Cantor's mistake.

[WM]

Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions. On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ. (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).) Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval. For this sake ℵo natural numbers are required. More are not available.

Regards, WM

[Greg Cunt]

On Friday, September 3, 2021 at 3:23:30 PM UTC+2, WM wrote:

> Cantor [...] is mistaken.

Of course. See:

Crank Dot Net | Cantor was wrong
| http://www.crank.net/cantor.html

[Serg io]

On 9/3/2021 8:23 AM, WM wrote:
> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions. On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ. (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).) Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval. For this sake ℵo natural numbers are required. More are not available.
>
> Regards, WM
>

you forgot the basics of infinity;

r * ℵo = ℵo where r is the number of your intervals

no worries, you will NEVER run out of numbers.

[WM]

Serg io schrieb am Sonntag, 5. September 2021 um 07:01:59 UTC+2:
> On 9/3/2021 8:23 AM, WM wrote:
> > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions. On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ. (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).) Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval. For this sake ℵo natural numbers are required. More are not available.
> >

> you forgot the basics of infinity;
> r * ℵo = ℵo where r is the number of your intervals

Here I discuss Cantor's claim. Cantor used 1*ℵo.

> no worries, you will NEVER run out of numbers.

That is a result of Cantor's claim. But this claim is nonsense, and so are its results.

Regards, WM

[Dan Christensen]

On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.

Wrong again, Mucke. Here again, is visual proof that ALL positive fractions can be enumerated: https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg

> On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ.

True. But this does not prove your claim.

> (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).)

Wrong again, Mucke. There is actually no need to invoke "transfinity" here.

> Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.

Pure gibberish!

> For this sake ℵo natural numbers are required. More are not available.
>

More gibberish! When will you learn, Mucke?

More absurd quotes from Wolfgang Muckenheim (WM):

“In my system, two different numbers can have the same value.”
-- sci.math, 2014/10/16

“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20

“1/9 has no decimal representation.”
-- sci.math, 2015/09/22

"0.999... is not 1."
-- sci.logic 2015/11/25

“Axioms are rubbish!”
-- sci.math, 2014/11/19

“Formal definitions have lead to worthless crap like undefinable numbers.”
-- sci.math 2017/02/05

“No set is countable, not even |N.”
-- sci.logic, 2015/08/05

“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05

Slipping ever more deeply into madness...

“There is no actually infinite set |N.”
-- sci.math, 2015/10/26

“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26

“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28

“Everything is in the list of everything and therefore everything belongs to a not uncountable set.”
-- sci.math, 2015/11/30

"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09

“The set of numbers will get empty after all have numbers been used.”
-- sci.math, 2016/08/24

“I need no set theory.”
-- sci.math, 2016/09/01

A special word of caution to students: Do not attempt to use WM's “system” (MuckeMath) in any course work in any high school, college or university on the planet. You will fail miserably. MuckeMath is certainly no shortcut to success in mathematics.

Using WM's “axioms” for the natural numbers, he cannot even prove that 1=/=2. His goofy little system is truly a dead-end.


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

[Serg io]

On 9/5/2021 5:55 AM, WM wrote:
> Serg io schrieb am Sonntag, 5. September 2021 um 07:01:59 UTC+2:
>> On 9/3/2021 8:23 AM, WM wrote:
>>> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions. On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ. (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).) Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval. For this sake ℵo natural numbers are required. More are not available.
>>>
>> you forgot the basics of infinity;
>> r * ℵo = ℵo where r is the number of your intervals
>
> Here I discuss Cantor's claim. Cantor used 1*ℵo.

Here I discuss your obvious error with your claim "More are not available "

r * ℵo = ℵo where r is the number of your intervals

do you deny this fact ?

This shows you are wrong, and do not know the basics of ℵo, no worries, just google for it.

>
>> no worries, you will NEVER run out of numbers.
>
> That is a result of Cantor's claim. But this claim is nonsense, and so are its results.

your opinion is nonsense,

under examination you fail to understand there are plenty of numbers, an infinity of numbers, multiple infinities

r * ℵo = ℵo where r is the number of your intervals





>

> Regards, WM
>

[WM]

Serg io schrieb am Sonntag, 5. September 2021 um 17:52:00 UTC+2:
> On 9/5/2021 5:55 AM, WM wrote:
> > Serg io schrieb am Sonntag, 5. September 2021 um 07:01:59 UTC+2:
> >> On 9/3/2021 8:23 AM, WM wrote:
> >>> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions. On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ. (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).) Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval. For this sake ℵo natural numbers are required. More are not available.
> >>>
> >> you forgot the basics of infinity;
> >> r * ℵo = ℵo where r is the number of your intervals
> >
> > Here I discuss Cantor's claim. Cantor used 1*ℵo.
> Here I discuss your obvious error with your claim "More are not available "
> r * ℵo = ℵo where r is the number of your intervals
> do you deny this fact ?

No, but it would not help your case. Indexing the first fractions in every interval will require ℵo natnumbers. Indexing the second fraction in every interval will require ℵo natnumbers again. And so on.

Hiwever, Cantor uses only ℵo natnumbers - and no matheologian has ever seen this blatant shortcoming. Most are even too stupid to see it now.

Regards, WM

[WM]

Dan Christensen schrieb am Sonntag, 5. September 2021 um 16:58:13 UTC+2:
> On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.

> Here again, is visual proof that ALL positive fractions can be enumerated: https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg

Wrong. The lower part of the matrix will never be hit.

> > On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ.
> True. But this does not prove your claim.

It proves that all natural numbers are needed to index only one fraction in each interval.

> > Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.
> Pure gibberish!

Really? When all fractions have been enumerated, not even one single fraction must have been enumerated first in every interval?

Regards, WM

[Dan Christensen]

On Sunday, September 5, 2021 at 3:58:53 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> Dan Christensen schrieb am Sonntag, 5. September 2021 um 16:58:13 UTC+2:
> > On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.

> > Here again, is visual proof that ALL positive fractions can be enumerated: https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg

> Wrong. The lower part of the matrix will never be hit.

Wrong again, Mucke. Every positive rational number is accounted for.

> > > On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ.

> > True. But this does not prove your claim.

> It proves that all natural numbers are needed to index only one fraction in each interval.

Gibberish.

> > > Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.

> > Pure gibberish!

> Really? When all fractions have been enumerated, not even one single fraction must have been enumerated first in every interval?
>

Yup, PURE gibberish! All fractions, in every interval on the Q+, can be enumerated as indicated. Deal with it, Mucke.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com

[Python]

Crank Wolfgang Mueckenheim, aka WM wrote:
> Dan Christensen schrieb am Sonntag, 5. September 2021 um 16:58:13 UTC+2:
>> On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
>>> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
>> Here again, is visual proof that ALL positive fractions can be enumerated: https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg
>
> Wrong. The lower part of the matrix will never be hit.

it is it at step 5.

>>> On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ.
>> True. But this does not prove your claim.
>
> It proves that all natural numbers are needed to index only one fraction in each interval.
>
>>> Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.
>> Pure gibberish!
>
> Really? When all fractions have been enumerated, not even one single fraction must have been enumerated first in every interval?

Specify any interval, then compute which index enumerate a
first fraction there. Even somebody as fool and idiot as
you, Crank Wolfgang Mueckenheim, from Wolfgang Mueckenheim,
should be able to do so.

[Serg io]

On 9/5/2021 2:44 PM, WM wrote:
> Serg io schrieb am Sonntag, 5. September 2021 um 17:52:00 UTC+2:
>> On 9/5/2021 5:55 AM, WM wrote:
>>> Serg io schrieb am Sonntag, 5. September 2021 um 07:01:59 UTC+2:
>>>> On 9/3/2021 8:23 AM, WM wrote:
>>>>> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions. On the other hand it is easy to prove that he is mistaken: There is an actual infinity of ℵo positive unit intervals (n-1, n], n ∈ ℕ. (Without actual infinity transfinity is a long twaddle about nothing (Alexander Zenkin).) Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval. For this sake ℵo natural numbers are required. More are not available.
>>>>>
>>>> you forgot the basics of infinity;
>>>> r * ℵo = ℵo where r is the number of your intervals
>>>
>>> Here I discuss Cantor's claim. Cantor used 1*ℵo.
>> Here I discuss your obvious error with your claim "More are not available "
>> r * ℵo = ℵo where r is the number of your intervals
>> do you deny this fact ?
>
> No, but it would not help your case. Indexing the first fractions in every interval will require ℵo natnumbers. Indexing the second fraction in every interval will require ℵo natnumbers again. And so on.

which adds up to ℵo for all your intervals

> Hiwever, Cantor uses only ℵo natnumbers

so, Cantor and I agree, ℵo enumerated it all

>- and no matheologian has ever seen this blatant shortcoming. Most are even too stupid to see it now.

no, it is a hole in your knowledge base, a well known property of ℵo

study this for a while, google and verify it, it has not sunk in yet, time to get started...

r * ℵo = ℵo where r is the number of your intervals


>

> Regards, WM
>

[Serg io]

take the interval 12379^123456 to 12379^123456 + 1
Since I have not written down all its digits of both they are not defined numbers, nor an interval IAW WM.

[WM]

Do you accept that *before* having enumerated all positive fractions at least one fraction must be enumerated in every unit interval (n-1, n], n ∈ ℕ?

Regards, WM

[Gus Gassmann]

I don't. *After* all positive rational numbers have been enumerated, of course one or more (in fact, all aleph_0) rational numbers in every interval (n-1, n], n ∈ ℕ must have been enumerated. But *before* all positive rational numbers have been enumerated, only a finite number of them have been enumerated, hence only a finite number of unit intervals can contain an enumerated rational. I repeat that you have *NO* clue about infinity and should not be let into a bathroom unattended.

[WM]

Dan Christensen schrieb am Montag, 6. September 2021 um 00:43:50 UTC+2:
> On Sunday, September 5, 2021 at 3:58:53 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > Dan Christensen schrieb am Sonntag, 5. September 2021 um 16:58:13 UTC+2:
> > > On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > > > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
>
> > > Here again, is visual proof that ALL positive fractions can be enumerated: https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg
>
> > Wrong. The lower part of the matrix will never be hit.

> Every positive rational number is accounted for.

But hardly before the first rational of every interval is accounted for.

> > > > Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.
>
> > > Pure gibberish!
>
> > Really? When all fractions have been enumerated, not even one single fraction must have been enumerated first in every interval?
> >

> ll fractions, in every interval on the Q+, can be enumerated as indicated.

