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 3:23:03 PM UTC+2, WM wrote:

> Cantor [...] is mistaken.

Of course. See:

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

[Eram semper recta]

On Friday, 3 September 2021 at 16:23:03 UTC+3, WM wrote:
> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.

I think they are systematically enumerable, but not "ALL".

If Cantor wrote "ALL", then he was simply boasting about something as completely unremarkable as a radix system.

A set is countable iff its members can be systematically named. This is why |N is countable, not because |N has any magical properties, but that the young, wily Jew Cantor realised this fact before he went insane.

> 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).)

They can't be "unit intervals" because there is nothing longer than a unit between any two consecutive natural numbers.

[Gus Gassmann]

On Friday, 3 September 2021 at 10:23:03 UTC-3, 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* [my emphasis] 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.

The very fact that Muckenheim is using the notion of "before" here shows that he has no clue about infinity and marks him as a crank.

[WM]

Can you understand that 2 comes before 6 in the sequence of natural numbers as Cantor used them?
Can you understand that 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, ...
1/3 is indexed before 5/1?

Regards, WM

[WM]

Eram semper recta schrieb am Freitag, 3. September 2021 um 15:39:00 UTC+2:
> On Friday, 3 September 2021 at 16:23:03 UTC+3, WM wrote:
> > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
> I think they are systematically enumerable, but not "ALL".
>
> If Cantor wrote "ALL", then he was simply boasting about something as completely unremarkable as a radix system.

Without enumerating all, ℵo is nonsense.

> > 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).)
> They can't be "unit intervals" because there is nothing longer than a unit between any two consecutive natural numbers.

(n-1, n] is a called unit interval because n - (n-1) = 1.

Regards, WM

[William]

Galileo's mistake.

Galileo claimed that all natural numbers can be enumerated. But consider the perfect squares. Clearly there are aleph_0 of them and so enumerating them uses aleph_0 natural numbers. More are not available to enumerate the natural numbers which are not perfect squares.

--
William Hughes

[Gus Gassmann]

As I said, you have no clue about the issues. I have some doubts that you even comprehend your own drivel any more. Your description of the diagonal method certainly has *nothing* to do with the statement in your previous post.

[Serg io]

On 9/3/2021 8:22 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
>

wrong.

Proof;

Assume there is a rational a/b in (n-1, n] n ∈ ℕ
Cantors Enumeration includes a/b
therefore Cantors Enumeration includes all rationals in (n-1, n] n ∈ ℕ



(“There are so many different kinds of stupidity, and cleverness is one of the worst.”
― Thomas Mann, )




[ABBA REUNION !!!]

[Gus Gassmann]

On Friday, 3 September 2021 at 14:07:22 UTC-3, Serg io wrote:
> On 9/3/2021 8:22 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
> >
> wrong.
>
> Proof;
>
> Assume there is a rational a/b in (n-1, n] n ∈ ℕ
> Cantors Enumeration includes a/b
> therefore Cantors Enumeration includes all rationals in (n-1, n] n ∈ ℕ

Even though this is of course correct, it does not address Muckenheim's blind spot. The truth is that every rational a/b in (n-1, n] has a position /k(a,b)/ in the sequence of enumerations and that there is an interval (M-1, M] whose *first* enumerated rational has a position that *exceeds* k(a,b). (In fact, there are infinitely many of these intervals.) That is the whole point of Cantor's enumeration. Muckenheim truly has no clue about infinity.

[Greg Cunt]

On Friday, September 3, 2021 at 7:06:16 PM UTC+2, Gus Gassmann wrote:
> On Friday, 3 September 2021 at 13:36:20 UTC-3, WM wrote:
>

> [bla bla]

>
> As I said, you have no clue about the issues. I have some doubts that you even comprehend your own drivel any more.

Agree.

[Jim Burns]

On 9/3/2021 9:22 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.

Cantor (and others) begin with claims that are we already know
are true of each positive rational. Claims like these
|
| x is a positive rational iff
| two FISONable naturals p,q exist such that q*x = p, q>0, p>0.
|
| k is a FISONable natural iff
| a steppable set {0,...,k} of FISONable naturals exists
| in which, for all adjacent i,j, j = i+1.

Cantor (and others) extend what we already know by truth-
-preserving inference to further claims. Because we only use
truth-preserving inferences, for anything for which it is _true_
that p and q exist, q*x = p, it will also be _true_ (for example)
that a unique index k exists for x, k being a FISONable natural,
and that no other x' has the same index k.

In this sense, all positive rationals can be enumerated,
and this can be proven for all positive rationals.

> 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).)

Do you (WM) agree with Zenkin that potential infinity
is a long twaddle about nothing?

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

No.
Before all rationals have been enumerated, no more than
finitely-many rationals have been enumerated.
There are infinitely-many unit intervals, so your claim
cannot be true.

Your claim is the same as you made a little while ago.
We went around and around and, in order to continue asserting
your claim, you ended up, in effect, declaring arithmetic
and/or geometry inconsistent.

Consider the much simpler hypothesis that your claim here
is wrong. Before all rationals have been enumerated there,
at least one unit interval has zero rationals enumerated in it.

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

I claim that all the integers Z = {...,-3,-2,-1,0,1,2,3,...}
can be enumerated by the naturals N = {0,1,2,3,...}.

Proof: We can re-order the integers this way
j 0, 1, -1, 2, -2, 3, -3, ...
k 0, 1, 2, 3, 4, 5, 6, ...

I can't list all the integers or all the naturals,
but, for each integer j, I can tell you index(j).
index(j) =
{ 2*j-1 for j >= 1
{ -2*j for j =< 0

For each integer j, a unique natural k exists, k = index(j).
For each natural k, a unique integer j exists, index(j) = k.
index(j) = index(j') iff j = j'.

Perhaps you (WM) will object to my claim that _all_ the integers
can be enumerated. But you might not. It's fairly easy to _imagine_
the number line being _folded_ at 1/3, giving us our new order.

I also claim that _all_ the positive rationals can be enumerated.
Any objection to this claim has a corresponding objection to
enumerating the integers.

There are many ways to enumerate the positive rationals.
My personal favorite folds _positive and negative_ powers of
each prime to correspond to _only positive_ powers of that prime.

2^j 1, 2, 1/2, 4, 1/4, 8, 1/8, ...
2^k 1, 2, 4, 8, 16, 32, 64, ...

3^j 1, 3, 1/3, 9, 1/9, 27, 1/27, ...
3^k 1, 3, 9, 27, 81, 243, 729, ...

...

The powers of primes be folded the same way as
the integers were folded onto the naturals.
j 0, 1, -1, 2, -2, 3, -3, ...
k 0, 1, 2, 3, 4, 5, 6, ...

Because each index and each numerator and denominator
have unique prime factorizations, each prime power can be
folded separately and then multiplied together.

index(2/7) = index(2)*index(1/7) = 2*49 = 98
index(11/3) = index(1/3)*index(11) = 9*11 = 99
index(1/10) = index(1/2)*index(1/5) = 4*25 = 100

We can also reverse the process and calculate, for a given
index, what rational has that index.

[WM]

William schrieb am Freitag, 3. September 2021 um 18:53:53 UTC+2:
> Galileo's mistake.
>
> Galileo claimed that all natural numbers can be enumerated. But consider the perfect squares. Clearly there are aleph_0 of them and so enumerating them uses aleph_0 natural numbers. More are not available to enumerate the natural numbers which are not perfect squares.

In this case the enumeration happened rather clumsy. Cantor's enumeration could be improved too. But I discuss what Cantor did.

Regards, WM

[WM]

Jim Burns schrieb am Freitag, 3. September 2021 um 21:27:39 UTC+2:
> On 9/3/2021 9:22 AM, WM wrote:

> > Before all fractions have been enumerated there, at least one
> > fraction must have been enumerated in every unit interval.
> No.

So all rational numbers in all intervals can be enumerated without having enumerated at least one rationale number in every interval?

> Before all rationals have been enumerated, no more than
> finitely-many rationals have been enumerated.
> There are infinitely-many unit intervals, so your claim
> cannot be true.

It is Cantor's claim that all rationals can be enumerated.
My claim is minimalistic: If all ℵo rationals of each of all the ℵo intervals are enumerated, at least one rational in each interval has to be enumerated.

Proof: If not every interval has at least one enumerated rational, then there exists at least one interval without any enumerated rational. This interval cannot have all its rationals enumerated.

> Your claim is the same as you made a little while ago.

No, it much better and sharper.

> Consider the much simpler hypothesis that your claim here
> is wrong. Before all rationals have been enumerated there,
> at least one unit interval has zero rationals enumerated in it.

No, not "before" but after all rationals have been enumerated, one interval has no enumerated rationals in it. That is what you claim.

