The Matheological Explosion

[WM]

In oder to index all positive fractions

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, ...
...

by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the first column, covering the fractions n/1. All other fractions get no indexes. This is expressed by an O's.

For brevity the presence of an index is represented by an X since its particular numerical value is irrelevant. So we get, as the start position:

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
...

Now every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... . At this place there appears an X and where the index has been taken from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 have lost their indexes:

XXXXOOOOO...
XOXOOOOOO...
XXOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
...

As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!

If, after all, the X's will cover the whole matrix and all the O's will have disappeared, this can only happen during the so-called Matheologoial Explosion, happening after all finite steps according to the Matheologial Syllogism: The O's will have disappeared never. Never we get into the infinite. ==> The O's will have diappeared in the infinite.

Regards, WM

[sergio]

On 1/15/2022 6:52 AM, WM wrote:
> In oder to index all positive fractions
>
> 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, ...
> ...
>
> by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the first column, covering the fractions n/1. All other fractions get no indexes. This is expressed by an O's.

However, Cantor used diagonal threading, which converts the above 2 dimensional matrix, into a 1 dimensional array.

>
> For brevity the presence of an index is represented by an X since its particular numerical value is irrelevant. So we get, as the start position:
>
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
>

> Now every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... . At this place there appears an X and where the *index has been taken* from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 *have lost their indexes*:

your mistakes are highlighted.

>
> XXXXOOOOO...
> XOXOOOOOO...
> XXOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>
> As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!

look, google core memory

It was early memory for computers using tiny ferrite rings, structured in a large matrix, with the reset line threaded diagonally through the matrix
just like Cantors Enumeration. no bits were "lost", or "taken from"

it is radiation hardened memory, and was used in outer space up till about 1980

>
> If, after all, the X's will cover the whole matrix and all the O's will have disappeared, this can only happen during the so-called Matheologoial Explosion, happening after all finite steps according to the Matheologial Syllogism: The O's will have disappeared never. Never we get into the infinite. ==> The O's will have diappeared in the infinite.

no, you *disappeared them* in your construct.

>
> Regards, WM

[William]

For the sake of argument we will say that the fractions cannot be indexed. This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P. The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark, can be considered and is not larger than every element of N_P, changes nothing.

--
William Hughes

[WM]

William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> For the sake of argument we will say that the fractions cannot be indexed.

I agree.

> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.

I agree.

> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,

This infinite number however cannot be a completed infinity. In a linear order completion implies last.

Complete, infinite, order cannot exist together.

Regards, WM

[William]

On Saturday, January 15, 2022 at 2:23:42 PM UTC-4, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:

> > ... no element of N_P is larger than every element of N_P.
> I agree.

So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.

--

William Hughes

[sergio]

no.

count the naturals 1,2,3,4,... it is in linear order and there is no last

>
> Complete, infinite, order cannot exist together.

you forget you can prove completeness by showing no element is missing...

>
> Regards, WM

[WM]

sergio schrieb am Samstag, 15. Januar 2022 um 19:45:00 UTC+1:
> On 1/15/2022 12:23 PM, WM wrote:
> > William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> >> For the sake of argument we will say that the fractions cannot be indexed.
> >
> > I agree.
> >
> >> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.
> >
> > I agree.
> >
> >> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,
> >
> > This infinite number however cannot be a completed infinity. In a linear order completion implies last.
> no.
>
> count the naturals 1,2,3,4,... it is in linear order and there is no last

You cannot know till to meet it.

> >
> > Complete, infinite, order cannot exist together.
> you forget you can prove completeness by showing no element is missing...

I showed that no O is missing in

> XXXXOOOOO...
> XOXOOOOOO...
> XXOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...

even after all possible transpositions.

Regards, WM

[WM]

But omega is the limit, the border before which all must happen. omega would be the last ordinal unless a smaller one next to omega could be found (if all was in linear order). You cannot deny it before you have reached it.

Regards, WM

[mitchr...@gmail.com]

The explosion started at math zero...
Mathematical God creates gravity...

[sergio]

On 1/15/2022 1:20 PM, WM wrote:
> sergio schrieb am Samstag, 15. Januar 2022 um 19:45:00 UTC+1:
>> On 1/15/2022 12:23 PM, WM wrote:
>>> William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
>>>> For the sake of argument we will say that the fractions cannot be indexed.
>>>
>>> I agree.
>>>
>>>> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.
>>>
>>> I agree.
>>>
>>>> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,
>>>
>>> This infinite number however cannot be a completed infinity. In a linear order completion implies last.
>> no.
>>
>> count the naturals 1,2,3,4,... it is in linear order and there is no last
>
> You cannot know till to meet it.

If you meet it, it is finite.

The set of Natural number is infinite, there is no last.

>>>
>>> Complete, infinite, order cannot exist together.
>> you forget you can prove completeness by showing no element is missing...
>
> I showed that no O is missing in
>
>> XXXXOOOOO...
>> XOXOOOOOO...
>> XXOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> XOOOOOOOO...
>> ...
>
> even after all possible transpositions.
>
> Regards, WM

you can show completion by proving there are no missing numbers,

take the rationals, n/k where n and k are natural numbers.
take the set composed of "for any n, and any k, the rational n/k exists"

all rationals are included, none are missing, it is complete.

"Complete, infinite, order cannot exist together." is bogus

all natural numbers are in N, there are none missing, it is complete, they are ordered, and infinite...

so...

[Jim Burns]

On 1/15/2022 2:25 PM, WM wrote:

> But omega is the limit, the border before which all must happen.

If there exists a totally-ordered collection {0,...,k}
for which each cut C is a step
( max(C) = i min({0,...,k}\C = j )
and each step is a count
( j = i+1 )
and which begins at 0 and ends at k,
then k is before omega.

That's what omega is.

> omega would be the last ordinal unless a smaller one
> next to omega could be found (if all was in linear order).

| Suppose psi was a smaller one next to omega.
| A totally-ordered collection {0,...,psi} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at psi.
|
| psi+1 exists and
| a totally-ordered collection {0,...,psi+1} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at psi+1.
|
| Because "smaller one next to omega" psi+1 = omega
|
| omega exists and
| a totally-ordered collection {0,...,omega} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at omega.
|
| omega+1 exists and
| a totally-ordered collection {0,...,omega+1} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at omega+1.
|
| Therefore omega+1 is before omega,
| by definition of omega.
| However, omega+1 is not in {0,...,omega}
| Contradiction.

Therefore,
there is no smaller one next to omega.

> You cannot deny it before you have reached it.

We can deny that omega is something other than
what we have defined omega to be.
We don't need to reach omega to do that.

From that, by reliable statement-steps,
we can learn things about omega, without ever reaching it.
For example: there is no smaller one next to omega.

[William]

On Saturday, January 15, 2022 at 3:25:41 PM UTC-4, WM wrote:

> William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:

> > So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
> But omega

We do not need omega. We have

(i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
larger than n,

(ii) There is no element of N_P that is larger than every element of N_P

[(i) you have repeated many times. (ii) is a statement that you agree with]
--

William Hughes

[sergio]

and you will never reach omega.

assume your at the largest number k

there is always a k + 1

conflict. so k is not the largest number.