Would you also think so if we change the bijection slightly such that all fractions n/1 for n > 1 come first? You velieven that all are in the bijection. Why not first index them?

Regards, WM

[WM]

Gus Gassmann schrieb am Montag, 6. September 2021 um 13:10:27 UTC+2:
> On Monday, 6 September 2021 at 07:17:01 UTC-3, WM wrote:

> > Do you accept that *before* having enumerated all positive fractions at least one fraction must be enumerated in every unit interval (n-1, n], n ∈ ℕ?
> I don't.

So all of all intervals are enumerated, but not a first one of each interval???

> *After* all positive rational numbers have been enumerated, of course one or more (in fact, all aleph_0) rational numbers in every interval (n-1, n], n ∈ ℕ must have been enumerated. But *before* all positive rational numbers have been enumerated, only a finite number of them have been enumerated, hence only a finite number of unit intervals can contain an enumerated rational.

You believe that all fractions including all n/1 for n > 1 will become enumerated. They belong to the bijection, don't they? Would you also think so if we changed the bijection slightly such that all fractions n/1 for n > 1 come first? If all are get indexed anhow, why not first index them?

Regards, WM

[jvr]

On Monday, September 6, 2021 at 1:51:49 PM UTC+2, WM wrote:
> Dan Christensen schrieb am Montag, 6. September 2021 um 00:43:50 UTC+2:
> > On Sunday, September 5, 2021 at 3:58:53 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > > Dan Christensen schrieb am Sonntag, 5. September 2021 um 16:58:13 UTC+2:
> > > > On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:

[...]

> > Every positive rational number is accounted for.
> But hardly before the first rational of every interval is accounted for.

[...]
OK, Mückenheim, here we have your error in pure and uncharacteristically unambiguous form.
Pay attention and you will see the light:

In Cantor's enumeration let p/q be any fraction and let n = (p+q)^2.
Then p/q is 'accounted for' before n intervals contain at least one fraction that
occurred earlier in the enumeration.
This is a simple kindergarten exercise in arithmetic.
This is true of the exact mapping that Cantor used to demonstrate the fact that you
find so difficult to understand.

[Gus Gassmann]

Oh great one. Can't you fathom that modifications to a perfectly well-defined mapping may result in something incorrect? If *YOU* make the modifications, failure is guaranteed.

[WM]

Gus Gassmann schrieb am Montag, 6. September 2021 um 15:15:13 UTC+2:
> On Monday, 6 September 2021 at 08:58:33 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 6. September 2021 um 13:10:27 UTC+2:

> > You believe that all fractions including all n/1 for n > 1 will become enumerated. They belong to the bijection, don't they? Would you also think so if we changed the bijection slightly such that all fractions n/1 for n > 1 come first? If all are get indexed anhow, why not first index them?

> Can't you fathom that modifications to a perfectly well-defined mapping may result in something incorrect? If *YOU* make the modifications, failure is guaranteed.

Not for a real bijection. Only for Cantor's fake bijections.

Regards, WM

[Dan Christensen]

On Monday, September 6, 2021 at 7:51:49 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Montag, 6. September 2021 um 00:43:50 UTC+2:
> > On Sunday, September 5, 2021 at 3:58:53 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > > Dan Christensen schrieb am Sonntag, 5. September 2021 um 16:58:13 UTC+2:
> > > > On Friday, September 3, 2021 at 9:23:30 AM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > > > > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
> >
> > > > Here again, is visual proof that ALL positive fractions can be enumerated: https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg
> >
> > > Wrong. The lower part of the matrix will never be hit.
> > Every positive rational number is accounted for.
> But hardly before the first rational of every interval is accounted for.
> > > > > Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.
> >
> > > > Pure gibberish!
> >
> > > Really? When all fractions have been enumerated, not even one single fraction must have been enumerated first in every interval?
> > >

> > All fractions, in every interval on the Q+, can be enumerated as indicated.

> Would you also think so if we change the bijection slightly such that all fractions n/1 for n > 1 come first?

It would make no difference. A set X is countable iff there exists an injective (1-1) function f: X --> N. Any such mapping will do.

[Python]

So you are claiming that if a bijection exists between two sets then ALL
applications between these sets should be a bijection? Are you really
that dumb, Crank Wolfgang Mueckenheim, from Hochschule Augsburg?

[WM]

jvr schrieb am Montag, 6. September 2021 um 14:46:30 UTC+2:
> On Monday, September 6, 2021 at 1:51:49 PM UTC+2, WM wrote:

> > > Every positive rational number is accounted for.
> > But hardly before the first rational of every interval is accounted for.
>

> In Cantor's enumeration let p/q be any fraction and let n = (p+q)^2.
> Then p/q is 'accounted for' before n intervals contain at least one fraction that
> occurred earlier in the enumeration.

That is nice but every p and every q have finitely many predecessors but infinitely many successors. Therefore your contemplation is irrelevant. Relevant is this: The enumeration cannot be complete before every interval is deflowered. Alas it cannot be completed after every interval is deflowered. Hence it cannot be completed at all.

Can you understand this simple kindergarten exercise in logic?

Regards, WM

[WM]

Dan Christensen schrieb am Montag, 6. September 2021 um 17:03:46 UTC+2:
> On Monday, September 6, 2021 at 7:51:49 AM UTC-4, WM wrote:

> > > All fractions, in every interval on the Q+, can be enumerated as indicated.
> > Would you also think so if we change the bijection slightly such that all fractions n/1 for n > 1 come first?
> It would make no difference.

Nice to see your expertise.

Regards, WM

[Gus Gassmann]

As I said, if you make the modifications, failure is guaranteed. You wouldn't know a bijection if two of them bit you in the gonads.

[Gus Gassmann]

Ahem. It seems that you put the word "kindergarten" in the wrong position. What you are talking about and using is kindergarten logic.

[WM]

Python schrieb am Montag, 6. September 2021 um 17:06:02 UTC+2:
>
> So you are claiming that if a bijection exists between two sets then ALL
> applications between these sets should be a bijection?

All transpositions are allowed. They will not change the bijection, if a bijection exists.

Regards, WM

[jvr]

Yes I can - I would ascribe it either to a slightly retarded 5-year-old or to
a seriously retarded retired pseudo-professor.

[Ross A. Finlayson]

And, "Cantor proves the line is drawn"?

I find it seriously comforting that a theory with line-drawing as a
fundamental primitive just like time, or at least one unidirectional
constant motion, allows for both that as a non-Cartesian function
between naturals and only points on a unit line segment, while then
in all the non-Cartesian functions there are thusly various models of
real numbers, each what all suffice for all sorts of usual things.

Because such a notion to be thorough requires actually addressing
theory writ large or the foundations of logic and mathematics, it
is to be expected that the bulwark of modern formalism as the giant
edifice, makes for a unified theory altogether with Cantor satisfied as
"Cantor proves the line is drawn, and the existence of uncountable
models of reals after Dedekind and Eudoxus, in terms of the Cartesian
and non-Cartesian in functions, allowing elements to be interchanged
besides only segments of continuous domains".

[Ross A. Finlayson]

That's perhaps a usual Cartesian notion and quite grand, of course
Des Cartes is widely known as grandiloquent, in the space of functions,
that instead some usual non-Cartesian function is defined by its endpoints,
middle, and monotonicity (thusly continuity).

Yes, Des Cartes was a premier thinker, and it's relevant to look at functions
in model theory in descriptive set theory about topology with the existence
of the outer space only contingent the existence of being the inner space,
of a function or projection.

[jvr]

Is this text generated by a program? It reminds me of the spoof papers
that got submitted to fake journals.

[Serg io]

that is called nardjection or de-nardjection

[Serg io]

who the heck is Des Cartes ?


it is not Rene Descartes...

[Serg io]

On 9/6/2021 10:59 AM, WM wrote:
> jvr schrieb am Montag, 6. September 2021 um 14:46:30 UTC+2:
>> On Monday, September 6, 2021 at 1:51:49 PM UTC+2, WM wrote:
>
>>>> Every positive rational number is accounted for.
>>> But hardly before the first rational of every interval is accounted for.
>>
>> In Cantor's enumeration let p/q be any fraction and let n = (p+q)^2.
>> Then p/q is 'accounted for' before n intervals contain at least one fraction that
>> occurred earlier in the enumeration.
>
> That is nice but every p and every q have finitely many predecessors but infinitely many successors.

that is irrelevant.

> Therefore your contemplation is irrelevant.

wrong. your ego is bruised.

> Relevant is this: The enumeration cannot be complete before every interval is deflowered.

define "deflowered" in terms of math.

enumeration is not dependent in any way on your intervals.

If you disagree, explain why.

> Alas it cannot be completed after every interval is deflowered. Hence it cannot be completed at all.

Fail. you need to show or prove an interval does not have a enumerated rational in it... If you were a real math professor, you could try that.

Instead, it is easy to show any interval has a rational that has been enumerated.



I will leave that as an excersize for you and the students

>
> Can you understand this simple kindergarten exercise in logic?

show any interval has a rational that has been enumerated.


>
> Regards, WM
>

[Dan Christensen]

On Monday, September 6, 2021 at 12:01:42 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Montag, 6. September 2021 um 17:03:46 UTC+2:
> > On Monday, September 6, 2021 at 7:51:49 AM UTC-4, WM wrote:
>
> > > > All fractions, in every interval on the Q+, can be enumerated as indicated.
> > > Would you also think so if we change the bijection slightly such that all fractions n/1 for n > 1 come first?

> > It would make no difference. A set X is countable iff there exists an injective (1-1) function f: X --> N. Any such mapping will do.

> Nice to see your expertise.

Yes, you really have to be careful about your definitions. You can't just make them up willy-nilly as you so often do, Mucke. Anyone with any training in math will see right through you. Must be frustrating as hell for you.

Dan

[Ross A. Finlayson]

I complement Uncle Rene's vortices with a spiral space-filling curve.
(Which was probably inherited from ancient theories after atomism.)

[Ross A. Finlayson]

https://en.wikipedia.org/wiki/Sokal_affair

There's a really great book by Anthony O'Hear on Karl Popper.
I much mostly agree with it and it's very good.

Have you read it?


And what about Shelah?

I think that most people after set theory's usual definition of function
as an indicator in the space of the Cartesian product, for bijections,
didn't think to consider that there are functions that instead are
built from lesser structure, what so results as non- or sub-Cartesian.