> > For this sake ℵo natural numbers are required.
> > More are not available.
> I claim that all the integers Z = {...,-3,-2,-1,0,1,2,3,...}
> can be enumerated by the naturals N = {0,1,2,3,...}.

This error is harder to detect, but it is not an excuse for the blatant mistake that I have discovered.

Regards, WM

[Dan Christensen]

On Friday, September 3, 2021 at 9:23:03 AM UTC-4, WM (aka Wolfgang Muckenheim (WM))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

[Dan Christensen]

STUDENTS BEWARE: Don't be a victim of JG's fake math

On Friday, September 3, 2021 at 9:39:00 AM UTC-4, I am Super Rectum (aka John Gabriel (JG), Troll Boy) wrote:
> On Friday, 3 September 2021 at 16:23:03 UTC+3, WM wrote:
> > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
> I think they are systematically enumerable, but not "ALL".
>

Yes, ALL of them, Mr. Rectum.

JG here claims to have a discovered a shortcut to mastering calculus without using limits. Unfortunately for him, this means he has no workable a definition of the derivative of a function. It blows up for functions as simple f(x)=|x|. Or even f(x)=0. As a result, he has had to ban 0, negative numbers and instantaneous rates of change rendering his goofy little system quite useless. What a moron!

Forget calculus. JG has also banned all axioms because he cannot even derive the most elementary results of basic arithmetic, e.g. 2+2=4. Such results require the use of axioms, so he must figure he's now off the hook. Again, what a moron!

Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:

"There are no points on a line."
--April 12, 2021

"Pi is NOT a number of ANY kind!"
--July 10, 2020

"1/2 not equal to 2/4"
--October 22, 2017

“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015

"3 =< 4 is nonsense.”
--October 28, 2017

"Zero is not a number."
-- Dec. 2, 2019

"0 is not required at all in mathematics, just like negative numbers."
-- Jan. 4, 2017

“There is no such thing as an empty set.”
--Oct. 4, 2019

“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
--Oct. 22, 2019

No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.

Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10, 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ

Dan

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

[Dan Christensen]

On Friday, September 3, 2021 at 12:53:53 PM UTC-4, William wrote:
> Galileo's mistake.
>
> Galileo claimed that all natural numbers can be enumerated. But consider the perfect squares. Clearly there are aleph_0 of them and so enumerating them uses aleph_0 natural numbers. More are not available to enumerate the natural numbers which are not perfect squares.

There are many infinite subsets of N. The set of perfect squares. The set of even numbers. The set of numbers greater than 2. All have the same cardinality as N. Deal with it.

>
> --
> William Hughes

An impostor???

Dan

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

Visit my Math Blog at http://www.dcproof.wordpress.com

[Eram semper recta]

On Friday, 3 September 2021 at 19:39:47 UTC+3, WM wrote:
> Eram semper recta schrieb am Freitag, 3. September 2021 um 15:39:00 UTC+2:
> > On Friday, 3 September 2021 at 16:23:03 UTC+3, WM wrote:
> > > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
> > I think they are systematically enumerable, but not "ALL".
> >
> > If Cantor wrote "ALL", then he was simply boasting about something as completely unremarkable as a radix system.
> Without enumerating all, ℵo is nonsense.

I know. My point is I don't think he meant "ALL", but rather systematic enumeration.

> > > 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).)
> > They can't be "unit intervals" because there is nothing longer than a unit between any two consecutive natural numbers.
> (n-1, n] is a called unit interval because n - (n-1) = 1.

Sorry, I think I misunderstood what you meant.

>
> Regards, WM

[Serg io]

On 9/3/2021 11:53 AM, William wrote:
> Galileo's mistake.
>
> Galileo claimed that all natural numbers can be enumerated. But consider the perfect squares.

having trouble ? here you go;

1 1
2 4
3 9
4 16
. .
. .
. .

EZ Cheesy

> Clearly there are aleph_0 of them and so enumerating them uses aleph_0 natural numbers.

what do you mean by "uses" ? Used Up? an infinite set ?

> More are not available to enumerate the natural numbers which are not perfect squares.

bullshit.

and you will say since all naturals are USED UP, then you cant number all the even numbers one to one,

double bullshit.


advice: try to be more clear about what you are stating by using MATH statements

[WM]

Dan Christensen schrieb am Samstag, 4. September 2021 um 05:43:48 UTC+2:
> On Friday, September 3, 2021 at 9:23:03 AM UTC-4, WM (aka Wolfgang Muckenheim (WM))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

Two statemenets contradicting each other. This is called a contradiction.

> > 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.

This proves that if every interval is hit at least once, then all indices have been exhausted.

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

Simplest logic. The contrary would be: All fractions have been enumerated, but in at least one interval no first one has been enumerated.

Regards, WM

[WM]

Eram semper recta schrieb am Samstag, 4. September 2021 um 07:46:59 UTC+2:
> On Friday, 3 September 2021 at 19:39:47 UTC+3, WM wrote:
> > Eram semper recta schrieb am Freitag, 3. September 2021 um 15:39:00 UTC+2:
> > > On Friday, 3 September 2021 at 16:23:03 UTC+3, WM wrote:
> > > > Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
> > > I think they are systematically enumerable, but not "ALL".
> > >
> > > If Cantor wrote "ALL", then he was simply boasting about something as completely unremarkable as a radix system.
> > Without enumerating all, ℵo is nonsense.
> I know. My point is I don't think he meant "ALL", but rather systematic enumeration.

"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" vollständig = complete means all.

Regards, WM

[markus...@gmail.com]

fredag 3 september 2021 kl. 15:23:03 UTC+2 skrev 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

The product of two countable set is countable. So your argument does not hold.

[Gus Gassmann]

On Saturday, 4 September 2021 at 09:18:08 UTC-3, WM wrote:
> "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" vollständig = complete means all.

What an arsehole you are! "eindeutig und vollständig, Element für Element" is simply Cantor's way of describing a bijection. (Did he have the term when he wrote that passage?) Every element in M has a single associated element in N, and vice versa. And, yes, "all" means that none are left behind, either in N or in M.

So, nothing at all to see here, folks.

[WM]

markus...@gmail.com schrieb am Samstag, 4. September 2021 um 14:40:02 UTC+2:
> fredag 3 september 2021 kl. 15:23:03 UTC+2 skrev 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.

> The product of two countable set is countable. So your argument does not hold.

Here we have not two countable sets but only one set of indices mapped to the rationals. One could use many infinite subsets of ℕ but Cantor didn't. (And it would not help, because we could ask for the point when at least two fractions were indexed in every unit interval and so on.)

Regards, WM

[WM]

Gus Gassmann schrieb am Samstag, 4. September 2021 um 14:45:01 UTC+2:
> On Saturday, 4 September 2021 at 09:18:08 UTC-3, WM wrote:
> > "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" vollständig = complete means all.

> "eindeutig und vollständig, Element für Element" is simply Cantor's way of describing a bijection.

Of course.

> (Did he have the term when he wrote that passage?)

Of course not.

> Every element in M has a single associated element in N, and vice versa. And, yes, "all" means that none are left behind, either in N or in M.

That is what I said. It means "complete" or "all".

Regards, WM

[markus...@gmail.com]

Sure we do? The usual argument shows that rationals in (0, 1] is in bijection to N. Since Q is a disjoint union of (n, n+1], Q is bijective to NxN, which is just N.

[WM]

markus...@gmail.com schrieb am Samstag, 4. September 2021 um 15:09:24 UTC+2:
> lördag 4 september 2021 kl. 15:02:05 UTC+2 skrev WM:
> > markus...@gmail.com schrieb am Samstag, 4. September 2021 um 14:40:02 UTC+2:
> > > fredag 3 september 2021 kl. 15:23:03 UTC+2 skrev 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.
> > > The product of two countable set is countable. So your argument does not hold.
> > Here we have not two countable sets but only one set of indices mapped to the rationals. One could use many infinite subsets of ℕ but Cantor didn't. (And it would not help, because we could ask for the point when at least two fractions were indexed in every unit interval and so on.)
> >

> Sure we do? The usual argument shows that rationals in (0, 1] is in bijection to N. Since Q is a disjoint union of (n, n+1], Q is bijective to NxN, which is just N.

We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.

Regards, WM

[NaCl]

Dan Christensen wrote:

> Yes, ALL of them, Mr. Rectum.
> JG here claims to have a discovered a shortcut to mastering calculus
> without using limits. Unfortunately for him, this means he has no
> workable a definition of the derivative of a function. It blows up for

amazed to realize, the grade of imbecility in this forum

eram recta, Dan Christensen, Michael Morony, FromtheRafters, Quantum
Bubbles and certainly a few other "mathematicians". "science" believers
and lethal injections promoters.

all guilty of premeditated *grave_indifference_to_human_life*, murder,
mass murder and conspiracy to commit *mass_murder*.

