Jim Burns schrieb am Dienstag, 10. Januar 2023 um 22:17:11 UTC+1:
> On 1/10/2023 2:28 PM, WM wrote:

> > The Peano set has no *fixed* end.
> ℕ has no end at all, fixed or otherwise.

Try the sequence (1/n). It ends at 0. Since there is no hole, it ends immediately before 0.
>
> It is not-possibly-false that
> an element of ℕ is its end...

There cannot be an end. What ends is a field of bunch numbers.
>
> > According to Cantor ℕ fits between 0 and ω.
> Yes.
> What we mean by 'ω' is the first infinite ordinal.
> Each element n ∈ ℕ is a finite ordinal.
> In the ordinals, 0 ≤ₒ n <ₒ w
> ℕ fits between 0 and ω
> > So it ends before ω.
> No.

It ends not after ω.
>
> Yes,
> ℕ fits between 0 and ω
> and,
> if an element of ℕ was its end,
> then that end would be before ω

There is no elements its end but a bunch of infinitely many indistinguishable elements.
>
> What is the set of FISON-ends?
> ℕ, by definition.

The potentially infinite collection ℕ.

Regards, WM

On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:

> The Peano set has no *fixed* end.

A Peano set is a set. It does not change. It has no end, "fixed" or otherwise

On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:

> The sequence (1/n) ends

Nope. It does not have an end.

On 1/10/2023 4:35 PM, WM wrote:
> William schrieb am Dienstag, 10. Januar 2023 um 20:42:03 UTC+1:
>> On Tuesday, January 10, 2023 at 3:28:34 PM UTC-4, WM wrote:
>>
>>> The Peano set has no *fixed* end.
>> A Peano set is a set. It does not change. It has no end "fixed" or otherwise.
>>
>> <snip>
>>> According to Cantor ℕ fits between 0 and ω.
>> Correct
>>> So it ends before ω.
>> Nope. it has no end.
>
> ω is a point at the ordinal axis.
> The sequence (1/n) ends before 0. But since there is no gap, it ends close before 0.

the sequence 1/n does not end. the sequence n does not end, nor does n^n, nor 1/n^n.

this is a hole in your knowledge, not knowing how to handle infinite sequences.

it is a dark hole.

>
> Regards, WM

On 1/10/2023 4:16 PM, Jim Burns wrote:
> On 1/10/2023 2:54 PM, Sergi o wrote:
>> On 1/10/2023 1:28 PM, WM wrote:
>>> Jim Burns schrieb am Montag,
>>> 9. Januar 2023 um 21:04:37 UTC+1:
>
>>>> and ℕ doesn't.
>>>
>>> According to Cantor ℕ fits between 0 and ω.
>>> So it ends before ω.
>>
>> infinite sets do not have an end
>
> Nitpicking,
> _some_ infinite sets do not have an end.

ah, now you see the inherent vagueness in language,
where high complexity of things is distilled down into a few words
Math is a language that avoids that, which WM does not use.

>
> However,
> all infinite ordered-sets
> have subsets which not not have an end.
> (finite == double-well-ordered)
>
>> you have asked for help in this newsgroup
>> too many times on this topic.
>> what about that dude that lost to his turtle ?
>
> He's on his way.
> Not quite halfway here.
> We'll check back later,
> and see how he's doing.

poor framed Achilles losing a race to a racing turtle

>
>

On 1/10/2023 4:43 PM, WM wrote:
> Jim Burns schrieb am Dienstag, 10. Januar 2023 um 22:17:11 UTC+1:
>> On 1/10/2023 2:28 PM, WM wrote:
>
>>> The Peano set has no *fixed* end.
>> ℕ has no end at all, fixed or otherwise.
>
> Try the sequence (1/n). It ends at 0. Since there is no hole, it ends immediately before 0.
>>
>> It is not-possibly-false that
>> an element of ℕ is its end...
>
> There cannot be an end. What ends is a field of bunch numbers.

Bunch Ants, yes they are rare...

>>
>>> According to Cantor ℕ fits between 0 and ω.
>> Yes.
>> What we mean by 'ω' is the first infinite ordinal.
>> Each element n ∈ ℕ is a finite ordinal.
>> In the ordinals, 0 ≤ₒ n <ₒ w
>> ℕ fits between 0 and ω
>>> So it ends before ω.
>> No.
>
> It ends not after ω.

yes the ωaffle-ωaffle Ants

>>
>> Yes,
>> ℕ fits between 0 and ω
>> and,
>> if an element of ℕ was its end,
>> then that end would be before ω
>
> There is no elements its end but a bunch of infinitely many indistinguishable elements.