It wouldn't be so much material except being so utterly fundamental.

I think it's better to have a replete theory with continuity,
a more complete and not more inconsistent theory,
that addresses for modern formalism why various effects in
continuum mechanics are real, what is otherwise lacking.

I think that anybody else should too if they were conscientious
mathematicians, in foundations the foundation.

Yes, this text was generated by a progam, 100% wet-ware, as
has been trained over about fifty years a usual corpus of canon.

Please excuse me I have my failings.

[Jim Burns]

You have returned to Hilbert's Hotel.

The problem with your rule
(and it is only your (WM's) rule, not logic's)
is that, if you try to describe infinite things,
your description will be either be incorrect or
it will break your rule.

Describe a successor '+1' operation.
|
| For each k which '+1' can be applied to,
| there is k+1, to which '+1' can also be applied.
|
| '+1' can be applied to 0.
| There is no k for which k+1 = 0.
|
| j = k iff j+1 = k+1.

'+1' is a bijection between _things '+1' can be applied to_
and _successors of those things_

Successors of those things is a proper subset of things
'+1' can be applied to. For example, 0 is the latter and
not the former. And yet, '+1' is a bijection.
Choose between your rule and describing the natural numbers.

----
Note that, although my description of '+1' is true of
natural numbers, it can also be true of some versions of
what could fairly be called "dark numbers".

We can add to that description, so that the longer description
is only of the natural numbers.
|
| As well as being a thing '+1' can be applied to,
| k can be counted to, in principle.

k can be counted to, in principle, iff
steppable {0,...,k} exists in which,
for adjacent i,j, j = i+1.

{0,...,k} is steppable iff
for each _split_ B,A of {0,...,k},
a unique adjacent pair i,j exists, i in B, j in A.

This is enough description to allow us to derive the familiar
natural-number induction.
|
| Let j and k refer to things countable-to, in principle.
|
| If exists k, P(0) and ~P(k),
| then exists j, P(j) and ~P(j+1).

And so on.

----
You (WM) make skeptic-noises about induction.
Those noises don't matter.
Your rule got broken back when we defined '+1'.

(Back then, the description included some kind of
dark numbers!)

A variable allows us to refer to an individual and
partially describe it, even if there are infinitely-many
individuals the variable is possibly referring to.

That's what allows us to break your rule, variables.
They allow us to describe infinitely-many, partially but
correctly, so far as it goes. Infinity breaks your rule.

You have a problem with infinity. Your only solution
is to avoid infinity. You try to dress it up, but
there isn't anything more to it than that.

For as long as we continue to describe infinite things
correctly, we will continue to break your rule.

For as long as you continue to obey your rule, you will
continue to describe infinite things incorrectly.

[Ross A. Finlayson]

On Monday, September 6, 2021 at 12:33:27 PM UTC-7, Jim Burns wrote:
> For as long as we continue to describe infinite things
> correctly, we will continue to break your rule.
>
> For as long as you continue to obey your rule, you will
> continue to describe infinite things incorrectly.

A usual effort then to address finite things correctly,
might see some assistance in addressing particularly [0,1],
that the unbounded or infinite divisibility of it, results in a
model of a function aka generalized function that presumes
to maintain scale, in the sense toward analyticity, what results
that the numbers partition [0,1] correctly.

Or, "regularly" as it were usually.

Excuse me, I'm to get back to stacking my books.

[Python]

What you describe is not a transposition (which means bijection btw),
you didn't dare to express it formally because you know it is not.

Aren't you tired of using the same sophistries again and again, crank

[Python]

The point is that he is NOT retired, he's still about to teach the same
fallacies at Hochschule Augsburg this academic year.

[Khong Dong]

What's the difference?

Why worrying about anything coming out of Hochschule Augsburg, when in other "ordinary" universities mathematics student must learn the _incorrectness_ of PA syntactically proving SSS0?

[Dan Christensen]

On Monday, September 6, 2021 at 8:51:03 PM UTC-4, khongdo...@gmail.com wrote:

> the _incorrectness_ of PA syntactically proving SSS0?

Huh??? Please explain.

Dan

[Khong Dong]

Sorry for the typo: it was meant "the _incorrectness_ of PA syntactically proving prime(SSS0)?'.

Of course (Shoenfield's) PA doesn't syntactically prove prime(SSS0).

>
> Dan

[Khong Dong]

> Why worrying about anything coming out of Hochschule Augsburg, when in other "ordinary" universities mathematics student must learn the _incorrectness_ of PA syntactically proving prime(SSS0)?

(Typo correction in the above).

[WM]

Jim Burns schrieb am Montag, 6. September 2021 um 21:33:27 UTC+2:
> On 9/6/2021 12:17 PM, WM wrote:

> > All transpositions are allowed.
> > They will not change the bijection, if a bijection exists.
> You have returned to Hilbert's Hotel.

Note that Hilbert's hotel will accommodate the new giest in romm number 1.

>
> The problem with your rule
> (and it is only your (WM's) rule, not logic's)

This is logic: Before (with respect to the well-order of the natural numbers) all fractions have been indexed, all unit intervals must have been indexed. This all must happen with finite distinguishable indices. Therefore it is impossible that the completion of intervals is simultaneous to the completion of fractions. According to logic the fractions within any interval (n-1, n] will be indexed *after* n/1. This holds for all intervals, whether or not there is a last one. All existing naturals are indexing all existing intervals. More is not possible, as soon as aleph_0 intervals have been indexed, i.e., as soon as all existing naturals have been issued.

Regards, WM

[WM]

Python schrieb am Dienstag, 7. September 2021 um 01:33:21 UTC+2:
> Crand Wolfgang Mueckenheim, aka WM wrote:

Do you mean Grand?

> > All transpositions are allowed. They will not change the bijection, if a bijection exists.
> What you describe is not a transposition (which means bijection btw),

When Cantor indexes all fractions

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...

then it will not destroy the surjectivity, when some fractions will be put to the front

1/1, 2/1, 3/1, 4/1, 5/1, 1/2, 1/3, 1/4, 2/3, 3/2, 1/5, 1/6, ....

Only when those fractions n/1 which never have been in the sequence are put to the front, then the failure of Cantor's claim becomes obvious:

1/1, 2/1, 3/1, 4/1, 5/1, ..., 1/2, 1/3, 1/4, 2/3, 3/2, 1/5, 1/6, ....

This is what I show with my remark that all fractions n/1 must have been enumerated before the enumeration of all fractions can be completed. Only if someone despises logic, he will refuse my claim.

Regards, WM

[Gus Gassmann]

Utter bullshit from you, as usual. You can't even apply "before" and "after" correctly, which is entirely unsurprising, because it involves the use of "all". In every unit interval (n-1,n] the rational number n (=n/1) is the first one to be enumerated. Nonetheless, n+1 gets enumerated before n/1000, etc. The indexing of aleph_0 intervals is *NOT* completed before, e.g., 1/2 is enumerated. Whatever the *FINITE* position of a rational number in the enumeration sequence, it is *FINITE*, and so there can only be at most finitely many intervals up to that point in the sequence that contain an enumerated rational. Your continued ignorance of these numbers is absolutely astounding.

[Gus Gassmann]

Reordering a mapping turns it into a *DIFFERENT* mapping, you fucking imbecile. Some reorderings work, others don't. Anything you get your hands on, breaks. Quel surprise.

[Serg io]

On 9/8/2021 11:21 AM, WM wrote:
> Jim Burns schrieb am Montag, 6. September 2021 um 21:33:27 UTC+2:
>> On 9/6/2021 12:17 PM, WM wrote:
>
>>> All transpositions are allowed.
>>> They will not change the bijection, if a bijection exists.
>> You have returned to Hilbert's Hotel.
>
> Note that Hilbert's hotel will accommodate the new giest in romm number 1.
>>
>> The problem with your rule
>> (and it is only your (WM's) rule, not logic's)
>
> This is logic: Before (with respect to the well-order of the natural numbers) all fractions have been indexed, all unit intervals must have been indexed.

wrong, if one fraction is not indexed, then one unit interval has not bee indexed.

>This all must happen with finite distinguishable indices.

nope.

> Therefore it is impossible that the completion of intervals is simultaneous to the completion of fractions.

no. you are intentionally misleading people.

> According to logic

according to WM's logic

> the fractions within any interval (n-1, n] will be indexed *after* n/1.

provide a proof of you imaginational idea

> This holds for all intervals,

nope, you fail to specify how many intervals, what size etc etc.

Provide a proof, or it remains WM's Spoof.

> whether or not there is a last one.

there are no last ones.

> All existing naturals are indexing all existing intervals.

who told them to do that ? seems unnatural that you would RUN OUT of an infinite number of numbers...

> More is not possible, as soon as aleph_0 intervals have been indexed, i.e., as soon as all existing naturals have been issued.

again, you flail away at infinity, because you have no training in it at all.



>
> Regards, WM
>

[Greg Cunt]

On Wednesday, September 8, 2021 at 6:30:37 PM UTC+2, WM wrote:

> When Cantor indexes all fractions
>
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...

Look, idiot, we are talking about a certain sequence here. This sequence can be (colloquially) referred to with the expression

(1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...)

Actually, there's a certain FUNCTION i |-> f(i)/g(i) which just _is_ this "sequence".

See: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function

> then it will not destroy the surjectivity, when some fractions will be put to the front
>
> 1/1, 2/1, 3/1, 4/1, 5/1, 1/2, 1/3, 1/4, 2/3, 3/2, 1/5, 1/6, ...

Right. Though this means that we are refering to a NEW function/sequence with now, say,

(1/1, 2/1, 3/1, 4/1, 5/1, 1/2, 1/3, 1/4, 2/3, 3/2, 1/5, 1/6, ...)

Actually, you woud have to be more specific concerning your "fractions ... put to the front" if doing real math.

> Only when those fractions n/1 which never have been in the sequence

Huh? There are no fractions "which never have been in the sequence", since ALL (pos.) fractions are terms of the sequence.

You might try to learn the meaning of a SIMPLE FORMULA, you silly asshole.

See: https://en.wikipedia.org/wiki/Pairing_function#Cantor_pairing_function

You know: "Since the Cantor pairing function is invertible, it must be one-to-one and onto."

When doing STUPID things (as usual)

> then [YOUR] failures become obvious:

>
> 1/1, 2/1, 3/1, 4/1, 5/1, ..., 1/2, 1/3, 1/4, 2/3, 3/2, 1/5, 1/6, ...

