Greg Cunt schrieb am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:

> 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.
>
What is its place in the sequence of ordinals?

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 15. September 2021 um 08:36:39 UTC+2:
> 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.
> >
> empty assertion, they all have infinite cardinality.

A sequence of ordinal oemga cannot have aleph_0 elements which are followed by aleph_0 elements. But almost all contents of all endsegments follows after the indices of endsegments.

As long as an endsegment is infinite, it cannot belong to an infinite set of endsegments.

Regards, WM

WM presented the following explanation :
> zelos...@gmail.com schrieb am Mittwoch, 15. September 2021 um 08:36:39 UTC+2:
>> 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.
>>>
>> empty assertion, they all have infinite cardinality.
>
> A sequence of ordinal oemga cannot have aleph_0 elements which are followed
> by aleph_0 elements. But almost all contents of all endsegments follows after
> the indices of endsegments.
>
> As long as an endsegment is infinite, it cannot belong to an infinite set of
> endsegments.

With all due respect, stop lying.

On Wednesday, 15 September 2021 at 08:18:20 UTC-3, WM wrote:
> Greg Cunt schrieb am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:
>
> > 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.
> >
> What is its place in the sequence of ordinals?

It follows its immediate predecessor and is followed by its successor S(n).

Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 12:51:50 UTC+2:
> On Tuesday, 14 September 2021 at 17:19:44 UTC-3, WM wrote:
> > 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)?

|N.

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

Of course. The limit of the sequence of endsegments is

lim{n -> oo} E(n) = lim{n -> oo} {n, n+1, n+2, ...} = lim{n -> oo} {n} = {ω}

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

Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

would be broken.

Regards, WM

FromTheRafters schrieb am Mittwoch, 15. September 2021 um 13:33:50 UTC+2:
> WM presented the following explanation :

> > As long as an endsegment is infinite, it cannot belong to an infinite set of
> > endsegments.
> With all due respect, stop lying.

The set of endsgments {E(n) | n ∈ ℕ} absorbs all natural numbers. None are remaining for subsequent contents of the endsegments. But an infinite endsegment can only use the first number of its contents as index enumerating its position. Between its position and ω there remain infinitely many natnumbers enumerating finite endsegments. If you are not convinced, you should confess that you despise mathematics, namely its basic definition of endsegments:

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Regards, WM

Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 14:03:00 UTC+2:
> On Wednesday, 15 September 2021 at 08:18:20 UTC-3, WM wrote:
> > Greg Cunt schrieb am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:
> >
> > > 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.
> > >
> > What is its place in the sequence of ordinals?
> It follows its immediate predecessor and is followed by its successor S(n).

Here is the sequence: 1, 2, 3, 4, 5, 6, ...

Please point to the position of n.

Regards, WM

On Wednesday, September 15, 2021 at 3:08:54 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 14:03:00 UTC+2:
> > On Wednesday, 15 September 2021 at 08:18:20 UTC-3, WM wrote:
> > > Greg Cunt schrieb am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:
> > > >
> > > > 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.
> > > >
> > > What is its place in the sequence of ordinals?
> > >
> > It follows its immediate predecessor [if it has one --GC] and is followed by its successor S(n).

Right.

> Here is the sequence: (1, 2, 3, 4, 5, 6, ...)
>
> Please point to the position of n.

No problem: The position of n in the sequence (1, 2, 3, 4, 5, 6, ...) is n.

Now THAT was easy!

On Wednesday, 15 September 2021 at 09:57:18 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 12:51:50 UTC+2:
> > On Tuesday, 14 September 2021 at 17:19:44 UTC-3, WM wrote:
> > > 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)?
> |N.
> > The beauty, of course, is that lim{n -> oo} [ |N \ FISON(n) ] *IS* |N \ [ lim{n -> oo} FISON(n) ].
> Of course. The limit of the sequence of endsegments is
>
> lim{n -> oo} E(n) = lim{n -> oo} {n, n+1, n+2, ...} = lim{n -> oo} {n} = {ω}

Are you *NUTS*? (No, you obviously *ARE* insane.) This limit is a set limit.. And you used to know how to deal with set limits.

> lim{n -> oo} E(n) = lim{n -> oo} {n, n+1, n+2, ...}

This is correct, based on your (in your own admission) not fully considered definition of E(n).

> lim{n -> oo} {n, n+1, n+2, ...} = lim{n -> oo} {n}

This is highly questionable. You might be able to justify it by first computing each limit and verifying that the limits coincide, but as a stand-alone assertion, it is garbage.

