FromTheRafters schrieb am Montag, 18. Oktober 2021 um 20:36:23 UTC+2:

> A sign outside the Hilbert says "No Double Occupancy" so you're out of
> luck there.

No, that is just my simple point. The hotel like every infinite set is a fixed quantity:

"In spite of significant difference between the notions of the potential and actual infinite, where the former is a variable finite magnitude, growing above all limits, the latter a constant quantity fixed in itself but beyond all finite magnitudes, it happens deplorably often that the one is confused with the other." [Cantor, p. 374]

"By the actual infinite we have to understand a quantity that on the one hand is not variable but fixed and definite in all its parts, a real constant, but at the same time, on the other hand, exceeds every finite size of the same kind by size." [G. Cantor, letter to A. Eulenburg (28 Feb 1886)]

Is it bad take thim at his word? Then Hilbert's hotel is nonsense. There are no rooms to let.

Regards, WM

Greg Cunt schrieb am Montag, 18. Oktober 2021 um 20:49:51 UTC+2:

> you explicitly claimed: "n isn't a natural number" and "n+1 is not a natural number too."
>
> Hence the claim
>
> "There are natnumbers in {n, n+1}"
>
> implies a contradiction.

Thank you for your attention. But I told you already that this phrase "the natural number n has property X" is merely an abbreviation for "every natural number which can be inserted in place of n has property X". For instance if you insert 4711 in place of n in E(n), then you must also insert 4711 in place of n in {n, n+1, n+2, ...}
>
Regards, WM

On Monday, 18 October 2021 at 15:58:15 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 19:01:36 UTC+2:
>
> > There you go again with your idiocies. The pigeonhole principle does *NOT* apply to infinite sets.
> It applies to all FISONs which shall be distinguished, because otherwise it is not possible to distinguish them. Only by different number of natnumbers or symbols like
> o
> oo
> ooo
> oooo
> ...
> it is possible to distinguish FISONs. If it does not apply, then there is no distinction possible. But even in infinite sets it is necessary to distinguish elements if any useful result shall be obtained. What fool has taught you, without considering this simple fact, that Dirichlet's principle does not apply?

Look. Are you talking about the pigeonhole principle now or Dirichlet's principle? Are you too far gone to understand what you are babbling about? Or do you think by changing the subject you can cover your tracks? Maybe you should start by actually giving *your* description of the pigeonhole principle, if you are even still able to do that.

On Monday, October 18, 2021 at 12:16:18 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Sonntag, 17. Oktober 2021 um 22:43:25 UTC+2:
>
> > E(1) = {1, 2, 3, 4, ... }
> > E(2) = {2, 3, 4, 5, ... }
> > E(3) = {3, 4, 5, 6, ... }
> > ...
> > E(n) = {n, n+1, n+2, ... }
> > ...
> >
> > Contrary to your outrageous claim here, each of these infinitely many end-segments has in it infinitely many elements.

> There are only potentially infinitely many endsegments.

There are infinitely many end-seements. There's nothing "potential" about it, Mucke.

> If the set of infinite endsegments had the cardinality aleph_0, then all natnumbers were used as indices and none would remain as contents.
>

Same gibberish again, Mucke. When will you learn?

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

On 10/18/2021 11:38 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 18:26:38 UTC+2:
>> On Monday, 18 October 2021 at 12:59:36 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 17:41:08 UTC+2:
>>>
>>>> If you cannot see the difference between "for every n in find the set difference N \ {1, 2, 3, ..., n}" and "find the set difference N \ *ALL* sets {1, 2, 3, ..., n}" then you are truly beyond hope.
>>> I see the difference.
>> You cannot fathom that there can be infinitely many finite objects
>
> Why should I accept nonsense? I can prove that there are not infinitely many FISONs

nope, you can't.

just index the natural numbers, k, to FISON(k)

How many natural numbers are there ?

> because of the simple pigeon hole principle which holds for all sets the elements of which must be distinguished - irrespective of whether the sets are finite ot infinite.
>
>> In particular, your insistence that N =/= union (n in N: {n}) is breathtaking in its stupidity.
>
> I did not say so. So the stupidity is not mine. Of course ℕ = {1, 2, 3, ...} is the union of the singletons of natural numbers. But most of them are dark.

no, none of them are dark.

