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.

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.)

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

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

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.

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.

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

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. 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

that is nonsense.

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.

> 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

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

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.

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!

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

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!

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

FromTheRafters schrieb am Montag, 25. April 2022 um 15:47:21 UTC+2:
> 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!

Is the contents of the first column an infinite sequence?

Regards, WM

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.

WM was thinking very hard :
> FromTheRafters schrieb am Montag, 25. April 2022 um 15:47:21 UTC+2:
>> 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!
>
> Is the contents of the first column an infinite sequence?

No, it isn't.

On 4/25/2022 12:39 PM, WM wrote:
> FromTheRafters schrieb am Montag, 25. April 2022 um 15:47:21 UTC+2:
>> 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!
>
> Is the contents of the first column an infinite sequence?
>
> Regards, WM

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

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

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

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

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

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.

[...]

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.

>
>> 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