> lim{n -> oo} {n} = {ω}

And this, of course, is the height of lunacy. I won't dignify it with further elaboration.

> > Note: No dark numbers. Put that in your pipe and smoke it!
> Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (*)
>
> would be broken.

Bullshit!! You have no clue how to compute set limits, and you have no idea what your equation involving the set difference in (*) actually tells you.

On Wednesday, September 15, 2021 at 2:57:18 PM UTC+2, WM wrote:

> Between every infinite endsegment and {ω} there <bla>

Huh?! How is "betweenness" defined in this case. In respect to which order relation?

After stating a proper definition, please prove your claim.

Hint: We may use "c" to order the endsegements. Then

.... c E_3 c E_2 c E_1.

But {w} isn't an endsegment; moreover for no endsegment E: E c {ω} or {ω} c E.

Please elaborate.

On Wednesday, September 15, 2021 at 2:57:18 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 12:51:50 UTC+2:
> >
> > What is lim{n -> oo} FISON(n)?
> >
> IN.

Right.

> > The beauty, of course, is that lim{n -> oo} [ IN \ FISON(n) ] *IS* IN \ [ lim{n -> oo} FISON(n) ].
> >
> Of course.

Hence we get IMMEDIATELY:

lim{n -> oo} E'(n) = lim{n -> oo} {n+1, n+2, n+3, ...} = lim{n -> oo} [ IN \ FISON(n) ] = IN \ [ lim{n -> oo} FISON(n) ] = IN \ IN = {}.

In other words,

> The [set] limit of the [set] sequence of [slightly modified] endsegments is
>
> lim{n -> oo} E'(n) = lim{n -> oo} {n+1, n+2, n+3, ...} = [...]

{} .

See: https://en.wikipedia.org/wiki/Set-theoretic_limit

Hint: We might call these slightly modified endsegments "tail segments". :-)

Definitions:

TEIL(n) := {m e IN : m > n} (n e IN)

or recursively:

TAIL(1) = IN \ {1}
TAIL(n+1) =TAIL(n) \ {n+1} .

Then TAIL(1) = {2, 3, 4, ...}, TAIL(2) = {3, 4, 5, ...}, TAIL(3) = {4, 5, 6, ...}.

This /tail segments/ have the following property:

An e IN: n !e TAIL(n).

Moreover, for all n e IN:

FISON(n) n TAIL(n) = {} and FISON(n) u TAIL(n) = IN .

Hence, for all n e IN:

TAIL(n) = IN \ FISON(n) and IN \ TAIL(n) = FISON(n).

Finally:

INTERSECTION {TAIL(n) : n e IN} = {}.

Proof: For all n e IN: n !e TAIL(n).

On Wednesday, 15 September 2021 at 10:51:21 UTC-3, Greg Cunt wrote:
> On Wednesday, September 15, 2021 at 2:57:18 PM UTC+2, WM wrote:
>
> > Between every infinite endsegment and {ω} there <bla>
>
> Huh?! How is "betweenness" defined in this case. In respect to which order relation?
>
> After stating a proper definition, please prove your claim.
>
> Hint: We may use "c" to order the endsegements. Then
>
> ... c E_3 c E_2 c E_1.
>
> But {w} isn't an endsegment; moreover for no endsegment E: E c {ω} or {ω} c E.
>
> Please elaborate.

What for? You know that all he can spout is nonsense, and you know what nonsense he is going to spout.

On Wednesday, September 15, 2021 at 4:14:54 PM UTC+2, Gus Gassmann wrote:
> On Wednesday, 15 September 2021 at 10:51:21 UTC-3, Greg Cunt wrote:
> > On Wednesday, September 15, 2021 at 2:57:18 PM UTC+2, WM wrote:
> > >
> > > Between every infinite endsegment and {ω} there <bla>
> > >
> > Huh?! How is "betweenness" defined in this case. In respect to which order relation?
> >
> > After stating a proper definition, please prove your claim.
> >
> > Hint: We may use "c" to order the endsegements. Then
> >
> > ... c E_3 c E_2 c E_1.
> >
> > But {w} isn't an endsegment; moreover for no endsegment E: E c {ω} or {ω} c E.
> >
> > Please elaborate.
> >
> What for?

Just for fun!

> You know that all he can spout is nonsense,

Right.

> and you know what nonsense he is going to spout.

Not really, that's the "funny" part. He's constantly cumming up with new nonsense.