>Therefore ℕ is not the union of FISONs.

Which FISONs ?

>> On the contrary, for all FISONs we have ℕ \ {1, 2, 3, ..., n} is an infinite endsegment.

that and a doughnut, gets you a doughnut hole.

>
> Regards, WM
>

On 10/18/2021 2:09 PM, WM wrote:
> Greg Cunt schrieb am Montag, 18. Oktober 2021 um 20:49:51 UTC+2:
>
>> you explicitly claimed: "n isn't a natural number" and "n+1 is not a natural number too."
>>
>> Hence the claim
>>
>> "There are natnumbers in {n, n+1}"
>>
>> implies a contradiction.
>
> Thank you for your attention. But I told you already that this phrase "the natural number n has property X" is merely an abbreviation for "every natural number which can be inserted in place of n has property X". For instance if you insert 4711 in place of n in E(n), then you must also insert 4711 in place of n in {n, n+1, n+2, ...}
>>
> Regards, WM
>

a few baby steps toward "variable" (?)

Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 22:59:09 UTC+2:
> On Monday, 18 October 2021 at 15:58:15 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 19:01:36 UTC+2:
> >
> > > There you go again with your idiocies. The pigeonhole principle does *NOT* apply to infinite sets.
> > It applies to all FISONs which shall be distinguished, because otherwise it is not possible to distinguish them. Only by different number of natnumbers or symbols like
> > o
> > oo
> > ooo
> > oooo
> > ...
> > it is possible to distinguish FISONs. If it does not apply, then there is no distinction possible. But even in infinite sets it is necessary to distinguish elements if any useful result shall be obtained. What fool has taught you, without considering this simple fact, that Dirichlet's principle does not apply?
> Are you talking about the pigeonhole principle now or Dirichlet's principle?

Both is the same. Look at Dirichlet 1834.

> Maybe you should start by actually giving *your* description of the pigeonhole principle, if you are even still able to do that.

Please look at the figure above. There are FISONs denoted by finite strings of o's. If there are more FISONs than finite strings of o'2, then at least two of them have the same string, They cannot be distinguished. They are not different. There are not more FISONs than finitely many.

Regards, WM

Dan Christensen schrieb am Montag, 18. Oktober 2021 um 23:03:37 UTC+2:
> On Monday, October 18, 2021 at 12:16:18 PM UTC-4, WM wrote:
> > Dan Christensen schrieb am Sonntag, 17. Oktober 2021 um 22:43:25 UTC+2:
> >
> > > E(1) = {1, 2, 3, 4, ... }
> > > E(2) = {2, 3, 4, 5, ... }
> > > E(3) = {3, 4, 5, 6, ... }
> > > ...
> > > E(n) = {n, n+1, n+2, ... }
> > > ...
> > >
> > > Contrary to your outrageous claim here, each of these infinitely many end-segments has in it infinitely many elements.
>
> > There are only potentially infinitely many endsegments.
> There are infinitely many end-seements. There's nothing "potential" about it, Prof. Dr. Mueckenheim.
> > If the set of infinite endsegments had the cardinality aleph_0, then all natnumbers were used as indices and none would remain as contents.
> >
> Same gibberish

Sorry for inconvenience. But do you really believe that an infinite sequence of indices 1, 2, 3, ... of endsegments can be followed by an infinite sequence of members of endsegments?

Regards, WM

måndag 18 oktober 2021 kl. 11:23:05 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 18. Oktober 2021 um 08:28:25 UTC+2:
>
> > ALL natural numbers have aleph_0 successors
> Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .
>
> Regards, WM

That is not a refutation of it you fucking imbecile. It has 0 to do with it..

How are you this fucking retarded?

