WM Challange, Dark property existence

[zelos...@gmail.com]

Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"

Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?

If your proof shows it applies to every natural number, it is invalid.
If your definition is not crystal clear FOL, it is invalid.
If your sets are not disjoint, it is invalid as you constnatly claim they are mutually exclusive.

Let's see you try.

No, your proof of
An e N: n e [0,n]=F_n
Aka, "all natural is in in a finite set" is instantly invalid as that applies to all natural numbers, thus one set is non-empty.

[Gus Gassmann]

On Thursday, 21 April 2022 at 07:34:25 UTC-3, zelos...@gmail.com wrote:
> Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
>
> Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?

That's actually not very difficult, nor is it original.

A number is called "instantiated" if it has been used by someone, somewhere. There is an obvious time dependence, so we call IN(t) the set of natural numbers that have been instantiated at time t. This is a finite set, and its complement could be called "dark", if you want. Ed Nelson talked in terms of definable numbers, and I accused WM previously of having stolen his ideas --- completely misunderstanding everything. (Of course he denied the stealing, but that he doesn't have a clue is axiomatic.)

Now, of course this definition is problematic and leads to countless difficulties down the road. I believe Nelson thought that he could somehow derive an inconsistency in ZFC from it, but he was a good enough mathematician to retract his proof attempt when it was pointed out that there was a fatal flaw. (Good luck getting WM to admit any fallibility.)

[WM]

zelos...@gmail.com schrieb am Donnerstag, 21. April 2022 um 12:34:25 UTC+2:
> Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
>
> Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?

This task is more difficult than the proof that there are dark fractions. If you have understood that it is impossible to enumerate all fractions, because starting from

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

every attempt to collect all fractions in the first column will fail:

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

When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.

If you have understood this, then we can proceed to understand, that even most places in the first column, all of which have been occupied by integer fractions or natural numbers, are undefinable too.

> No, your proof of
> An e N: n e [0,n]=F_n
> Aka, "all natural is in in a finite set" is instantly invalid as that applies to all natural numbers, thus one set is non-empty.

By induction we prove that every definable natural number, i.e., every natural number which is subject to induction, belongs to a finite set F_n but has ℵo successors which cannot be removed whatever n you consider. That means they cannot be used as individuals.

Regards, WM

[WM]

horand....@gmail.com schrieb am Donnerstag, 21. April 2022 um 13:22:54 UTC+2:

> A number is called "instantiated" if it has been used by someone, somewhere.

A number is called instantiatable if it can be instantiated. By induction we prove that every instantiatable natural number, i.e., every natural number which is subject to induction, belongs to a finite set F_n but has ℵo successors which cannot be removed whatever n you consider. That means they cannot be instantiated or used as individuals.

Regards, WM

[Gus Gassmann]

On Thursday, 21 April 2022 at 09:06:03 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 21. April 2022 um 13:22:54 UTC+2:
>
> > A number is called "instantiated" if it has been used by someone, somewhere.

> A number is called instantiatable if it can be instantiated. By induction we prove ...

In order to prove *ANYTHING* you'd actually have to have a brain function. Your verbal diarrhea does not count as proof, of anything. Also look up "inductive set" some time.

[Gus Gassmann]

On Thursday, 21 April 2022 at 09:03:43 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 21. April 2022 um 12:34:25 UTC+2:
> > Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
> >
> > Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?
> This task is more difficult than the proof that there are dark fractions. If you have understood that it is impossible to enumerate all fractions, because starting from
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
>
> every attempt to collect all fractions in the first column will fail:
>
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
> ... ... ... ...
>

> When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, ...

Your usual bullshit. "When all ... have transferred" indicates a *LIMIT*. It is not hard to figure out what that limit looks like -- provided one has a brain function, of course.

[sergio]

On 4/21/2022 7:03 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 21. April 2022 um 12:34:25 UTC+2:
>> Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
>>
>> Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?
>
> This task is more difficult than the proof that there are dark fractions. If you have understood that it is impossible to enumerate all fractions, because starting from
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...

that is called the matrix of rational numbers which is infinite number of rows and infinite number of columns

>
> every attempt to collect all fractions in the first column will fail:
>
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
> ... ... ... ...

that is your failed attempt to confuse people, using your swaparoo mixerupper technique, you still have two indexes, and you need one index to go one to
one with the naturals.

Instead, try converting the two indexes to one index like Cantor does. Simple and easy.


<snip>

>
>> No, your proof of
>> An e N: n e [0,n]=F_n
>> Aka, "all natural is in in a finite set" is instantly invalid as that applies to all natural numbers, thus one set is non-empty.
>
> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction, belongs to a finite set F_n but has ℵo successors which cannot be removed whatever n you consider. That means they cannot be used as individuals.

nonsense. Your "definable", "cannot be removed" and "used as individuals" are bogus, they are not mathematical operations.

>
> Regards, WM

[sergio]

that is nonsense.

[WM]

But there is no limit. All of Cantor's sequence are all that have ℵo successors. You cannot remove them like you cannot remove all terms of the sequence (1/n) to *reach* the limit. The limit exists, it *is* but you cannot approach it.

> It is not hard to figure out what that limit looks like -- provided one has a brain function, of course.

The limit looks like omega, but between n and omega there are ℵo natnumbers which never can be passed one by one. And in the matrix, at *every* place outside the first column, there are ℵo fractions waiting before omega - and cannot be passed.

Regards, WM

[sergio]

On 4/21/2022 8:58 AM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 21. April 2022 um 15:00:42 UTC+2:
>> On Thursday, 21 April 2022 at 09:03:43 UTC-3, WM wrote:
>>> zelos...@gmail.com schrieb am Donnerstag, 21. April 2022 um 12:34:25 UTC+2:
>>>> Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
>>>>
>>>> Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?
>>> This task is more difficult than the proof that there are dark fractions. If you have understood that it is impossible to enumerate all fractions, because starting from
>>>
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>>
>>> every attempt to collect all fractions in the first column will fail:
>>>
>>> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
>>> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
>>> ... ... ... ...
>>>
>>> When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, ...
>>
>> Your usual bullshit. "When all ... have transferred" indicates a *LIMIT*.
>
> But there is no limit. All of Cantor's sequence are all that have ℵo successors. You cannot remove them like you cannot remove all terms of the sequence (1/n) to *reach* the limit. The limit exists, it *is* but you cannot approach it.

*this is very simple math*, limits, infinite sequences, enumeration,

sorry you are having so much extreme difficulty understanding any of it.

>
>> It is not hard to figure out what that limit looks like -- provided one has a brain function, of course.
>
> The limit looks like omega, but between n and omega there are ℵo natnumbers which never can be passed one by one.

By counting ? No one has that many sheeps. However if you tried to count them on Earth, it would never end, there would be infinite # of sheeps
to you, as more are born in flocks, as you are counting other flocks.

> And in the matrix, at *every* place outside the first column, there are ℵo fractions waiting before omega - and cannot be passed.

No. your matrix is a failed idea you had. It underscores over and over your limited understanding of math.

>
> Regards, WM

[Gus Gassmann]

On Thursday, 21 April 2022 at 10:58:19 UTC-3, WM wrote:
[...]

> But there is no limit.

[...]
That would be a whole lot more convincing if you had a clue what a limit actually is. In the present case, you are spouting irrelevant, nonsensical shit.

[zelos...@gmail.com]

torsdag 21 april 2022 kl. 14:03:43 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 21. April 2022 um 12:34:25 UTC+2:
> > Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
> >
> > Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?
> This task is more difficult than the proof that there are dark fractions. I

Go ahead and define it.