Look dumbo, in contrast to the expressions above THIS expression

(1/1, 2/1, 3/1, 4/1, 5/1, ..., 1/2, 1/3, 1/4, 2/3, 3/2, 1/5, 1/6, ...)

DOESN'T denote a sequence. It's just meaningless undefined gibberish.

Hint:

"Cranks who contradict some mainstream opinion in some highly technical field, such as mathematics or physics, frequently [...] misunderstand or fail to use standard notation and terminology" (Wikipedia)

[Khong Dong]

So, if one says of arithmetic truths (Gödel used in his 1931 paper) as being malleable and deformable, would one be a crank, misunderstanding or failing to "use standard notation and terminology"?

[WM]

Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 19:11:53 UTC+2:
> On Wednesday, 8 September 2021 at 13:21:29 UTC-3, WM wrote:
> >
> > This is logic: Before (with respect to the well-order of the natural numbers) all fractions have been indexed, all unit intervals must have been indexed. This all must happen with finite distinguishable indices. Therefore it is impossible that the completion of intervals is simultaneous to the completion of fractions. According to logic the fractions within any interval (n-1, n] will be indexed *after* n/1. This holds for all intervals, whether or not there is a last one. All existing naturals are indexing all existing intervals. More is not possible, as soon as aleph_0 intervals have been indexed, i.e., as soon as all existing naturals have been issued.

> You can't even apply "before" and "after" correctly,

Of course. For natural numbers n, k: n comes before n + k.

> In every unit interval (n-1,n] the rational number n (=n/1) is the first one to be enumerated. Nonetheless, n+1 gets enumerated before n/1000, etc.

That shows that many fractions come after 1/n for every natural n.

> The indexing of aleph_0 intervals is *NOT* completed before, e.g., 1/2 is enumerated.

The exhaustion of aleph_0 naturals is completed when all intervals are deflowered and before all rationals in all intervals have been indexed.

> Whatever the *FINITE* position of a rational number in the enumeration sequence, it is *FINITE*, and so there can only be at most finitely many intervals up to that point in the sequence that contain an enumerated rational.

So it is. But there are infinitely many intervals. Your argument reaffirms what I have shown (thank you) namely that not all rationals can be enumerated. Because there are not only finitely many intervals.

Regards, WM

[WM]

That doesn't matter if it is surjective before, it will remain surjective afterwards.

> Some reorderings work, others don't.

But wyh? What is the reason? Never pondered about? I can tell you: Because Cantor's swindle comes to light.

Regards, WM

[Python]

Crank Wolfgang Mueckenheim, aka WM wrote:
> Python schrieb am Dienstag, 7. September 2021 um 01:33:21 UTC+2:

>> CranK Wolfgang Mueckenheim, aka WM wrote:
>
> Do you mean Grand?

I meant Crank, disgusting shameful crank abusing students. Criminal
pseudo-teacher that deserve jail, Crank Wolfgang Mueckenheim, from
Hochschule Augsburg.

>[snip bla no answer]

[WM]

Greg Cunt schrieb am Mittwoch, 8. September 2021 um 20:28:17 UTC+2:
> On Wednesday, September 8, 2021 at 6:30:37 PM UTC+2, WM wrote:

> > Only when those fractions n/1 which never have been in the sequence
> Huh? There are no fractions "which never have been in the sequence", since ALL (pos.) fractions are terms of the sequence.

Why then is it prohibited to reorder them?

>
> You might try to learn the meaning of a SIMPLE FORMULA,

Your formulas are mainly shit. Listen to Gus Gassmann: "Whatever the *FINITE* position of a rational number in the enumeration sequence, it is *FINITE*, and so there can only be at most finitely many intervals up to that point in the sequence that contain an enumerated rational."

But there are infinitelymany intervals according to the bijection n <--> 1/n.

> You know: "Since the Cantor pairing function is invertible, it must be one-to-one and onto."

Pairing with rationals requires defloration of their intervals first. But Infinitely many deflorations disable further pairing.

Regards, WM

[Python]

crank Wolfgang Mueckenheim, aka WM wrote:
>...

> Pairing with rationals requires defloration of their intervals first.

Holy shit! You are ready for the psychiatric ward, crank Wolfgang
Mueckenheim, from Hochschule Augsburg. And you shouldn't be allowed
BY LAW to approach any student anymore.

[Greg Cunt]

How do you like

"Erst nach der Defloration kann die Paarung richtig einsetzen." (W. Mückenheim, de.sci.mathematik)

[Jim Burns]

On 9/8/2021 12:21 PM, WM wrote:
> Jim Burns schrieb
> am Montag, 6. September 2021 um 21:33:27 UTC+2:
>> On 9/6/2021 12:17 PM, WM wrote:

>>> All transpositions are allowed.
>>> They will not change the bijection, if a bijection exists.
>>
>> You have returned to Hilbert's Hotel.
>
> Note that Hilbert's hotel will accommodate the new giest
> in romm number 1.

Hilbert's Hotel is infinite.
Bijections exist between its rooms and proper subsets of
its rooms.

The same (infinite) roster of guests can either be accommodated
one to each room or accommodated one to every prime-numbered room,
leaving composite-numbered rooms empty.

How can finite beings know "infinite" facts like this?
We can partially describe one of infinitely-many guests or rooms,
and reason from that description to further claims about the
same infinitely-many guests or rooms.
Our reasoning preserves the truth of the original description of
a guest or of a room in the further claims about that guest or that
room, whether or not which guest or room is referred to is part of
the description.

>> The problem with your rule
>> (and it is only your (WM's) rule, not logic's)
>
> This is logic:
> Before (with respect to the well-order of the natural numbers)
> all fractions have been indexed,
> all unit intervals must have been indexed.

Consider the collection
{ k | {0,...,k} is doubly-well-ordered }

{0,...,k} is doubly-well-ordered iff
each non-empty subset of {0,...,k} contains a first and
a last, including {0,...,k}.
0 is first of and k is last of {0,...,k}.

{ k | {0,...,k} is doubly-well-ordered }
is NOT doubly-well-ordered.

For each j in { k | {0,...,k} is doubly-well-ordered },
j+1 is in { k | {0,...,k} is doubly-well-ordered }.
Thus, { k | {0,...,k} is doubly-well-ordered } does not
have a last and so is not doubly-well-ordered.

----
Before all fractions have been indexed,
there is some index m such that {0,...,m}
contains all the indexes used so far.

Let Firsts = { first index of a rational in (n,n+1] }
Firsts is NOT doubly-well-ordered,
because it does not contain a last first index.

If this is possible
|
| Before all fractions have been indexed,

| all unit intervals must have been indexed.
|

then Firsts is a subset of doubly-well-ordered {0,...,m}
for some m.
If it were a subset, it would have a last.
It doesn't have a last.
Contradiction.

| Before all fractions have been indexed,

| all unit intervals must have been indexed.
|

is impossible.

> This all must happen with finite distinguishable indices.

No.
For any m, Firsts cannot be a subset of {0,...,m}.

> Therefore it is impossible that the completion of intervals
> is simultaneous to the completion of fractions.

All the firsts-in-an-interval do not have a completing index
If you describe them that way, you will describe them incorrectly.
All the rationals do not have a completing index
If you describe them that way, you will describe them incorrectly.

It is impossible for these non-existent indexes to be
simultaneous. That isn't because they're dark. It's because
they're non-existent.

> According to logic the fractions within any interval (n-1, n]
> will be indexed *after* n/1. This holds for all intervals,
> whether or not there is a last one. All existing naturals are
> indexing all existing intervals.
> More is not possible, as soon as aleph_0 intervals have been
> indexed, i.e., as soon as all existing naturals have been issued.

More is not necessary.
As soon as aleph_0 intervals have been indexed, all
finitely-indexed rationals have been indexed.
That's all of them.

We know that's all of them because,
if
steppable {0,...,j} exists with adjacent h, i = h+1, and
steppable {0,...,k} exists with adjacent h, i = h+1,
then
steppable {0,...,j+k} exists with adjacent h, i = h+1, and
steppable {0,...,j*k} exists with adjacent h, i = h+1, and
...
steppable {0,...,j+(j+k-1)*(j+k-2)/2} exists
with adjacent h, i = h+1.

[Gus Gassmann]

On Wednesday, 8 September 2021 at 15:54:24 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 19:11:53 UTC+2:
> > On Wednesday, 8 September 2021 at 13:21:29 UTC-3, WM wrote:
> > >
> > > This is logic: Before (with respect to the well-order of the natural numbers) all fractions have been indexed, all unit intervals must have been indexed. This all must happen with finite distinguishable indices. Therefore it is impossible that the completion of intervals is simultaneous to the completion of fractions. According to logic the fractions within any interval (n-1, n] will be indexed *after* n/1. This holds for all intervals, whether or not there is a last one. All existing naturals are indexing all existing intervals. More is not possible, as soon as aleph_0 intervals have been indexed, i.e., as soon as all existing naturals have been issued.
> > You can't even apply "before" and "after" correctly,
> Of course. For natural numbers n, k: n comes before n + k.
> > In every unit interval (n-1,n] the rational number n (=n/1) is the first one to be enumerated. Nonetheless, n+1 gets enumerated before n/1000, etc.
> That shows that many fractions come after 1/n for every natural n.

Well. Not to burst your bubble, or your attempt at misdirection: For every natural number n there are infinitely many natural numbers m > n and there are infinitely many fractions mappable beyond 1/n, namely beyond the integer k that maps to the rational 1/n.

I repeat that you have *NO* clue about infinity. All you are still capable of doing is make obscene similes about rationals and intervals. Logic is clearly far beyond you at this point.

> > The indexing of aleph_0 intervals is *NOT* completed before, e.g., 1/2 is enumerated.
> The exhaustion of aleph_0 naturals is completed when all intervals are deflowered and before all rationals in all intervals have been indexed.

Nope. Aleph_0 natural numbers are sufficient to map not just one, but all aleph_0 rationals in *EVERY* interval. Clearly you are incapable of fathoming this, so why don't you call it a life and shut the *FUCK* up.

> > Whatever the *FINITE* position of a rational number in the enumeration sequence, it is *FINITE*, and so there can only be at most finitely many intervals up to that point in the sequence that contain an enumerated rational.
> So it is. But there are infinitely many intervals. Your argument reaffirms what I have shown (thank you) namely that not all rationals can be enumerated. Because there are not only finitely many intervals.