The swine-flu "vaccine", for instance, was canceled by the FDA after 24
deaths. The covid_19 lethal injection, said vaccine, already counts
deaths in millions, and still not canceled by those conspirators.

[Greg Cunt]

On Saturday, September 4, 2021 at 3:02:05 PM UTC+2, WM wrote:
> markus...@gmail.com schrieb am Samstag, 4. September 2021 um 14:40:02 UTC+2:
> >
> > The product of two countable set is countable. So your argument does not hold.
> >

> Here we have not two countable sets but <bla bla>

Sure we do, you silly asshole.

Hint: The set of numerators is {1, 2, 3, 4, ...}
The set of denumerators is {1, 2, 3, 4, ...}

Hence the set of fractions is IN x IN. Where a fraction (n,m) usually is written as n/m.

Now Cantor's pairing function is a bijection from IN x IN onto IN.

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

[Greg Cunt]

On Saturday, September 4, 2021 at 3:19:53 PM UTC+2, WM wrote:

> We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.

Huh?!

No, dumbo, IN is "indexing" ALL fractions n/m with n e IN and m e IN.

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

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

[NaCl]

Serg io wrote:

> On 9/3/2021 11:53 AM, William wrote:
>> Galileo's mistake.
>> Galileo claimed that all natural numbers can be enumerated. But
>> consider the perfect squares.
>
> having trouble ? here you go; 1 1 2 4 3 9 4 16 . .

Biological Weapons Law Author Explains How to Arrest Fauci & Friends For
Unleashing Bioweapon on U.S
https://seed306.bitchute.com/SWtD5kO0Ia5X/E2XI2INdtyRo.mp4

[Dan Christensen]

On Saturday, September 4, 2021 at 8:14:57 AM UTC-4, WM (Wolfgang Muckenheim, aka Mucke) wrote:
> Dan Christensen schrieb am Samstag, 4. September 2021 um 05:43:48 UTC+2:
> > On Friday, September 3, 2021 at 9:23:03 AM UTC-4, WM (aka Wolfgang Muckenheim (WM))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

> Two statemenets contradicting each other. This is called a contradiction.

And if you could disprove a theorem of ZFC using only the axioms of ZFC itself, you would have an inconsistency, not to mention instant fame and fortune. How about it, Mucke? Not today??? Oh, well....

> > > 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.
> This proves that if every interval is hit at least once, then all indices have been exhausted.
> > > Before all fractions have been enumerated there, at least one fraction must have been enumerated in every unit interval.
> > Pure gibberish!

> Simplest logic. The contrary would be: All fractions have been enumerated, but in at least one interval no first one has been enumerated.
>

Requires a proof using the axioms of ZFC, Mucke. Sorry, but MuckeMath simply won't cut it here. You cannot even prove 1=/=2 in your goofy little system.

Dan

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

[Gus Gassmann]

Excellent work, Herr Professor. Note, however, and this is important and probably *WAY* over your head, that he does *NOT* say that the mapping has to be determined sequentially. For instance, the mapping from IR to IR given by x |-> 2*x for every x in IR is a bijection.

[Dan Christensen]

On Saturday, September 4, 2021 at 9:40:36 AM UTC-4, NaCl wrote:
> Dan Christensen wrote:
>
> > Yes, ALL of them, Mr. Rectum.
> > JG here claims to have a discovered a shortcut to mastering calculus
> > without using limits. Unfortunately for him, this means he has no
> > workable a definition of the derivative of a function. It blows up for
> amazed to realize, the grade of imbecility in this forum
>
> eram recta, Dan Christensen, Michael Morony, FromtheRafters, Quantum
> Bubbles and certainly a few other "mathematicians". "science" believers
> and lethal injections promoters.
>

So, how are things down on the Putin "farm" these days, comrade NaCl?

[markus...@gmail.com]

You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.

[Serg io]

Putin is giving away land! 1 hector to even forieners who apply for naturalization.


https://www.rt.com/business/341892-far-east-land-infrastructure/


President Vladimir Putin has ordered free land handouts in Russia’s Far East to be provided with minimal infrastructure. On Monday, he signed a law
offering land plots of 1 hectare (2.5 acres) to Russians and naturalized citizens for free.

[Serg io]

red herring from the Miss Leader.


When you chop a carrot into pieces, you don't need a knife for each piece.

[Serg io]

I am formally revoking you of all your Math duties, except for counting rocks.

[FromTheRafters]

Serg io brought next idea :

> On 9/4/2021 9:38 AM, Dan Christensen wrote:
>> On Saturday, September 4, 2021 at 9:40:36 AM UTC-4, NaCl wrote:
>>> Dan Christensen wrote:
>>>
>>>> Yes, ALL of them, Mr. Rectum.
>>>> JG here claims to have a discovered a shortcut to mastering calculus
>>>> without using limits. Unfortunately for him, this means he has no
>>>> workable a definition of the derivative of a function. It blows up for
>>> amazed to realize, the grade of imbecility in this forum
>>>
>>> eram recta, Dan Christensen, Michael Morony, FromtheRafters, Quantum
>>> Bubbles and certainly a few other "mathematicians". "science" believers
>>> and lethal injections promoters.
>>>
>>
>> So, how are things down on the Putin "farm" these days, comrade NaCl?
>>
>>
>>
>
>
>
> Putin is giving away land! 1 hector to even forieners who apply for
> naturalization.

Sounds like a communist plot.

[Jim Burns]

On 9/3/2021 5:29 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 3. September 2021 um 21:27:39 UTC+2:

>> On 9/3/2021 9:22 AM, WM wrote:

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

>> No.
>
> So all rational numbers in all intervals can be enumerated
> without having enumerated at least one rationale number
> in every interval?

Before all rationals have been enumerated, at least one
unit interval does not have any rationals enumerated in it.

>> Consider the much simpler hypothesis that your claim here
>> is wrong. Before all rationals have been enumerated there,
>> at least one unit interval has zero rationals enumerated in it.
>
> No, not "before" but after all rationals have been enumerated,
> one interval has no enumerated rationals in it.
> That is what you claim.

No, this is what I claim:

Before all rationals have been enumerated, at least one
unit interval does not have any rationals enumerated in it.

Your confusion is probably caused by assuming that I also
claim that there is a last natural number.
I don't claim that.

I claim...

For each k in N+, the set of positive naturals,
a steppable {1,...,k} exists in which,
for adjacent i,j, j = i+1.

For each p in Q+, the set of positive rationals,
j and k exist in N+ such that k*p = j.

( One way to define this multiplication '*' is geometrically.
( Construct similar right triangles K0J and U0P with sides
( d(0K) = k, d(0J) = j, d(0U) = 1, d(0P) = p.
( Because they're similar, p/1 = j/k, k*p = j.

I claim we have enumerated all the elements of Q+ with
the elements of N+. By that, I mean
I claim that index: Q+ --> N+ can be shown such that
for each p in Q+, a unique i in N+ exists, i = index(p),
index(p) = index(q) iff p = q,
and (extra credit, for k*k*j*j/rad(j))
for each k in N+,a unique p in Q+ exists, index(q) = k,


One consequence of these claims is that,
before all rationals have been enumerated, at least one
unit interval does not have any rationals enumerated in it.


"Before all rationals have been enumerated", there is
at least one rational p which has not been enumerated.
p has an index, i = index(p). i is FISONable.

A steppable set {0,...,i-1} exists in which, for all
adjacent j,k, k = j+1.

Let Q[i-1] be the set of rationals indexed by
an element of {0,...,i-1}.
Let r be the maximum rational in Q[i-1]
(Lemma: maximum r exists. [1])

r is in Q+
j,k exist in N+, r = j/k
The set { i =< j/k } is a subset of {0,...,j},
is also steppable,
is not empty,
has a last member m.
m =< j/k and m+1 > j/k.

Each rational in unit interval (m+1,m+2] is larger than
each rational indexed by one of {1,...,i-1}.

The unit interval (m+1,m+2] has no rationals in it
which are indexed before p, index(p) = i.

Therefore,
before all rationals have been enumerated, at least one
unit interval does not have any rationals enumerated in it.

[1]
The set Q[i-1] of rationals indexed by one of {1,...,i-1}
contains a maximum r.

Consider BestYet, a subset of {1,...,i-1}.
k is in BestYet iff
the rational indexed by k is larger than
each rational indexed by one of 1,...,k-1

Each subset of {1,...,i-1} contains a first and a last,
unless it's empty.
BestYet is a subset of {1,...,i-1}
1 is in BestYet. It's not empty.
BestYet contains a last. Call it m.

m is the index for some rational r, index(r) = m

Consider EvenBetter, a subset of {m+1,...,i-1} of
indexes of rationals following r that are larger than r.