In the real number number set (0,1) there are infinitely many numbers after every real number, but |(0,1)\(0,1)|=0

So your shit proves nothing!

måndag 18 oktober 2021 kl. 11:31:25 UTC+2 skrev WM:
> Gus Gassmann schrieb am Sonntag, 17. Oktober 2021 um 22:18:16 UTC+2:
> > On Sunday, 17 October 2021 at 15:51:47 UTC-3, WM wrote:
> > > This simple fact cannot be remedied by quantifier tricks. Infinite endsegments have an infinite intersection and there are not aleph_0 infinite endsegments. There are not aleph_0 natural numbers having aleph_0 successors.
> > Of course there are! *EVERY* one of the aleph_0 natural numbers has aleph_0 successors.
> That would require two consecutive sets in omega. Impossible.
> > > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> > Finite intersections of end segments (as you also wrote!) have cardinality aleph_0.
> Yes.
> > To extend that to infinite intersections you are switching quantifiers.
> No, I simply apply mathematics. It says truly
> |ℕ \ {1, 2, 3, ...}| = 0 .
>
> Regards, WM

>That would require two consecutive sets in omega. Impossible.

It requires no suck thing you imbecile. This is an empty assertion.

>No, I simply apply mathematics. It says truly

No, you aren't! You are abandoning logic and making shit up!

zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:18:46 UTC+2:
> måndag 18 oktober 2021 kl. 11:23:05 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 18. Oktober 2021 um 08:28:25 UTC+2:
> >
> > > ALL natural numbers have aleph_0 successors
> > Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .
>
> That is not a refutation of it

It is obviosuly a refutation. And here is another one:

The axiom of choice is equivalent to Zorn's lemma: If a partially ordered set S has the property that every chain has an upper bound in S, then the set S contains at least one maximal element. Without this feature, every element would have another next element. This successorship would not stop at natural indices but would run through all ordinal numbers.

How could and why should it enter larger than natural numbers if there was not a last natural one?

> In the real number number set (0,1) there are infinitely many numbers after every real number, but |(0,1)\(0,1)|=0

Sorry that is as wrong and cannot serve as an argument. There are infinitely many real numbers beyond every definable one. But there is a point where the cursor encounters zero. Why should it, if always infinitely many points would lie between the cursor and zero? Can you imagine a reason?

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:19:52 UTC+2:
> måndag 18 oktober 2021 kl. 11:31:25 UTC+2 skrev WM:

> >That would require two consecutive sets in omega. Impossible.
>
> It requires no suck thing

If there is an infinite initial segment in |N which is followed by an infinite endsegment, then we have two consecutive infinite sets in order omega.

> You are abandoning logic and making shit up!

Logic says that an infinite sequence followed by an infinite sequence with no or at most finite overlap yields two infinite sequences. What do you think it says?

Regards, WM

tisdag 19 oktober 2021 kl. 07:53:34 UTC+2 skrev WM:
> Dan Christensen schrieb am Montag, 18. Oktober 2021 um 23:03:37 UTC+2:
> > On Monday, October 18, 2021 at 12:16:18 PM UTC-4, WM wrote:
> > > Dan Christensen schrieb am Sonntag, 17. Oktober 2021 um 22:43:25 UTC+2:
> > >
> > > > E(1) = {1, 2, 3, 4, ... }
> > > > E(2) = {2, 3, 4, 5, ... }
> > > > E(3) = {3, 4, 5, 6, ... }
> > > > ...
> > > > E(n) = {n, n+1, n+2, ... }
> > > > ...
> > > >
> > > > Contrary to your outrageous claim here, each of these infinitely many end-segments has in it infinitely many elements.
> >
> > > There are only potentially infinitely many endsegments.
> > There are infinitely many end-seements. There's nothing "potential" about it, Prof. Dr. Mueckenheim.
> > > If the set of infinite endsegments had the cardinality aleph_0, then all natnumbers were used as indices and none would remain as contents.
> > >
> > Same gibberish
> Sorry for inconvenience. But do you really believe that an infinite sequence of indices 1, 2, 3, ... of endsegments can be followed by an infinite sequence of members of endsegments?
>
> Regards, WM
therei s no follow because that is ALL of them you imbecile!

tisdag 19 oktober 2021 kl. 11:32:12 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:18:46 UTC+2:
> > måndag 18 oktober 2021 kl. 11:23:05 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 18. Oktober 2021 um 08:28:25 UTC+2:
> > >
> > > > ALL natural numbers have aleph_0 successors
> > > Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .
> >
> > That is not a refutation of it
> It is obviosuly a refutation. And here is another one:
>
> The axiom of choice is equivalent to Zorn's lemma: If a partially ordered set S has the property that every chain has an upper bound in S, then the set S contains at least one maximal element. Without this feature, every element would have another next element. This successorship would not stop at natural indices but would run through all ordinal numbers.
>
> How could and why should it enter larger than natural numbers if there was not a last natural one?
> > In the real number number set (0,1) there are infinitely many numbers after every real number, but |(0,1)\(0,1)|=0
> Sorry that is as wrong and cannot serve as an argument. There are infinitely many real numbers beyond every definable one. But there is a point where the cursor encounters zero. Why should it, if always infinitely many points would lie between the cursor and zero? Can you imagine a reason?
>
> Regards, WM

>It is obviosuly a refutation.