Oh, po' Wolfi. There are infinitely many intervals, and after every rational number has been mapped there are still infinitely many natural numbers left to map them. It's really not my fault that you are too dumb and too stupid to realize that; it is quite a simple idea, and it is very elegant, in contrast to your clumsy rape phantasies.

[Gus Gassmann]

Rape fantasies have no place in newsgroups.

[Greg Cunt]

Why rape? The sexual innuendos as such are bad enough.

What's the matter with this guy? Old age bullshit?

[Khong Dong]

On Wednesday, 8 September 2021 at 15:05:45 UTC-6, Jim Burns wrote:
> On 9/8/2021 12:21 PM, WM wrote:

> > Note that Hilbert's hotel will accommodate the new giest
> > in romm number 1.

> Hilbert's Hotel is infinite.
>

> The same (infinite) roster of guests can either be accommodated
> one to each room or

> accommodated one to every prime-numbered room,
> leaving composite-numbered rooms empty.

Right on! All these guests are named after prime numbers -- Two, Three, Five, Seven ... -- as it happens in this example!

> How can finite beings know "infinite" facts like this?

I'll bet in this example finite beings (including Gödel) can't know how to _enumerate_ the fact-list of pairs (room-number, guest's name). Would you agree Jim?

[WM]

Jim Burns schrieb am Mittwoch, 8. September 2021 um 23:05:45 UTC+2:
> On 9/8/2021 12:21 PM, WM wrote:

> > This is logic:
> > Before (with respect to the well-order of the natural numbers)
> > all fractions have been indexed,
> > all unit intervals must have been indexed.

Diversion deleted.

> | Before all fractions have been indexed,
> | all unit intervals must have been indexed.
> |
> is impossible.

Of course it is impossible. Why do you think that I have devised this example?

> > This all must happen with finite distinguishable indices.
> No.

Yes! Every rational has to be indexed with a finite natural number.

> For any m, Firsts cannot be a subset of {0,...,m}.

Whatever your excuse is. My point stants.

> > Therefore it is impossible that the completion of intervals
> > is simultaneous to the completion of fractions.
> All the firsts-in-an-interval do not have a completing index

No completion index will be available. There are no completion indexes for infinite bijections. But we know that aleph_0 natural indexes are required to index the intervals. This fact is unavoidable and cannot be circumvented.

> If you describe them that way, you will describe them incorrectly.
> All the rationals do not have a completing index
> If you describe them that way, you will describe them incorrectly.

I do not describe them that way. Whatever the definition of completion is: All intervals will be completed before all rationals can be completed. And fact is in set theory tat completing the enumeration will require all aleph_0 finite indexes.

>
> It is impossible for these non-existent indexes to be
> simultaneous. That isn't because they're dark. It's because
> they're non-existent.

Of course they are not simultaneous as you erroneously proposed.

> > According to logic the fractions within any interval (n-1, n]
> > will be indexed *after* n/1. This holds for all intervals,
> > whether or not there is a last one. All existing naturals are
> > indexing all existing intervals.
> > More is not possible, as soon as aleph_0 intervals have been
> > indexed, i.e., as soon as all existing naturals have been issued.
> More is not necessary.

More would be necessary if all intervals had aleph_0 successors of n/1.

> As soon as aleph_0 intervals have been indexed, all
> finitely-indexed rationals have been indexed.

Again simultaneously? Try to adhere to logic. All intervals are indexed by all finite indexes. But every interval requires infinitely more indexes. That is impossible.

> That's all of them.
>
> We know that's all of them because,

because you despise logic.

> if
> steppable {0,...,j} exists

My proof shows that one of your premises is wrong. Perhaps it is this one.

Regards, WM

[WM]

Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 23:52:16 UTC+2:
> On Wednesday, 8 September 2021 at 15:54:24 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 19:11:53 UTC+2:
> > > On Wednesday, 8 September 2021 at 13:21:29 UTC-3, WM wrote:
> > > >
> > > > This is logic: Before (with respect to the well-order of the natural numbers) all fractions have been indexed, all unit intervals must have been indexed. This all must happen with finite distinguishable indices. Therefore it is impossible that the completion of intervals is simultaneous to the completion of fractions. According to logic the fractions within any interval (n-1, n] will be indexed *after* n/1. This holds for all intervals, whether or not there is a last one. All existing naturals are indexing all existing intervals. More is not possible, as soon as aleph_0 intervals have been indexed, i.e., as soon as all existing naturals have been issued.
> > > You can't even apply "before" and "after" correctly,
> > Of course. For natural numbers n, k: n comes before n + k.
> > > In every unit interval (n-1,n] the rational number n (=n/1) is the first one to be enumerated. Nonetheless, n+1 gets enumerated before n/1000, etc.
> > That shows that many fractions come after 1/n for every natural n.
> Well. Not to burst your bubble, or your attempt at misdirection: For every natural number n there are infinitely many natural numbers m > n and there are infinitely many fractions mappable beyond 1/n, namely beyond the integer k that maps to the rational 1/n.

But before they all can be mapped, all intervals must have been mapped once at least. This exhausts all finite indices.

>
> I repeat that you have *NO* clue about infinity.

That will not erase your mistake.

> Aleph_0 natural numbers are sufficient to map not just one, but all aleph_0 rationals in *EVERY* interval.

Here we talk about Cantor's way. Here every fraction n/1 comes before all other fractions of the infinite set of fractions of the interval (n-1, n]. As this is true for every n/1, it is irrelevant how complition is defined. In any case the intervals are completed, by the complete set of indices, before their contents can be completed. No escape and no diversionpossible.

> Clearly you are incapable of fathoming this,

but I am able to disprove it. The sequence 1/1, 1/2, 1/3, ... must be completely indexed before all other fractions can be completely indexed.

> > > Whatever the *FINITE* position of a rational number in the enumeration sequence, it is *FINITE*, and so there can only be at most finitely many intervals up to that point in the sequence that contain an enumerated rational.
> > So it is. But there are infinitely many intervals. Your argument reaffirms what I have shown (thank you) namely that not all rationals can be enumerated. Because there are not only finitely many intervals.

> There are infinitely many intervals, and after every rational number has been mapped there are still infinitely many natural numbers left to map them.

Not according Cantor's approach! Only that is discussed here
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... will exhaust all natural numbers as indices, but before 1/1, 2/1, 3/1, 4/1, ... will exhaust them. Note: Not simultaneously, but *before* all fractions have been mapped.

I could also disprove your approach but probably it would stretch you too far.

Regards, WM

[WM]

Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 23:53:42 UTC+2:

> > "Erst nach der Defloration kann die Paarung richtig einsetzen." (W. Mückenheim, de.sci.mathematik)
> Rape fantasies have no place in newsgroups.

Chuckle. Never heard of technical terms in mathematics? Unterkörper, Paarungsfunktion? Defloration here means that a unit interval receives an index for the first time.

Regards, WM

[WM]

Greg Cunt schrieb am Donnerstag, 9. September 2021 um 02:37:21 UTC+2:
> The sexual innuendos as such are bad enough.
>

Never realized that technical terms have meanings different from common language? Gruppe, Ring, Körper, Unterkörper, Paarungsfunktion?

Regards, WM

[Greg Cunt]

Go away, you disgusting asshole full of shit.

[Greg Cunt]

On Thursday, September 9, 2021 at 2:30:54 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Donnerstag, 9. September 2021 um 02:37:21 UTC+2:
> >
> > The sexual innuendos as such are bad enough.
> >

> Never realized that <bla>

[Serg io]

On 9/9/2021 7:25 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 23:52:16 UTC+2:
>> On Wednesday, 8 September 2021 at 15:54:24 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Mittwoch, 8. September 2021 um 19:11:53 UTC+2:
>>>> On Wednesday, 8 September 2021 at 13:21:29 UTC-3, WM wrote:
>>>>>
>>>>> This is logic: Before (with respect to the well-order of the natural numbers) all fractions have been indexed, all unit intervals must have been indexed. This all must happen with finite distinguishable indices. Therefore it is impossible that the completion of intervals is simultaneous to the completion of fractions. According to logic the fractions within any interval (n-1, n] will be indexed *after* n/1. This holds for all intervals, whether or not there is a last one. All existing naturals are indexing all existing intervals. More is not possible, as soon as aleph_0 intervals have been indexed, i.e., as soon as all existing naturals have been issued.
>>>> You can't even apply "before" and "after" correctly,
>>> Of course. For natural numbers n, k: n comes before n + k.
>>>> In every unit interval (n-1,n] the rational number n (=n/1) is the first one to be enumerated. Nonetheless, n+1 gets enumerated before n/1000, etc.
>>> That shows that many fractions come after 1/n for every natural n.
>> Well. Not to burst your bubble, or your attempt at misdirection: For every natural number n there are infinitely many natural numbers m > n and there are infinitely many fractions mappable beyond 1/n, namely beyond the integer k that maps to the rational 1/n.
>
> But before they all can be mapped, all intervals must have been mapped once at least. This exhausts all finite indices.

indices are not finite, boy.

>>
>> I repeat that you have *NO* clue about infinity.
>
> That will not erase your mistake.

give it up WM, you have no clue about oo, you try to mush your darkies in to explain it to yourself.

>
>> Aleph_0 natural numbers are sufficient to map not just one, but all aleph_0 rationals in *EVERY* interval.
>
> Here we talk about Cantor's way. Here every fraction n/1 comes before all other fractions of the infinite set of fractions of the interval (n-1, n]. As this is true for every n/1, it is irrelevant how complition is defined. In any case the intervals are completed, by the complete set of indices, before their contents can be completed. No escape and no diversionpossible.

that is flawed thinking.

>
>> Clearly you are incapable of fathoming this,
>
> but I am able to disprove it. The sequence 1/1, 1/2, 1/3, ... must be completely indexed before all other fractions can be completely indexed.

no it does not. This simply underscores the fact you do not know Math at all.

>
>>>> Whatever the *FINITE* position of a rational number in the enumeration sequence, it is *FINITE*, and so there can only be at most finitely many intervals up to that point in the sequence that contain an enumerated rational.
>>> So it is. But there are infinitely many intervals. Your argument reaffirms what I have shown (thank you) namely that not all rationals can be enumerated. Because there are not only finitely many intervals.
>
>> There are infinitely many intervals, and after every rational number has been mapped there are still infinitely many natural numbers left to map them
>

> Not according Cantor's approach! Only that is discussed here

Wrong.