EvenBetter is a subset of {1,...,i-1}.
Either it is empty or it contains a first and a last.

Suppose n is the first element of EvenBetter,
and index(s) = n.
s > r, everything between s and r is =< r,
everything before r is =< r
s is larger than each rational indexed by 1,...,n-1.
But then n = index(s) should be in BestYet.
But then, m = index(r) is NOT last in BestYet.
Contradiction.
Therefore, EvenBetter has no first.
EvenBetter is empty.

So. We have m = index(r).
m is last in BestYet, so r > each rational indexed in {1,...,m-1}
EvenBetter is empty, so r > each rational indexed in {m+1,...,i-1}
r > each rational indexed in {1,...,i-1} other than itself.

Therefore,
r is the maximum rational indexed by one of {1,...,i-1}.

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

>> I claim that all the integers Z = {...,-3,-2,-1,0,1,2,3,...}
>> can be enumerated by the naturals N = {0,1,2,3,...}.
>
> This error is harder to detect, but it is not an excuse for
> the blatant mistake that I have discovered.

The bijection between Z and N is simpler than the bijection
between Q+ and N+. Why would simplicity and clarity
make this "blatant mistake" *harder* to detect?

I think that this is because your "detection process"
requires the generation of large quantities of fog
to obscure what you are saying.

This is why my counter-argument is mostly clarification of
what a natural is, what a rational is, what an enumeration is,
the significance of truth-preservation.

The points I make are intended to be simple points.
It is their simplicity which opposes your obfuscation.

[Serg io]

ugh... but very true!

missed that one entirely!

[WM]

Jim Burns schrieb am Samstag, 4. September 2021 um 21:21:26 UTC+2:
> On 9/3/2021 5:29 PM, WM wrote:
> > Jim Burns schrieb
> > am Freitag, 3. September 2021 um 21:27:39 UTC+2:
> >> On 9/3/2021 9:22 AM, WM wrote:
>
> >>> Before all fractions have been enumerated there, at least one
> >>> fraction must have been enumerated in every unit interval.
> >>
> >> No.
> >
> > So all rational numbers in all intervals can be enumerated
> > without having enumerated at least one rationale number
> > in every interval?
>

> > No, not "before" but after all rationals have been enumerated,
> > one interval has no enumerated rationals in it.
> > That is what you claim.
>
> No, this is what I claim:
>
> Before all rationals have been enumerated, at least one
> unit interval does not have any rationals enumerated in it.

That is nonsense, because the rationals of this unit interval cannot become enumerated in one step simultaneously with the completion of this interval.

>
> Your confusion is probably caused by assuming that I also
> claim that there is a last natural number.
> I don't claim that.

But you claim that no natural number and no rational is missing from the complete enumeration.

>
> One consequence of these claims is that,
> before all rationals have been enumerated, at least one
> unit interval does not have any rationals enumerated in it.
>

Immediately before all rationals have been enumerated, at least one unit interval does not have any rationals enumerated in it? I say: Between the complete enumeration of all rationals of all intervals and the enumeration of at least one rational in every interval there are infinitely many further steps.

Regards, WM

[WM]

Gus Gassmann schrieb am Samstag, 4. September 2021 um 16:30:12 UTC+2:
> On Saturday, 4 September 2021 at 10:04:37 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Samstag, 4. September 2021 um 14:45:01 UTC+2:
> > > On Saturday, 4 September 2021 at 09:18:08 UTC-3, WM wrote:
> > > > "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" vollständig = complete means all.
> > > "eindeutig und vollständig, Element für Element" is simply Cantor's way of describing a bijection.
> > Of course.
> > > (Did he have the term when he wrote that passage?)
> > Of course not.
> > > Every element in M has a single associated element in N, and vice versa. And, yes, "all" means that none are left behind, either in N or in M.
> > That is what I said. It means "complete" or "all".

> he does *NOT* say that the mapping has to be determined sequentially.

The natural numbers are applied as a sequence.

Regards, WM

[WM]

markus...@gmail.com schrieb am Samstag, 4. September 2021 um 16:49:16 UTC+2:
> lördag 4 september 2021 kl. 15:19:53 UTC+2 skrev WM:

> > We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.
> >

> You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.

You believe that you can construct the bijection. But you forget or have never realized that most nunbers are dark. This is proven here:

Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.

Regards, WM

[Dan Christensen]

On Saturday, September 4, 2021 at 5:15:29 PM UTC-4, WM wrote:
> markus...@gmail.com schrieb am Samstag, 4. September 2021 um 16:49:16 UTC+2:
> > lördag 4 september 2021 kl. 15:19:53 UTC+2 skrev WM:
>
> > > We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.
> > >
> > You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.
> You believe that you can construct the bijection. But you forget or have never realized that most nunbers are dark.

There are no "dark numbers," Mucke. Recall that, by YOUR OWN formal definition, EVERY natural number is defined. So, after decades of fruitless effort, you STILL have no inconsistency in ZFC Must be frustrating as hell for you!

[Serg io]

Ill have the Cursor Function Apply Natural Numbers in sequence to all rocks on all beaches after they are put in a single line, without markings.

[Gus Gassmann]

OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3, ...} = IN. The whole point of the diagonal argument (obviously utterly lost on you and clearly no longer within reach of your demented mind) is that the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1, n] will have been mapped, and more is not required.

But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient", obviously can't be expected to process any of the previous paragraph. You'd be better off putting all your poofs into a pipe and smoke them...

[Chris M. Thomasson]

Here it is:

https://youtu.be/xrsbjjuDTzU

lol!

[Jim Burns]

On 9/4/2021 5:07 PM, WM wrote:
> Jim Burns schrieb

> am Samstag, 4. September 2021 um 21:21:26 UTC+2:

>> One consequence of these claims is that,
>> before all rationals have been enumerated, at least one
>> unit interval does not have any rationals enumerated in it.
>
> Immediately before all rationals have been enumerated,

I see that you have accidentally typed "Immediately" in front of
what I actually claim. A side effect of "academic freedom",
most likely.

Before all rationals have been enumerated, *anywhere* before
all rationals have been enumerated, there are infinitely-many
rationals which aren't enumerated. There is no "immediately before".

This is a consequence of there being infinitely-many rationals,
each with a finite index.

> Immediately before all rationals have been enumerated,
> at least one unit interval does not have any rationals
> enumerated in it?

No. There is no "immediately before".

Your confusion is probably caused by assuming that I also
claim that there is a last natural number.
I don't claim that.

> I say:
> Between the complete enumeration of all rationals of all intervals
> and the enumeration of at least one rational in every interval
> there are infinitely many further steps.

I'll take a moment to point out that, while you say things,
I and others prove things.

There are infinitely many unit intervals (k,k+1].

If at least one rational in each of infinitely-many intervals
has been indexed, each finitely-indexed rational has been
indexed.

All rationals are finitely-indexed rationals.

If at least one rational in each of infinitely-many intervals
has been indexed, each rational has been indexed.

There are no steps, zero, none between the complete enumeration of

all rationals of all intervals and the enumeration of at least

one rational in every interval.
Such a step would require a finitely-indexed rational to
come after infinitely-many rationals.

Please review "finite", "infinite", and "index".

[Serg io]

big Fail. r * ℵ0 = ℵ0 for r =/= 0 where r is your number of intervals, more is always available.


you do not understand infinity, nor the properties of infinity




WM, Go Study => https://en.wikipedia.org/wiki/Aleph_number

[Serg io]

the follow up where they discover they are parked
https://www.youtube.com/watch?v=ZEavqjHWOv0

[Chris M. Thomasson]

ROFL!!!!

[Dan Christensen]

On Saturday, September 4, 2021 at 5:50:50 PM UTC-4, Dan Christensen wrote:
> On Saturday, September 4, 2021 at 5:15:29 PM UTC-4, WM wrote:
> > markus...@gmail.com schrieb am Samstag, 4. September 2021 um 16:49:16 UTC+2:
> > > lördag 4 september 2021 kl. 15:19:53 UTC+2 skrev WM:
> >
> > > > We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.
> > > >
> > > You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.
> > You believe that you can construct the bijection. But you forget or have never realized that most nunbers are dark.
> There are no "dark numbers," Mucke. Recall that, by YOUR OWN formal definition, EVERY natural number is defined.