>f you have understood that it is impossible to enumerate all fractions, because starting from

You still haven't defined what "dark" means

>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
>
> every attempt to collect all fractions in the first column will fail:

No one is doing that

>
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
> ... ... ... ...
>
> When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.

N^2 and N are not the same sets so this is invalid. You ahve STILL NOT DEFINED DARK!

>
> If you have understood this, then we can proceed to understand, that even most places in the first column, all of which have been occupied by integer fractions or natural numbers, are undefinable too.
> > No, your proof of
> > An e N: n e [0,n]=F_n
> > Aka, "all natural is in in a finite set" is instantly invalid as that applies to all natural numbers, thus one set is non-empty.
> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction, belongs to a finite set F_n but has ℵo successors which cannot be removed whatever n you consider. That means they cannot be used as individuals.
>
> Regards, WM

It means nothing of the sort.

You failed to define what "dark" is, you failed to demonstrate their existence, you failed EVERYTHING!

[WM]

zelos...@gmail.com schrieb am Montag, 25. April 2022 um 07:33:44 UTC+2:
> torsdag 21 april 2022 kl. 14:03:43 UTC+2 skrev WM:

> >f you have understood that it is impossible to enumerate all fractions, because starting from
> You still haven't defined what "dark" means

Dark natural numbers cannot be used individually. They can only be used collectively. They can be removed to isolate omega
{1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
Definable natural numbers can be used individually, but omega will never be isolated
∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .

> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, 3/2, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> >
> > every attempt to collect all fractions in the first column will fail:
> No one is doing that

Everyone claiming that all fractions can be enumerated claims that all fractions can be collected in the first column.

> > When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.
> N^2 and N are not the same sets so this is invalid.

According to Cantor N^2 can be linearized.

Regards, WM

[FromTheRafters]

WM laid this down on his screen :

> zelos...@gmail.com schrieb am Montag, 25. April 2022 um 07:33:44 UTC+2:
>> torsdag 21 april 2022 kl. 14:03:43 UTC+2 skrev WM:
>
>>> f you have understood that it is impossible to enumerate all fractions,
>>> because starting from
>> You still haven't defined what "dark" means
>
> Dark natural numbers cannot be used individually. They can only be used
> collectively. They can be removed to isolate omega {1, 2, 3, ..., omega} \
> {1, 2, 3, ...} = {omega} Definable natural numbers can be used individually,
> but omega will never be isolated ∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2,
> 3, ..., n} = {n+1, n+2, ..., omega} .
>
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>>
>>> every attempt to collect all fractions in the first column will fail:
>> No one is doing that
>
> Everyone claiming that all fractions can be enumerated claims that all
> fractions can be collected in the first column.

Liar!

[sergio]

On 4/25/2022 5:15 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 25. April 2022 um 07:33:44 UTC+2:
>> torsdag 21 april 2022 kl. 14:03:43 UTC+2 skrev WM:
>
>>> f you have understood that it is impossible to enumerate all fractions, because starting from
>> You still haven't defined what "dark" means
>
> Dark natural numbers cannot be used individually. They can only be used collectively. They can be removed to isolate omega
> {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}

Wrong. 1,2,3,4,... are all natural numbers. Where in the equation above is your "collectively removed darkies" ? Which term is it ?

> Definable natural numbers can be used individually, but omega will never be isolated
> ∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .

Wrong again. you stopped at n. when you stop at n, you only create a FISON and an ENDSEGMENT, no darkies there.

Where in the equation above is your "collectively removed darkies" ? Point out that term.

>
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>>
>>> every attempt to collect all fractions in the first column will fail:
>> No one is doing that
>
> Everyone claiming that all fractions can be enumerated claims that all fractions can be collected in the first column.

BS. WM is the only one claiming that, then denies it.

>
>>> When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.
>> N^2 and N are not the same sets so this is invalid.
>
> According to Cantor N^2 can be linearized.

since every single one of your posts has several falsehoods in it, why should anyone believe you on this statement ?

>
> Regards, WM

[WM]

Is the contents of the first column an infinite sequence?

Regards, WM

[Jim Burns]

On 4/25/2022 1:39 PM, WM wrote:

> Is the contents of the first column
> an infinite sequence?

It is an infinite sequence as "infinite sequence"
is usually meant.

No one can be sure of what you mean by
your question, though.
I include you (WM) among those not sure of
what you (WM) mean.

The contents of the first column are
totally-ordered such that,
there is a first entry but no last entry,
and,
for each entry k/1,
for each split BEFORE and AFTER of
the entries from 1/1 to k/1,
BEFORE contains a last entry and
AFTER contains a first entry.

That's an infinite sequence.

[FromTheRafters]

WM was thinking very hard :

No, it isn't.

[sergio]

is the above matrix first column showing "1,2,3,4,5..." ?

[zelos...@gmail.com]

måndag 25 april 2022 kl. 12:15:16 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 25. April 2022 um 07:33:44 UTC+2:
> > torsdag 21 april 2022 kl. 14:03:43 UTC+2 skrev WM:
>
> > >f you have understood that it is impossible to enumerate all fractions, because starting from
> > You still haven't defined what "dark" means
> Dark natural numbers cannot be used individually.

Define that in FOL because "individually" is meaningless.

>They can only be used collectively.

Define that in FOL

>They can be removed to isolate omega

Omega is an ordinal, it is not part of natural numbers so ti is irrelevant here.

> {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
> Definable natural numbers can be used individually, but omega will never be isolated

Define in FOL

> ∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .

And?

> > > 1/1, 1/2, 1/3, 1/4, ...
> > > 2/1, 2/2, 2/3, 2/4, ...
> > > 3/1, 3/2, 3/3, 3/4, ...
> > > 4/1, 4/2, 4/3, 4/4, ...
> > > 5/1, 5/2, 5/3, 5/4, ...
> > > ...
> > >
> > > every attempt to collect all fractions in the first column will fail:
> > No one is doing that
> Everyone claiming that all fractions can be enumerated claims that all fractions can be collected in the first column.

No one is claiming that, what one claim is that there is a function from N to Q+, that is a different claim you imbecile.

> > > When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.
> > N^2 and N are not the same sets so this is invalid.
> According to Cantor N^2 can be linearized.

We have a bijection between them, but they are not the same fucking set.

>
> Regards, WM

[WM]

Jim Burns schrieb am Montag, 25. April 2022 um 20:48:48 UTC+2:
> On 4/25/2022 1:39 PM, WM wrote:
>
> > Is the contents of the first column
> > an infinite sequence?
> It is an infinite sequence as "infinite sequence"
> is usually meant.

So it is. 1/1, 2/1, 3/1, ...
By values this is same as 1, 2, 3, ...
Therefore here are all indices which Cantor wants to spread over all positive fractions.
But it is obvious that by exchanging X's and O's as Cantor prescribed

XOOO...
XOOO...
XOOO...
XOOO...
...

where the O's indicate not indexed fractions and the X indicate indexed fractions

XXOO...
OOOO...
XOOO...
XOOO...
...

XXOO...
XOOO...
OOOO...
XOOO...
...

XXXO...
XOOO...
OOOO...
OOOO...
...

never any O will disappear, let alone all.

> No one can be sure of what you mean by
> your question, though.

Everybody knowing Cantor's approach

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

can understand it, if he is not too stupid.

> The contents of the first column are
> totally-ordered such that,
> there is a first entry but no last entry,

It is the set of indices.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Dienstag, 26. April 2022 um 06:50:32 UTC+2:
> måndag 25 april 2022 kl. 12:15:16 UTC+2 skrev WM:

> > > You still haven't defined what "dark" means
> > Dark natural numbers cannot be used individually.
> Define that in FOL because "individually" is meaningless.
> >They can only be used collectively.
> Define that in FOL

Why should I use the language FOOL?