On Wednesday, September 15, 2021 at 4:14:54 PM UTC+2, Gus Gassmann wrote:

> You know that all he can spout is nonsense, and you know what nonsense he is going to spout.

| "Without an actually infinite term there is no actually infinite number of terms." [WM]

On 9/15/2021 7:18 AM, WM wrote:
> Greg Cunt schrieb
> am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:

>> 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.
>
> What is its place in the sequence of ordinals?

We don't know. At least, we apparently haven't been told yet.

However, we already know other things about n.

There is an order-relation '<' between n and other individuals
such that, for any collection B of these (ordinal) individuals,
B contains a first element, unless B is empty.
('<' is a well-order.)

We know facts that can be derived from this fact about n
It's interesting that there are a lot of these derived facts,
but, interesting or not, we know them.

Not-knowing which individual n is does not remove from our
brains anything we know about n.
"You don't know X" is not in itself an argument for
"You don't know Y".

On Wednesday, September 15, 2021 at 4:32:37 PM UTC+2, Jim Burns wrote:
> On 9/15/2021 7:18 AM, WM wrote:
> > Greg Cunt wrote:
> > >
> > > 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.
> > >
> > What is its place in the sequence of ordinals?
> >
> We don't know.

Oh, but we *do* know! It's n (!).

Really!

I mean, it's not n-k for all k e {m e {1, 2, 3, ...} : m <= n} and it is not n+k for all k e IN = {1, 2, 3, ...}, it's just n.

Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 15:46:10 UTC+2:
> On Wednesday, 15 September 2021 at 09:57:18 UTC-3, WM wrote:
>
> > Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition
> >
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (*)
> >
> > would be broken.
>
> you have no idea what your equation involving the set difference in (*) actually tells you.

It tells me and every person able to read math text, that no empty set can result unless finite endsegments are passed by.

Regards, WM

Greg Cunt schrieb am Mittwoch, 15. September 2021 um 15:51:21 UTC+2:

> But {w} isn't an endsegment; moreover for no endsegment E: E c {ω} or {ω} c E.
>
> Please elaborate.

ω is not a natural number, but it is the limit of the sequence of natural numbers: it is called limit ordinal. For endsegments it depends whether the limit is implied internally (for every endsegment) or only natural numbers are admitted. In the latter case the limit of the sequence (E(n)) is the empty set.

Regards, WM

Greg Cunt schrieb am Mittwoch, 15. September 2021 um 15:44:34 UTC+2:
> On Wednesday, September 15, 2021 at 3:08:54 PM UTC+2, WM wrote:

> > Here is the sequence: (1, 2, 3, 4, 5, 6, ...)
> >
> > Please point to the position of n.
> No problem: The position of n in the sequence (1, 2, 3, 4, 5, 6, ...) is n.
>
There is no n visible yet. Try again.

Regards, WM

Jim Burns schrieb am Mittwoch, 15. September 2021 um 16:32:37 UTC+2:
> On 9/15/2021 7:18 AM, WM wrote:
> > Greg Cunt schrieb
> > am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:
>
> >> 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.
> >
> > What is its place in the sequence of ordinals?
> We don't know. At least, we apparently haven't been told yet.

But you have been told, long ago, the position of 3 and 17 and even 10^10.
>
> However, we already know other things about n.

Yes, it should be the placeholder for a natural number which satisfies the following conditions:
>
> There is an order-relation '<' between n and other individuals
> such that, for any collection B of these (ordinal) individuals,
> B contains a first element, unless B is empty.
> ('<' is a well-order.)
>
> We know facts that can be derived from this fact about n
> It's interesting that there are a lot of these derived facts,
> but, interesting or not, we know them.

Of course.
>
> Not-knowing which individual n is does not remove from our
> brains anything we know about n.

But it should remove the silly idea that it is a natural number. On the other hand this simple example explains how fanatical matheologians can defend the obviously wrong results of set theory. To mention only the most striking ones which every sober mind can identify as nonsense:
- All fractions can be enumerated.
- There are uncountably many real numbers although only countably many can be distinguished.
- There are more paths in the Binary Tree than nodes.
- The bankruptcy of McDuck.
- aleph_0 rational points separate uncountably many irrational points such that never two are existing without rational between them.

It is painful to see so much fool's crap blocking your minds. I will really enjoy my lectures next semester. It will be relieving to address an audience of unbiased thinkers.

Regards, WM

On Wednesday, September 15, 2021 at 5:52:04 PM UTC+2, WM wrote:

> There is no n visible yet.