It isn't, only a complete idiot thinks it is!

The fact that a A\A={} does not prove anything of what you fucking want!

>The axiom of choice is equivalent to Zorn's lemma: I

It is but this won't help you.

>f a partially ordered set S has the property that every chain has an upper bound in S, then the set S contains at least one maximal element. Without this feature, every element would have another next element. This successorship would not stop at natural indices but would run through all ordinal numbers.

And it does not. It doesn't apply to natural numbers as a whole because not every chain has it. So Zorns lemma is not applicable to natural numbers.

>How could and why should it enter larger than natural numbers if there was not a last natural one?

You are now taking in ordinal numbers which are different things together and no, ordinal numbers do not need there to be a last one by their definition. This is a result of you thinking in finites and trying to go to infinite.

If you read up on how ordinal numbers are constructed you can and will see there is no need for a last.

tisdag 19 oktober 2021 kl. 11:36:43 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:19:52 UTC+2:
> > måndag 18 oktober 2021 kl. 11:31:25 UTC+2 skrev WM:
>
> > >That would require two consecutive sets in omega. Impossible.
> >
> > It requires no suck thing
> If there is an infinite initial segment in |N which is followed by an infinite endsegment, then we have two consecutive infinite sets in order omega..
> > You are abandoning logic and making shit up!
> Logic says that an infinite sequence followed by an infinite sequence with no or at most finite overlap yields two infinite sequences. What do you think it says?
>
> Regards, WM

>If there is an infinite initial segment in |N which is followed by an infinite endsegment,

There are no such things so your point is moot.

>Logic says that an infinite sequence followed by an infinite sequence with no or at most finite overlap yields two infinite sequences. What do you think it says?

There is nothing in mathematics that says that is so in natural numbers. You are also conflating sequence with ordinal. Sequences are constructed using cardinals and not ordinals so you are once again dishonest.

On Tuesday, 19 October 2021 at 02:49:44 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 22:59:09 UTC+2:
> > On Monday, 18 October 2021 at 15:58:15 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 19:01:36 UTC+2:
> > >
> > > > There you go again with your idiocies. The pigeonhole principle does *NOT* apply to infinite sets.
> > > It applies to all FISONs which shall be distinguished, because otherwise it is not possible to distinguish them. Only by different number of natnumbers or symbols like
> > > o
> > > oo
> > > ooo
> > > oooo
> > > ...
> > > it is possible to distinguish FISONs. If it does not apply, then there is no distinction possible. But even in infinite sets it is necessary to distinguish elements if any useful result shall be obtained. What fool has taught you, without considering this simple fact, that Dirichlet's principle does not apply?
> > Are you talking about the pigeonhole principle now or Dirichlet's principle?
> Both is the same. Look at Dirichlet 1834.

Well, when I look for "Dirichlet's principle", I find something like

"In mathematics, and particularly in potential theory, Dirichlet's principle is the assumption that the minimizer of a certain energy functional is a solution to Poisson's equation.

Not to be confused with Pigeonhole principle."

Hence my question.

> > Maybe you should start by actually giving *your* description of the pigeonhole principle, if you are even still able to do that.
> Please look at the figure above. There are FISONs denoted by finite strings of o's. If there are more FISONs than finite strings of o'2, then at least two of them have the same string,

This, of course, is your usual bullshit. No two strings have the same length, every single one is finite, and yet there are aleph_0 of them. In fact, to every such string there exist aleph_0 of them that have greater length. The collections of those are the endsegments. You have no clue, never have, and never will.

On Monday, October 18, 2021 at 9:09:58 PM UTC+2, WM wrote:

> Thank you for <bla>

Look, you silly crank, you explicitly claimed: "n isn't a natural number" and "n+1 is not a natural number too."