> >They can be removed to isolate omega
> Omega is an ordinal, it is not part of natural numbers so ti is irrelevant here.

I use the set {1, 2, 3, ..., omega} which is existing in ZF.

> > {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
> > Definable natural numbers can be used individually, but omega will never be isolated

> > ∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
> And?

This is a difference.

> > > > 1/1, 1/2, 1/3, 1/4, ...
> > > > 2/1, 2/2, 2/3, 2/4, ...
> > > > 3/1, 3/2, 3/3, 3/4, ...
> > > > 4/1, 4/2, 4/3, 4/4, ...
> > > > 5/1, 5/2, 5/3, 5/4, ...
> > > > ...
> > > >
> > > > every attempt to collect all fractions in the first column will fail:
> > > No one is doing that
> > Everyone claiming that all fractions can be enumerated claims that all fractions can be collected in the first column.
> No one is claiming that, what one claim is that there is a function from N to Q+, that is a different claim

No, that is precisely this claim. All fractions can be collected in the first column since the line numbers are the indices of an infinite sequence.

> > > > When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.
> > > N^2 and N are not the same sets so this is invalid.
> > According to Cantor N^2 can be linearized.
> We have a bijection between them,

that establishs my claim.

Regards, WM

[zelos...@gmail.com]

tisdag 26 april 2022 kl. 13:15:57 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 26. April 2022 um 06:50:32 UTC+2:
> > måndag 25 april 2022 kl. 12:15:16 UTC+2 skrev WM:
>
> > > > You still haven't defined what "dark" means
> > > Dark natural numbers cannot be used individually.
> > Define that in FOL because "individually" is meaningless.
> > >They can only be used collectively.
> > Define that in FOL
> Why should I use the language FOOL?

FOL; First order logic.

Because then you cannot be a snake and evade things by using ambiguity of words.

> > >They can be removed to isolate omega
> > Omega is an ordinal, it is not part of natural numbers so ti is irrelevant here.
> I use the set {1, 2, 3, ..., omega} which is existing in ZF.

The infinite set in ZFC by AoI has no omega, but the ordinal you are on about, DOES exist. But it is not the natural numbers + omega for natural numbers.

> > > {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
> > > Definable natural numbers can be used individually, but omega will never be isolated
> > > ∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
> > And?
> This is a difference.

Nope, all it means is that your N_def=N

> > > > > 1/1, 1/2, 1/3, 1/4, ...
> > > > > 2/1, 2/2, 2/3, 2/4, ...
> > > > > 3/1, 3/2, 3/3, 3/4, ...
> > > > > 4/1, 4/2, 4/3, 4/4, ...
> > > > > 5/1, 5/2, 5/3, 5/4, ...
> > > > > ...
> > > > >
> > > > > every attempt to collect all fractions in the first column will fail:
> > > > No one is doing that
> > > Everyone claiming that all fractions can be enumerated claims that all fractions can be collected in the first column.
> > No one is claiming that, what one claim is that there is a function from N to Q+, that is a different claim
> No, that is precisely this claim. All fractions can be collected in the first column since the line numbers are the indices of an infinite sequence.

Nope, those are DIFFERENT claims. How is this hard for you to understand? How retarded are you?

> > > > > When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.
> > > > N^2 and N are not the same sets so this is invalid.
> > > According to Cantor N^2 can be linearized.
> > We have a bijection between them,
> that establishs my claim.

Nope, it doesn't

>
> Regards, WM

[Gus Gassmann]

On Tuesday, 26 April 2022 at 08:37:20 UTC-3, zelos...@gmail.com wrote:
> tisdag 26 april 2022 kl. 13:15:57 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 26. April 2022 um 06:50:32 UTC+2:
> > > måndag 25 april 2022 kl. 12:15:16 UTC+2 skrev WM:

[...]

> > > > Everyone claiming that all fractions can be enumerated claims that all fractions can be collected in the first column.
> > > No one is claiming that, what one claim is that there is a function from N to Q+, that is a different claim
> > No, that is precisely this claim. All fractions can be collected in the first column since the line numbers are the indices of an infinite sequence.
> Nope, those are DIFFERENT claims. How is this hard for you to understand? How retarded are you?

It is the same claim, or at least very closely related. The first column of the matrix has aleph_0 places. If Q+ is countable, then the positive fractions can be arranged to fill (exactly) the places in the first column. And the stepwise matrix process is a series of transformations that gets you there. It is constructed exactly so that the positive fractions line up in the column in the order described by the Cantor sequence. What WM does not understand/accept/whatever is that the process is infinite, and at no finite step is the process complete. In order to figure out the final process one has to take a limit, which is also not very difficult.

[...]

[sergio]

On 4/26/2022 6:10 AM, WM wrote:
> Jim Burns schrieb am Montag, 25. April 2022 um 20:48:48 UTC+2:
>> On 4/25/2022 1:39 PM, WM wrote:
>>
>>> Is the contents of the first column
>>> an infinite sequence?
>> It is an infinite sequence as "infinite sequence"
>> is usually meant.
>
> So it is. 1/1, 2/1, 3/1, ...

<snip crap>

>
> Everybody knowing Cantor's approach
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> can understand it, if he is not too stupid.

So you agree that Cantors Enumeration of the rationals is simple and easy.

In contrast, your approach using switcharoo swap-um + mutulating what was the matrix of rationals, and then declaring that you lost fractions along the
way, is clearly deception, bogus, fumbled, + bungled.

[FromTheRafters]

Gus Gassmann presented the following explanation :

He simply refuses to accept that |N|^2 = |N|

[sergio]

Face it, WM is Mathematically Retarded.

Would you trust this guy counting your change, since he made Scrooge McDuck go broke ?

[Gus Gassmann]

On Tuesday, 26 April 2022 at 11:52:05 UTC-3, sergio wrote:
[...]
> Face it, WM is Mathematically Retarded.
>
> Would you trust this guy counting your change, since he made Scrooge McDuck go broke ?

I wouldn't trust him with the time of day, but that's a different matter.

[WM]

FromTheRafters schrieb am Dienstag, 26. April 2022 um 16:09:00 UTC+2:

> He simply refuses to accept that |N|^2 = |N|

No I simply disprove it.

XOOO...
XOOO...
XOOO...
XOOO...
...

XXOO...
OOOO...
XOOO...
XOOO...
...

XXOO...
XOOO...
OOOO...
XOOO...
...

XXXO...
XOOO...
OOOO...
OOOO...
...

Regards, WM

[Gus Gassmann]

I think you mean |N^2| = |N|. This is not my area of expertise, but if I understand correctly, you may need the axiom of choice to infer equality.

But that's not really what I responded to in my previous post. WM has various ways to conceptualize for himself that the infinite cannot possibly be real. He is quite creative in his dementia at constructing new and ever more convoluted schemes to delude his own head. (Always brings to mind Bob Dylan's song "License To Kill" with the line "his brain has been mismanaged with great skill".) Whenever I see a new idiocy of his, I always try to figure out exactly where the mistake occurs and how to fix it.

[Tom Bola]

An unsavory Idiot WM drivels:

> FromTheRafters schrieb:

>
>> He simply refuses to accept that |N|^2 = |N|

> No I simply disprove it.

ROFL.

> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> ...
>
> XXOO...
> XOOO...
> OOOO...
> XOOO...
> ...
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> ...

ROFL - Ingenious for a clown.

[sergio]

you are doing it all wrong, My turn to disprove it;


XXOO...
OOXO...

XOOO...
XOOO...
...

XXOO...
XOOO...