Look, you psychotic asshole, NO ONE has ever s e e n a number. (Well except you, it seems.)

On Wednesday, September 15, 2021 at 6:08:36 PM UTC+2, WM wrote:

> But you have been told, long ago, the position of 3 and 17 and even 10^10.

Yeah, the position of 3 is 3, the position of 17 is 17, the position of 10^10 is 10^10, and the position of n is n.

> > it should be <bla>

Shut up, you psychotic asshole full of shit.

Since we "introduced" the term (actually a constant) "n" the following way:

| Let n e IN.

n is a natural number (i.e. an element in IN).

Hence we know quite a lot about the natural number n (since we know quite a lot about "the natural numbers").

> But it should remove the silly idea that it is a natural number.

Only if suffering from a psychosis, just like you.

On Wednesday, 15 September 2021 at 12:42:51 UTC-3, WM wrote:
> Greg Cunt schrieb am Mittwoch, 15. September 2021 um 15:51:21 UTC+2:
>
> > But {w} isn't an endsegment; moreover for no endsegment E: E c {ω} or {ω} c E.
> >
> > Please elaborate.
> ω is not a natural number, but it is the limit of the sequence of natural numbers: it is called limit ordinal. For endsegments it depends whether the limit is implied internally (for every endsegment) or only natural numbers are admitted. In the latter case the limit of the sequence (E(n)) is the empty set.

*NO*, you fucking idiot. Everything depends on what you take the limit of. The limit lim{n -> oo} n is the limit of a sequence of natural numbers, and it requires a metric to measure distances from ω, but lim{n -> oo} {n} is a limit of *SETS*, which is computed in a completely different way. (You even knew that way, and, I believe, you copied a proof in your dime store novel.) Only a complete idiot can think that there should be anything linking these two limits.

On 9/15/2021 12:08 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 15. September 2021 um 16:32:37 UTC+2:
>> On 9/15/2021 7:18 AM, WM wrote:
>>> Greg Cunt schrieb
>>> am Mittwoch, 15. September 2021 um 07:46:55 UTC+2:

>>>> 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.
>>>
>>> What is its place in the sequence of ordinals?
>>
>> We don't know. At least, we apparently haven't been told yet.
>
> But you have been told, long ago,
> the position of 3 and 17 and even 10^10.

The difference between
"n is 3", "n is 17", "n is 10^10"
and
"n is one of these well-ordered things"
is that we know
"n is one of these well-ordered things".

From things we know, we can derive more things we know,
by truth-preservation.

From things we don't know, not so much.

( An exception pops into my mind.
( https://en.wikipedia.org/wiki/Sum_and_Product_Puzzle
( Two perfect logicians S and P reason to the value of
( a pair of numbers starting from what the other
( _doesn't_know.
( ( Still, the opposite is pretty rare.

>> However, we already know other things about n.
>
> Yes, it should be the placeholder for a natural number which
> satisfies the following conditions:

We know that that number satisfies the following conditions.
And, from that, we can derive a lot more.

>> There is an order-relation '<' between n and other individuals
>> such that, for any collection B of these (ordinal) individuals,
>> B contains a first element, unless B is empty.
>> ('<' is a well-order.)
>>
>> We know facts that can be derived from this fact about n
>> It's interesting that there are a lot of these derived facts,
>> but, interesting or not, we know them.
>
> Of course.
>
>> Not-knowing which individual n is does not remove from our
>> brains anything we know about n.
>
> But it should remove the silly idea that it is a natural number.

| Something which cannot be counted to, even in principle,
| is not a natural number.
| If it is a natural number, it can be counted to, in principle.

Is that silly, in your opinion?

WM was thinking very hard :
> FromTheRafters schrieb am Mittwoch, 15. September 2021 um 13:33:50 UTC+2:
>> WM presented the following explanation :
>
>>> As long as an endsegment is infinite, it cannot belong to an infinite set
>>> of endsegments.
>> With all due respect, stop lying.
>
> The set of endsgments {E(n) | n ∈ ℕ} absorbs all natural numbers. None are
> remaining for subsequent contents of the endsegments. But an infinite
> endsegment can only use the first number of its contents as index enumerating
> its position. Between its position and ω there remain infinitely many
> natnumbers enumerating finite endsegments. If you are not convinced, you
> should confess that you despise mathematics, namely its basic definition of
> endsegments:
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

You misunderstand. I said *STOP* lying.