On Friday, December 30, 2022 at 12:07:17 PM UTC-4, WM wrote:
> William schrieb am Freitag, 30. Dezember 2022 um 16:16:01 UTC+1:
> > On Friday, December 30, 2022 at 4:59:33 AM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 28. Dezember 2022 um 18:23:44 UTC+1:
> > > > On Friday, December 23, 2022 at 5:01:00 AM UTC-4, WM wrote:
> > > > > William schrieb am Donnerstag, 22. Dezember 2022 um 19:17:29 UTC+1:
> > > > > > On Thursday, December 22, 2022 at 5:39:38 AM UTC-4, WM wrote:
> > > > > > > William schrieb am Donnerstag, 22. Dezember 2022 um 02:11:35 UTC+1:
> > > > > > > > On Wednesday, December 21, 2022 at 3:01:14 PM UTC-4, WM wrote:
> > > > > > > >
> > > > > > > > > But all infinite sets which have not a fixed number of members are potentially infinite.
> > > > > > > > Potentially infinite sets are nonsense. All sets are fixed.
> > > > > > > Then call it collection.
> > > > > > Nope, N is a set.
> > > > > Well, may be.
> > > >
> > > >
> > > > No "may be" about it. N is a Peano set.
> > > Peano's axioms derfine a potentially infinite collection.
> > Nope Peano's axioms define a set
> A set in the sense of ZFC must be actually infinite. That means it must have a successor proving that the set is complete.

Nope, it means that the set has no largest element.

> In case of the natural numbers this is ω.

Piffle. ω is not a natural number. Despite this the natural numbers are a complete set (no element is missing or can be added)

William schrieb am Freitag, 30. Dezember 2022 um 17:29:32 UTC+1:
> On Friday, December 30, 2022 at 12:07:17 PM UTC-4, WM wrote:

> > A set in the sense of ZFC must be actually infinite. That means it must have a successor proving that the set is complete.
> Nope, it means that the set has no largest element.

That is not suitable to distinguish potentail and actual infinity.

> > In case of the natural numbers this is ω.
> Piffle. ω is not a natural number.

No, but it limits them.

> Despite this the natural numbers are a complete set (no element is missing or can be added)

The definable numbers are not completed. Definable numbers can be written down. Can you write down them such that none is missing? Could you write down them with infinite ressources? No! Always ℵo would be missing.

Regards, WM

On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> William schrieb am Freitag, 30. Dezember 2022 um 17:29:32 UTC+1:
> > On Friday, December 30, 2022 at 12:07:17 PM UTC-4, WM wrote:
>
> > > A set in the sense of ZFC must be actually infinite. That means it must have a successor proving that the set is complete.
> > Nope, it means that the set has no largest element.
> That is not suitable to distinguish potentail and actual infinity.

Nope. A potentially infinite collection has a largest element but it is something that changes. An "actually" infinite set does not change and does not have a largest element.

On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> Could you write down them with infinite ressources? No! Always ℵo would be missing.

So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.

On Friday, 30 December 2022 at 14:01:59 UTC-4, William wrote:
> An "actually" infinite set does not change and does not [*necessarily*] have a largest element.

The set {..., -3, -2, -1, 0} has --- in the usual order --- a largest element (but not a smallest).

On Friday, December 30, 2022 at 8:11:34 PM UTC+1, Gus Gassmann wrote:
> On Friday, 30 December 2022 at 14:01:59 UTC-4, William wrote:
> >
> > An "actually" infinite set does not change and does not [*necessarily*] have a largest element.
> >
> The set {..., -3, -2, -1, 0} has --- in the usual order --- a largest element (but not a smallest).

We may also consider the intervall (a set) [0, 1] = {x e IR : 0 <= x <= 1} which has a smallest and a largest element (but is "actually" infinite). :-P

C'mon, man, you know that William meant! :-)

[ William is no Mückenheim. ]

William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.

And the reason is: Most elements of ℕ are dark. Otherwise you could write down all of them because up to every writable element it is only a FISON.

Regards, WM

On Saturday, 31 December 2022 at 10:16:18 UTC+2, WM wrote:
> William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> And the reason is: Most elements of ℕ are dark.