OOOX...
XOOO...
...
XOOX...
XOOO...
XOXO...
XOOO...
...


Bonus Question:

is
XXXO...
OXOO...
XOXO...
XOOX... ?

[sergio]

XXOO!! OOXO, and XOOO!

XOOX ?? "XXOX."

[Jim Burns]

On 4/26/2022 7:10 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 25. April 2022 um 20:48:48 UTC+2:
>> On 4/25/2022 1:39 PM, WM wrote:

>>> Is the contents of the first column
>>> an infinite sequence?
>>
>> It is an infinite sequence as "infinite sequence"
>> is usually meant.
>
> So it is. 1/1, 2/1, 3/1, ...
> By values this is same as 1, 2, 3, ...
> Therefore here are all indices which
> Cantor wants to spread over all positive fractions.
>
> But it is obvious that
>

[...]

>> The contents of the first column are
>> totally-ordered such that,
>> there is a first entry but no last entry,
>
> It is the set of indices.

You haven't described an infinite sequence.
You've only given a name, "indices", as though
that name is the answer to the question.

You're trying to hide your dark numbers in vagueness.

But we reject your vagueness,
so you call us "matheologians".

[FromTheRafters]

WM pretended :

> FromTheRafters schrieb am Dienstag, 26. April 2022 um 16:09:00 UTC+2:
>
>> He simply refuses to accept that |N|^2 = |N|
>
> No I simply disprove it.

Hugs and kisses are not proof.

[FromTheRafters]

Gus Gassmann expressed precisely :

What is N^2 in that equation?

> This is not my area of expertise, but if I
> understand correctly, you may need the axiom of choice to infer equality.

ZFC has AoC, but I refer here to cardinal arithmetic.

The cardinality of the natural numbers times the cardinality of the
natural numbers equals the cardinality of the natural numbers. He won't
accept it because he denies or does not understand cardinal arithmetic.

> But that's not really what I responded to in my previous post. WM has various
> ways to conceptualize for himself that the infinite cannot possibly be real.
> He is quite creative in his dementia at constructing new and ever more
> convoluted schemes to delude his own head. (Always brings to mind Bob Dylan's
> song "License To Kill" with the line "his brain has been mismanaged with
> great skill".) Whenever I see a new idiocy of his, I always try to figure out
> exactly where the mistake occurs and how to fix it.

His mistake is always his ego.

[Gus Gassmann]

Very sloppy notation for NxN.

[FromTheRafters]

Then yes, it sort of describes WM's matrix's dimensions and that the
cardinality of a single row or column can be the same as the
cardinality of the whole thing. It is exactly this which he denies
because it doesn't "feel" right to him.

[zelos...@gmail.com]

You don't disprove it, all you prove is that you are an idiot.

[zelos...@gmail.com]

it is not sloppy, it is very standard

[zelos...@gmail.com]

tisdag 26 april 2022 kl. 13:15:57 UTC+2 skrev WM:

so face it VM, you failed EVERYTHING here

[WM]

Jim Burns schrieb am Dienstag, 26. April 2022 um 19:51:50 UTC+2:
> On 4/26/2022 7:10 AM, WM wrote:

> > So it is. 1/1, 2/1, 3/1, ...
> > By values this is same as 1, 2, 3, ...

> > It is the set of indices.
> You haven't described an infinite sequence.
> You've only given a name, "indices", as though
> that name is the answer to the question.

Liar.

I have described the first column of the matrix:

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

Regards, WM

[WM]

FromTheRafters schrieb am Dienstag, 26. April 2022 um 22:05:48 UTC+2:

> Then yes, it sort of describes WM's matrix's dimensions and that the
> cardinality of a single row or column can be the same as the
> cardinality of the whole thing. It is exactly this which he denies
> because it

it can be disproved by the fact that every permutation of the X's fails to cover the whole matrix

XOOO...
XOOO...
XOOO...
XOOO...
...

Regards, WM

[zelos...@gmail.com]

None of this is relevant to you proving your darknumbers.

You couldn't even fucking DEFINE THEM USING FIRST ORDER LOGIC!

You are a failure!

[Gus Gassmann]

More unreflected garbage from you. What the fuck do you think a "permutation of the X's" might look like? And where do the O's fit into your nonthinking? Please get lost.

[sergio]

very sloppy math, WM

take your first column, write it as its own series, assuming your last three dots ... means it goes on to infinity, {1/1,2/1,3/1,4/1...}, this is the
set of natural numbers, however it is not in reduced form.


Bonus Question:

what are the indices to this new series ?

[sergio]

nope. wrong. try again.

[David Petry]

On Thursday, April 21, 2022 at 5:03:43 AM UTC-7, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 21. April 2022 um 12:34:25 UTC+2:
> > Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
> >
> > Define it in in terms of FOL and then prove that there exists two disjoint non-empty sets that has the union of natural numbers, where one set is your supposed claimed "dark" and the other is not, undark?
> This task is more difficult than the proof that there are dark fractions. If you have understood that it is impossible to enumerate all fractions, because starting from

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

> every attempt to collect all fractions in the first column will fail:
>

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

>
> When all definable fractions of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... will have transferred into the first column, nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone.
>

> If you have understood this, then we can proceed to understand, that even most places in the first column, all of which have been occupied by integer fractions or natural numbers, are undefinable too.
> > No, your proof of
> > An e N: n e [0,n]=F_n
> > Aka, "all natural is in in a finite set" is instantly invalid as that applies to all natural numbers, thus one set is non-empty.
> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction, belongs to a finite set F_n but has ℵo successors which cannot be removed whatever n you consider. That means they cannot be used as individuals.
>
> Regards, WM


The argument that I like to make is that mathematics has an important purpose, which is to provide a conceptual framework that facilitates reasoning about the real world (i.e. mathematics is the language of science). And Cantor's theory of infinite sets is simply irrelevant to that important purpose.

What I'm wondering is, how is your (Mueckenheim) argument relevant to the very important purpose of mathematics? Perhaps more importantly, why should people care about your argument? What's the point?

Need I remind you that any theory of infinity is a fiction? Why is your fiction more important than Cantor's?

[FromTheRafters]

WM formulated on Thursday :

How many permutations are there for an infinite set? Are you sure you
have covered them all?

[sergio]

sequence, not series

[sergio]

No, During his life Georg Cantor proved several concepts in mathematics that other mathematicians ahead of him were unable to prove. Cantor was
encouraged by his friend at Halle who was working on trigonometric series to work on the uniqueness of infinite series. In 1873 he was able to prove
that rational numbers are countable, he added that algebraic numbers that are roots, squares and square roots of polynomial equation with integer
coefficients are countable (Dauben, pp. 23-78). He published his first paper on theory of sets in 1874 where he proved that the set of integers had an
equal number of members.

He also came up with the argument that real numbers are not countable which he proved, he said that transcendental numbers are irrational numbers that
are not root, square or square root of any polynomial equation having integer coefficients (Ruker, pp. 27-79). Georg was able to show that the interval
between zero and one is uncountable. He is the only mathematician who was able to show that almost all numbers are transcendental by proving that real
numbers are not countable while proving that algebraic numbers were countable, he also showed that the set of all subsets of a given set are larger than
the original set. The introduction of the concept of the first derived set was his initiative (Dauben, pp. 23-78). Cantor also showed that union of two
countable sets should also be countable and brought across the existence of uncountable numbers.

Georg Cantor was the first person to discuss the continuum hypothesis which states that there exists a set of numbers whose power is greater than that
of the naturals and less than that of real, he tried it but all was in vain as he was able to prove and disprove it. Golden and Paul Cohen in 1963 said
that the hypothesis can be proved or disproved (Ruker, pp. 27-79).