In case you forgot, you wrote:

"ℕ_def = {k ∈ ℕ : |∩{E(1), E(2), ..., E(k)}|= ℵo}"
--sci.math 2021-08-11

[WM]

Gus Gassmann schrieb am Sonntag, 5. September 2021 um 00:20:51 UTC+2:
> On Saturday, 4 September 2021 at 18:15:29 UTC-3, WM wrote:

> > Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.
> OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3, ...} = IN.

That is irrelevant since Cantor here uses only {1, 2, 3, ...}.

> The whole point of the diagonal argument is that the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1, n] will have been mapped, and more is not required.

Firstly, this is wrong, since the lower part of the matrix is never reached, but secondly it is irrelevant, since I show that Cantor would have used up all the indices for the first rationals of all intervals already.

>
> But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient",

Galilei had no grasp of the difference between potential and actual infinity. His treatment of the problem is as insufficient as would be Newton's treatment of a quantum computer.

Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 5. September 2021 um 00:58:57 UTC+2:
> On 9/4/2021 5:07 PM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 4. September 2021 um 21:21:26 UTC+2:
> >> One consequence of these claims is that,
> >> before all rationals have been enumerated, at least one
> >> unit interval does not have any rationals enumerated in it.
> >
> > Immediately before all rationals have been enumerated,
> I see that you have accidentally typed "Immediately" in front of

> what I actually claim.wrote you that

So you did not mean immediately? Why then wrote you that sentence? Of course there are many intervals not enumeretaed at the beginning. That is trivial and not of interest.

>
> Before all rationals have been enumerated, *anywhere* before
> all rationals have been enumerated, there are infinitely-many
> rationals which aren't enumerated. There is no "immediately before".

Of course there are many rationals not enumerated. But Cantor and you claim that there is an instance when all rationals have been enumerated. Hence there must be an instance way before, where all intervals have at least one rational enumerated.

>
> This is a consequence of there being infinitely-many rationals,
> each with a finite index.

Cantor's and your claim are wrong. This is a consequence of there being infinitely many rationals. Neverftheless you will continue to claim that all will be enumerated at some instance. But you will deny that before that has happened, all intervals will have received at least one hit? Then my "immediately" was appropriate.

> > Immediately before all rationals have been enumerated,
> > at least one unit interval does not have any rationals
> > enumerated in it?

> No. There is no "immediately before".

There is a sequence. Each index can well be distinguished.

> Your confusion is probably caused by assuming that I also
> claim that there is a last natural number.
> I don't claim that.

I don't claim that. But you claim that the indexing is completed somewhere.

> > I say:
> > Between the complete enumeration of all rationals of all intervals
> > and the enumeration of at least one rational in every interval
> > there are infinitely many further steps.

> I'll take a moment to point out that, while you say things,
> I and others prove things.

My statement is easily proved: Every interval has aleph_0 rationals. So for every interval between enumerating the first rational and completing it, there are aleph_0 steps of the sequence.

>
> There are infinitely many unit intervals (k,k+1].
>
> If at least one rational in each of infinitely-many intervals
> has been indexed, each finitely-indexed rational has been
> indexed.

So it is.

>
> All rationals are finitely-indexed rationals.

That is refuted by my proof.

>
> If at least one rational in each of infinitely-many intervals
> has been indexed, each rational has been indexed.

Impossible, because then aleph_0 indices have been applied, but at least one interval has been not enumerated completely.

>
> There are no steps, zero, none between the complete enumeration of
> all rationals of all intervals and the enumeration of at least
> one rational in every interval.

Herewith you prove that it is impossible to enumerate all rationals which lie between the first enumerated of every interval and the completion.

> Such a step would require a finitely-indexed rational to
> come after infinitely-many rationals.

Now you've got it! Yes, that is the core of my proof. in order to avoid it you prove that it is impossible to enumerate all rationals which lie between the first enumerated of every interval and the completion.

>
> Please review "finite", "infinite", and "index".

Why? I know that. Otherwise I would not have been able to refute Cantor's claim. But you can't believe it, can you?

Regards, WM

[WM]

Serg io schrieb am Sonntag, 5. September 2021 um 01:36:34 UTC+2:

> r * ℵ0 = ℵ0 for r =/= 0 where r is your number of intervals, more is always available.

Cantor applied 1*ℵo. That is what I refuted.

Regrads, WM

[Gus Gassmann]

On Sunday, 5 September 2021 at 09:07:53 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 5. September 2021 um 00:20:51 UTC+2:
> > On Saturday, 4 September 2021 at 18:15:29 UTC-3, WM wrote:
>
> > > Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.
> > OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3, ...} = IN.
> That is irrelevant since Cantor here uses only {1, 2, 3, ...}.

As usual you have zero understanding of the issues. A positive fraction is a ratio of two integers, and the correspondence between n/m (with n,m > 0) and the ordered pair (n,m) (which *is* an element of IN x IN ought to be even within your limited grasp.

> > The whole point of the diagonal argument is that the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1, n] will have been mapped, and more is not required.
>
> Firstly, this is wrong, since the lower part of the matrix is never reached,

I have no idea what image you have in your birdsized brain when you talk about the "lower" (and presumably "upper") part of a matrix not previously mentioned.

> but secondly it is irrelevant, since I show that Cantor would have used up all the indices for the first rationals of all intervals already.

This idiotic remark is exactly why I said that you have no clue how the diagonal method works. The unit intervals are *NOT* enumerated one at a time. I believe you even knew this at a time when you made your density argument. So you are either too dense to remember that or lying in the hopes of getting away with a knowingly false statement. (So maybe you *DO* know how the diagonal method works but pretend not to; it makes no difference.) Either way, this point is no longer worth discussing.

> > But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient",
> Galilei had no grasp of the difference between potential and actual infinity.

He may not have stated it explicitly, but this passage from "Two New Sciences" (taken from Wikipedia) is pretty clear on the subject:

Salviati: So far as I see we can only infer that the totality of all numbers is infinite, that the number of squares is infinite, and that the number of their roots is infinite; neither is the number of squares less than the totality of all the numbers, nor the latter greater than the former; and finally the attributes "equal," "greater," and "less," are not applicable to infinite, but only to finite, quantities. When therefore Simplicio introduces several lines of different lengths and asks me how it is possible that the longer ones do not contain more points than the shorter, I answer him that one line does not contain more or less or just as many points as another, but that each line contains an infinite number.

There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity. (Note also that this is the *LAST* book that he was working on and the ideas expressed therein presumably supersede any previous contradictory opinions.) That you, as a self-proclaimed "expert" on the History of the Infinite, do not understand this, is telling. At this point you should not be left into a toilet unsupervised; you might be inclined to think it is for drinking.

[Serg io]

On 9/5/2021 7:32 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 00:58:57 UTC+2:
>> On 9/4/2021 5:07 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 4. September 2021 um 21:21:26 UTC+2:
>>>> One consequence of these claims is that,
>>>> before all rationals have been enumerated, at least one
>>>> unit interval does not have any rationals enumerated in it.
>>>
>>> Immediately before all rationals have been enumerated,
>> I see that you have accidentally typed "Immediately" in front of
>> what I actually claim.wrote you that
>
> So you did not mean immediately? Why then wrote you that sentence? Of course there are many intervals not enumeretaed at the beginning. That is trivial and not of interest.
>>
>> Before all rationals have been enumerated, *anywhere* before
>> all rationals have been enumerated, there are infinitely-many
>> rationals which aren't enumerated. There is no "immediately before".
>
> Of course there are many rationals not enumerated. But Cantor and you claim that there is an instance when all rationals have been enumerated. Hence there must be an instance way before, where all intervals have at least one rational enumerated.

wrong, that is your conjecture, you do not specify an interval size either.

You need to Prove that first, "there must be an instance way before, where all intervals have at least one rational enumerated."

[Serg io]

no, you stated that "More are not available." which is false on its face.


This is what you snipped and are trying to hide;

"... at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available."

[Greg Cunt]

> "... at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available." [WM]

I guess, the reason for this is that his claim is nonsensical. For example the fractions n/1 might be "enumerated" by, say, the natural numbers 2, 4, 6, 8, ... Then at least one fraction is enumerated in every unit fraction. And right, for this sake "ℵo natural numbers are required". But there are STILL ℵo natural numbers "available", namely 1, 3, 5, 7, .... Hence WM's claim is WRONG.

[Jim Burns]

On 9/5/2021 8:32 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 5. September 2021 um 00:58:57 UTC+2:
>> On 9/4/2021 5:07 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 4. September 2021 um 21:21:26 UTC+2:

>>>> One consequence of these claims is that,
>>>> before all rationals have been enumerated, at least one
>>>> unit interval does not have any rationals enumerated in it.
>>>
>>> Immediately before all rationals have been enumerated,
>>
>> I see that you have accidentally typed "Immediately" in front of
>> what I actually claim.wrote you that
>
> So you did not mean immediately? Why then wrote you that
> sentence?