Perhaps "You cannot deny it before you have reached it", but I deny that you will ever reach it as I say there is always a k+1, and you never will get
to omega. Simple

[Gus Gassmann]

What a crock. WM does not know what a linear order is. Never has, never will. A linear order does not imply the existence of a smallest element, nor a largest, nor an element "next to" omega. The only fact is that every natural number n is smaller than omega, with an infinite number of other natural numbers k satisfying n < k < omega.

[WM]

William schrieb am Samstag, 15. Januar 2022 um 21:48:05 UTC+1:
> On Saturday, January 15, 2022 at 3:25:41 PM UTC-4, WM wrote:
> > William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
>
> > > So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
> > But omega
> We do not need omega.

Then we have no finished infinity and no infinite set.

We have
>
> (i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
> larger than n,
>
> (ii) There is no element of N_P that is larger than every element of N_P

So it is. For every element of N_P that you choose there is a larger element of N_P. But that does not mean completed infinity.

>
> [(i) you have repeated many times. (ii) is a statement that you agree with]

Of course. I call it potential infinity.

Regards, WM

[WM]

sergio schrieb am Samstag, 15. Januar 2022 um 20:45:17 UTC+1:
> On 1/15/2022 1:20 PM, WM wrote:
>
> >> you forget you can prove completeness by showing no element is missing...
> >
> > I showed that no O is missing in
> >
> >> XXXXOOOOO...
> >> XOXOOOOOO...
> >> XXOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> XOOOOOOOO...
> >> ...
> >
> > even after all possible transpositions.

> you can show completion by proving there are no missing numbers,

I have proved that after all transpositions, there are infinitely many fractions without index.

>
> all rationals are included, none are missing, it is complete.

How do you bring that into line with the fact that there are infinitely many fractions without index?

Regards, WM

[WM]

Jim Burns schrieb am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> On 1/15/2022 2:25 PM, WM wrote:

> > omega would be the last ordinal unless a smaller one
> > next to omega could be found (if all was in linear order).
> | Suppose psi was a smaller one next to omega.
> | A totally-ordered collection {0,...,psi} exists

No.

> | for which each cut is a step and each step is a count
> | and which begins at 0 and ends at psi.
> |
> | psi+1 exists and
> | a totally-ordered collection {0,...,psi+1} exists
> | for which each cut is a step and each step is a count
> | and which begins at 0 and ends at psi+1.
> |
> | Because "smaller one next to omega" psi+1 = omega

Therefore psi is not accessible.

> |
> Therefore,
> there is no smaller one next to omega.

Then it is not next to omega. Where is the first smaller one?
What is between it an omega?

Regards, WM

[sergio]

he already told you psi+1, or try psi, or psi-1

these are very very simple proofs there is no number "next to" omega

and yet again you fail, which convinces everyone that you never took algebra I.

Can you convince people you are right ? no, if they had Algebra I, then no.

[sergio]

On 1/15/2022 4:30 PM, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 21:48:05 UTC+1:
>> On Saturday, January 15, 2022 at 3:25:41 PM UTC-4, WM wrote:
>>> William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
>>
>>>> So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
>>> But omega
>> We do not need omega.
>
> Then we have no finished infinity and no infinite set.

so you stopped, stopped at k again. because you can only deal with the finite.

>
>
> We have
>>
>> (i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
>> larger than n,
>>
>> (ii) There is no element of N_P that is larger than every element of N_P
>
> So it is. For every element of N_P that you choose there is a larger element of N_P. But that does not mean completed infinity.

sure it does, as you can show all numbers are in the set, it is complete.

>>
>> [(i) you have repeated many times. (ii) is a statement that you agree with]
>
> Of course. I call it potential infinity.

which is finite, because you stopped at k.

>
> Regards, WM

[sergio]

On 1/15/2022 4:34 PM, WM wrote:
> sergio schrieb am Samstag, 15. Januar 2022 um 20:45:17 UTC+1:
>> On 1/15/2022 1:20 PM, WM wrote:
>>
>>>> you forget you can prove completeness by showing no element is missing...
>>>
>>> I showed that no O is missing in
>>>
>>>> XXXXOOOOO...
>>>> XOXOOOOOO...
>>>> XXOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> XOOOOOOOO...
>>>> ...
>>>
>>> even after all possible transpositions.
>
>> you can show completion by proving there are no missing numbers,
>
> I have proved that after all transpositions, there are infinitely many fractions without index.

it is your bad math, you deleted them.

>>
>> all rationals are included, none are missing, it is complete.
>
> How do you bring that into line with the fact that there are infinitely many fractions without index?

you deleted them, your construct is just bad math.

"At this place there appears an X and where the *index has been taken from* there appears an O. After eight steps for instance we see the following
picture: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 *have lost their indexes*"

Highlighted is where you deleted numbers.

your missing Os are deception.

>
> Regards, WM

[Jim Burns]

On 1/15/2022 5:38 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
>> On 1/15/2022 2:25 PM, WM wrote:

>>> omega would be the last ordinal unless a smaller one
>>> next to omega could be found (if all was in linear order).
>>
>> | Suppose psi was a smaller one next to omega.
>> | A totally-ordered collection {0,...,psi} exists
>
> No.

{0,...,psi} exists because,
for every k smaller than omega, {0,...,k} exists.

omega is the first infinite ordinal.
Anything before omega is a finite ordinal.
Any finite ordinal is before omega.

Being a finite ordinal k is equivalent to
{0,...,k} existing, starting at 0, ending at k,
with a counting-order
(each cut is a step, each step is a count).

"No" means that what you intend by your claim is not
what is broadly meant by those words in your claim.
Since we are not mind-readers we must depend upon
what is broadly meant. So, "no" means that you
have picked yourself up by the scruff of your neck
and thrown yourself in the dumpster.
Carry on. Don't let me interrupt your self-dumping.

>> | for which each cut is a step and each step is a count
>> | and which begins at 0 and ends at psi.
>> |
>> | psi+1 exists and
>> | a totally-ordered collection {0,...,psi+1} exists
>> | for which each cut is a step and each step is a count
>> | and which begins at 0 and ends at psi+1.
>> |
>> | Because "smaller one next to omega" psi+1 = omega
>
> Therefore psi is not accessible.

If psi < omega then {0,...,psi} exists.
When you aren't rigging the game, that means that
psi is accessible.

>> |
>> Therefore,
>> there is no smaller one next to omega.
>
> Then it is not next to omega.
> Where is the first smaller one?

How could the first smaller than omega not be
next to omega? Do you know what "next" means?

> What is between it an omega?

There is no first smaller.
For each smaller there are infinitely-many between
-- which is why there is no first smaller.

[William]

On Saturday, January 15, 2022 at 6:30:12 PM UTC-4, WM wrote:

> William schrieb am Samstag, 15. Januar 2022 um 21:48:05 UTC+1:


> We have
> >
> > (i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
> > larger than n,
> >
> > (ii) There is no element of N_P that is larger than every element of N_P

> So it is.

N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.

--
William Hughes

[WM]

William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:

> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.

Construct the largest number you can. It is the the largest one, but only as long as you do not construct a larger one. And after that one, you can construct an even larger one. That is potential infinity.

Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> On 1/15/2022 5:38 PM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:

> There is no first smaller.
> For each smaller there are infinitely-many between
> -- which is why there is no first smaller.

But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.

[JVR]

The man is completely tireless, repeating the same nonsense now for more than 20 years.

\pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

Now shut up until you have understood this fact.