[Jim Burns]

On 9/9/2021 8:11 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 8. September 2021 um 23:05:45 UTC+2:
>> On 9/8/2021 12:21 PM, WM wrote:

>>> This is logic:
>>> Before (with respect to the well-order of the natural numbers)
>>> all fractions have been indexed,
>>> all unit intervals must have been indexed.
>
> Diversion deleted.

A variable allows us to refer to an individual and
partially describe it, even if there are infinitely-many
individuals the variable is possibly referring to.

That's what allows us to break your rule, variables.
They allow us to describe infinitely-many, partially but
correctly, so far as it goes. Infinity breaks your rule.

You have a problem with infinity. Your only solution
is to avoid infinity. You try to dress it up, but
there isn't anything more to it than that.

>> | Before all fractions have been indexed,
>> | all unit intervals must have been indexed.
>> |
>> is impossible.
>
> Of course it is impossible.
> Why do you think that I have devised this example?

It would appear that you are arguing that "set theory"
is wrong because it agrees with you, and you are notorious
for being wrong.

That is cleverer than I expected from you.
Or luckier.

>>> This all must happen with finite distinguishable indices.
>>
>> No.
>
> Yes! Every rational has to be indexed with
> a finite natural number.

"This all must happen" is impossible. Therefore, no.

Before all finite indexes have been used,
no more than finitely-many indexes have been used.

One rational in each interval: more than finitely-many.
All rationals in one interval: more than finitely-many.

It is impossible for either of those to happen before
all finite indexes have been used.

>> For any m, Firsts cannot be a subset of {0,...,m}.
>
> Whatever your excuse is. My point stants.

If each interval (n,n+1] has had at least one rational
indexed, then there is no index m after all used indexes.

Before all rationals have been indexed, there is some
index m after all used indexes.

Contradiction.

| Before all fractions have been indexed,
| all unit intervals must have been indexed.
|
is impossible.

>>> Therefore it is impossible that the completion of intervals
>>> is simultaneous to the completion of fractions.
>>
>> All the firsts-in-an-interval do not have a completing index
>
> No completion index will be available.

Available or unavailable, it doesn't exist.

> There are no completion indexes for infinite bijections.
> But we know that aleph_0 natural indexes are required to
> index the intervals.

We know that aleph_0 indexes are required to index the primes.
Is arithmetic inconsistent?

>> if
>> steppable {0,...,j} exists
>
> My proof shows that one of your premises is wrong.
> Perhaps it is this one.

Not a premise, a lemma.

Shorter:
if j and k are finite naturals,
then j+k and j*k are finite naturals.

Is arithmetic inconsistent?

[Greg Cunt]

On Thursday, September 9, 2021 at 7:20:47 PM UTC+2, Jim Burns wrote:

> if j and k are finite naturals,
> then j+k and j*k are finite naturals.

Since each and every natural number is "finite" (i.e. a finite set, if so), we may simplify your claim:

| if j and k are natural numbers,
| then j+k and j*k are natural numbers.

Cool!

[Jim Burns]

I certainly am not objecting to this.
I agree, all natural numbers have finitely-many
(eventual) predecessors.

I am arguing against the claim
|
| We need more than the finitely-preceded naturals
| to index all the rationals.

We have ways to index all the rationals. For example,
index(j/k) = j + (j+k-2)*(j+k-2)/2

In order to demonstrate that index(j/k) is finite,
I need start with some sort of expression of
"j and k are finite".

It's not different from claiming that right triangle ABC
has a right angle as one of its angles. It seems too
obvious to need claiming. Perhaps it is. The point, though
is to show what follows from that obvious claim.
Sometimes, what follows is far from obvious.

> Cool!
>

[Greg Cunt]

On Thursday, September 9, 2021 at 10:16:47 PM UTC+2, Jim Burns wrote:

> I am arguing against the claim
>
> | We need more than the finitely-preceded naturals to index all the rationals.

which, of course, is pure Mückenheim nonsense.

> We have ways to index all the rationals. For example,
> index(j/k) = j + (j+k-2)*(j+k-2)/2

Exactly!

Or, alternatively, i/k |-> 2^i * 3^k would do too. Showing that a subset of IN already suffice.

> In order to demonstrate that index(j/k) is finite,
> I need start with some sort of expression of
> "j and k are finite".

Well... Again, if i and k are natural numbers, then j + (j+k-2)*(j+k-2)/2 is a natural number too. And hence "finite". :-P

> [...] It seems too obvious to need claiming. Perhaps it is. [...]

Once "invented" such (or a similar?) an approach myself.

There's a common claim that the notion of a "set" cannot be defined (say in ZFC). Well, technically that's no quite true.

Actually, we may just state the follwing _definition_ (in the context of ZFC):

set(x) :<-> x = x /
"x is a set"

This would allow to PROVE that "everything is a set" in ZFC:

Ax set(x) .
"Everything is a set."

(Since Ax(x = x) by identity logic in the context of ZFC.)

Yeah, such a definition is superfluous (in ZFC). But... :-)

[Jim Burns]

On 9/9/2021 4:48 PM, Greg Cunt wrote:
> On Thursday, September 9, 2021 at 10:16:47 PM UTC+2,
> Jim Burns wrote:

>> I am arguing against the claim
>> |
>> | We need more than the finitely-preceded naturals
>> | to index all the rationals.
>
> which, of course, is pure Mückenheim nonsense.

Oh, absolutely.

>> We have ways to index all the rationals. For example,
>> index(j/k) = j + (j+k-2)*(j+k-2)/2
>
> Exactly!
>
> Or, alternatively, i/k |-> 2^i * 3^k would do too.
> Showing that a subset of IN already suffice.

That would work. It's also simpler than j + (j+k-2)*(j+k-2)/2

Aesthetically, I prefer i/k |-> k*k*i*i/rad(i)

with gcd(i,k) = 1

It's invertible from N+ to Q+ and that makes some of my
arguments clearer, without having to deal with largely
irrelevant cases in which one rational maps to multiple
indexes.

I just enjoy the idea of a unique prime factorization
_of a rational_ something that I hadn't heard of before.

WM has, in the past, been accepting of unique prime
factorization of the naturals (though, who knows? today),
and, starting with UPF, I think it's simpler.

One reason I like having a chance to bring Unique Prime
Factorization into the conversation with WM is that I think
he has some very odd ideas about how definitions work.
This gives me a reason to address those ideas.

>> In order to demonstrate that index(j/k) is finite,
>> I need start with some sort of expression of
>> "j and k are finite".
>
> Well... Again, if i and k are natural numbers,
> then j + (j+k-2)*(j+k-2)/2 is a natural number too.
> And hence "finite". :-P

Your comment is fair.
However, what if you were addressing someone who did not
accept that all naturals are finitely-preceded?
What would you say then?

[Greg Cunt]

On Friday, September 10, 2021 at 12:49:09 AM UTC+2, Jim Burns wrote:

> However, what if you were addressing someone who did not
> accept that all naturals are finitely-preceded?
> What would you say then?

Well, if not dealing with a crank, I guess I'd try a proof!

Claim: For all n e IN: {m e IN : m < n} is finite. (N = {1, 2, 3, ...}}

Proof (by induction):

If n = 1 then {m e IN : m < n} is = {} and hence is finite.
Assume that for some arbitrary n e IN the set {m e IN : m < n} is finite. We want to show that {m e IN : m < n + 1} is finite too. But {m e IN : m < n + 1} = {m e IN : m < n} u {n} and {m e IN : m < n} is finite, hence {m e IN : m < n + 1} (obviously) is finite too. qed. [Ok, there are some small "gaps" in this proof, but it should be rather clear how to fill them.]

[Ross A. Finlayson]

If X is a set of all finite ordinals, it contains an infinite ordinal.

Diagonalize the set, there's thusly an element in it.

I.e., collecting the sets that don't contain themselves from the set,
in a little universe of only finite ordinals you just proved,
structurally the result's an ordinal and contains itself.

So, in a way, you are saying that omega isn't an ordinal or set.

Or that it contains itself.


Here the point that perhaps you can read and so so established via
a usual course of inference, is, if there are infinite numbers there are
(infinite numbers).


Not that counting one-at-a-time in finite time gets to them, or,
that counting down one-at-a-time leaves them, there's still though
for infinite divisibility that it solves Zeno's paradoxes directly, and,
there either is or isn't well-ordering uncountably many elements
according to their normal ordering. (Or, elements of a continuous
domain as it were, or that "Cantor proves the line is drawn".)

Anyways, though, you might enjoy kicking trolls with such pretty
usual monotone progressions, and making us suffer your dysnonymous
quack, but there really is something to think about the infinite that
really is just sitting around in all these usual example pieces, that
both of course recognizes such usual trivia and fills in such vacua.

I.e. kicking the troll is fundamentally boring!

[Khong Dong]

So your set {m e IN : m < n} is interesting: you haven't specified the underlying set theory, and haven't defined the binary relation denoted by '<' in _that_ theory.

Suppose then the underlying set theory is ZFCU

http://www.pgrim.org/philosophersannual/34articles/menzelwide.pdf

in which

"ZFCU is ZFC modified to allow for the existence of urelements, or atoms, i.e., things that can be members of sets but are not themselves sets and do not themselves have members"

and there is (Soa) Set-of-atoms axiom:

"SoA ∃x∀y(y ∈ x ↔ ~Set(y))"

and either your m or n is a prime and ~Set(m) or ~Set(n), then how would '<' or "m < n" get defined in _that_ theory ZFCU?

[WM]

> disgusting asshole full of shit.

Another technical term? What is its meaning?

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 9. September 2021 um 19:20:47 UTC+2:
> On 9/9/2021 8:11 AM, WM wrote:

> >>> This is logic:
> >>> Before (with respect to the well-order of the natural numbers)
> >>> all fractions have been indexed,
> >>> all unit intervals must have been indexed.

Diversion deleted.

> You have a problem with infinity.

No, I know that matheologians have a problem with infinity, and I have explained above the simplest example.

> Your only solution
> is to avoid infinity.

No, I assume that it exists and that it can be exhausted by indexing all fractions n/1. This proves that it is also exhausted when these terms are hidden in
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .
Hence there is nothing avaliable to index the remaining fractions within the unit intervals succeeding all first rationals which are n/1. (Note that I use this variable like you have recommended.)