>
> What I'm wondering is, how is your (Mueckenheim) argument relevant to the very important purpose of mathematics? Perhaps more importantly, why should people care about your argument? What's the point?

WM's pointless non mathematical post is to make himself feel important.

>
> Need I remind you that any theory of infinity is a fiction? Why is your fiction more important than Cantor's?

You write fiction too, finitely.


>
>
>
>

[Jim Burns]

On 4/28/2022 8:45 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 26. April 2022 um 19:51:50 UTC+2:
>> On 4/26/2022 7:10 AM, WM wrote:

>>> So it is. 1/1, 2/1, 3/1, ...
>>> By values this is same as 1, 2, 3, ...
>
>>> It is the set of indices.
>>
>> You haven't described an infinite sequence.
>> You've only given a name, "indices", as though
>> that name is the answer to the question.
>
> Liar.

'...' is not a description.

This is a description:
|
| In discussion A, each collection
| is empty or contains a first element.
|
| In discussion B, each _bounded-in-B_ collection
| is empty or contains a first _and a last_ element.
|
| _omega_ is defined to be the first thing in
| discussion A which is not in discussion B.


But '...' is not a description.
'...' is _at best_ a way to say
"You don't need to be told the description here".

At best.
'...' can also be used to avoid giving a description.
That's the way you (wM) use "..."

> I have described the first column of the matrix:
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...

No, you haven't.
'...' is not a description.

[zelos...@gmail.com]

You are however wrong, now GTFO

[WM]

I am sure that none will change the sets of O's and X's. That is all I need.

Regards, WM

[WM]

> The argument that I like to make is that mathematics has an important purpose, which is to provide a conceptual framework that facilitates reasoning about the real world (i.e. mathematics is the language of science). And Cantor's theory of infinite sets is simply irrelevant to that important purpose.
>
> What I'm wondering is, how is your (Mueckenheim) argument relevant to the very important purpose of mathematics? Perhaps more importantly, why should people care about your argument? What's the point?

The point is to show that Cantor's theory is ninsense and should be dropped.

>
> Need I remind you that any theory of infinity is a fiction? Why is your fiction more important than Cantor's?

I am not improving mathematics. That would go beyond my power. I am only proving, that Cantor's countability and uncountability are ill concepts.

Regards, WM

[WM]

Jim Burns schrieb am Freitag, 29. April 2022 um 04:16:56 UTC+2:
> On 4/28/2022 8:45 AM, WM wrote:

> '...' is not a description.

1/1, 2/1, 3/1, ...

is a description.

>

> > I have described the first column of the matrix:
> >
> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, 3/2, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> No, you haven't.
> '...' is not a description.

Liar. As anybody if he knows the next terms.

Regards, WM

[FromTheRafters]

WM has brought this to us :

That's easy because each row is countable and each column is countable
and they exhibit an easy to follow pattern.

[Jim Burns]

On 4/29/2022 12:07 PM, FromTheRafters wrote:
> WM has brought this to us :
>> Jim Burns schrieb
>> am Freitag, 29. April 2022 um 04:16:56 UTC+2:

>>> '...' is not a description.
>>
>> 1/1, 2/1, 3/1, ...
>> is a description.
>>
>>>> I have described the first column of the matrix:
>>>> 1/1, 1/2, 1/3, 1/4, ...
>>>> 2/1, 2/2, 2/3, 2/4, ...
>>>> 3/1, 3/2, 3/3, 3/4, ...
>>>> 4/1, 4/2, 4/3, 4/4, ...
>>>> 5/1, 5/2, 5/3, 5/4, ...
>>>> ...
>>>
>>> No, you haven't. '...' is not a description.
>>
>> Liar. As anybody if he knows the next terms.
>
> That's easy because
> each row is countable and each column is countable
> and they exhibit an easy to follow pattern.

Yes.
Everyone knows the next terms, but
knowing the next terms is not having a description.

WM deletes my description.
Not-having a description is clearly central to
his "argument" == bait-and-switch.

This is a description:
|
| In discussion A, each collection
| is empty or contains a first element.
|
| In discussion B, each _bounded-in-B_ collection
| is empty or contains a first _and a last_ element.
|
| _omega_ is defined to be the first thing in
| discussion A which is not in discussion B.

Discussion B is of the natural numbers.
Discussion A is of the finite and transfinite ordinals.

[WM]

Jim Burns schrieb am Freitag, 29. April 2022 um 19:38:47 UTC+2:
> On 4/29/2022 12:07 PM, FromTheRafters wrote:
> > WM has brought this to us :
> >> Jim Burns schrieb
> >> am Freitag, 29. April 2022 um 04:16:56 UTC+2:
> >>> '...' is not a description.
> >>
> >> 1/1, 2/1, 3/1, ...
> >> is a description.
> >>
> >>>> I have described the first column of the matrix:
> >>>> 1/1, 1/2, 1/3, 1/4, ...
> >>>> 2/1, 2/2, 2/3, 2/4, ...
> >>>> 3/1, 3/2, 3/3, 3/4, ...
> >>>> 4/1, 4/2, 4/3, 4/4, ...
> >>>> 5/1, 5/2, 5/3, 5/4, ...
> >>>> ...
> >>>
> >>> No, you haven't. '...' is not a description.
> >>

> >> Liar. Ask anybody if he knows the next terms.

> >
> > That's easy because
> > each row is countable and each column is countable
> > and they exhibit an easy to follow pattern.
> Yes.
> Everyone knows the next terms, but
> knowing the next terms is not having a description.

Wrong. You try to blur the discussion by nonensical words.

Regards, WM

[sergio]

I am sure he did not intentionally step into your area of expertise.

>
> Regards, WM

[Jim Burns]

On 4/29/2022 4:25 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 29. April 2022 um 19:38:47 UTC+2:
>> On 4/29/2022 12:07 PM, FromTheRafters wrote:
>>> WM has brought this to us :
>>>> Jim Burns schrieb
>>>> am Freitag, 29. April 2022 um 04:16:56 UTC+2:

>>>>>> I have described the first column of the matrix:
>>>>>> 1/1, 1/2, 1/3, 1/4, ...
>>>>>> 2/1, 2/2, 2/3, 2/4, ...
>>>>>> 3/1, 3/2, 3/3, 3/4, ...
>>>>>> 4/1, 4/2, 4/3, 4/4, ...
>>>>>> 5/1, 5/2, 5/3, 5/4, ...
>>>>>> ...
>>>>>
>>>>> No, you haven't. '...' is not a description.
>>>>
>>>> Liar. Ask anybody if he knows the next terms.
>>>
>>> That's easy because
>>> each row is countable and each column is countable
>>> and they exhibit an easy to follow pattern.
>>
>> Yes.
>> Everyone knows the next terms, but
>> knowing the next terms is not having a description.
>
> Wrong.
> You try to blur the discussion by nonensical words.

Ask.

In discussion A, each collection
is empty or contains a first element.

In discussion B, each _bounded-in-B_ collection
is empty or contains a first _and a last_ element.

_omega_ is defined to be the first thing in
discussion A which is not in discussion B.

B is the discussion of the natural numbers.
A is the discussion of the finite and transfinite ordinals.

[FromTheRafters]

WM brought next idea :

Slogging through some thick irony here, but it all makes ense.

[zelos...@gmail.com]

All you manage ot show is that you are an idiot.