[WM]

JVR schrieb am Sonntag, 16. Januar 2022 um 11:22:38 UTC+1:
> On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
> > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> > > On 1/15/2022 5:38 PM, WM wrote:
> > > > Jim Burns schrieb
> > > > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> > > There is no first smaller.
> > > For each smaller there are infinitely-many between
> > > -- which is why there is no first smaller.
> > But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> > That is potential infinity.
> >
>

> \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

How can you reconcile this with the fact that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!

Note that the so-called Matheologial Explosion is not acceptable in rational mathematics.

> Now shut up until you have understood this fact.

This fact is easily understood and reconciled with the constancy of the numbers of O's by recognizing dark numbers. Do you have a better idea?

Regards, WM

[sergio]

being 'chosen' is an 'external event'. Each 'chosen one' is the last one YOU 'chose'.

this has nothing to do with sets or numbers at all.

[JVR]

Yes, I have a better idea. Learn to understand the obvious facts such as

\pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

Then find the errors in your simplistic argument yourself.

[sergio]

following your instructions;

Ok, I construct the number 17.

it is the largest one, I have not constructed a larger one.

after that, I can construct an even larger one, Ok how about 19.

that is 'potential infinity' (??)


no, that is silly. I have a card with 19 on it.

that is called "observer dependent Math", the math changes depending upon the observer.


you are missing some lines to get from picking an number, to a statement about (non) infinity.
that is where Math comes in, use math to be clear.

[WM]

JVR schrieb am Sonntag, 16. Januar 2022 um 16:36:20 UTC+1:
> On Sunday, January 16, 2022 at 4:28:49 PM UTC+1, WM wrote:
> > JVR schrieb am Sonntag, 16. Januar 2022 um 11:22:38 UTC+1:
> > > On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
> > > > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> > > > > On 1/15/2022 5:38 PM, WM wrote:
> > > > > > Jim Burns schrieb
> > > > > > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> > > > > There is no first smaller.
> > > > > For each smaller there are infinitely-many between
> > > > > -- which is why there is no first smaller.
> > > > But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> > > > That is potential infinity.
> > > >
> > >
> > > \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.
> > How can you reconcile this with the fact that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!
> >
> > Note that the so-called Matheologial Explosion is not acceptable in rational mathematics.
> > > Now shut up until you have understood this fact.
> > This fact is easily understood and reconciled with the constancy of the numbers of O's by recognizing dark numbers. Do you have a better idea?
> >

> Yes, I have a better idea. Learn to understand the obvious facts such as
> \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

That is not an idea how to reconcile complete indexing with the constancy of the numbers of O's.

> Then find the errors in your simplistic argument yourself.

Unfortunately there are no errors. The O's could only decrease during transpositions at finite steps. But they don't. Any idea?

Regards, WM

[sergio]

wrong, you made several intentional mistakes, you *disappeared them* in your construct.

Quote your snippage, which shows your 2 errors;

"Now every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... . At this place
there appears an X and where the *index has been taken* from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2,
2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 *have lost their indexes*: "

[William]

On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>
> > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> Construct the largest number you can.

It is an element of N_P. And it was an element of N_P before you "constructed" it. N_P does not change.

> It is the the largest one, but only as long as you do not construct a larger one.

It may be the largest element that you have "constucted" but it is not the largest element of N_P.
N_P does not change. N_P is not something you have to "construct".

--
William Hughes

[Jim Burns]

On 1/16/2022 3:46 AM, WM wrote:
> Jim Burns schrieb

> am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:

>> There is no first smaller.
>> For each smaller there are infinitely-many between
>> -- which is why there is no first smaller.
>
> But not all can be chosen.

All k which end some two-ended-counting-ordered
{0,...,k} beginning at 0
end some two-ended-counting-ordered
{0,...,k} beginning at 0

All k which end some two-ended-counting-ordered
{0,...,k} beginning at 0
are before omega.

No k which ends some two-ended-counting-ordered
{0,...,k} beginning at 0
is first smaller than omega.

> Each chosen one is the last one only as long as
> you do not construct a larger one.
> That is potential infinity.

Then potential infinity has nothing to say about
k which end some two-ended-counting-ordered
{0,...,k} beginning at 0

[WM]

William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> >
> > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > Construct the largest number you can.
> It is an element of N_P. And it was an element of N_P before you "constructed" it.

Who told you, if you are the first to construct it? Nobody. Therefore you cannot be sure. N_P may be potentially infinite.

> N_P does not change.

If you are sure and right, then there are many, many elements of N_P which never will be used. That means they cannot not have been used at the end of times. That means they are dark.

Whatever you choose, the alternatives are potential infinity or dark numbers.

> N_P does not change. N_P is not something you have to "construct".

Let's assume you are right. Then N_P is something most elements of which cannot be used. They are dark.

Regards, WM

[WM]

Jim Burns schrieb am Sonntag, 16. Januar 2022 um 19:18:56 UTC+1:
> On 1/16/2022 3:46 AM, WM wrote:
> > Jim Burns schrieb
> > am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> >> There is no first smaller.
> >> For each smaller there are infinitely-many between
> >> -- which is why there is no first smaller.
> >
> > But not all can be chosen.
>
> All k which end some two-ended-counting-ordered
> {0,...,k} beginning at 0

> are before omega.

But not all can be chosen and used. Almost all will not have been used at the end of times. That means they cannot have been used. That means they are dark.

Regards, WM

[sergio]

No, you are using "Observer Dependent" Math, which is totally unworkable.

Your chooser is some external person, you require time for your math operations, two more flaws.

[sergio]

On 1/16/2022 3:24 PM, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
>> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>>>
>>>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>>> Construct the largest number you can.
>> It is an element of N_P. And it was an element of N_P before you "constructed" it.
>
> Who told you, if you are the first to construct it? Nobody. Therefore you cannot be sure. N_P may be potentially infinite.

Observer Dependent math again. Fail.

>
>> N_P does not change.
>
> If you are sure and right, then there are many, many elements of N_P which never will be used. That means they cannot not have been used at the end of times. That means they are dark.

Used Numbers ? Yuck! (same with used Ants) There is no end of time.

>
> Whatever you choose, the alternatives are potential infinity or dark numbers.

false choice.

>
>> N_P does not change. N_P is not something you have to "construct".
>
> Let's assume you are right. Then N_P is something most elements of which cannot be used. They are dark.

your assumption, not fact.


>
> Regards, WM

[Jim Burns]

On 1/16/2022 4:29 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 16. Januar 2022 um 19:18:56 UTC+1:
>> On 1/16/2022 3:46 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:

>>>> There is no first smaller.
>>>> For each smaller there are infinitely-many between
>>>> -- which is why there is no first smaller.
>>>
>>> But not all can be chosen.
>>
>> All k which end some two-ended-counting-ordered
>> {0,...,k} beginning at 0
>> are before omega.
>
> But not all can be chosen and used.
> Almost all will not have been used at the end of times.
> That means they cannot have been used.
> That means they are dark.

We have a technique for reasoning about infinitely-many
individuals. Start with some claim we know is true of
each individual of the infinitely-many. Take only
reliable statement-steps from there.

Which infinitely-many? Which starting-claims?
The answers to these two questions are tightly linked.

We can make a claim about one of infinitely-many
right triangles that one of its angles is 90 degrees,
without knowing which right triangle our claim is
about.