<WM>

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

<JB>
| Before all rationals have been enumerated, at least one

| unit interval does not have any rationals enumerated in it.

Do you want me to explain the difference between these claims?

> Of course there are many intervals not enumeretaed at the
> beginning. That is trivial and not of interest.

Before all rationals have been enumerated, *anywhere* before
all rationals have been enumerated, there are infinitely-many
rationals which aren't enumerated. There is no "immediately before".

>> Before all rationals have been enumerated, *anywhere* before
>> all rationals have been enumerated, there are infinitely-many
>> rationals which aren't enumerated. There is no "immediately before".
>
> Of course there are many rationals not enumerated.

For each p in Q+, there are j,k in N+, j/k = p.
For each j,k in N+, there are steppable {0,...,j}, {0,...,k}
in which, for adjacent h,i, i = h+1.

Each j, each k in N+ is countable to, in principle.

_All the positive rationals_ are in Q+

----
There are many schemes for indexing _all the positive rationals_

Cantor mapped each pair j,k to a single natural m.
m = j + (j+k-1)*(j+k-2)/2

That scheme maps each rational p to infinitely-many indexes m,
which seems more than sufficient for our purposes here.

Each j and each k in N+ is countable to, in principle.
Two naturals which are countable to, in principle,
have a sum which is a natural countable to, in principle.
Two naturals which are countable to, in principle,
have a product which is a natural countable to, in principle.
Since m = j + (j+k-1)*(j+k-2)/2
m is a natural countable to, in principle.

In Cantor's scheme, _all the positive rationals_ are
enumerated by an index countable to, in principle.
(Actually, each rational is enumerated infinitely-many times.)

> But Cantor and you claim that there is an instance when
> all rationals have been enumerated.

That's a point of contention between us. I have been trying to
work my way around that point to the conclusions I want.

Can we agree that there are instances when NOT all rationals
have been enumerated? Can we agree that that is what
"before all rationals have been enumerated" means?

> Hence there must be an instance way before, where all
> intervals have at least one rational enumerated.

You keep saying that, but it doesn't get any more correct.

Each rational has a finite index, countable to, in principle.
For each m, there is a steppable {0,...,m} in which,
for adjacent h,i, i = h+1.

For each rational, there are no more than finitely-many
rationals before it in the enumeration.
Finitely-many aren't enough for one in all unit intervals.
Finitely-many aren't enough to "complete" any unit interval.

( Cantor's way, each rational has infinitely-many indexes,
( which can make expressing things a bit tricky, a bit messy.
(
( This is why I prefer mapping the _unique prime factorization_
( of a rational to the unique prime factorization of its
( index. For each rational, a unique index, and, for each index
( a unique rational.
(
( Of course, Cantor's scheme has more historical resonance,
( for those who care about that sort of thing.

Before all the rationals have been enumerated, no more than
finitely-many rationals have been enumerated

>> This is a consequence of there being infinitely-many rationals,
>> each with a finite index.
>
> Cantor's and your claim are wrong.

Do you mean this claim?
| If steppable {0,...,j} and {0,...,k} exist in which,
| for adjacent h,i, i = h+1,
| then steppable {0,...,j+k} and {0,...,j*k} exist in which,
| for adjacent h,i, i = h+1.

Does your "intuition" disagree with that claim?
Do you want me to prove it for you?

> Cantor's and your claim are wrong.
> This is a consequence of there being infinitely many rationals.
> Neverftheless you will continue to claim that all will be
> enumerated at some instance. But you will deny that before
> that has happened, all intervals will have received at least
> one hit?

Before that has happened, no more than finitely-many rationals
have been enumerated. So, yes, I deny it.

> Then my "immediately" was appropriate.

Your "immediately" is incorrect. There is no "immediately".
You claim there is.

What was _inappropriate_ was your putting _your_ claim in
_my_ mouth. One might think that someone holding onto his
classroom because of "academic freedom" would understand that.

Of course that assumes a lot about the honesty and intelligence
of such a person.

>>> Immediately before all rationals have been enumerated,
>>> at least one unit interval does not have any rationals
>>> enumerated in it?
>
>> No. There is no "immediately before".
>
> There is a sequence. Each index can well be distinguished.

Only a last enumerated rational can be immediately before
all the rationals being enumerated. There isn't any.

There is no "immediately before".

>> Your confusion is probably caused by assuming that I also
>> claim that there is a last natural number.
>> I don't claim that.
>
> I don't claim that.
> But you claim that the indexing is completed somewhere.

I've avoided claiming that.
Whatever it means for the indexing to be completed somewhere,
it should be clear what it means for the indexing to NOT-be
complete. As in: "before all rationals have been enumerated".

Before all rationals have been enumerated, some rational and
all the rationals enumerated after it have not been enumerated.

Before all rationals have been enumerated, no more than
finitely-many rationals have been enumerated.

Before all rationals have been enumerated, no unit interval
has had all of its rationals enumerated.

Before all rationals have been enumerated, not all
unit intervals have at least one rational enumerated.

[WM]

Jim Burns schrieb am Sonntag, 5. September 2021 um 20:32:21 UTC+2:

> Before all rationals have been enumerated, not all
> unit intervals have at least one rational enumerated.

So, do you claim that all rationals have been enumerated before every interval has at least one rational enumerated?

Regards, WM

[WM]

Greg Cunt schrieb am Sonntag, 5. September 2021 um 16:32:42 UTC+2:

> > "... at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available." [WM]
>
> I guess, the reason for this is that his claim is nonsensical. For example the fractions n/1 might be "enumerated" by, say, the natural numbers 2, 4, 6, 8, ...

Cantor enumerates by 1, 2, 3, ...

Do you claim that all rationals have been enumerated by 1, 2, 3, ... before every interval has at least one rational enumerated? Yes or no?

Regards, WM

[WM]

Gus Gassmann schrieb am Sonntag, 5. September 2021 um 14:53:55 UTC+2:

> There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity.

Well, fine, irrelevant.

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

Regards, WM

[Jim Burns]

No infinite subset of finite indexes is all enumerated before
any other infinite subset of finite indexes is all enumerated.

For example,
is all of Primes = {2,3,5,7,11,...} enumerated
before all of Squares = {1,4,9,16,25,...} ?

No, because there is no square in Squares after all
the primes in Primes.

Is all of Squares = {1,4,9,16,25,...} enumerated
before all of Primes = {2,3,5,7,11,...} ?

No, because there is no prime in Primes after all
the squares in Squares.

Therefore, no,
all (infinitely-many) rationals will NOT have been enumerated
before every (infinitely-many) interval has at least one
rational enumerated.

Notice that "there is no ... after all ..." does not
refer to anything infinite, just primes and squares,
finite things which are all followed by more finite things.

[Gus Gassmann]

On Sunday, 5 September 2021 at 17:20:56 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 5. September 2021 um 14:53:55 UTC+2:
>
> > There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity.
> Well, fine, irrelevant.

After you removed the context, of course. Can't admit even once that you may have been even a smidgen wrong? What a pathetic, narcissistic prick!

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

Why should *anyone* subscribe to your deluded version of reality?

[Python]

This is not Jim's claim. This small exchange is, nevertheless, a
blatant illustration of both a) your inability to deal with any
kind of simple logical statements and b) your profound intellectual
dishonesty, Crank Wolfgang Mueckenheim, from Hochschule Augsburg.

[Serg io]

Why did you snip out the word "Before", Miss Leader ? Munging peoples replies! Shame!

[Serg io]

An Enumerated Fractional Ant in Every unit interval Pot

[zelos...@gmail.com]

>Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.

We can prove it for ALL rational numbers. So you are wrong.

[WM]

zelos...@gmail.com schrieb am Montag, 6. September 2021 um 07:07:17 UTC+2:
> >Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
>
> We can prove it for ALL rational numbers.

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?

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

[WM]

Jim Burns schrieb am Sonntag, 5. September 2021 um 22:42:10 UTC+2:
> On 9/5/2021 4:14 PM, WM wrote:
> > Jim Burns schrieb
> > am Sonntag, 5. September 2021 um 20:32:21 UTC+2:
>
> >> Before all rationals have been enumerated, not all
> >> unit intervals have at least one rational enumerated.
> >
> > So, do you claim that all rationals have been enumerated
> > before every interval has at least one rational enumerated?
> No infinite subset of finite indexes is all enumerated before
> any other infinite subset of finite indexes is all enumerated.

So you cannot analyze the enumeration between having indexed every interval at least once and having enumerated all intervals completely? That is the more deplorable as only finite natural numbers are used for indexing. Where does the visible part of the sequence 1, 2, 3, ... cease?