> > Yes! Every rational has to be indexed with
> > a finite natural number.
> "This all must happen" is impossible. Therefore, no.

It is Cantor's claim as the basis of set theory.

>
> Before all finite indexes have been used,
> no more than finitely-many indexes have been used.

But they are used for all fractions n/1.

>
> All rationals in one interval: more than finitely-many.

Yes, therefore they are not indexed finitely.

> > Whatever your excuse is. My point stants.
> If each interval (n,n+1] has had at least one rational
> indexed, then there is no index m after all used indexes.
>
> Before all rationals have been indexed, there is some
> index m after all used indexes.
>
> Contradiction.

Try logic. All indices are used here: 1/1, 2/1, 3/1, ...
This cannot change when these fractions are distributed in
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

> | Before all fractions have been indexed,
> | all unit intervals must have been indexed.
> |
> is impossible.

Yes. Therefore not all fractions can be indexed.

> We know that aleph_0 indexes are required to index the primes.
> Is arithmetic inconsistent?

No, it is not inconsistent. But set theory is inconsistent.

Regards, WM

[WM]

Jim Burns schrieb am Freitag, 10. September 2021 um 00:49:09 UTC+2:
> On 9/9/2021 4:48 PM, Greg Cunt wrote:
> > On Thursday, September 9, 2021 at 10:16:47 PM UTC+2,
> > Jim Burns wrote:

> >> We have ways to index all the rationals. For example,
> >> index(j/k) = j + (j+k-2)*(j+k-2)/2
> >
> > Exactly!

All rationals indexed in that way can be put to the front of

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

by transpositions. Nothing will change.

> >
> > Or, alternatively, i/k |-> 2^i * 3^k would do too.
> > Showing that a subset of IN already suffice.
> That would work. It's also simpler than j + (j+k-2)*(j+k-2)/2

All rationals indexed in that way can be put to the front of

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

or reordered in any arbitrary way by transpositions. Nothing will change. Or can you find any such rational that cannot be transposed? But this will not work

1/1, 2/1 3/1, ..., 1/2, 1/3, 1/4, 2/3, 3/2, 1/5,1/6, ... .

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 9. September 2021 um 22:16:47 UTC+2:

> I am arguing against the claim
> |
> | We need more than the finitely-preceded naturals
> | to index all the rationals.

But you are not arguing against the claim that there is a bijection n <--> n/1? Note that a bijection does not leave any index n unemployed.

>
> We have ways to index all the rationals. For example,
> index(j/k) = j + (j+k-2)*(j+k-2)/2

All rationals defined in this way can be indexed by finite natnumbers before the infinite remainder. No problem. But all n/1 cannot be indexed before 2/3, say.

Regards, WM

[Greg Cunt]

On Friday, September 10, 2021 at 11:53:22 AM UTC+2, WM wrote:
> Jim Burns schrieb am Donnerstag, 9. September 2021 um 22:16:47 UTC+2:
> >
> > We have ways to index all the rationals. For example,
> > index(j/k) = j + (j+k-2)*(j+k-2)/2
> >

> All rationals [...] can be indexed by finite natnumbers [in this way]. No problem.

Indeed!

> But all n/1 cannot be indexed before 2/3, say.

So what? No one claimed that this is possible.

Hint: The fraction 2/3 has a certain index. Hence only finitely many fractions have an smaller index. Hence ist's not possible for all fractions of the form "n/1" (n e IN) to have a smaller index (since there are infinitely many such fractions).

[Greg Cunt]

On Friday, September 10, 2021 at 7:46:56 AM UTC+2, khongdo...@gmail.com wrote:

> So your set {m e IN : m < n} is interesting: you haven't specified the underlying set theory,

That's not necessary in this case, the context of the whole discussion is ZF(C) or some similar set theory. The "meaning (definitions) of IN and < are known in this context (as well as the set builder notation, etc.)

> and haven't defined the binary relation denoted by '<' in _that_ theory.

We all know (well except you it seems) what is meant by "<" in this context. This means, we all know the formal properties of "<" in this context.

IF "+" is defined the usual way on IN "<" might be defined (for n,m e IN) the following way

n < m <-> Ek e IN\{0}: m = n + k .

If in our system (of ZF(C)) the natural numbers are define due to von Neumann. We might as well define "<" for n,m e IN the following way:

n < m <-> n e m .

It can be shown that for all n,m e IN:

n e m <-> Ek e IN\{0}: m = n + k

of "+" is defined the usual way on IN.

Hence both definitions are "equivalent".

> Suppose then the underlying set theory is ZFCU
>
> http://www.pgrim.org/philosophersannual/34articles/menzelwide.pdf
>
> in which
>
> "ZFCU is ZFC modified to allow for the existence of urelements, or atoms, i.e., things that can be members of sets but are not themselves sets and do not themselves have members"

Ok.

> [...] then how would '<' or "m < n" get defined in _that_ theory ZFCU?

Depends. One approach might be to add additionals axioms to ZFCU - concerning the natural numbers (considered as atoms in this case) and the set IN (as well as a successor function). Actually, a variant of the Peano axioms would do, I guess.

In this case, we might define "+" (on IN) as usual, i.e. "recursively", and then "<" (again) the following way:

n < m <-> Ek e IN\{0}: m = n + k .

[Jim Burns]

On 9/10/2021 5:02 AM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 9. September 2021 um 19:20:47 UTC+2:
>> On 9/9/2021 8:11 AM, WM wrote:

>>>>> This is logic:
>>>>> Before (with respect to the well-order of
>>>>> the natural numbers) all fractions have been indexed,
>>>>> all unit intervals must have been indexed.
>
> Diversion deleted.
>
>> You have a problem with infinity.
>
> No, I know that matheologians have a problem with infinity,

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

A positive rational p for which there are no positive
naturals j,k such that j/k = p is not a positive rational.

As a consequence of a natural being a natural,
there are examples of schemes to match all of them to
fewer than all of them, one-to-one, leaving none out.

As a consequence of a rational being a rational,
there are examples of schemes to match all of them to
fewer than all of them, one-to-one, leaving none out.

The problem that you imagine we have with infinity is
only the "problem" of having correctly described naturals,
of having correctly described rationals.

> and I have explained above the simplest example.
>
>> Your only solution
>> is to avoid infinity.
>
> No, I assume that it exists and that it can be exhausted by
> indexing all fractions n/1.

The naturals can be matched to 1/1, 2/1, 3/1, ...

> This proves that it is also exhausted when these terms are
> hidden in
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

The naturals can be matched to all of

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

and the naturals can be matched to a proper subset of

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

This is not a contradiction.
This is a consequence of there being infinitely-many.

> Hence there is nothing avaliable to index the remaining
> fractions within the unit intervals succeeding all first
> rationals which are n/1.

There are indexing schemes under which each rational is
assigned a finite index.
There are indexing schemes under which that's not true.

The claim was never "all schemes index all rationals",
so showing a scheme that doesn't is NOT a counter-example
to the claim actually made.
Please review "counter-example".

> (Note that I use this variable like you have recommended.)

If you use variables, you can describe a successor '+1'
operation correctly.

| If '+1' can be applied to k, then k+1 exists and
| '+1' can be applied to k+1.
|
| '+1' can be applied to 0.
| If '+1' can be applied to k, k+1 is not 0.
|
| If '+1' can be applied to j and to k, then
| j = k iff j+1 = k+1.

If you describe a successor '+1' operation correctly,
then you can describe a scheme for matching all of
the things '+1' operates on to fewer than all of them,
one-to-one, leaving none out: k to k+1.

----
There's your "problem with infinity".
If you don't want to describe or reason about naturals
or rationals or other things that ship in the Extra Large
Infinity-Pack, then don't do that. Arrange flowers. Hike.

If you describe infinitely-many things as though they were
finitely-many, you describe them incorrectly. If you teach
your students at Hochschule Augsburg to describe
infinitely-many things as though they are finitely-many,
you teach them lies. Because "academic freedom". Hurray.

[Khong Dong]

That -- your '<' being defined by your 'e' -- is a total mix-up between two concepts (paradigms), at least according to Terreence Tao:

<quote>
we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set
</quote>

Until you (JB, some others ...) acknowledge this is a conceptual mix-up, you wouldn't understand _the main underlying theme_ behind WM's posts -- here.

(Caveat: I'm not saying I understand everything WM has said and perhaps some of the things he has elaborated might not be that clear, and I'm not speaking for him. Otoh, if you don't acknowledge this mix-up, what would be the point of arguing about something that's _already_ a mixed-up on your side?).

[Khong Dong]

My apology: I meant "Terence Tao".

[Greg Cunt]

On Saturday, September 11, 2021 at 4:17:47 AM UTC+2, khongdo...@gmail.com wrote:

> Terreence Tao:
>
> <quote>
> we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set
> </quote>

This may be true in *his* framework (see my comments concerning ZFCU in the other post).

It's false if the background theory is ZFC, since THERE *everything* is a set.

Since it seems that you dind't even read that part - a repost:

===============================================================

If in our system (of ZF(C)) the natural numbers are define due to von Neumann, we might as well define "<" for n,m e IN the following way:

n < m <-> n e m .

It can be shown that for all n,m e IN:

n e m <-> Ek e IN\{0}: m = n + k

of "+" is defined the usual way on IN.

Hence both definitions are "equivalent".

> Suppose then the underlying set theory is ZFCU
>
> http://www.pgrim.org/philosophersannual/34articles/menzelwide.pdf
>
> in which
>
> "ZFCU is ZFC modified to allow for the existence of urelements, or atoms, i.e., things that can be members of sets but are not themselves sets and do not themselves have members"

Ok.

> [...] then how would '<' or "m < n" get defined in _that_ theory ZFCU?

Depends. One approach might be to add additionals axioms to ZFCU - concerning the natural numbers (considered as atoms in this case) and the set IN (as well as a successor function). Actually, a variant of the Peano axioms would do, I guess.

In this case, we might define "+" (on IN) as usual, i.e. "recursively", and then "<" (again) the following way:

n < m <-> Ek e IN\{0}: m = n + k .

===============================================================

Here Tao's claim would be (partially) true:

| "we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set"

Since in this case (ZFCU), 3 ∈ 4 would not be "meaningless", but simply false. Actually, we could PROVE in this framework the theorem "Ax(~(x ∈ 4))" and hence, especially, ~(3 ∈ 4) [with 4 e IN.]