We _can'_t make that claim about one of infinitely-many
_triangles_ not without knowing more. Some have a
90 degree angle, but some do not.

You tell us that "k has been chosen and used" is not
a claim that can be made for each one of the k which
end some two-ended-counting-ordered {0,...,k} which
begins at 0.

Okay, then. "k has been chosen and used" is not the
sort of starting-claim which we should use to reason
about one of the k which end some two-ended-counting-
-ordered {0,...,k} which begins at 0.

Since we wouldn't do that anyway, it's very easy
to follow that recommendation.

What else is there to say about this?
You got what you wanted didn't you?

[William]

On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > >
> > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > Construct the largest number you can.
> > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> Who told you

It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,

--
William Hughes

[zelos...@gmail.com]

your "dark" is irrelevant here and serves no purpose.

[WM]

Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> On 1/16/2022 4:29 PM, WM wrote:

> > But not all can be chosen and used.
> > Almost all will not have been used at the end of times.
> > That means they cannot have been used.
> > That means they are dark.

> We have a technique for reasoning about infinitely-many
> individuals.

Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.

Regards, WM

[WM]

By induction we can show that almost all elements are dark and will never have been used:
∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That is tantamount to they cannot be used. They can only be used collectively:
|ℕ \ {1, 2, 3, ...}| = 0 .

Regards, WM

[Gus Gassmann]

Logic really is not your strong suit. There is no way to say of any particular number that it will never be used in the future. Any one of them can be reasoned about, any one of them is linkable by a chain of successor operations back to 0, and any one of them may turn out not only to be used, but to be very useful indeed. So stop your inane whining.

[WM]

horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
> On Monday, 17 January 2022 at 06:56:35 UTC-4, WM wrote:
> > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> > > We have a technique for reasoning about infinitely-many
> > > individuals.
> > Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.

> There is no way to say of any particular number that it will never be used in the future.

Of course not. Thar one would have been used.

> Any one of them can be reasoned about,

Almost all cannot. Proof: Each used one has ℵo succesors, ℵo of which will be successors of every used number.

Regards, WM

[Fritz Feldhase]

On Monday, January 17, 2022 at 2:31:38 PM UTC+1, WM wrote:
> horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
> >
> > Any one of them can be reasoned about
> >

> Almost all cannot.

Which one has been left out in this case:

An e IN: n + n = 2*n

?

> Proof: Each [...] one has ℵo succesors,

Yeah: An e IN: card({m e IN : m > n}) = aleph_0.

> ℵo of which will be successors of every [...] number.

Mückenheim's rule?

An e IN: EM c IN (card(M) = aleph_0): [Am e M: n < m]

implies

EM c IN (card(M) = aleph_0): An e IN: [Am e M: n < m]

?

Fascinating! Hint: This rule boils down to the following "principle":

An e IN: Em e M: n < m

implies

Em e M: An e IN: n < m.

A truism in mückenmath.

[sergio]

wrong. That is observer dependent math, Fail.

Any number may or can be used, which directly conflicts with your personal decision *they cannot be used*

[sergio]

that is spoof, not a proof.

the number of successors has no bearing at all on weather a number is "used" (by an external observer)

You idea that numbers cannot be used, and therefore you can discount, remove, or turn them dark, is failure.

Numbers are not an extension of a persons personality

>
> Regards, WM
>
>

[sergio]

On 1/17/2022 4:59 AM, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
>> On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
>>>> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
>>>>> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>>>>>
>>>>>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>>>>> Construct the largest number you can.
>>>> It is an element of N_P. And it was an element of N_P before you "constructed" it.
>>> Who told you
>> It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
>
> By induction we can show that almost all elements are dark and will never have been used:
> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

wrong. all that says is if you remove a finite set from an infinite set, you have an infinite set. Fail.

> That is tantamount to they cannot be used. They can only be used collectively:
> |ℕ \ {1, 2, 3, ...}| = 0 .

wrong. all that says is if you remove all the elements of a set there are no elements left. Fail.

>
> Regards, WM

[Jim Burns]

I continue...

Start with some claim we know is true of
each individual of the infinitely-many. Take only
reliable statement-steps from there.

However, the step you make from
| For each j, there are infinitely-many k > j
to
| There are infinitely-many k, for each j, k > j
is not a reliable statement-step.

By "reliable", I mean that, in any case in which
the first claim(s) is (are) true, the second
claim is true. (AKA "valid")

There are cases in which the second claim
(AKA "conclusion") is provably true. But the
quantifier shift is not used in the proof in those
cases.

There are cases in which the second claim is
provably false. It is because of these cases that
we cannot use a quantifier shift in the proof
even in case in which the second claim is true.
A reliable step cannot be otherwise: that's what
justifies the second claim.

An example of what looks like a quantifier shift
but isn't:
From:
| For each inductive set, there is an inductive set
| set which is contained in each of its inductive subsets.
To:
| There is an inductive set which, for each inductive
| set, is contained in each of its inductive subsets.

That's true, that's provable, but not by using a quantifier
shift. We use the fact that the intersection of inductive
sets is an inductive set.

An example showing why a quantifier shift is not reliable:
From:
| For each natural j, there is a natural k, k > j
To:
| There is NOT a natural j, for each natural k, k =< j

THIS step is reliable, because k > j xor k =< j

Also, this is why we can't justify a claim by
quantifier shift.

Also also, the "To" contradicts
| There are infinitely-many k, for each j, k > j

That's NOT a case of proofs contradicting each other.
There is one proof, ours, and one not-proof using a
quantifier shift, yours.

[William]

On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > >
> > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > Construct the largest number you can.
> > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > Who told you
> > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> By induction we can show that almost all elements are dark

Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.

> and will never have been used:

"used" is another of the many ways you have of saying "can be written down" Before you try to prove it at again let me specify that almost all elements of N_P will never be "used". The fact that an element of N_P has the property that it "cannot be written down" does not mean it does not have other properties, E,g is the largest element of a FISON.

--
William Hughes

.

[WM]

Jim Burns schrieb am Montag, 17. Januar 2022 um 18:37:16 UTC+1:

> An example showing why a quantifier shift is not reliable:

Here I show you a proof that cannot be refuted by shouting "quantifier shift". Indexing the fractions

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, ...
...

by natural numbers according to
k = (m + n - 1)(m + n - 2)/2 + m (*)
is called Cantor's first diagonl argument.

We put the indexes X in the first column and mark all not yet indexed fractions by O.

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
...

Then every index k will be attached to a fraction m/n according to (*). The O's will be removed from the target element but will appear where the index has taken from. I call this change of places between X and O a transposition. Here is the configuration according to (*) for k = 8:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
...

Note that every transposed O is dropped above the X to be applied next. Therefore it is impossible that an O disappears from the board as long as X's are available. Indexing all fractions of the matrix would require all O's to disappear. That is impossible as long as indexes X are in the first column below the dropped O's.

No try to find a way to revert the positions of O's and X's in the first column or to get rid of the O's in another way.

Regards, WM

[WM]

William schrieb am Montag, 17. Januar 2022 um 19:15:09 UTC+1:
> On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> > William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > > >
> > > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > > Construct the largest number you can.
> > > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > > Who told you
> > > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> > By induction we can show that almost all elements are dark
> Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.

No, that is only provable for n+1 if n has been named.

> > and will never have been used:
> "used" is another of the many ways you have of saying "can be written down"

Yes.