> >
> > Need I remind you that any theory of infinity is a fiction? Why is your fiction more important than Cantor's?
> I am not improving mathematics. That would go beyond my power. I am only proving, that Cantor's countability and uncountability are ill concepts.
>
> Regards, WM

Those concepts work very fine, the issue is you are an idiot

[Gus Gassmann]

On Sunday, 1 May 2022 at 12:11:27 UTC-3, zelos...@gmail.com wrote:
> fredag 29 april 2022 kl. 14:41:13 UTC+2 skrev WM:

> > I am not improving mathematics. That would go beyond my power. I am only proving, that [...]

Proving *anything* goes well beyond your powers (other than that you are an idiot, at which you are very good).

[sergio]

On 4/29/2022 7:41 AM, WM wrote:
> david...@gmail.com schrieb am Donnerstag, 28. April 2022 um 16:29:42 UTC+2:

>
> I am not improving mathematics. That would go beyond my power.

totally agree, in fact you are trying to destroy vast areas of math with your Dark Army.

> I am only proving, that Cantor's countability and uncountability are ill concepts.

so where is your proof ? and what type of proof is it ?

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


>
> Regards, WM

[mitchr...@gmail.com]

On Thursday, April 21, 2022 at 3:34:25 AM UTC-7, zelos...@gmail.com wrote:
> Hey WM, let's start fresh. You keep claiming there are such things as "dark numbers"
>

They are below the finite.

[zelos...@gmail.com]

Stay out of this, you are too stupid to understand even basic negatives

[zelos...@gmail.com]

Notice WM, how you failed defining your "dark numbers"

[WM]

zelos...@gmail.com schrieb am Montag, 9. Mai 2022 um 06:53:40 UTC+2:

> Notice WM, how you failed defining your "dark numbers"

No, you have failed to understand. That's a difference.

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

Every attempt to collect all fractions in the first column of this matrix will fail. But all definable fractions of Cantor's sequence

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

will be transferred into the first column. Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.

Regards, WM

[Gus Gassmann]

On Monday, 9 May 2022 at 09:58:49 UTC-3, WM wrote:
[...]

> No, you have failed to understand. That's a difference.
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
> Every attempt to collect all fractions in the first column of this matrix will fail.

That brain-dead garbage again? Please take your dump elsewhere.

[sergio]

On 5/9/2022 7:58 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 9. Mai 2022 um 06:53:40 UTC+2:
>
>> Notice WM, how you failed defining your "dark numbers"
>
> No, you have failed to understand. That's a difference.
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
>

OK, a matrix of the rationals

> Every attempt to collect all fractions in the first column of this matrix will fail.

Wrong on 2 counts.

Mistake 1. Overwrite the first column, you destroy the matrix of rationals, just write out a new sequence as its own set.
Mistake 2. You say you fail doing this.

remember you are only doing a simple re-indexing.

Google "Cantor's Enumeration" for further help and guidance.

> But all definable fractions of Cantor's sequence
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...

no need for "definable", your failed daffynitions of "defined,definable" are the silly sand your math is based upon, and floundering.

you leave no room for you to be correct.

> will be transferred into the first column.

that is the wrong way to do it, so you fail again. no need to transfer, "swapparoobees" that is diversion.

> Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.

wrong, your math degenerates on its own.

>
> Regards, WM

Use equations and math to prove your dark ones exist, can you do that ?

[sergio]

On 4/21/2022 7:05 AM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 21. April 2022 um 13:22:54 UTC+2:
>
>> A number is called "instantiated" if it has been used by someone, somewhere.
>
> A number is called instantiatable if it can be instantiated.

A Post is called Bullshit if it has bullshit in it.

>By induction

where is your induction ?

> we prove

no, you have never proved anything.

that every instantiatable

what does that mean, you make them magically appear ?

> natural number, i.e., every natural number which is subject to induction

which is all of N

>, belongs to a finite set F_n

no. infinite cannot fit in finite. Go get a bucket, and put it under the tap for 12 hours and report back.

>but has ℵo successors which cannot be removed

you need the removal squad to do that

> whatever n you consider. That means they cannot be instantiated

hocus pocus, you a circus dude! make that rabbit appear !

>or used as individuals.

no one wants "Used individuals", that is sick, dude.

>
> Regards, WM

[zelos...@gmail.com]

måndag 9 maj 2022 kl. 14:58:49 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 9. Mai 2022 um 06:53:40 UTC+2:
>
> > Notice WM, how you failed defining your "dark numbers"
> No, you have failed to understand. That's a difference.

You are the one failing to define it.

> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
> Every attempt to collect all fractions in the first column of this matrix will fail.

No one is "collecting" them in the first column

>But all definable fractions of Cantor's sequence
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> will be transferred into the first column. Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.
>
> Regards, WM

You have not defined "definable" or "dark", you just say "they exist" based on misunderstandings by you. This is not a fucking definition

[WM]

zelos...@gmail.com schrieb am Dienstag, 10. Mai 2022 um 07:02:43 UTC+2:
> måndag 9 maj 2022 kl. 14:58:49 UTC+2 skrev WM:

> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, 3/2, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> > Every attempt to collect all fractions in the first column of this matrix will fail.

> No one is "collecting" them in the first column.

That is the way to index them because by definition the indices are the positions in the first column.

> >But all definable fractions of Cantor's sequence
> > 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> > will be transferred into the first column. Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.

> You have not defined "definable" or "dark", you just say "they exist" based on

the fact that most cannot be indexed by tranferring them into the first column.

Regards, WM

[Gus Gassmann]

On Tuesday, 10 May 2022 at 10:26:20 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 10. Mai 2022 um 07:02:43 UTC+2:

[...]

> > You have not defined "definable" or "dark", you just say "they exist" based on
> the fact that

the great perfosser desperately wishes them to exist, against reason, and without any shred of evidence. Piss off, moron.

[sergio]

On 5/10/2022 8:26 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 10. Mai 2022 um 07:02:43 UTC+2:
>> måndag 9 maj 2022 kl. 14:58:49 UTC+2 skrev WM:
>
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>> Every attempt to collect all fractions in the first column of this matrix will fail.
>> No one is "collecting" them in the first column.
>
> That is the way to index them because by definition the indices are the positions in the first column.

wrong. your "by definition" is your mistake.

as well as intentionally mutilating the matrix of rationals.

>

>
>> You have not defined "definable" or "dark", you just say "they exist" based on
>
> the fact that most cannot be indexed by tranferring them into the first column.

Liar.

>
> Regards, WM

[zelos...@gmail.com]

tisdag 10 maj 2022 kl. 15:26:20 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 10. Mai 2022 um 07:02:43 UTC+2:
> > måndag 9 maj 2022 kl. 14:58:49 UTC+2 skrev WM:
>
> > > 1/1, 1/2, 1/3, 1/4, ...
> > > 2/1, 2/2, 2/3, 2/4, ...
> > > 3/1, 3/2, 3/3, 3/4, ...
> > > 4/1, 4/2, 4/3, 4/4, ...
> > > 5/1, 5/2, 5/3, 5/4, ...
> > > ...
> > > Every attempt to collect all fractions in the first column of this matrix will fail.
> > No one is "collecting" them in the first column.
>
> That is the way to index them because by definition the indices are the positions in the first column.

There is no such definition, "indicing" just means you have a function f: N->Q+

> > >But all definable fractions of Cantor's sequence
> > > 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> > > will be transferred into the first column. Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.
> > You have not defined "definable" or "dark", you just say "they exist" based on
> the fact that most cannot be indexed by tranferring them into the first column.