Bunch Ants indistinguishable with badges on their uniforms, all in a row

>>
>> What is the set of FISON-ends?
>> ℕ, by definition.
>
> The potentially infinite collection ℕ.

you were doing so well, quaaaaaccckkk!

>
> Regards, WM
>

On 1/10/2023 4:37 PM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 10. Januar 2023 um 20:42:10 UTC+1:
>> On Tuesday, January 10, 2023 at 8:33:08 PM UTC+1, WM wrote:
>>> William schrieb am Dienstag, 10. Januar 2023 um 19:16:35 UTC+1:
>>>>
>>>> IN is a Peano set.
>>>>
>>> The Peano set can be collected individually. Every Peano-number has a FISON. Every Peano-number is accessible by induction, proving that
>>>
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>> So what?! Actually, we have:
>>
>> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>>
> Try to think harder. The never vanishing rest ℵo cannot appear on the left-hand side.

wrong, it is simply an endsegment which is defined as ℵo

>
> Regards WM

On 1/8/2023 2:35 AM, WM wrote:
> "It never happens that the (always finite number of) pathbunches split into 2^ℵo pathbunches. However, if we posit a time after an infinite number of steps, we can say that the split happened. This 'it never happens but it happened' is unique to processes with no last step. For a process with no last step, you cannot conclude something cannot have happened, because there is no step where it happens."
>
> "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;"
>
> "At every level we have more lamettas than nodes in the binary tree. But through each of these nodes pass infinitely many paths. Maybe you can not distinguish them just by looking at the nodes but they do exist."
>
> "Unfortunately you have not provided a way of moving this from 'true for all finite' to 'true for infinite n', and induction won't take you from 'all finite n' to 'infinite n'."
>
> "If all nodes of a path have been deleted, that does not mean you ever deleted the path!!! You could delete all and only the finite paths from the tree, and that would delete all the nodes, yet you would never have deleted any infinite path, and they would all still exist in any case, regardless of what had been deleted!"
>
> Gruß, WM

Tonto says: "WM grub-um, try push-um darkies on us-ums. this White man say darkies good, but cant see-um, try to make me pay-um big wampum for nothing,
next time we see him, we get him marry him to Dofus, fat squaw, keep him in her teepee with his darkies, make-um nookie"

tisdag 10 januari 2023 kl. 08:08:28 UTC+1 skrev Eram semper recta:
> On Tuesday, 10 January 2023 at 08:08:42 UTC+2, zelos...@gmail.com wrote:
> > söndag 8 januari 2023 kl. 09:35:15 UTC+1 skrev WM:
> > > "It never happens that the (always finite number of) pathbunches split into 2^ℵo pathbunches. However, if we posit a time after an infinite number of steps, we can say that the split happened. This 'it never happens but it happened' is unique to processes with no last step. For a process with no last step, you cannot conclude something cannot have happened, because there is no step where it happens."
> > >
> > > "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;"
> > >
> > > "At every level we have more lamettas than nodes in the binary tree. But through each of these nodes pass infinitely many paths. Maybe you can not distinguish them just by looking at the nodes but they do exist."
> > >
> > > "Unfortunately you have not provided a way of moving this from 'true for all finite' to 'true for infinite n', and induction won't take you from 'all finite n' to 'infinite n'."
> > >
> > > "If all nodes of a path have been deleted, that does not mean you ever deleted the path!!! You could delete all and only the finite paths from the tree, and that would delete all the nodes, yet you would never have deleted any infinite path, and they would all still exist in any case, regardless of what had been deleted!"
> > >
> > > Gruß, WM
> > your dark numbers do not exist as you cannot define them to be distinct from natural number.
> They don't exist anymore than your trailing "infinite 3s" do in 0.333...
>
> What a moron.

False, those are valid and rigorously defined. His is not as he cannot even define it.