Dark in the sense that they "cannot be known" even if one could live forever and do nothing else but write them down.

> Otherwise you could write down all of them because up to every writable element it is only a FISON.
>
> Regards, WM

Eram semper recta schrieb am Samstag, 31. Dezember 2022 um 10:24:57 UTC+1:
> On Saturday, 31 December 2022 at 10:16:18 UTC+2, WM wrote:
> > William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> > And the reason is: Most elements of ℕ are dark.
> Dark in the sense that they "cannot be known" even if one could live forever and do nothing else but write them down.

Yes. Because for every known number we have:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

And you need not write them all. Start at any desired number and leave out as many as you like. The dark realm will never shrink below ℵo (if we assume actual infinity).

Regards, WM

On Friday, 30 December 2022 at 15:58:09 UTC-4, Fritz Feldhase wrote:
> On Friday, December 30, 2022 at 8:11:34 PM UTC+1, Gus Gassmann wrote:
> > On Friday, 30 December 2022 at 14:01:59 UTC-4, William wrote:
> > >
> > > An "actually" infinite set does not change and does not [*necessarily*] have a largest element.
> > >
> > The set {..., -3, -2, -1, 0} has --- in the usual order --- a largest element (but not a smallest).
> We may also consider the intervall (a set) [0, 1] = {x e IR : 0 <= x <= 1} which has a smallest and a largest element (but is "actually" infinite). :-P
>
> C'mon, man, you know that William meant! :-)
>
> [ William is no Mückenheim. ]

Of course not. But every inaccuracy and mistake someone makes has the potential for WM to discredit the author. One should strive for correctness. You correct me, too. and I am no Mückenheim, either. (Thank god!)

On 12/31/2022 2:16 AM, WM wrote:
> William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
>> On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
>>> Could you write down them with infinite ressources? No! Always ℵo would be missing.
>> So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
>
> And the reason is: bla bla...

>
> Regards, WM

On Saturday, December 31, 2022 at 4:16:18 AM UTC-4, WM wrote:
> William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> And the reason is: Most elements of ℕ are dark

Nope. |N is a Peano se. A Peano set cannot have a dark element.

On 12/31/2022 1:24 AM, Eram semper recta wrote:
> On Saturday, 31 December 2022 at 10:16:18 UTC+2, WM wrote:
>> William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
>>> On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
>>>> Could you write down them with infinite ressources? No! Always ℵo would be missing.
>>> So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
>> And the reason is: Most elements of ℕ are dark.
>
> Dark in the sense that they "cannot be known" even if one could live forever and do nothing else but write them down.

[...]

Dark. So, is n + 1 dark? I guess so, because I have not defined what n
is: Is that your line of thinking? So, if I said, n = 41, then 42 is
implied right? What, what about 43 being _automatically_ implied if n =
42? See the infinite chain? Notice the pattern? All of the naturals are
automatically implied, right? And the use of the word, All, here does
not mean finite in any way!

https://youtu.be/HEYrRNMr2kg

;^)

On 12/20/2022 4:17 AM, WM wrote:
> FromTheRafters schrieb am Dienstag, 20. Dezember 2022 um 10:16:30 UTC+1:
>
>> For every guest there is a next guest, for every room there is a next
>> room.
>
> And none is without a guest.

| (እያንዳንዱ ልውውጥ ተከስቷል) ⇒ ¬(ቦብ ከሴሎች በአንዱ ውስጥ አለ።)

>
> Regards, WM

do the guests have toilets ?

torsdag 22 december 2022 kl. 11:19:41 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 22. Dezember 2022 um 10:49:48 UTC+1:
>
> > no one is saying any is added
> The set of guests = natural numbers ℕ is complete, i.e., none can be added.
> The set ℕ of room numbers of Hilbert's hotel is complete too.
> The bijection between guests and room numbers has |ℕ| pairs.
> If another guest, 0, arrives and is accomodated, then the bijection has one more pair. The natural room numbers contained therein have one more element. Contradiction.
>
> Regards, WM

Saying "one more" does not work in infinite cardinal arithmetic you idiot!