Which is nothing but fallacious so you again fail

>
> Regards, WM

[WM]

zelos...@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 06:44:52 UTC+2:
> tisdag 10 maj 2022 kl. 15:26:20 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 10. Mai 2022 um 07:02:43 UTC+2:
> > > måndag 9 maj 2022 kl. 14:58:49 UTC+2 skrev WM:
> >
> > > > 1/1, 1/2, 1/3, 1/4, ...
> > > > 2/1, 2/2, 2/3, 2/4, ...
> > > > 3/1, 3/2, 3/3, 3/4, ...
> > > > 4/1, 4/2, 4/3, 4/4, ...
> > > > 5/1, 5/2, 5/3, 5/4, ...
> > > > ...
> > > > Every attempt to collect all fractions in the first column of this matrix will fail.
> > > No one is "collecting" them in the first column.
> >
> > That is the way to index them because by definition the indices are the positions in the first column.
> There is no such definition,

It is: 11, 21, 31, ...

> "indicing" just means you have a function f: N->Q+

11, 21, 31, ... --> 1/1, 1/2, 2/1, ...

Voila, you are too stupid to understand this?

> > > >But all definable fractions of Cantor's sequence
> > > > 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> > > > will be transferred into the first column. Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.
> > > You have not defined "definable" or "dark", you just say "they exist" based on
> > the fact that most cannot be indexed by tranferring them into the first column.
> Which is nothing but fallacious

You are too stupid to understand that exchanging will never reduce anything? Your fault.

Regards, WM

[WM]

XOOO...
XOOO...
XOOO...
XOOO...
...

XXOO...
OOOO...
XOOO...
XOOO...
...

XXOO...
XOOO...
OOOO...
XOOO...
...

XXXO...
XOOO...
OOOO...
OOOO...
...

You are too stupid to understand this? Your fault.

Regards, WM

[Gus Gassmann]

What is entirely on you is to think that the stuff you wrote about is a proof of dark numbers. In fact, as I repeatedly demonstrated, it is exactly the opposite.

And now fuck off, you silly crank.

[WM]

You repeatedly demonstrated that you cannot think. Your matrix with infinitely many oo is impossible because when the first position has reached an oo, then there is no continuation possible.

Regards, WM

[Gus Gassmann]

On Wednesday, 11 May 2022 at 09:44:34 UTC-3, WM wrote:
[...]

> You repeatedly demonstrated that you cannot think. Your matrix with infinitely many oo is impossible because when the first position has reached an oo, then there is no continuation possible.

Do you have to announce your utter ignorance quite so blatantly?!? The limiting value oo is *NOT* "reached" --- it is a *LIMIT*. Now FUCK OFF, MORON.

[sergio]

Wrong, all columns are infinite, all rows are too.


FAIL. you get an F, for Fake Math

[sergio]

insults will not correct your bad math.

[WM]

But enumerating all fractions must be reached. As soon as infinitely many have been reached at one position of the matrix, the counting is finished.

Regards, WM

[sergio]

your thoughts require a 'finish', an end, a finite time, a termination. However, that is not math.

You took a novice approach to enumerating the rationals, and failed.


That is why Cantor's enumeration is brilliant, he starts in the corner and goes diagonally back and forth.

[Gus Gassmann]

I can only repeat what I said before: Do you really have to announce your utter ignorance quite so blatantly?!? The limiting value oo is *NOT* "reached" --- it is a *LIMIT*. It seems that over the years you learned how to spell the word "limit", but sadly you still have no clue what it means in mathematics.

[Jim Burns]

On 5/11/2022 1:51 PM, WM wrote:

> But enumerating all fractions must be reached.

All fractions are described.
For top and for bottom,
each BEFORE ends and each AFTER begins.

We describe the index for each fraction.
One consequence of that description is that
each BEFORE ends and each AFTER begins.

And so, Bob is not conserved.

> As soon as infinitely many have been reached

*BZZZZZT*
Sorry, thanks for playing.

Things-that-can-be-reached are
reallyreallyreallyreallyreallyreally large numbers.

Infinity is not a reallyreallyreallyreallyreallyreally
large number. Infinity is a different kind of thing.

[mitchr...@gmail.com]

Zero isn't a quantity. That is the proof of its dark.

Mitchell Raemsch

[sergio]

On 5/11/2022 6:55 PM, mitchr...@gmail.com wrote:
> Zero isn't a quantity. That is the proof of its dark.
>
> Mitchell Raemsch

nope, you have to be able to see that you have nothing first.

[zelos...@gmail.com]

onsdag 11 maj 2022 kl. 13:32:05 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 06:44:52 UTC+2:
> > tisdag 10 maj 2022 kl. 15:26:20 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Dienstag, 10. Mai 2022 um 07:02:43 UTC+2:
> > > > måndag 9 maj 2022 kl. 14:58:49 UTC+2 skrev WM:
> > >
> > > > > 1/1, 1/2, 1/3, 1/4, ...
> > > > > 2/1, 2/2, 2/3, 2/4, ...
> > > > > 3/1, 3/2, 3/3, 3/4, ...
> > > > > 4/1, 4/2, 4/3, 4/4, ...
> > > > > 5/1, 5/2, 5/3, 5/4, ...
> > > > > ...
> > > > > Every attempt to collect all fractions in the first column of this matrix will fail.
> > > > No one is "collecting" them in the first column.
> > >
> > > That is the way to index them because by definition the indices are the positions in the first column.
> > There is no such definition,
> It is: 11, 21, 31, ...

That is a function N to whatever set you have, no column in sight.

> > "indicing" just means you have a function f: N->Q+
> 11, 21, 31, ... --> 1/1, 1/2, 2/1, ...
>
> Voila, you are too stupid to understand this?

Are you to stupid to understand that this is not your fucking column?

> > > > >But all definable fractions of Cantor's sequence
> > > > > 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> > > > > will be transferred into the first column. Nevertheless all matrix places remain occupied by fractions which are undefinable, because all definable fractions have gone. That means dark.
> > > > You have not defined "definable" or "dark", you just say "they exist" based on
> > > the fact that most cannot be indexed by tranferring them into the first column.
> > Which is nothing but fallacious
> You are too stupid to understand that exchanging will never reduce anything? Your fault.

Are you too stupid to understand that all this of yours is irrelevant and proves NOTHING of what you want?
Yes, yes you are

>
> Regards, WM

[zelos...@gmail.com]

We can think consideraby better than you.

[WM]

horand....@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 21:12:10 UTC+2:
> On Wednesday, 11 May 2022 at 14:51:09 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 15:53:39 UTC+2:
> > > On Wednesday, 11 May 2022 at 09:44:34 UTC-3, WM wrote:
> > > [...]
> > > > You repeatedly demonstrated that you cannot think. Your matrix with infinitely many oo is impossible because when the first position has reached an oo, then there is no continuation possible.
> > > Do you have to announce your utter ignorance quite so blatantly?!? The limiting value oo is *NOT* "reached" --- it is a *LIMIT*.
> > But enumerating all fractions must be reached. As soon as infinitely many have been reached at one position of the matrix, the counting is finished.
> The limiting value oo is *NOT* "reached" --- it is a *LIMIT*.

Therefore it is irrelevant. "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen (was, wenn es auf eine Art möglich ist, immer auch noch auf viele andere Weisen geschehen kann), so möge für das Folgende die Ausdrucksweise gestattet sein, daß diese Mannigfaltigkeiten gleiche Mächtigkeit haben, oder auch, daß sie äquivalent sind." [Cantor, p. 119]

For infinite set completeness means infinitely many elements are treated.

Regards, WM

[WM]

Jim Burns schrieb am Mittwoch, 11. Mai 2022 um 21:39:09 UTC+2:
> On 5/11/2022 1:51 PM, WM wrote:
>
> > But enumerating all fractions must be reached.
> All fractions are described.

No. All fractions
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
described by indices k = (m + n - 1)(m + n - 2)/2 + m
are collected in the first column.
But all other columns are filled too.