> Before you try to prove it at again let me specify that almost all elements of N_P will never be "used".

That is true.

> The fact that an element of N_P has the property that it "cannot be written down" does not mean it does not have other properties, E,g is the largest element of a FISON.

Maybe. But we cannot know that FISON. It remains dark to us. Otherwise the element is visible, defined, and used.

Regards, WM

[sergio]

On 1/17/2022 1:31 PM, WM wrote:
> Jim Burns schrieb am Montag, 17. Januar 2022 um 18:37:16 UTC+1:
>
>> An example showing why a quantifier shift is not reliable:
>
> Here I show you a proof that cannot be refuted by shouting "quantifier shift". Indexing the fractions
>
> 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, ...
> ...
>
> by natural numbers according to
> k = (m + n - 1)(m + n - 2)/2 + m (*)
> is called Cantor's first diagonl argument.

wrong. when m=1 and n=1 k should = 1, but you have 0 Fail.

also, in your equation m and n are symetrical, which means when they are switched you get the same k. Fail.

this is not enumeration at all. It is fake math.

>
> We put the indexes X in the first column and mark all not yet indexed fractions by O.

this is not Cantors enumeration at all, which diagonally threads

<snip rest of misleading garf>

[sergio]

On 1/17/2022 1:34 PM, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 19:15:09 UTC+1:
>> On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
>>> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
>>>> On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
>>>>> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
>>>>>> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
>>>>>>> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>>>>>>>
>>>>>>>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>>>>>>> Construct the largest number you can.
>>>>>> It is an element of N_P. And it was an element of N_P before you "constructed" it.
>>>>> Who told you
>>>> It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
>>> By induction we can show that almost all elements are dark
>> Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.
>
> No, that is only provable for n+1 if n has been named.

No.

>
>>> and will never have been used:
>> "used" is another of the many ways you have of saying "can be written down"
>
> Yes.
>
>> Before you try to prove it at again let me specify that almost all elements of N_P will never be "used".
>
> That is true.
>
>> The fact that an element of N_P has the property that it "cannot be written down" does not mean it does not have other properties, E,g is the largest element of a FISON.
>
> Maybe. But we cannot know that FISON. It remains dark to us. Otherwise the element is visible, defined, and used.

which is Observer Dependent Math, a farce.

>
> Regards, WM

[Fritz Feldhase]

On Monday, January 17, 2022 at 8:34:47 PM UTC+1, WM wrote:
>
> No, that is only provable for n+1 if n has [a] name[.]

If you say so. Ok, then I will supply all n e IN with names: The name for n (where n is any natural number) is "|...|" where "|..|" consist of n strokes. Hence the name of the first natural number is just "|", that of the second is "||", etc.

We can do this "recursively" too: 1 has the name "|". For all n e IN: n+1 has the name of n appended by "|".

Hence each and every natural number has a name.

And no, I can't personally "batize" them all, one after the other. :-)

Only Chuck Norris could do that!

[William]

On Monday, January 17, 2022 at 3:34:47 PM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 19:15:09 UTC+1:
> > On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > > > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > > > >
> > > > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > > > Construct the largest number you can.
> > > > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > > > Who told you
> > > > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> > > By induction we can show that almost all elements are dark
> > Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.
> No, that is only provable for n+1 if n has been named.

Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)

... know[n]... visible, defined, used.

You have a lot of ways of say "can be written down"

--
William Hughes

[Gus Gassmann]

That is not a proof that any particular number cannot be used, you fucking imbecile. You are up to your old quantifier switching. Man, you are dense!

[zelos...@gmail.com]

it is entirely irrelevant

[WM]

William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:

> Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
>
> ... know[n]... visible, defined, used.
>
> You have a lot of ways of say "can be written down"

To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions

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, ...
...

can be represented by diagrams. When we put all indexes X in the first column and apply them as Cantor prescribed, then we get first, as the starting position:

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
...

where the X's represent the indexes 1, 2, 3, ...

The index 1 remains at fraction 1/1. The first move, by 2, indexes 1/2:

XXOOOOOOO...
OOOOOOOOO...

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
...

and so on: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ...

After eight steps for instance we get

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
...

Important is, when continuing, that no O can disappear as long as any index X blocks the possible drain, i.e., the first column.

X
X
X
.
.
.
O
O
O
.
.
.
X
X
X
.
.
.

As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.

This proves there remain many fractions without indexes. But we cannot find them. They are dark.

In this example I have assumed that all indexes are visible and can be used. But why should they, if there are dark fractions like n/1?

Regards, WM

[zelos...@gmail.com]

proves nothing of the sort and your stupid attempt at this proves you are an idiot.

The original bijection is all that matters, not your stupid shit.

[WM]

zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:

> > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions

> > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> >
> > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> >

> proves nothing of the sort

Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?

Regards, WM

[sergio]

On 1/18/2022 4:41 AM, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
>
>> Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
>>
>> ... know[n]... visible, defined, used.
>>
>> You have a lot of ways of say "can be written down"
>
> To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
>
> 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, ...
> ...
>
> can be represented by diagrams. When we put all indexes X in the first column and apply them as Cantor prescribed, then we get first, as the starting position:
>
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
>
> where the X's represent the indexes 1, 2, 3, ...
>

that is not Cantor's mapping at all. You associate an index to an entire row, Fail.

[sergio]

you listed the k's as a column associating it directly with m.


An intentional misdirection.

[Jim Burns]

On 1/17/2022 2:31 PM, WM wrote:
> Jim Burns schrieb
> am Montag, 17. Januar 2022 um 18:37:16 UTC+1:

>> An example showing why a quantifier shift is
>> not reliable:
>
> Here I show you a proof that cannot be refuted by
> shouting "quantifier shift".

I did not intend to suggest that quantifier shifts
are the only fallacies you commit.

> Indexing the fractions
> 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, ...
> ...
>
> by natural numbers according to
> k = (m + n - 1)(m + n - 2)/2 + m (*)
> is called Cantor's first diagonl argument.
>
> We put the indexes X in the first column and
> mark all not yet indexed fractions by O.

Step 0.

> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
>
> Then every index k will be attached to a fraction m/n
> according to (*).

Steps 1 through k.

Steps 1 through k have a counting-order
(total order, each cut is a step, each step is a count)
which begins at 1 and ends at k, whatever is.

Without ending somewhere, they're not the first kind.

> The O's will be removed from the target element but
> will appear where the index has taken from.
> I call this change of places between X and O
> a transposition. Here is the configuration according
> to (*) for k = 8:

Steps 1 through 8 have a counting-order
(total order, each cut is a step, each step is a count)
which begins at 1 and ends at 8.

Without ending somewhere, they're not the first kind.

> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>
> Note that every transposed O is dropped above the X
> to be applied next. Therefore it is impossible that
> an O disappears from the board as long as X's are
> available.

As long as X's are available,
the transpositions have a stepping-order
(total order, each cut is a step)
which begins somewhere and ends somewhere.

Without ending somewhere, they're not the first kind.

> Indexing all fractions of the matrix

All fractions of the matrix do NOT have a stepping-
-order (total order, each cut is a step)
which begins somewhere and ends somewhere.

Without ending somewhere, they're not the first kind.

> Indexing all fractions of the matrix would require
> all O's to disappear.

All fractions, the collection of, is the second kind.
d/n gets an x from i/1 where i = (d+n-1)*(d+n-2)/2 + n
All O's disappear.