> Therefore, no,
> all (infinitely-many) rationals will NOT have been enumerated
> before every (infinitely-many) interval has at least one
> rational enumerated.
>
> Notice that "there is no ... after all ..." does not
> refer to anything infinite, just primes and squares,
> finite things which are all followed by more finite things.

For intervals this is very clear. All are finite and all are in a well-order.
And here we have an "after all": It is impossible to complete the enumeration of all intervals unless all intervals have been deflowered before. That is a very simple chain of cause and result.

Bet perhaps you can imagine a slight correction of Cantor's sequence

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

Since you believe that all fractions are indexed, also all fractions n/1 for n > 0 will be indexed. Therefore it would do no harm if we indexed first als fractions of the form n/1 and then the remaining fractions. Note that this is what Cantor does, if he covers all these fractions. He does it only slightly later in order to confuse his audience. Every set theorist knows that my proposal will fail. But when these fractions are better hidden in the sequence, nobody will notice.

Of course your agument with primes and squares is of same kind. If he really indexed all primes, then indexing them first would not change the whole story.

Fight under the banner of confusion!

Regards, WM

[Serg io]

On 9/6/2021 7:56 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 22:42:10 UTC+2:
>> On 9/5/2021 4:14 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 5. September 2021 um 20:32:21 UTC+2:
>>
>>>> Before all rationals have been enumerated, not all
>>>> unit intervals have at least one rational enumerated.
>>>
>>> So, do you claim that all rationals have been enumerated
>>> before every interval has at least one rational enumerated?
>> No infinite subset of finite indexes is all enumerated before
>> any other infinite subset of finite indexes is all enumerated.
>
> So you cannot analyze the enumeration between having indexed every interval at least once and having enumerated all intervals completely?

Since you say are a Math Professor, you should be the one providing answers along with the showing or Proof. But you do not.

> That is the more deplorable as only finite natural numbers are used for indexing.

100% wrong, did you stop at k again ?

> Where does the visible part of the sequence 1, 2, 3, ... cease?

at the 'last' one, silly.

>
>> Therefore, no,
>> all (infinitely-many) rationals will NOT have been enumerated
>> before every (infinitely-many) interval has at least one
>> rational enumerated.
>>
>> Notice that "there is no ... after all ..." does not
>> refer to anything infinite, just primes and squares,
>> finite things which are all followed by more finite things.
>
> For intervals this is very clear. All are finite and all are in a well-order.

wrong , you have brain fart. Intervals are infinite

> And here we have an "after all": It is impossible to complete the enumeration of all intervals unless all intervals have been deflowered before. That is a very simple chain of cause and result.

wrong, and wrong. Your flailing is noted

>
> Bet perhaps you can imagine a slight correction of Cantor's sequence
>
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...
>
> Since you believe that all fractions are indexed, also all fractions n/1 for n > 0 will be indexed. Therefore it would do no harm if we indexed first als fractions of the form n/1 and then the remaining fractions. Note that this is what Cantor does, if he covers all these fractions. He does it only slightly later in order to confuse his audience. Every set theorist knows that my proposal will fail. But when these fractions are better hidden in the sequence, nobody will notice.

no. Go back to the math and study, avoid jumping to conclusions

>
> Of course your agument with primes and squares is of same kind. If he really indexed all primes, then indexing them first would not change the whole story.

red herring

[Serg io]

On 9/6/2021 7:40 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 6. September 2021 um 07:07:17 UTC+2:
>>> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
>>
>> We can prove it for ALL rational numbers.
>
> 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?

go ahead, its your idea, show us how...

>
> 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?

no, you have not provided a proof. where is it ?

>
> Regards, WM
>

[zelos...@gmail.com]

>You believe that all fractions

We KNOW it because it can be PROVEN.

>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?

You need to prove that the new function is a bijection then.

When it comes to the question of cardinality the actual bijection is not relevant only that ONE exists.

>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?

It is entirely fucking irrelevant. All that matters is that the bijection exist.

If we can prove that there is a surjection Q->N and another surjection N->Q we know there exists a bijection N->Q and nothing else is needed and we can EASILY create a surjection in either direction.

[WM]

zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
> >You believe that all fractions
> We KNOW it because it can be PROVEN.

The contrary can be proven too. That means we have found an inconsistency.

> >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?
> You need to prove that the new function is a bijection then.

A bijection cannot lose its character by transpositions. That's proof enough.

>
> >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?
> It is entirely fucking irrelevant.

No, it shows an inconsistency. Don't you like inconsistencies?

Regards, WM

[Greg Cunt]

On Tuesday, September 7, 2021 at 9:04:52 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
> >
> > We KNOW it because it can be PROVEN.
> >
> The contrary can be proven too.

No, it can't, you lying sack of shit!

[Python]

Crank Wolfgang Mueckenheim, WM a écrit :

> zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
>>> You believe that all fractions
>> We KNOW it because it can be PROVEN.
>
> The contrary can be proven too. That means we have found an inconsistency.
>
>>> 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?
>> You need to prove that the new function is a bijection then.
>
> A bijection cannot lose its character by transpositions.

You are not using a transposition (i.e. bijection) but a re-ordering.

> That's proof enough.

So NO, Crank Wolfgang Mueckenheim, from Hochschule Augsburg, it is
not a proof.

[Ross A. Finlayson]

That "transpositions of elements of a bijection is still a bijection" is
the _Cartesian_ notion of function. That there are _non-Cartesian_
functions in the sense that transpositions result in "not a function",
is a key understanding what results that something like the Equivalency
Function, or sweep, is unique and so detailed as a function only as
by its bounds and the course-of-passage of elements of the domain
with respect to the range.

So, with sweep or f, where, f(0) = 0 and lim d->oo f(d) = 1, there
is only that there is one antidiagonal, and, it's only at the end,
where, it exists. Combined with other results that establish
extent, density, completeness, and measure, makes for the
modern example, a unique counterexample to uncountability.

So, to extend modern mathematics for continuity's sake,
first is the emplacement _underneath_ of a countable
continuous domain [0,1], exemplified by the "natural/unit
equivalency function", then there is field continuity the
usual notion of the infinite divisibility of finites, _after_
the infinitely-divided finite 1, then, there's a third notion
of continuity the signal continuity, what with respect to
dense domains that exhaust the rational, result in a time-signal
component, that also helps explains how the rationals can
be dense in the reals while the irrationals are uncountable
according to _Cartesian_ functions, the space thereof,
because every neighborhood of rationals contains rationals.


So, Python et alia, we all know that WM is mostly Hodges' hopeless,
this "usual" approach to standard infinitesimals, and a "usual"
approach to the signal or information after the rational to the
continuous, makes for _at least three different definitions of
continuity that reflect its character_.

This way, and, largely no other way, it's possible to have a
mathematics _consistent in this way_, what both has the
uncountable after regular ordinary trans-finite cardinals,
and, what results as a particular countable continuous domain
after the naturals the line reals, and, a particular countable
continuous domain after the rationals ("which is at least two
copies of the naturals"), three different models of continuous
domains, mathemtical, thus, mathematics.

Otherwise a usual notion of ultrafilters and countable saturation
as for making countable additivity which is the analytical character
of the real analysis, ends up being a giant stack not cool, that instead
the "non-Archimedean", or "re-Archimedean" as it were, is teachable
as directly from Zeno's notions, and makes for a simple placement of
constant uniform motion, and explaining infinite divisibility, neatly.

So: Python, Zelos, Sergio, Gus, WM, Jim, Dan, Chris, Rafters, (Markus):
_this way they get along together_, and explaining why Weierstrass
is of course correct, that also Leibniz, Maclaurin, et cetera aren't wrong.

I suppose there were only two things in school instruction that
I rejected, or, had to accept to go along: that .999 wasn't a notation
for the line through 1.0, and, that space-time locally isn't flat,
or at most, that the cosmological constant wasn't an infinitesimal.

Then, later with respect to "all the naturals exhaust divisibility of
the unit", that didn't have to be defended because it didn't happen
to be part of instruction. It's pretty much the same thing, though.



Univalency, the illative even, supertasks, the anti-foundational,
modern rehabilitations of infinitesimals, ultrafilters, all these things
are simply approaches _after_ "Cartesian Cantorian set theory with
LUB and measure 1.0 as furthermore axioms", the missing link was
that it was actually _before_ DesCartes than _after_ Cantor, to care.


For then an actual entire theory about the powerset result more
generally, is for the theory with ubiquitous ordinals for the ordinal,
and what results as powerset as order type as successor: making
for that it's reduced in symbolic resources to the possibility of
_one, good, theory_.


So, amigos or amigas as it were, rest assured the continuum's replete.