tisdag 10 januari 2023 kl. 08:25:35 UTC+1 skrev Eram semper recta:
> On Tuesday, 10 January 2023 at 09:21:55 UTC+2, Eram semper recta wrote:
> > On Tuesday, 10 January 2023 at 09:08:28 UTC+2, Eram semper recta wrote:
> > > On Tuesday, 10 January 2023 at 08:08:42 UTC+2, zelos...@gmail.com wrote:
> > > > söndag 8 januari 2023 kl. 09:35:15 UTC+1 skrev WM:
> > > > > "It never happens that the (always finite number of) pathbunches split into 2^ℵo pathbunches. However, if we posit a time after an infinite number of steps, we can say that the split happened. This 'it never happens but it happened' is unique to processes with no last step. For a process with no last step, you cannot conclude something cannot have happened, because there is no step where it happens."
> > > > >
> > > > > "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;"
> > > > >
> > > > > "At every level we have more lamettas than nodes in the binary tree. But through each of these nodes pass infinitely many paths. Maybe you can not distinguish them just by looking at the nodes but they do exist."
> > > > >
> > > > > "Unfortunately you have not provided a way of moving this from 'true for all finite' to 'true for infinite n', and induction won't take you from 'all finite n' to 'infinite n'."
> > > > >
> > > > > "If all nodes of a path have been deleted, that does not mean you ever deleted the path!!! You could delete all and only the finite paths from the tree, and that would delete all the nodes, yet you would never have deleted any infinite path, and they would all still exist in any case, regardless of what had been deleted!"
> > > > >
> > > > > Gruß, WM
> > > > your dark numbers do not exist as you cannot define them to be distinct from natural number.
> > > They don't exist anymore than your trailing "infinite 3s" do in 0.333....
> > Lots of dark 3s. 🤣🤣🤣🤣
> Proof:
> 1. There are ℵ dark numbers always as pointed out by WM.

Nope, he has never once been able to, his supposed "dark numbers" are always the empty set and his "defined numbers" are always equivalent with N. That is what "non-existence" means.

> 2. For each 3 in the sequence 0.333..., there is an index by a natural number.
> 3. We cannot know the dark numbers that are indexes for ℵ threes, therefore those ℵ threes are dark threes.
> Q.E.D.
>
>
> >
> > >
> > > What a moron.

tisdag 10 januari 2023 kl. 23:36:03 UTC+1 skrev WM:
> William schrieb am Dienstag, 10. Januar 2023 um 20:42:03 UTC+1:
> > On Tuesday, January 10, 2023 at 3:28:34 PM UTC-4, WM wrote:
> >
> > > The Peano set has no *fixed* end.
> > A Peano set is a set. It does not change. It has no end "fixed" or otherwise.
> >
> > <snip>
> > > According to Cantor ℕ fits between 0 and ω.
> > Correct
> > >So it ends before ω.
> > Nope. it has no end.
> ω is a point at the ordinal axis.
> The sequence (1/n) ends before 0. But since there is no gap, it ends close before 0.
>
> Regards, WM
it doesn't end you retard

William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
> On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
>
> > The sequence (1/n) ends
>
> Nope. It does not have an end.

It streches to the negative axis?
Surprising.

Regards, WM

WM submitted this idea :
> Fritz Feldhase schrieb am Dienstag, 10. Januar 2023 um 20:38:41 UTC+1:
>> On Tuesday, January 10, 2023 at 8:28:34 PM UTC+1, WM wrote:
>>
>>> According to Cantor ℕ fits between 0 and ω. So it ends before ω.
>> Oh, really?! Were exactly does it end?
>
> It ends immediately before ω. There are no holes in the ordinals.

What? You're still confused about this!?

https://mathworld.wolfram.com/LimitOrdinal.html

On Tuesday, January 10, 2023 at 11:33:05 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 10. Januar 2023 um 20:38:41 UTC+1:
> > On Tuesday, January 10, 2023 at 8:28:34 PM UTC+1, WM wrote:
> > >
> > > According to Cantor ℕ fits between 0 and ω. So it ends before ω.
> > >
> > Oh, really?! Were exactly does it end?
> >
> It ends immediately before ω.

False.

Hint: Any ordinal smaller than ω is _a natural number_ and (hence) has a natural number (i. e. not ω) as successor.

Formally: Ax e ORD: x < ω -> x e IN & x' e IN & x < x' < ω.

So there is no end "before ω" (immediately or not).

After all: "There are no holes in the ordinals." (WM)

On Tuesday, January 10, 2023 at 11:36:03 PM UTC+1, WM wrote:
> William schrieb am Dienstag, 10. Januar 2023 um 20:42:03 UTC+1:
> > On Tuesday, January 10, 2023 at 3:28:34 PM UTC-4, WM wrote:
> > >
> > > The Peano set has no *fixed* end. [WM]
> > >
> > A Peano set is a set. It does not change. It has no end "fixed" or otherwise.
> >
> > > According to Cantor ℕ fits between 0 and ω. So it ends before ω.
> > >
> > Nope. it has no end.
> >
> ω is a point at the ordinal axis.

So what?