You should be aware of the fact that Tao is a highly capable mathematician, but not really recognized as an expert concerning "foundational questions". :-)

Hint: In a set theoretic framework (even if it allows for atoms) a stetement like "3 ∈ 4" would certainly be a wff (if "3" and "4" are -possibly defined- terms in it). And clearly a reasonable "set theory" should allow to derive either "3 e 4" or "~(3 e 4)". Otherwise "3 e 4" would be "undecidable". :-)

[Khong Dong]

On Friday, 10 September 2021 at 20:36:41 UTC-6, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 4:17:47 AM UTC+2, khongdo...@gmail.com wrote:
>
> > Terreence Tao:
> >
> > <quote>
> > we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set
> > </quote>
> This may be true in *his* framework (see my comments concerning ZFCU in the other post).
>
> It's false if the background theory is ZFC, since THERE *everything* is a set.
>
> Since it seems that you dind't even read that part - a repost:
>
> ===============================================================
>
> If in our system (of ZF(C)) the natural numbers are define due to von Neumann, we might as well define "<" for n,m e IN the following way:
> n < m <-> n e m .

You missed the point entirely. The so-called "natural numbers" _are_ _not_ sets. Period.

[Out of von Neumann sets, you can construct something that you can _subjectively_ -- but don't have to -- interpret as integers, or Presburger numbers, too.]

[Khong Dong]

On Friday, 10 September 2021 at 20:48:03 UTC-6, Khong Dong wrote:
> On Friday, 10 September 2021 at 20:36:41 UTC-6, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 4:17:47 AM UTC+2, khongdo...@gmail.com wrote:
> >
> > > Terreence Tao:
> > >
> > > <quote>
> > > we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set
> > > </quote>
> > This may be true in *his* framework (see my comments concerning ZFCU in the other post).
> >
> > It's false if the background theory is ZFC, since THERE *everything* is a set.
> >
> > Since it seems that you dind't even read that part - a repost:
> >
> > ===============================================================
> >
> > If in our system (of ZF(C)) the natural numbers are define due to von Neumann, we might as well define "<" for n,m e IN the following way:
> > n < m <-> n e m .

> You missed the point entirely. The so-called "natural numbers" _are_ _not_ sets. Period.

Which also means the statement like "A natural number is finite" is also meaningless.

[Greg Cunt]

On Saturday, September 11, 2021 at 4:36:41 AM UTC+2, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 4:17:47 AM UTC+2, khongdo...@gmail.com wrote:

> You should be aware of the fact that Tao is a highly capable mathematician, but not really recognized as an expert concerning "foundational questions". :-)

Let's check his statement again:

| we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set

Actually, the fact (if so) that 4 is not a set does not suffice to justify that "3 ∈ 4" is "meaningless" (in general), since in, say, ZFCU 4 is indeed not a set, but there "3 ∈ 4" is NOT "meaningless"; actually, there "~3 ∈ 4" is a theorem.

Still, Tao may consider things differently. I mean if he's able to specify a framework (comprising some sort of set theory) for that (where "the natural number's" aren't sets), why not.

Just an idea. If his system has (infinitely many) constants especially for "natural numbers", say, I, II, III, IIII, ... then he may define the notion of a /wff/ in his system in a way such that not expression of the form t ∈ Z, where t is any term and Z is a constant for a natural number, is a wff. In other words, then, "III ∈ IIII" would be a "meaningless" expression in his system. (Technically: no wff.) I guess it's possible to extend this approach in a way such that all constants referring to "natural numbers" are "affected", I mean, even defined constants, like, say, 4 := IIII and 3 := III.

Then his statement

| we consider the statement 3 ∈ 4 to neither be true or false, but simply meaningless, since 4 is not a set

would make sense, at least for me.

Of course, some sort of "type theory" may be involved here too. Who knows?

[Greg Cunt]

On Saturday, September 11, 2021 at 5:00:10 AM UTC+2, khongdo...@gmail.com wrote:

> Which also means the statement like "A natural number is finite" is also meaningless.

Look, you silly crank: In the contect of, say, ZFC, where the "natural numbers" are defined due to von Neumann, it's not "meaningless", but a provable statement (at least if formulated in set theoretic language). Of course assuming THE USUAL definition of the notion /finite/.

There we can prove, say, An e IN: finite(n). ("All natural numbers are finite.")

See: https://en.wikipedia.org/wiki/Dedekind-infinite_set

So, please shut up, idiot!

[Greg Cunt]

On Saturday, September 11, 2021 at 4:48:03 AM UTC+2, khongdo...@gmail.com wrote:

> The so-called "natural numbers" _are_ _not_ sets. Period.

In the context of a pure set theory, like ZFC, they are, you silly crank.

Holy shit!

EOD.

(I'm sorry, but you are too dumb for any reasonable discussion.)

[Khong Dong]

On Friday, 10 September 2021 at 21:18:46 UTC-6, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 5:00:10 AM UTC+2, khongdo...@gmail.com wrote:
>
> > Which also means the statement like "A natural number is finite" is also meaningless.

> n the contect of, say, ZFC, where the "natural numbers" are defined due to von Neumann

Look, Greg Crank, "4 is not a set" -- von Neumann's set or otherwise. Go and learn the basics of foundation of mathematics.

[Greg Cunt]

On Saturday, September 11, 2021 at 5:23:55 AM UTC+2, khongdo...@gmail.com wrote:
> On Friday, 10 September 2021 at 21:18:46 UTC-6, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 5:00:10 AM UTC+2, khongdo...@gmail.com wrote:
> > >
> > > Which also means the statement like "A natural number is finite" is also meaningless.
> > >

> > In the contect of, say, ZFC, where the "natural numbers" are defined due to von Neumann
> >
> Look <bla>

Try to learn at least SOME math, crank:

https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers
https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory

Hint: "In the area of mathematics called set theory, a specific construction due to John von Neumann defines the natural numbers as follows:

[...]

With this definition, a natural number n is a particular set with n elements, and n ≤ m if and only if n is a subset of m. The standard definition, now called definition of von Neumann ordinals, is: "each ordinal is the well-ordered set of all smaller ordinals."

I don't know what's wrong with you, but it's clear that you do not know ANY math.

Hint: The claim "the natural number are this and that" is completely meaningless without referring to a certain context/background framework. (The simple question "How would you know?" usually would suffice to make that clear. But it certainly won't work when dealing with a crank.)

[Khong Dong]

On Friday, 10 September 2021 at 21:39:56 UTC-6, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 5:23:55 AM UTC+2, khongdo...@gmail.com wrote:
> > On Friday, 10 September 2021 at 21:18:46 UTC-6, Greg Cunt wrote:
> > > On Saturday, September 11, 2021 at 5:00:10 AM UTC+2, khongdo...@gmail.com wrote:
> > > >
> > > > Which also means the statement like "A natural number is finite" is also meaningless.
> > > >
> > > In the contect of, say, ZFC, where the "natural numbers" are defined due to von Neumann
> > >
> > Look <bla>
>

> Try <blah>

Look, Greg Crank, "4 is not a set" -- von Neumann's or otherwise. Go and learn the basics of foundation of mathematics.

[Ross A. Finlayson]

Who made you the crank police?

Hyperbole or overgeneralization if not for its own sake,
is almost always incorrect in a usual formal sense.

In "pure set theory" "4" is the equivalence class the set of all things that "are" 4.

And simple tiny models like implicit or ubiquitous ordinals are only
the easiest abstraction of number sense.

Arithmetization and all what follows in groups of them is rather strong.

" I don't know what's wrong with you, but it's clear that you do not know ANY math.

-"Greg C"



See, here I am putting down "Greg C" for insulting intelligence.

Because, if someone was having a conversation, and, along comes
some guy berating the guy, thinking it was kind of him to yell at it,
like a dog, or for whatever other would be un-kind whatever motivates him,
an animal that treats people like dogs doesn't deserve respect,
"Greg C The Crank Police" without having any input but to kick the troll,
isn't but a troll to kick.


I am glad though you've heard about von Neumann ordinals now.


There's pretty much only one convention of von Neumann ordinals
but several conventions are after him, always for arithmetization's
sake, whether the set is an ordered pair convention or simple nesting
another convention, the "ordinal".

I.e., the inductive set the simple "infinite" nesting, the ordinal,
it's among "ubiquitous ordinals" in higher set theory.

Try to learn _all_ the math, troll. Then get back to me about _some_.

[Ross A. Finlayson]

One's actually obliged to interpret a model of ordinals as they are.

[Greg Cunt]

On Saturday, September 11, 2021 at 6:56:28 AM UTC+2, khongdo...@gmail.com wrote:
> On Friday, 10 September 2021 at 21:39:56 UTC-6, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 5:23:55 AM UTC+2, khongdo...@gmail.com wrote:
> > > On Friday, 10 September 2021 at 21:18:46 UTC-6, Greg Cunt wrote:
> > > > On Saturday, September 11, 2021 at 5:00:10 AM UTC+2, khongdo...@gmail.com wrote:
> > > > >
> > > > > Which also means the statement like "A natural number is finite" is also meaningless.
> > > > >
> > > > In the contect of, say, ZFC, where the "natural numbers" are defined due to von Neumann
> > > >
> > > Look <bla>

Greg Cunt's profile photo
Greg Cunt
5:39 AM (2 hours ago)
to

On Saturday, September 11, 2021 at 5:23:55 AM UTC+2, khongdo...@gmail.com wrote:
> On Friday, 10 September 2021 at 21:18:46 UTC-6, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 5:00:10 AM UTC+2, khongdo...@gmail.com wrote:
> > >
> > > Which also means the statement like "A natural number is finite" is also meaningless.
> > >
> > In the contect of, say, ZFC, where the "natural numbers" are defined due to von Neumann
> >
> Look <bla>

[Greg Cunt]

On Saturday, September 11, 2021 at 6:59:38 AM UTC+2, Ross A. Finlayson wrote:

> In "pure set theory" "4" is the equivalence class the set of all things that "are" 4.

Nonsens. This approach does NOT work in ZFC, for example.

"This definition works in type theory, and in set theories that grew out of type theory, such as New Foundations and related systems. However, it does not work in the axiomatic set theory ZFC nor in certain related systems, because in such systems the equivalence classes under equinumerosity are proper classes rather than sets."

Source: https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers

Holy shit! What's wrong with you cranks?