.
> > As soon as infinitely many have been reached

> Things-that-can-be-reached are
> reallyreallyreallyreallyreallyreally large numbers.

If two infinite sets "can be related completely, element by element, to each other [...], then for the following the expression may be permitted that these manifolds have the same cardinality or that they are equivalent." [Cantor, p. 119]
>
Of course it is nonsense, but it is the definition of what you are defending.

Regards, WM

[Gus Gassmann]

In other words, you are now stating that your matrix sequence was irrelevant. Good job! (This also gives me the opportunity to point out to you again that "element for element" does not mean "one after the other". Why, oh why, do you insist on constantly misreading Cantor?)

[WM]

horand....@gmail.com schrieb am Donnerstag, 12. Mai 2022 um 15:42:01 UTC+2:
> On Thursday, 12 May 2022 at 10:05:38 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 21:12:10 UTC+2:
> > > On Wednesday, 11 May 2022 at 14:51:09 UTC-3, WM wrote:
> > > > horand....@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 15:53:39 UTC+2:
> > > > > On Wednesday, 11 May 2022 at 09:44:34 UTC-3, WM wrote:
> > > > > [...]
> > > > > > You repeatedly demonstrated that you cannot think. Your matrix with infinitely many oo is impossible because when the first position has reached an oo, then there is no continuation possible.
> > > > > Do you have to announce your utter ignorance quite so blatantly?!? The limiting value oo is *NOT* "reached" --- it is a *LIMIT*.
> > > > But enumerating all fractions must be reached. As soon as infinitely many have been reached at one position of the matrix, the counting is finished.
> > > The limiting value oo is *NOT* "reached" --- it is a *LIMIT*.
> > Therefore it is irrelevant. "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen (was, wenn es auf eine Art möglich ist, immer auch noch auf viele andere Weisen geschehen kann), so möge für das Folgende die Ausdrucksweise gestattet sein, daß diese Mannigfaltigkeiten gleiche Mächtigkeit haben, oder auch, daß sie äquivalent sind." [Cantor, p. 119]
> >
> > For infinite set completeness means infinitely many elements are treated.
> In other words, you are now stating that your matrix sequence was irrelevant.

No one matrix for one step of enumerating.

> Good job! (This also gives me the opportunity to point out to you again that "element for element" does not mean "one after the other".

"unter beidseitiger Wahrung der Rangfolge ihrer Elemente auf einander beziehen, abbilden lassen" [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

"So stellt uns beispielsweise eine veränderliche Größe x, die nacheinander die verschiedenen endlichen ganzen Zahlwerte 1, 2, 3, ..., ν, ... anzunehmen hat, ein potentiales Unendliches vor, während die durch ein Gesetz begrifflich durchaus bestimmte Menge (y) aller ganzen endlichen Zahlen y das einfachste Beispiel eines actual unendlichen Quantums darbietet." [1886, Mai. Cantor to S. Giulio Vivanti in Mantua]

Do you know this law? If you have n, then go to n+1. One after the other:

We have 1 = |E0|. Now combine another thing e1 with E0. The sum set be called E1, such that E1 = (E0, e1) = (e0, e1). The cardinal number |E1| of E1 is called "two" and is denoted by 2 = |E1|. By adding new elements we obtain the sequence of sets E2 = (E1, e2), E3 = (E2, e3), ... which in an unbounded sequence successively supply the other so-called finite cardinal numbers denoted by 3, 4, 5, ... . [G. Cantor: "Beiträge zur Begründung der transfiniten Mengenlehre 1", Math. Annalen 46 (1895) § 5]

Denn wir haben hier ein dem Rang nach niedrigstes Element, die kleinste Kardinalzahl 1 und eine auf jede endliche Kardinalzahl  dem Range, d. h. hier der Größe nach nächstfolgende endliche Kardinalzahl  + 1. So erhalten wir die Gesamtheit aller endlichen Kardinalzahlen in der sogenannten natürlichen endlosen Folge: 1, 2, 3, ..., , ..., in welcher Folge sie eine wohlgeordnete Menge vom Ordnungstypus  darstellt.

Regards, WM

[sergio]

those quotes do not save your bad math, they underscore the fact that you do not fully understand them.

[sergio]

On 5/12/2022 8:05 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 11. Mai 2022 um 21:39:09 UTC+2:
>> On 5/11/2022 1:51 PM, WM wrote:
>>
>>> But enumerating all fractions must be reached.
>> All fractions are described.
>
> No. All fractions
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
> described by indices k = (m + n - 1)(m + n - 2)/2 + m
> are collected in the first column.
> But all other columns are filled too.

nonsense.

> .
>>> As soon as infinitely many have been reached
>
>> Things-that-can-be-reached are
>> reallyreallyreallyreallyreallyreally large numbers.
>
> If two infinite sets "can be related completely, element by element, to each other [...], then for the following the expression may be permitted that these manifolds have the same cardinality or that they are equivalent." [Cantor, p. 119]
>>
> Of course it is nonsense, but it is the definition of what you are defending.

wrong, it is only one to one mapping.

>
> Regards, WM

keep trying to prove Cantor wrong, it is good use of your time.

[Gus Gassmann]

On Thursday, 12 May 2022 at 11:08:32 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 12. Mai 2022 um 15:42:01 UTC+2:
> > On Thursday, 12 May 2022 at 10:05:38 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Mittwoch, 11. Mai 2022 um 21:12:10 UTC+2:

[...]

> > Good job! (This also gives me the opportunity to point out to you again that "element for element" does not mean "one after the other".
> "unter beidseitiger Wahrung der Rangfolge ihrer Elemente auf einander beziehen, abbilden lassen" [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

That does not say or mean "one after the other".

> "So stellt uns beispielsweise eine veränderliche Größe x, die nacheinander die verschiedenen endlichen ganzen Zahlwerte 1, 2, 3, ..., ν, ... anzunehmen hat, ein potentiales Unendliches vor, während die durch ein Gesetz begrifflich durchaus bestimmte Menge (y) aller ganzen endlichen Zahlen y das einfachste Beispiel eines actual unendlichen Quantums darbietet." [1886, Mai. Cantor to S. Giulio Vivanti in Mantua]
>
> Do you know this law? If you have n, then go to n+1.

That emphatically is *NOT* stated here.

> We have 1 = |E0|. Now combine another thing e1 with E0. The sum set be called E1, such that E1 = (E0, e1) = (e0, e1). The cardinal number |E1| of E1 is called "two" and is denoted by 2 = |E1|. By adding new elements we obtain the sequence of sets E2 = (E1, e2), E3 = (E2, e3), ... which in an unbounded sequence successively supply the other so-called finite cardinal numbers denoted by 3, 4, 5, ... . [G. Cantor: "Beiträge zur Begründung der transfiniten Mengenlehre 1", Math. Annalen 46 (1895) § 5]
>
> Denn wir haben hier ein dem Rang nach niedrigstes Element, die kleinste Kardinalzahl 1 und eine auf jede endliche Kardinalzahl  dem Range, d. h. hier der Größe nach nächstfolgende endliche Kardinalzahl  + 1. So erhalten wir die Gesamtheit aller endlichen Kardinalzahlen in der sogenannten natürlichen endlosen Folge: 1, 2, 3, ..., , ..., in welcher Folge sie eine wohlgeordnete Menge vom Ordnungstypus  darstellt.

OK. This is the process of induction. Please give a proper citation.

You seem to be really, really efficient at dredging up citations and so frightfully ill-equipped to actually understand them.

[sergio]

WM doesnt have the background, nor the knowledge structure, to understand.

If he cannot follow or do a Proof, he is pre-algebra at best. Cantor subject matter is out of reach for him.