Hint: Any ordinal smaller than ω is _a natural number_ and (hence) has a natural number (i. e. not ω) as successor.

Formally: Ax e ORD: x < ω -> x e IN & x' e IN & x < x' < ω.

So it _does not end befor ω_, since there is no end before ω.

> The sequence (1/n) ends before 0.

No, it doesn't end before 0.

Hint: Since there is no end (minimal term) before 0 it does not end before 0.

> But since there is no gap, it ends close before 0.

Ex falso quodlibet. Saudummer Scheißdreck, wieder mal, Mückenheim!

On Tuesday, January 10, 2023 at 11:43:05 PM UTC+1, WM wrote:
> Jim Burns schrieb am Dienstag, 10. Januar 2023 um 22:17:11 UTC+1:
> > On 1/10/2023 2:28 PM, WM wrote:
> > >
> > > The Peano set has no *fixed* end. [WM]
> > >
> > ℕ has no end at all, fixed or otherwise.
> >
> Try the sequence (1/n). It ends at 0.

No, it does not end at 0.

Since it does not have an end (smallest term) it does not end. Hence it does not end at 0 either.

> Since there is no hole, it ends immediately before 0.

Ex falso quodlibet.

Hint: Your claim is not even wrong, since there is no point on the real number line which is "immediately before 0".

You are dumb like shit, Mückenheim.

> > It is [...] false that an element of ℕ is its end...
> >
> [...] What ends is a field of bunch numbers.

Wo, now new type of numbers?

"bunch numbers" - how are they defined?

Even a _field_ of them?

"a field of bunch numbers" - how is a /field/ (of bunch numbers) defined?

You know: https://en.wikipedia.org/wiki/Field_(mathematics)

> > > According to Cantor ℕ fits between 0 and ω. So it ends before ω.
> > >
> > No.
> >
> It ends not after ω.

Wow! What an argument!

A: Mr. Mückeheim has won the Abel Prize before the year 2000.

B: No!

A: But Mr. Mückeheim has not won the Abel Prize after the year 2000!

B: Well...

> > Yes,
> > ℕ fits between 0 and ω
> > and,
> > if an element of ℕ was its end,
> > then that end would be before ω
> >
> There is no elements its end

Huh? What are you babbling?

> but a bunch of infinitely many indistinguishable elements.

"a bunch of infinitely many indistinguishable elements"? You mean something like a Bose-Einstein condensate? Wow, fascinating!

["Das Bose-Einstein-Kondensat ist ein extremer Aggregatzustand eines Systems ununterscheidbarer Teilchen ..." --Wikipedia]

"indistinguishable elements": "In quantum mechanics, identical particles (also called indistinguishable or indiscernible particles) are particles that cannot be distinguished from one another, even in principle."

See: https://en.wikipedia.org/wiki/Identical_particles

> The potentially infinite collection ℕ <bla>

Nope, you psychotic crank. In the context of the present discussion "IN" denotes the /infinite set/ IN, as defined in the context of set theory.

On Wednesday, January 11, 2023 at 12:48:34 PM UTC+1, WM wrote:
> William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
> > On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
> > >
> > > The sequence (1/n) ends
> > >
> > Nope. It does not have an end.
> >
> It streches to the negative axis?

At least in your psychotic realm it does, it seems.

On Wednesday, January 11, 2023 at 7:48:34 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
> > On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
> >
> > > The sequence (1/n) ends
> >
> > Nope. It does not have an end.
> It streches to the negative axis?

Nope. Every element is greater than 0. However, there is no last element. There is no end.

William schrieb am Mittwoch, 11. Januar 2023 um 15:50:09 UTC+1:
> On Wednesday, January 11, 2023 at 7:48:34 AM UTC-4, WM wrote:
> > William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
> > > On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
> > >
> > > > The sequence (1/n) ends
> > >
> > > Nope. It does not have an end.
> > It streches to the negative axis?
> Nope. Every element is greater than 0. However, there is no last element. There is no end.

The sequence does not stretch to 0, therefore it ends before 0 but there is no last element. How can these two diverging properties be reconciled?

The only way to describe an end without identifying a last element is the assumption of dark elements.

Regards, WM

On 1/11/2023 10:27 AM, WM wrote:
> William schrieb am Mittwoch, 11. Januar 2023 um 15:50:09 UTC+1:
>> On Wednesday, January 11, 2023 at 7:48:34 AM UTC-4, WM wrote:
>>> William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
>>>> On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
>>>>
>>>>> The sequence (1/n) ends
>>>>
>>>> Nope. It does not have an end.
>>> It streches to the negative axis?
>> Nope. Every element is greater than 0. However, there is no last element. There is no end.
>
> The sequence does not stretch to 0, therefore it ends before 0 but there is no last element. How can these two diverging properties be reconciled?