> That is impossible as long as indexes X are
> in the first column below the dropped O's.
>
> No try to find a way to revert the positions
> of O's and X's in the first column or
> to get rid of the O's in another way.

First kind ≠ second kind.

[William]

On Tuesday, January 18, 2022 at 6:41:22 AM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
>
> > Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
> >
> > ... know[n]... visible, defined, used.
> >
> > You have a lot of ways of say "can be written down"
> To explain it better I have to go a long way back

Nope, I have already said that for the sake of argument I will say the fractions cannot be indexed. If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.) So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.


--
William Hughes

[JVR]

I don't know. I can't bring myself to read any more of your repetitive nonsense.
The function k = (m + n - 1)(m + n - 2)/2 + m is biunique as every child can verify.
That's all there is to the countability of the rationals.

[Gus Gassmann]

It's his usual shit about thinking that somehow all integer values have to be processed to "unclog the brain" and allow the remaining fractional values to be enumerated. It's even clearer where he goes wrong when he uses the X's and O's, because there even he should see that some of the X's move off the first column before the entire column has been processed and that the vertical will not get exhausted. I am not holding my breath, however. Even the three dots he puts at the bottom are smeared there with zero comprehension on his part.

[WM]

JVR schrieb am Dienstag, 18. Januar 2022 um 19:50:24 UTC+1:
> On Tuesday, January 18, 2022 at 3:25:55 PM UTC+1, WM wrote:
> > zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
> >
> > > > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
> > > > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> > > >
> > > > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> > > >
> > > proves nothing of the sort
> > Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
> >

> I don't know. I can't bring myself to read any more of your repetitive nonsense.
> The function k = (m + n - 1)(m + n - 2)/2 + m is biunique as every child can verify.
> That's all there is to the countability of the rationals.

No, there is a contradiction. No O can leave before all indexes have been issued. No O can leave after all indexes have been issued. (Note that I don't claim that this will ever happen.)

Regards, WM

[WM]

William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:
> On Tuesday, January 18, 2022 at 6:41:22 AM UTC-4, WM wrote:
> > William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
> >
> > > Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
> > >
> > > ... know[n]... visible, defined, used.
> > >
> > > You have a lot of ways of say "can be written down"
> > To explain it better I have to go a long way back
> Nope, I have already said that for the sake of argument I will say the fractions cannot be indexed.

But those not indexed are either not existing or not available, i.e., dark. Wouldn't dark fractions hint to dark integers?

> If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)

I agree.

> So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.

But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's

o
oo
ooo
...

than are o's avaible. There are less than ℵo, i.e., less than more than all.

Regards, WM

[WM]

Jim Burns schrieb am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> On 1/17/2022 2:31 PM, WM wrote:

> Steps 1 through k have a counting-order
> (total order, each cut is a step, each step is a count)
> which begins at 1 and ends at k, whatever is.
>
> Without ending somewhere, they're not the first kind.

My model need not end (although Cantor claims all indexes are issued). But before ending, the O's cannot leave.

> Without ending somewhere, they're not the first kind.

My model need not end (although Cantor claims all indexes are issued). But before ending, no O can leave.

> As long as X's are available,
> the transpositions have a stepping-order
> (total order, each cut is a step)
> which begins somewhere and ends somewhere.

Of course.

>
> Without ending somewhere, they're not the first kind.

But without ending all O's will remain.

> Without ending somewhere, they're not the first kind.
> > Indexing all fractions of the matrix would require
> > all O's to disappear.
> All fractions, the collection of, is the second kind.
> d/n gets an x from i/1 where i = (d+n-1)*(d+n-2)/2 + n
> All O's disappear.

No, that is provably wrong.

> First kind ≠ second kind.

My model works for all kinds.

Regards, WM

[William]

Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available. By induction we can show that every element of the Peano set N_P is the largest element of a FISON.

--
William Hughes

[WM]

William schrieb am Dienstag, 18. Januar 2022 um 20:51:23 UTC+1:
> On Tuesday, January 18, 2022 at 3:21:53 PM UTC-4, WM wrote:
> > William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:
>
> > > If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)
> > I agree.
> > > So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.
> > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> >
> > o
> > oo
> > ooo
> > ...
> >
> > than are o's avaible.
> Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.

FISONs have less than aleph_0 o's.

> By induction we can show that every element of the Peano set N_P is the largest element of a FISON.

But aleph_0 = |ℕ| is not reached. ℕ \ U(FISONs) =/= { }.

Regards, WM

[sergio]

On 1/18/2022 1:26 PM, WM wrote:
> Jim Burns schrieb am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>> On 1/17/2022 2:31 PM, WM wrote:
>
>> Steps 1 through k have a counting-order
>> (total order, each cut is a step, each step is a count)
>> which begins at 1 and ends at k, whatever is.
>>
>> Without ending somewhere, they're not the first kind.
>
> My model need not end (although Cantor claims all indexes are issued). But before ending, the O's cannot leave.

your O's will not leave until you pay what you owe them!

>
>> Without ending somewhere, they're not the first kind.
>
> My model need not end (although Cantor claims all indexes are issued). But before ending, no O can leave.

your model is no good at all, because you stuck your index into the matrix as a column. Fail

>
>> As long as X's are available,
>> the transpositions have a stepping-order
>> (total order, each cut is a step)
>> which begins somewhere and ends somewhere.
>
> Of course.
>>
>> Without ending somewhere, they're not the first kind.
>
> But without ending all O's will remain

you stopped again!!

there is no ending an infinite matrix

>
>> Without ending somewhere, they're not the first kind.
>>> Indexing all fractions of the matrix would require
>>> all O's to disappear.
>> All fractions, the collection of, is the second kind.
>> d/n gets an x from i/1 where i = (d+n-1)*(d+n-2)/2 + n
>> All O's disappear.
>
> No, that is provably wrong.

no. It can be proven you have failed again.
you never prove anything.

>
>> First kind ≠ second kind.
>
> My model works for all kinds.

no. your model does not work.

>
> Regards, WM

[sergio]

No, your model is inherently flawed. Fix your mistake.

>
> Regards, WM

[William]

On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 18. Januar 2022 um 20:51:23 UTC+1:
> > On Tuesday, January 18, 2022 at 3:21:53 PM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:
> >
> > > > If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)
> > > I agree.
> > > > So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.
> > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > >
> > > o
> > > oo
> > > ooo
> > > ...
> > >
> > > than are o's avaible.
> > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> FISONs have less than aleph_0 o's.

Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.

--
William Hughes

[Jim Burns]

On 1/18/2022 2:26 PM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:

>> Without ending somewhere, they're not the first kind.
>
> But without ending all O's will remain.

If the O on n/d remains, the replacement has ended
before step i = (n+d-1)*(n+d-2)/2 + n

[zelos...@gmail.com]

It fails because you moving shit around is of no relevants.

[WM]

OK, if you believe in infinite FISONs, then I cannot win the argument.

Regards, WM

[WM]

Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> On 1/18/2022 2:26 PM, WM wrote:
> > Jim Burns schrieb
> > am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> >> Without ending somewhere, they're not the first kind.
> >
> > But without ending all O's will remain.
> If the O on n/d remains,

Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.

> the replacement has ended
> before step i = (n+d-1)*(n+d-2)/2 + n

No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.

Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!