[Ross A. Finlayson]

(Also delta-epsilonics works in most all models of infinitesimals
because it's basically topological, while there are basically notions of
the "standard" infinitesimals "all-countable" and "signal" infinitesimals
"all-uncountable", that establish analytical for topological character
besides as topologically, of course, for usual point-set topologies,
like geometry.)

Mathematics here has models of continuous domains or real numbers
what are "defining a continuous function f(x) = 0, in fixed numerical
resources", in real-valued formulae, with line, field, and signal continuity.


Then, about quasi-invariant measure theory, the hyperasymptotic,
results only in Ramsey theory, running constants and the invariants of
physical constants in physics and phase transition, parastatistics,
parametrization after symmetry-flex, results invoking Banach-Tarski,
and so on, is for clearing up replete continuity before stacking on
a trash-can fire.

Which of course is the relephant....

[Dan Christensen]

On Tuesday, September 7, 2021 at 3:04:52 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
> > >You believe that all fractions

> > We KNOW it because it can be PROVEN.

> The contrary can be proven too. That means we have found an inconsistency.

Not if it is only proven in your MuckeMath system. That would only prove that MuckeMath is inconsistent. If you want to demonstrate an inconsistency in ZFC, you would have to prove it using ONLY the axioms of ZFC. Good luck with that.

It seems you have failed once again, Mucke. Still NO INCONSISTENCIES in ZFC!!! Deal with it.

Dan

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

[Serg io]

On 9/7/2021 2:04 PM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
>>> You believe that all fractions
>> We KNOW it because it can be PROVEN.
>
> The contrary can be proven too.

Wrong!

[zelos...@gmail.com]

>The contrary can be proven too. That means we have found an inconsistency.

None of your so called "proofs" are valid however. That means you have NOT proven the contrary.

>A bijection cannot lose its character by transpositions. That's proof enough.

Prove that statement.

>No, it shows an inconsistency. Don't you like inconsistencies?

it doesn't prove that at all. You want it to but it doesn't.

[Ross A. Finlayson]

Which of course is the relephant!

[Ross A. Finlayson]

On Tuesday, September 7, 2021 at 7:34:48 PM UTC-7, Dan Christensen wrote:

My inconsistency in ZFC is it contains itself.

I.e., it's never deductively _not closed_.

Which breaks many or most Dan's usual theorems.

[Gus Gassmann]

On Wednesday, 8 September 2021 at 03:31:04 UTC-3, Ross A. Finlayson wrote:
[...]

> Which of course is the relephant!

"*A* relephant". If you quote Marx brothers without attribution, please do it right...

[Ross A. Finlayson]

This one's a kaleidoscope, the relephant including a flammable mathematical trash can.

[WM]

Dan Christensen schrieb am Mittwoch, 8. September 2021 um 04:34:48 UTC+2:
> On Tuesday, September 7, 2021 at 3:04:52 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
> > > >You believe that all fractions
>
> > > We KNOW it because it can be PROVEN.
>
> > The contrary can be proven too. That means we have found an inconsistency.
> Not if it is only proven in your MuckeMath system.

It is proven using basic logic: If all fractions of https://en.wikipedia.org/wiki/File:Diagonal_argument.svg are to be indexed, then also the fractions of the first column must habe been indexed before completing the task. But after indexing the fractions of the first column aleph_0 indexes have been issued. More are not available.

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

Regards, WM

[WM]

zelos...@gmail.com schrieb am Mittwoch, 8. September 2021 um 07:04:31 UTC+2:
> >The contrary can be proven too. That means we have found an inconsistency.
> None of your so called "proofs" are valid however. That means you have NOT proven the contrary.
> >A bijection cannot lose its character by transpositions. That's proof enough.
> Prove that statement.

A transposition changes two pairs (n, a) and (m, b) into (n, b) and (m, a). That does not change injectivity and surjectivity for these two pairs. All others remain unaltered. Hence injectivity and surjectivity remain intact.

Regards, WM

[Serg io]

On 9/8/2021 2:31 PM, WM wrote:
> Dan Christensen schrieb am Mittwoch, 8. September 2021 um 04:34:48 UTC+2:
>> On Tuesday, September 7, 2021 at 3:04:52 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
>>> zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
>>>>> You believe that all fractions
>>
>>>> We KNOW it because it can be PROVEN.
>>
>>> The contrary can be proven too. That means we have found an inconsistency.
>> Not if it is only proven in your MuckeMath system.
>
> It is proven using basic logic: If all fractions of https://en.wikipedia.org/wiki/File:Diagonal_argument.svg are to be indexed, then also the fractions of the first column must habe been indexed before completing the task.

Liar!

> But after indexing the fractions of the first column aleph_0 indexes have been issued. More are not available.

LIAR !


<snip crap>

[Serg io]

LIAR! your brain is f*ked up. Why do you Lie ?

trying to get even with someone ?

[Dan Christensen]

On Wednesday, September 8, 2021 at 3:32:00 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Mittwoch, 8. September 2021 um 04:34:48 UTC+2:
> > On Tuesday, September 7, 2021 at 3:04:52 PM UTC-4, WM (Wolfgang Muckenheim aka Mucke) wrote:
> > > zelos...@gmail.com schrieb am Dienstag, 7. September 2021 um 07:05:45 UTC+2:
> > > > >You believe that all fractions
> >
> > > > We KNOW it because it can be PROVEN.
> >
> > > The contrary can be proven too. That means we have found an inconsistency.
> > Not if it is only proven in your MuckeMath system.

> It is proven using basic logic: If all fractions of https://en.wikipedia.org/wiki/File:Diagonal_argument.svg are to be indexed, then also the fractions of the first column must habe been indexed before completing the task. But after indexing the fractions of the first column aleph_0 indexes have been issued. More are not available.

Pure gibberish.

We need proofs of two contradictory theorems from you, both using ONLY the axioms of ZFC. The axioms of MuckeMath are quite useless for this purpose (or any other AFAIK). Start by stating these theorems. (Hint: One theorem is the negation of the other.) How about it, Mucke? Instant fame and fortune awaits!

Theorem 1: _____________________

Theorem 2 ~_____________________

Fill in the blanks.

Don't hold your breath, folks! (Hee, hee!)

Dan

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

Visit my Math Blog at http://www.dcproof.wordpress.com

[zelos...@gmail.com]

Do a proper proof :) You are so lose with words I do not trust you using them like this!

[WM]

zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:06:27 UTC+2:
> onsdag 8 september 2021 kl. 21:35:28 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Mittwoch, 8. September 2021 um 07:04:31 UTC+2:
> > > >The contrary can be proven too. That means we have found an inconsistency.
> > > None of your so called "proofs" are valid however. That means you have NOT proven the contrary.
> > > >A bijection cannot lose its character by transpositions. That's proof enough.
> > > Prove that statement.
> > A transposition changes two pairs (n, a) and (m, b) into (n, b) and (m, a). That does not change injectivity and surjectivity for these two pairs. All others remain unaltered. Hence injectivity and surjectivity remain intact.
> >

> Do a proper proof :)

Abobe you see a proper proof.

Regards, WM

[WM]

Serg io schrieb am Mittwoch, 8. September 2021 um 23:08:20 UTC+2:

> LIAR! your brain is f*ked up. Why do you Lie ?

Let's split the proof in two steps.First step:
https://en.wikipedia.org/wiki/File:Diagonal_argument.svg After indexing the fractions of the first column aleph_0 indexes have been issued.
Understandable?

Regards, WM

[WM]

Dan Christensen schrieb am Mittwoch, 8. September 2021 um 23:40:27 UTC+2:

Let's split the proof in two steps.First step:

https://en.wikipedia.org/wiki/File:Diagonal_argument.svg After indexing the fractions of the first column aleph_0 indexes have been issued.
Understandable?

Regards, WM

[FromTheRafters]

WM expressed precisely :

No. Follow the arrows and 'index' in that order.

[Dan Christensen]

On Thursday, September 9, 2021 at 9:45:41 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Mittwoch, 8. September 2021 um 23:40:27 UTC+2:
>
> Let's split the proof in two steps.First step:
> https://en.wikipedia.org/wiki/File:Diagonal_argument.svg After indexing the fractions of the first column aleph_0 indexes have been issued.

I think I see your problem now. Every natural number could be used to index just those fractions in the first column. So you think it must then be impossible to also index all the other fractions as well. Correct?

This is similar to your problem in the past with the fact that there are just as many natural numbers as there are even numbers, i.e. the fact that there is a bijection f mapping {1, 2, 3, ... } to {2, 4, 6, ... }, namely f(x)=2x. You could say that the first even number is 2, the second is 4 and so on. If you cannot grasp this fundamental principle, you will never understand Cantor.