Hence the claim

"There are natnumbers in {n, n+1}"

implies a contradiction.

Hint: To express that for "every natural number ..." the _universal quantifier_ is used in mathematics.

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

On Monday, October 18, 2021 at 11:23:05 AM UTC+2, WM wrote:

> Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .

Look, you psychotic asshole, since "{1, 2, 3, ...}" denotes the very same set as "ℕ", you just stated that

|ℕ \ ℕ| = 0 ,

which is A TRIVIAL theorem, since for each and ever set A

A \ A = {}

and

|{}| = 0 .

Is this the refutation of the claim that you have a clue?

On Tuesday, October 19, 2021 at 1:53:34 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Montag, 18. Oktober 2021 um 23:03:37 UTC+2:
> > On Monday, October 18, 2021 at 12:16:18 PM UTC-4, WM wrote:
> > > Dan Christensen schrieb am Sonntag, 17. Oktober 2021 um 22:43:25 UTC+2:
> > >
> > > > E(1) = {1, 2, 3, 4, ... }
> > > > E(2) = {2, 3, 4, 5, ... }
> > > > E(3) = {3, 4, 5, 6, ... }
> > > > ...
> > > > E(n) = {n, n+1, n+2, ... }
> > > > ...
> > > >
> > > > Contrary to your outrageous claim here, each of these infinitely many end-segments has in it infinitely many elements.
> >

> > > There are only potentially infinitely many endsegments.

> > There are infinitely many end-segments. There's nothing "potential" about it, Prof. Dr. Mueckenheim.

> > > If the set of infinite endsegments had the cardinality aleph_0, then all natnumbers were used as indices and none would remain as contents.
> > >
> > Same gibberish
> Sorry for inconvenience. But do you really believe that an infinite sequence of indices 1, 2, 3, ... of endsegments can be followed by an infinite sequence of members of endsegments?
>

It is a fact that there are infinitely many end-segments {n, n+1, n+2, ... } in N. And each one of them is infinite. Clearly your goofy little system is broken if you have used it to convince yourself otherwise. It's back to the drawing board for you, Mucke! HA, HA, HA!!!

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Dan Christensen schrieb am Dienstag, 19. Oktober 2021 um 18:14:57 UTC+2:
> On Tuesday, October 19, 2021 at 1:53:34 AM UTC-4, WM wrote:

> > Sorry for inconvenience. But do you really believe that an infinite sequence of indices 1, 2, 3, ... of endsegments can be followed by an infinite sequence of members of endsegments?
> >
> It is a fact that there are infinitely many end-segments {n, n+1, n+2, ... } in N.

It is fact that this is impossible.

> And each one of them is infinite.

Try mathematics. Infinitely many endsegments exhaust all natnumbers 1, 2, 3, ... as indexes. Or could there be a smaller infinity? No. What remains a contents?

Regards, WM

Greg Cunt schrieb am Dienstag, 19. Oktober 2021 um 15:41:24 UTC+2:
> On Monday, October 18, 2021 at 11:23:05 AM UTC+2, WM wrote:
>
> >> ALL natural numbers have aleph_0 successors

> > Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .

> since "{1, 2, 3, ...}" denotes the very same set as "ℕ", you just stated that
>
> |ℕ \ ℕ| = 0 ,
>
> which is A TRIVIAL theorem,

Of course. It is trivial that there is nothing between ℕ, i.e., all natural numbers, and ω.
>
> Is this the refutation of the claim

Of course. It shows that nothing is between all natural numbers and ω. Not all natural numbers have ℵ₀ successors. But all definable numbers have ℵ₀ successors between themselves and omega:

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

Regards, WM

Greg Cunt schrieb am Dienstag, 19. Oktober 2021 um 15:36:28 UTC+2:

> you explicitly claimed: "n isn't a natural number" and "n+1 is not a natural number too."
> Hence the claim
>
> "There are natnumbers in {n, n+1}"
>
> implies a contradiction.