Regards, WM

[zelos...@gmail.com]

Again, your re-arranging is entirely fuckign IRRELEVANT

[FromTheRafters]

WM expressed precisely :

If it doesn't work the wrong way, try doing it the right way.

[sergio]

On 1/19/2022 3:38 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>> On 1/18/2022 2:26 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>>>> Without ending somewhere, they're not the first kind.
>>>
>>> But without ending all O's will remain.
>> If the O on n/d remains,
>
> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.

Using a boogered Tic-Tac-Toe as a substitute for Cantor's Enumeration is silly.

>
>> the replacement has ended
>> before step i = (n+d-1)*(n+d-2)/2 + n
>
> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.

because you are doing it wrong.

>
> Look, I did nothing else but ask where Cantor's indexes come from.

no, you put your index in a column of the matrix, wrong move.

>Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
>

both intentionally wrong and silly.

Just wiki Cantor's Enumeration and use that, and see where your mistakes are.


> Regards, WM

[sergio]

the # of FISONs is infinite, so you cannot win the argument.

[William]

Nope. I do not believe in an infinite FISON. Each line is finite. Each FISON is finite.
Does not change the fact that there are aleph_0 lines. Thus there are aleph_0 o's availiable.

--
William Hughes

[Gus Gassmann]

On Wednesday, 19 January 2022 at 09:40:38 UTC-4, sergio wrote:
> On 1/19/2022 3:25 AM, WM wrote:
> > OK, if you believe in infinite FISONs, then I cannot win the argument.
> >

> the # of FISONs is infinite, so you cannot win the argument.

The premise of WM's statement is, of course, entirely redundant. In order to win any argument, one has to be able to reason logically, and WM has proven himself utterly incapable of this for at least the last fourteen years. His premise is also ambiguous, likely intentionally so. Nobody, except perhaps WM himself, thinks that there could be FISONs that have infinite cardinality (the 'F' is there for a reason!). But, of course, there are infinitely many of them. And of course WM also believes that there cannot be infinitely many natural numbers unless omega is counted as a natural number.

[Jim Burns]

On 1/19/2022 4:38 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>> On 1/18/2022 2:26 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:

>>>> Without ending somewhere, they're not the first kind.
>>>
>>> But without ending all O's will remain.
>> If the O on n/d remains,
>
> Every O moves only once and then remains fixed ---
> in the first column --- above the never empty rest of X's.

No.

i = (n+d-1)*(n+2-2)/2 + n

The O at 1/2 moves to 2/1
because (1+2-1)(1+2-2)/2 + 1 = 2

The O at 2/1 moves to 3/1
because (2+1-1)*(2+1-2)/2 + 2 = 3

The O at 3/1 moves to 6/1
because (3+1-1)*(3+1-2)/2 + 3 = 6

The O at 6/1 moves to 21/1
because (6+1-1)*(6+1-2)/2 + 6 = 21

The O at 21/1 moves to 231/1
because (21+1-1)*(21+1-2)/2 + 21 = 231

...

No O remains fixed.
Other than 1/1 n/d moves to a location i/1 with an
index later than n/d. That later-indexed location
moves to a still later-indexed location, ad infinitum.


No definable O moves to a dark location.

i = (n+d-1)*(n+2-2)/2 + n

definable n, definable d
definable n+d-1, definable n+d-2
either 2 divides n+d-1 or 2 divides n+d-2
2 divides (n+d-1)*(n+d-2)
definable (n+d-1)*(n+d-2)/2 + n
i = (n+d-1)*(n+2-2)/2 + n
definable i

Define {1,...,k} to be a collection with a counting-order
which begins at 1 and ends somewhere. Name where it ends k.

Define ℕ⁺ = ⋃{{1,...,k}} union all the {1,...,k}

Define 𝔽⁺ = { n/d | n,d ∈ ℕ⁺ }

n/d ---> i/1
moves the X's and O's within the defined 𝔽⁺

You want the reason to be existence of dark numbers.
But i = (n+d-1)*(n+d-2)/2 + n
Whether or not dark numbers exist, no O moves to
a dark number.

Dark numbers don't explain how all of
𝔽⁺ = { n/d | n,d ∈ ⋃{{1,...,k}} }
fits in only its first column.

The reason that all of 𝔽⁺ can move to only its first column
is that 𝔽⁺ is NOT the first kind of collection,
the sheep-kind, the pebble-kind.

You:
Look! 𝔽⁺ does not behave like the first kind of collection.

Me:
Right. Because 𝔽⁺ is not the first kind of collection.

[WM]

> Again, your re-arranging is entirely fuckign IRRELEVANT

Why do you think so?

Of course nobody need be interested in my model. But for those beein interested I can state that my model applies Cantor's first diagonal arguments meticulously. Every fraction is indexed as he did. The only difference is that I ask where the indexes come from.

Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.

Regards, WM

[WM]

FromTheRafters schrieb am Mittwoch, 19. Januar 2022 um 13:23:08 UTC+1:
> WM expressed precisely :

> > Look, I did nothing else but ask where Cantor's indexes come from.
> > Matheologians never did so and thus remained in ignorance. It is clear that
> > the elements of one column will never flood the whole matrix because for
> > every taken X (necessarily in the finite) there opens a gap receiving an O
> > (necessarily in the finite). It is simply unbelievable that people could
> > believe that this could change!!!
> If it doesn't work the wrong way, try doing it the right way.

I did. My simple precondition is: Every Index is taken from a finite place. Hence every O lands at a finite place. No magic, no limits, simply application of
k = (m + n - 1)(m + n - 2)/2 + m.
The only violation of Cantor's way is to ask where the indexes come from.

If you don't like my result, simply don't ask, and you may stay happy with set theory.

Regards, WM

[WM]

sergio schrieb am Mittwoch, 19. Januar 2022 um 14:39:49 UTC+1:
> On 1/19/2022 3:38 AM, WM wrote:

> > Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
> Using a boogered Tic-Tac-Toe as a substitute for Cantor's Enumeration is silly.

It is allowed to use diagrams in order to get the overview, in particular to know where the indxes come from.

> >
> >> the replacement has ended
> >> before step i = (n+d-1)*(n+d-2)/2 + n
> >
> > No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
> because you are doing it wrong.

What is done wrong?

> Just wiki Cantor's Enumeration and use that,

I do: k = (m + n - 1)(m + n - 2)/2 + m

> and see where your mistakes are.

My "mistake" is to show where the indexes come from. That's the only deviation from Cantor-

Regards, WM

[WM]

William schrieb am Mittwoch, 19. Januar 2022 um 16:06:05 UTC+1:
> On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> > > On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> >
> > > > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > > > >
> > > > > > o
> > > > > > oo
> > > > > > ooo
> > > > > > ...
> > > > > >
> > > > > > than are o's avaible.
> > > > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> > > > FISONs have less than aleph_0 o's.
> > > Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
> > OK, if you believe in infinite FISONs,
> Nope. I do not believe in an infinite FISON. Each line is finite.

Each line is the union of all preceding lines + o. If there is an actually infinite number of o's, ℵo o's, then there is a preceding FISON with one o less than ℵo.

> Each FISON is finite.

The addition of finite FISONs is a fiite FISON.

> Does not change the fact that there are aleph_0 lines. Thus there are aleph_0 o's availiable.

o
oo
ooo
oooo
ooooo
...

Never. Width = Height.