On Thursday, December 22, 2022 at 11:19:41 AM UTC+1, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 22. Dezember 2022 um 10:49:48 UTC+1:

> The bijection between guests [IN] and room numbers [IN] has |ℕ| pairs.

This bijection is

f := {(n, n) : n e IN} = {(1, 1), (2, 2), (3, 3), ...} - a function from IN onto IN.

[ This function represents the occupancy of the hotel before the arrival and accommodation of the new guest: guest 1 occupies the room with nr. 1, guest 2 occupies the room with nr. 2, guest 3 occupies the room with nr. 3, etc. ]

> If another guest, 0, arrives and is accomodated, then the bijection

It is NOT THE SAME bijection, you silly idiot!

The NEW bijection is

g := {(n, n + 1) : n e IN u IN u {0}} = {(0, 1), (1, 2), (2, 3), ...} - a function from IN u {0} onto IN.

[ This function represents the occupancy of the hotel after the arrival and accommodation of the new guest: guest 0 occupies the room with nr. 1, guest 1 occupies the room with nr. 2, guest 2 occupies the room with nr. 3, etc.. ]

> has one more pair.

Actually, the pairs in g are completely different from the pairs in f: f and g are _disjoint_ sets.

Moreover card(f) = card(g) = aleph_0.

Hence "one more" (and/or "one less") is not applicable in this case.

> Contradiction.

Nonsense.

Have You tried to solve Hanoi towers ?
It is in my opinion the most beautiful mathematical problem.

WM kirjutas Laupäev, 31. detsember 2022 kl 11:33:42 UTC+2:
> Eram semper recta schrieb am Samstag, 31. Dezember 2022 um 10:24:57 UTC+1:
> > On Saturday, 31 December 2022 at 10:16:18 UTC+2, WM wrote:
> > > William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > > > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > > > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> > > And the reason is: Most elements of ℕ are dark.
> > Dark in the sense that they "cannot be known" even if one could live forever and do nothing else but write them down.
> Yes. Because for every known number we have:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> And you need not write them all. Start at any desired number and leave out as many as you like. The dark realm will never shrink below ℵo (if we assume actual infinity).
>
> Regards, WM

On Monday, 19 December 2022 at 15:45:49 UTC+1, Julio Di Egidio wrote:
> On Monday, 19 December 2022 at 11:04:58 UTC+1, WM wrote:
>
> > If the complete set ℕ of natural numbers exists, then it can be used to
> > populate Hilbert's Hotel. Is it possible to add a natural number to the set ℕ?
>
> Good question and the valid and correct answer indeed is NO!
>
> And here is an analysis:
> <http://seprogrammo.blogspot.com/2014/02/hilberts-impossible-hotel.html>
>
> HTH, somebody, somewhere...

Up. Spamming pieces of shit.

Julio

William schrieb am Samstag, 31. Dezember 2022 um 16:32:12 UTC+1:
> On Saturday, December 31, 2022 at 4:16:18 AM UTC-4, WM wrote:
> > William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> > And the reason is: Most elements of ℕ are dark
> Nope. |N is a Peano set.

ℕ_def with ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo is a Peano set.

> A Peano set cannot have a dark element.

Correct. But Cantor's set ℕ has.

Regards, WM

zelos...@gmail.com schrieb am Sonntag, 1. Januar 2023 um 09:02:16 UTC+1:
> torsdag 22 december 2022 kl. 11:19:41 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 22. Dezember 2022 um 10:49:48 UTC+1:
> >
> > > no one is saying any is added
> > The set of guests = natural numbers ℕ is complete, i.e., none can be added.
> > The set ℕ of room numbers of Hilbert's hotel is complete too.
> > The bijection between guests and room numbers has |ℕ| pairs.
> > If another guest, 0, arrives and is accomodated, then the bijection has one more pair. The natural room numbers contained therein have one more element. Contradiction.
> >
> > Regards, WM
>
> Saying "one more" does not work in infinite cardinal arithmetic

But it works in sets. That's why cardinal arithmetic is nonsense.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 1. Januar 2023 um 10:09:45 UTC+1:
> On Thursday, December 22, 2022 at 11:19:41 AM UTC+1, WM wrote:

> > If another guest, 0, arrives and is accomodated, then the bijection
> It is NOT THE SAME bijection

irrelevant. There is one more element. And the bijection.
>
> > has one more pair.
>
> Actually, the pairs in g are completely different from the pairs in f: f and g are _disjoint_ sets.