I corrected your statement => "The sequence does not include 0, there is no last element."

these problems were solved in the early 1700's, so try to keep up.

>

> Regards, WM

WM wrote on 1/10/2023 :
> FromTheRafters schrieb am Dienstag, 10. Januar 2023 um 20:53:14 UTC+1:
>> After serious thinking WM wrote :
>>> Every Peano-number has a FISON.
>> Where is the FISON associated with the initial element?
>
> {1}

How can this be if E(1) = |N ?

On Wednesday, January 11, 2023 at 5:35:00 PM UTC+1, FromTheRafters wrote:
> WM wrote on 1/10/2023 :
> > FromTheRafters schrieb am Dienstag, 10. Januar 2023 um 20:53:14 UTC+1:
> > >
> > > Where is the FISON associated with the initial element?

*** For WM ***

IN = {1, 2, 3, ...}

and

F_1 = {1}
F_2 = {1, 2}
etc.

> > {1}
> >
> How can this be if E(1) = IN ?

E_1 = {1, 2, 3, ...}
E_2 = {2, 3, 4, ...}
etc.

And right. Hence NOT E_n = IN \ F_1 (in this context).

Just deal with it. :-P

Not that elegant. On the other hand, not wrong either.

On Wednesday, January 11, 2023 at 5:27:58 PM UTC+1, WM wrote:
> William schrieb am Mittwoch, 11. Januar 2023 um 15:50:09 UTC+1:
> > On Wednesday, January 11, 2023 at 7:48:34 AM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
> > > > On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
> > > > <
> > > > > The sequence (1/n) ends
> > > > >
> > > > Nope. It does not have an end.
> > > >
> > > It streches to the negative axis?
> > >
> > Nope. Every element is greater than 0. However, there is no last element. There is no end.
> >
> The sequence does not stretch to 0, therefore it ends before 0

*** Nope. It does not have an end. ***

Can't you read?

> but there is no last element.

There is no last element - no "but".

> How can these two diverging properties be reconciled?

There are no "two diverging properties", only the false claim "[therefore] it ends before 0" and the correct claim "there is no last element."

So what's your problem, you psychotic asshole?

> The only way to

Go, fuck yourself, crank.

On Wednesday, January 11, 2023 at 12:27:58 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 11. Januar 2023 um 15:50:09 UTC+1:
> > On Wednesday, January 11, 2023 at 7:48:34 AM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 10. Januar 2023 um 23:48:11 UTC+1:
> > > > On Tuesday, January 10, 2023 at 6:36:03 PM UTC-4, WM wrote:
> > > >
> > > > > The sequence (1/n) ends
> > > >
> > > > Nope. It does not have an end.
> > > It streches to the negative axis?
> > Nope. Every element is greater than 0. However, there is no last element. There is no end.
> The sequence does not stretch to 0, therefore it ends before 0

Nope. The fact that there is a bound on the elements does not mean there is a last element. There is no end.

On 1/11/2023 12:11 PM, Fritz Feldhase wrote:
> On Wednesday, January 11, 2023 at 5:35:00 PM UTC+1, FromTheRafters wrote:
>> WM wrote on 1/10/2023 :
>>> FromTheRafters schrieb am Dienstag, 10. Januar 2023 um 20:53:14 UTC+1:
>>>>
>>>> Where is the FISON associated with the initial element?
>
> *** For WM ***
>
> IN = {1, 2, 3, ...}
>
> and
>
> F_1 = {1}
> F_2 = {1, 2}
> etc.
>
>>> {1}
>>>
>> How can this be if E(1) = IN ?
>
> E_1 = {1, 2, 3, ...}
> E_2 = {2, 3, 4, ...}
> etc.
>
> And right. Hence NOT E_n = IN \ F_1 (in this context).
>
> Just deal with it. :-P
>
> Not that elegant. On the other hand, not wrong either.

so, E_1 ∩ E_2 = F_1 = {1}

and E_2 ∩ E_3 = F_2 ∪ F_1 = {2}

and E_3 ∩ E_4 = F_3 ∪ F_4 = {3}

or
E_(k) ∩ E_(k+1) = F_(k) ∪ F_(k+1) = {k}

now we're countin !

(for completness F_0 = {}, so above E_1 ∩ E_2 = F_1 ∪ F_0 , so far no need for E_0)