Regards, WM

[WM]

Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:
> On 1/19/2022 4:38 AM, WM wrote:
> > Jim Burns schrieb
> > am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> >> On 1/18/2022 2:26 PM, WM wrote:

> > Every O moves only once and then remains fixed ---
> > in the first column --- above the never empty rest of X's.
> No.

It is fact by construction that all X are taken from finite places and O's are landing at finite places. The matrix has no other places.

>
> i = (n+d-1)*(n+2-2)/2 + n

Slightly corrected: i = (n+d-1)*(n+d-2)/2 + n

>
> The O at 1/2 moves to 2/1
> because (1+2-1)(1+2-2)/2 + 1 = 2

So it is.

>
> The O at 2/1 moves to 3/1
> because (2+1-1)*(2+1-2)/2 + 2 = 3
>
> The O at 3/1 moves to 6/1
> because (3+1-1)*(3+1-2)/2 + 3 = 6

After the move k = 8 we have:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...

...
>
> No O remains fixed.

Yes, you are right. My mistake. The X, when moved will remain fixed

> Other than 1/1 n/d moves to a location i/1 with an
> index later than n/d. That later-indexed location
> moves to a still later-indexed location, ad infinitum.

Of course. But since every X stems from a finite place, every transposed O willland at a finite place.

>
> No definable O moves to a dark location.
>

> i = (n+d-1)*(n+d-2)/2 + n

>
> definable n, definable d
> definable n+d-1, definable n+d-2
> either 2 divides n+d-1 or 2 divides n+d-2
> 2 divides (n+d-1)*(n+d-2)
> definable (n+d-1)*(n+d-2)/2 + n
> i = (n+d-1)*(n+2-2)/2 + n
> definable i

Right. Every move ends at a definable place.

Therefore the number of O's in the matrix does not change. Agreed?

Regards, WM

[WM]

Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:

> n/d ---> i/1
> moves the X's and O's within the defined 𝔽⁺

Yes. Every transposition where the X_i1 moves to O_nd and vice versa will preserve the numbers of X's and O's.

>
> You want the reason to be existence of dark numbers.
> But i = (n+d-1)*(n+d-2)/2 + n
> Whether or not dark numbers exist, no O moves to
> a dark number.

Right.

>
> Dark numbers don't explain how all of
> 𝔽⁺ = { n/d | n,d ∈ ⋃{{1,...,k}} }
> fits in only its first column.

They don't fit in the first column.

>
> The reason that all of 𝔽⁺ can move to only its first column
> is that 𝔽⁺ is NOT the first kind of collection,
> the sheep-kind, the pebble-kind.

Whatever. Even if all O's fit in the first column, they stay within the matrix. Therefore their number is preserved.

>
> You:
> Look! 𝔽⁺ does not behave like the first kind of collection.

I am not interested in the kind of collection. I am only interested in the fact that no O can leave, and hence not all fractions can be indeXed.

Regards, WM

[sergio]

On 1/19/2022 12:23 PM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 19. Januar 2022 um 13:23:08 UTC+1:
>> WM expressed precisely :
>
>>> Look, I did nothing else but ask where Cantor's indexes come from.
>>> Matheologians never did so and thus remained in ignorance. It is clear that
>>> the elements of one column will never flood the whole matrix because for
>>> every taken X (necessarily in the finite) there opens a gap receiving an O
>>> (necessarily in the finite). It is simply unbelievable that people could
>>> believe that this could change!!!
>> If it doesn't work the wrong way, try doing it the right way.
>
> I did. My simple precondition is: Every Index is taken from a finite place.

stealing #'s off the sides of sheeps, shame!

> Hence every O lands at a finite place.

you don't need a "place" to take numbers from or land in in Math.

> No magic, no limits, simply application of
> k = (m + n - 1)(m + n - 2)/2 + m.

wrong. Cantor does it clearly. Yours is boogered.

> The only violation of Cantor's way is to ask where the indexes come from.

wrong. you just do not know how to do math, so quit trying.

>
> If you don't like my result, simply don't ask, and you may stay happy with set theory.

set theory is just fine without your permanent confusion.

>
> Regards, WM

[sergio]

On 1/19/2022 12:27 PM, WM wrote:
> sergio schrieb am Mittwoch, 19. Januar 2022 um 14:39:49 UTC+1:
>> On 1/19/2022 3:38 AM, WM wrote:
>
>>> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
>> Using a boogered Tic-Tac-Toe as a substitute for Cantor's Enumeration is silly.
>
> It is allowed to use diagrams in order to get the overview, in particular to know where the indxes come from.

no, you use it to fool people. Indexes are NOT in the matrix. Fail.

>>>
>>>> the replacement has ended
>>>> before step i = (n+d-1)*(n+d-2)/2 + n
>>>
>>> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
>> because you are doing it wrong.
>
> What is done wrong?

you are manipulating the Matrix of fractions.

there is absolutely no need to do any of what you are doing except to intentionally arrive at a false conclusion.

>
>> Just wiki Cantor's Enumeration and use that,
>
> I do: k = (m + n - 1)(m + n - 2)/2 + m

on the wrong matrix.

>
>> and see where your mistakes are.
>
> My "mistake" is to show where the indexes come from. That's the only deviation from Cantor-

intentional mistakes, that fail. There is no need to show where indexes come from.

>
> Regards, WM

Your shell game is kaput.

[sergio]

On 1/19/2022 12:56 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:
>

<snip bad math>

> Regards, WM

Fail.

[sergio]

On 1/19/2022 12:17 PM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 19. Januar 2022 um 12:31:53 UTC+1:
>> onsdag 19 januari 2022 kl. 10:38:59 UTC+1 skrev WM:
>>> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>>>> On 1/18/2022 2:26 PM, WM wrote:
>>>>> Jim Burns schrieb
>>>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>>>>>> Without ending somewhere, they're not the first kind.
>>>>>
>>>>> But without ending all O's will remain.
>>>> If the O on n/d remains,
>>> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
>>>> the replacement has ended
>>>> before step i = (n+d-1)*(n+d-2)/2 + n
>>> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
>>>
>>> Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
>>>
>> Again, your re-arranging is entirely fuckign IRRELEVANT
>
> Why do you think so?

it is obvious it is irrelevant.

>
> Of course nobody need be interested in my model. But for those beein interested I can state that my model applies Cantor's first diagonal arguments meticulously.

Liar.

> Every fraction is indexed as he did. The only difference is that I ask where the indexes come from.

only because you do not know how to use numbers, and you failed.

>
> Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.
>
> Regards, WM

yoru shell game is uncovered, you manipulated the matrix, to deceive others.

You are a Deceiver.

[William]

On Wednesday, January 19, 2022 at 2:32:18 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 19. Januar 2022 um 16:06:05 UTC+1:
> > On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> > > > On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> > >
> > > > > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > > > > >
> > > > > > > o
> > > > > > > oo
> > > > > > > ooo
> > > > > > > ...
> > > > > > >
> > > > > > > than are o's avaible.
> > > > > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> > > > > FISONs have less than aleph_0 o's.
> > > > Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
> > > OK, if you believe in infinite FISONs,
> > Nope. I do not believe in an infinite FISON. Each line is finite.
> Each line is the union of all preceding lines + o.

Correct. Does not, nothing you say can, change the fact that there are an infinite number of lines.
---
William Hughes

[WM]

An infinite number of finite lines. That implies that the infinity is potential.

Regards, WM