> > >
> > 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
> And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible. Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite. In the limit n --> ω we get:
>
> Lim n = ω
> Lim E(n) = E(ω) = { }
> Lim |E(n)| = 0
>
> Regards, WM

>And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible.

Who says otherwise? Not me!

>Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite.

The intersection is still emtpy.

In the limit n --> ω we get:

>Lim n = ω

Can be said

>Lim E(n) = E(ω) = { }

E is not defined for omega.

>Lim |E(n)| = 0

incorrect, it is aleph_0

tisdag 5 oktober 2021 kl. 09:54:23 UTC+2 skrev WM:
> 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
no one says you can, I just cannot place it anywhere.

tisdag 5 oktober 2021 kl. 10:35:21 UTC+2 skrev WM:
> Jim Burns schrieb am Dienstag, 5. Oktober 2021 um 00:42:23 UTC+2:
>
> > A natural number which cannot be counted to, even in principle,
> > is not a natural number.
> Then there is no set of natural numbers but only a potentially infinite collection, namely the ends of FISONs. These are the only alternatives.
>
> Regards, WM

There is a set of natural numbers you moron

On Tuesday, 5 October 2021 at 05:06:49 UTC-3, WM wrote:
> From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on.

Didn't notice this lunacy before. From the fact that E(3) is not empty you try to deduce that E(3) is a subset of E(4), E(5), etc.??? Get a grip!

> If all endsegments have infinite contents, then infinite contents is in all of them.

This, of course, is equally untenable. You should open a gofundme page. Maybe that will enable you to get a clue.

On 10/3/2021 7:47 AM, WM wrote:
> 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
>

there are no empty endsegments by definition.

On 10/5/2021 8:34 AM, Gus Gassmann wrote:
> On Tuesday, 5 October 2021 at 05:06:49 UTC-3, WM wrote:
>> From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on.
>
> Didn't notice this lunacy before. From the fact that E(3) is not empty you try to deduce that E(3) is a subset of E(4), E(5), etc.??? Get a grip!
>
>> If all endsegments have infinite contents, then infinite contents is in all of them.
>
> This, of course, is equally untenable. You should open a gofundme page. Maybe that will enable you to get a clue.
>

WM has lost the definition/meaning of an endsegment, and the ability to reason using their characteristics.
Seems like WM thinks of them as bags of rocks, and infinity is some sort of special rock.

On 05.Oct.2021, Serg io wrote:

>> Only an empty endsegment can cause an emty intersection.
>> Regards, WM
>
> there are no empty endsegments by definition.

except is used thereafter in tensors and everything. Give me a break.

in america for instance, the CDC, a deeply evil private org, is using
Fluoridated water mostly in black communities, to keep their IQ low. The
same in Europe, ie in czech and polish communities. They have
concentration layers right now.

zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 15:25:50 UTC+2:
> tisdag 5 oktober 2021 kl. 10:03:10 UTC+2 skrev WM:

> > > 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
> > And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible. Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite. In the limit n --> ω we get:
> >
> > Lim n = ω
> > Lim E(n) = E(ω) = { }
> > Lim |E(n)| = 0
>
> >And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible.
> Who says otherwise? Not me!

Everyone who says that all natural numbers are used up as indices of endsegments but infinitely many are remaining as contents of all endsegments (which used up all already) say that there is an infinite set |N (of indices) which is followed by an infinite set |N (of contents).
> >Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite.
> The intersection is still emtpy.

Every endsegment has infinite contents. But all endsegments exhaust |N.
> In the limit n --> ω we get:
>
> >Lim n = ω
> Can be said
> >Lim E(n) = E(ω) = { }
> E is not defined for omega.

The limit is defined if Lim n = ω.
>
> >Lim |E(n)| = 0
>
> incorrect, it is aleph_0

This is impossible if the limit of E(n) is empty and

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Regards, WM

On 10/5/2021 9:21 AM, Hilton Blome wrote:
> On 05.Oct.2021, Serg io wrote:
>
>>> Only an empty endsegment can cause an emty intersection.
>>> Regards, WM
>>
>> there are no empty endsegments by definition.
>
> except is used thereafter in tensors and everything. Give me a break.
>
> in america for instance, the CDC, a deeply evil private org, is using
> Fluoridated water mostly in black communities, to keep their IQ low. The
> same in Europe, ie in czech and polish communities. They have
> concentration layers right now.
>

agree, in fact USA chemical companies get rid of their toxic infected industrial waste by spraying it from Airliners over Commie and Socialist-like
countries in a process known as Chemtrails

all know that commie endsegments are on the low rung of the ladder

On 05.Oct.2021, Serg io wrote:

>>> there are no empty endsegments by definition.
>>
>> except is used thereafter in tensors and everything. Give me a break.
>>
>> in america for instance, the CDC, a deeply evil private org, is using
>> Fluoridated water mostly in black communities, to keep their IQ low.
>> The same in Europe, ie in czech and polish communities. They have
>> concentration layers right now.
>
> agree, in fact USA chemical companies get rid of their toxic infected
> industrial waste by spraying it from Airliners

kiss my ass, this is incomplete, economic theories are not theories.

ROACH QUARANTINE https://www.bitchute.com/video/HP0JsBgE0sVl/

Polish MPs Protest Outside Australian Embassy "Australia Has Contracted
COVID Madness" https://www.bitchute.com/video/VpaxMVuFJz16/

how did this ugly bitch arrived in Australia?

Bye Bye Gladis Celebration and the peoples Reaction
https://www.bitchute.com/video/58N93D7zeKZB/