Thank you for your attention. But I told you already that this phrase "the natural number n has property X" is merely an abbreviation for "every natural number which can be inserted in place of n has property X". For instance if you insert 4711 in place of n in E(n), then you must also insert 4711 in place of n in {n, n+1, n+2, ...}
>
Regards, WM

Gus Gassmann schrieb am Dienstag, 19. Oktober 2021 um 12:24:25 UTC+2:
> On Tuesday, 19 October 2021 at 02:49:44 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 22:59:09 UTC+2:
> > > On Monday, 18 October 2021 at 15:58:15 UTC-3, WM wrote:
> > > > Gus Gassmann schrieb am Montag, 18. Oktober 2021 um 19:01:36 UTC+2:
> > > >
> > > > > There you go again with your idiocies. The pigeonhole principle does *NOT* apply to infinite sets.
> > > > It applies to all FISONs which shall be distinguished, because otherwise it is not possible to distinguish them. Only by different number of natnumbers or symbols like
> > > > o
> > > > oo
> > > > ooo
> > > > oooo
> > > > ...
> > > > it is possible to distinguish FISONs. If it does not apply, then there is no distinction possible. But even in infinite sets it is necessary to distinguish elements if any useful result shall be obtained. What fool has taught you, without considering this simple fact, that Dirichlet's principle does not apply?
> > > Are you talking about the pigeonhole principle now or Dirichlet's principle?
> > Both is the same. Look at Dirichlet 1834.
> Well, when I look for "Dirichlet's principle", I find something like
>
> "In mathematics, and particularly in potential theory, Dirichlet's principle is the assumption that the minimizer of a certain energy functional is a solution to Poisson's equation.
>
> Not to be confused with Pigeonhole principle."

Although the pigeonhole principle appears as early as 1624 in a book attributed to Jean Leurechon,[2] it is commonly called Dirichlet's box principle or Dirichlet's drawer principle after an 1834 treatment of the principle (Wikipedia)
>
> Hence my question.
> > > Maybe you should start by actually giving *your* description of the pigeonhole principle, if you are even still able to do that.

Look into Wikipedia: Pigeonhole principle.

> > Please look at the figure above. There are FISONs denoted by finite strings of o's. If there are more FISONs than finite strings of o'2, then at least two of them have the same string,
> This, of course, is your usual bullshit.

No, it is simply true.

> No two strings have the same length, every single one is finite,

Right.

> and yet there are aleph_0 of them.

Nonsense.

> In fact, to every such string there exist aleph_0 of them that have greater length.

No, there are far less than aleph_0. Simply adhere to mathematics:
For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
If you don't trust this simple and clear theorem, then I cannot help you.
Similarly
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
But ℕ \ {1, 2, 3, ...} = { }.
The last line shows that not all natnumbers have an infinite distance from ω.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:47:24 UTC+2:
> tisdag 19 oktober 2021 kl. 11:32:12 UTC+2 skrev WM:
> >
> > Sorry that is as wrong and cannot serve as an argument. There are infinitely many real numbers beyond every definable one. But there is a point where the cursor encounters zero. Why should it, if always infinitely many points would lie between the cursor and zero? Can you imagine a reason?

> >The axiom of choice is equivalent to Zorn's lemma: I
> It is but this won't help you.
> >If a partially ordered set S has the property that every chain has an upper bound in S, then the set S contains at least one maximal element. Without this feature, every element would have another next element. This successorship would not stop at natural indices but would run through all ordinal numbers.
> And it does not. It doesn't apply to natural numbers as a whole because not every chain has it.

But many chains have it. Note: This successorship would not stop at natural indices but would run through all ordinal numbers. That means there is a direct transition from natnumbers to omega and larger.

> So Zorns lemma is not applicable to natural numbers.

Wrong. Note: This successorship would not stop at natural indices but would run through all ordinal numbers.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:48:44 UTC+2:
> tisdag 19 oktober 2021 kl. 11:36:43 UTC+2 skrev WM:

> >If there is an infinite initial segment in |N which is followed by an infinite endsegment,
> There are no such things so your point is moot.

If there were infinitely many infinite endsegments, then this would happen. Of course it can't. Therefore there are not nfinitely many infinite endsegments.

> >Logic says that an infinite sequence followed by an infinite sequence with no or at most finite overlap yields two infinite sequences. What do you think it says?
> There is nothing in mathematics that says that is so in natural numbers.

If there were infinitely many infinite endsegments, then this would happen. Of course it can't.

Regards, WM