There is one more pair.
>
> Moreover card(f) = card(g) = aleph_0.

card is utter nonsense.
>
> Hence "one more" (and/or "one less") is not applicable in this case.

It is applicable in set theory. {0, 1, 2, 3, ...} has one element more than {1, 2, 3, ...}
The latter is a proper subset of the former.
>
> > Contradiction.
>
The set of pairs has one more element - but only on one side. Clash.

Regards, WM

On 1/1/2023 3:40 PM, WM wrote:
> William schrieb am Samstag, 31. Dezember 2022 um 16:32:12 UTC+1:
>> On Saturday, December 31, 2022 at 4:16:18 AM UTC-4, WM wrote:
>>> William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
>>>> On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
>>>>> Could you write down them with infinite ressources? No! Always ℵo would be missing.
>>>> So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
>>> And the reason is: Most elements of ℕ are dark
>> Nope. |N is a Peano set.
>
> ℕ_def with ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo is a Peano set.
>
>> A Peano set cannot have a dark element.
>
> Correct. But Cantor's set ℕ has.
>
> Regards, WM
>

wrong, there are no darkies in ℕ.

On Sunday, January 1, 2023 at 5:40:09 PM UTC-4, WM wrote:

> ℕ_def

is not a set. It is something that changes.

> with ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

You do not have to restrict things to |N_def, Replace |N_def with |N. The statement is still true

∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo

Every element of |N is finite. |N is infinite

söndag 1 januari 2023 kl. 22:40:09 UTC+1 skrev WM:
> William schrieb am Samstag, 31. Dezember 2022 um 16:32:12 UTC+1:
> > On Saturday, December 31, 2022 at 4:16:18 AM UTC-4, WM wrote:
> > > William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > > > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > > > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> > > And the reason is: Most elements of ℕ are dark
> > Nope. |N is a Peano set.
>
> ℕ_def with ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo is a Peano set.
> > A Peano set cannot have a dark element.
> Correct. But Cantor's set ℕ has.
>
> Regards, WM
nope, it has no "dark element", it is a made up thing you cannot define

On Saturday, 31 December 2022 at 11:33:42 UTC+2, WM wrote:
> Eram semper recta schrieb am Samstag, 31. Dezember 2022 um 10:24:57 UTC+1:
> > On Saturday, 31 December 2022 at 10:16:18 UTC+2, WM wrote:
> > > William schrieb am Freitag, 30. Dezember 2022 um 19:10:27 UTC+1:
> > > > On Friday, December 30, 2022 at 1:05:52 PM UTC-4, WM wrote:
> > > > > Could you write down them with infinite ressources? No! Always ℵo would be missing.
> > > > So what? You can write down M iff M is finite. Every element of |N is finite. |N is infinite.
> > > And the reason is: Most elements of ℕ are dark.
> > Dark in the sense that they "cannot be known" even if one could live forever and do nothing else but write them down.
> Yes. Because for every known number we have:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> And you need not write them all. Start at any desired number and leave out as many as you like. The dark realm will never shrink below ℵo (if we assume actual infinity).

Of course. The point is not about actually writing them all out, but rather that by so doing you actually identify them by NAME (A number is a name given to a measure of a ratio of magnitudes).

Simply put, no matter how large is the element of N that you identify, there are still (in Cantor world) ℵo elements that remain unidentified and without any possibility of ever being identified. Call them "dark" if you wish, but a better phrase is "cannot be known", that is, ℵo elements cannot be known.

> (if we assume actual infinity)

Cantor's idiots always assume infinity, whatever the fuck "infinity" means because it does not exist outside of a dysfunctional brain.

>
> Regards, WM