On 2/26/2022 7:02 AM, WM wrote:
> horand....@gmail.com schrieb am Freitag, 25. Februar 2022 um 17:25:27 UTC+1:
>
>> The sequences {m_{i,j}(k):k \in IN} are sequences of rational numbers with well-defined formation rules. (The definitions are recursive, but they are finite.) Does the sequence {1/2, 2/1, 3/1, 3/1, 6/1, ...} have a limit or not?
>
> The sequence does not have a limit like 1, 2, 3, ... does not have a limit.

answer the question.

> Regards, WM

It happens that WM formulated :
> horand....@gmail.com schrieb am Donnerstag, 24. Februar 2022 um 23:25:34
> UTC+1:
>> I am not interested in the sequence of fractions; I am interested in the
>> individual elements of the matrices M(k).
>>
>> 1/1, 1/2, 1/3, 1/4, ... 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, 6/1, 4/1, 1/4, ... 2/1, 2/2, 2/3, 2/4, ... 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, ... 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, ... 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, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4,
>> ... 6/1, 6/2, 6/3, 6/4, ... 6/1, 6/2, 6/3, 6/4, ... 6/1, 6/2, 6/3, 6/4, ...
>> 6/1, 6/2, 6/3, 6/4, ... 3/1, 6/2, 6/3, 6/4, ... ... ... ... ...
>>
>> The elements in the first column will, of course, be the positive fractions
>> in Cantor's order, and after step k, m_{k,1} will be constant, since the
>> k'th fraction will have been swapped into place. But look at m_{1,2}(k), for
>> instance. The sequence starts 1/2, 2/1, 3/1, 3/1, 6/1, .... How does it
>> continue, and does it have a limit? If it does have a limit, do you think it
>> is one of the fractions you swapped into the first column? (Which one?)
>
> It has no limit. It vanishes in darkness. Note that only fractions which have
> been there in the matrix are changing places. There were no "limits" in the
> first matrix and there are no limits later. But there are only potentially
> many lit fractions. Every place of the matrix has its sequence. The places in
> the first column get settled. The other places get different sequences of
> fractions with no end. You can extend 1/2, 2/1, 3/1, 3/1, 6/1, ... as far as
> you like. Somewhere darkness falls.

I see no pattern to continue here. How does one extend this sequence?

FromTheRafters schrieb am Samstag, 26. Februar 2022 um 15:55:22 UTC+1:
> It happens that WM formulated :
> > You can extend 1/2, 2/1, 3/1, 3/1, 6/1, ... as far as
> > you like. Somewhere darkness falls.
>
> I see no pattern to continue here. How does one extend this sequence?

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, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1 ... will appear in the first column. So 6/1 has to occupy line 21. Therefore 21/1 sitting there will be exchanged by 6/1 and extend the sequence in (1, 2):
1/2, 2/1, 3/1, 3/1, 6/1, 21/1.

Regards, WM

WM laid this down on his screen :
> FromTheRafters schrieb am Samstag, 26. Februar 2022 um 15:55:22 UTC+1:
>> It happens that WM formulated :
>>> You can extend 1/2, 2/1, 3/1, 3/1, 6/1, ... as far as
>>> you like. Somewhere darkness falls.
>>
>> I see no pattern to continue here. How does one extend this sequence?
>
> 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, 2/4,
> 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1 ... will appear in the first
> column. So 6/1 has to occupy line 21. Therefore 21/1 sitting there will be
> exchanged by 6/1 and extend the sequence in (1, 2): 1/2, 2/1, 3/1, 3/1, 6/1,
> 21/1.

Why does 3/1 appear twice?

FromTheRafters schrieb am Samstag, 26. Februar 2022 um 21:27:21 UTC+1:
> WM laid this down on his screen :
> > FromTheRafters schrieb am Samstag, 26. Februar 2022 um 15:55:22 UTC+1:
> >> It happens that WM formulated :
> >>> You can extend 1/2, 2/1, 3/1, 3/1, 6/1, ... as far as
> >>> you like. Somewhere darkness falls.
> >>
> >> I see no pattern to continue here. How does one extend this sequence?
> >
> > 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, 2/4,
> > 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1 ... will appear in the first
> > column. So 6/1 has to occupy line 21. Therefore 21/1 sitting there will be
> > exchanged by 6/1 and extend the sequence in (1, 2): 1/2, 2/1, 3/1, 3/1, 6/1,
> > 21/1.
> Why does 3/1 appear twice?

An error. In fact it appears thrice, and then 6/1 several times until 21/1 appears.

Regards, WM

On Saturday, 26 February 2022 at 17:41:27 UTC-4, WM wrote:
> FromTheRafters schrieb am Samstag, 26. Februar 2022 um 21:27:21 UTC+1:
> > WM laid this down on his screen :
> > > FromTheRafters schrieb am Samstag, 26. Februar 2022 um 15:55:22 UTC+1:
> > >> It happens that WM formulated :
> > >>> You can extend 1/2, 2/1, 3/1, 3/1, 6/1, ... as far as
> > >>> you like. Somewhere darkness falls.
> > >>
> > >> I see no pattern to continue here. How does one extend this sequence?
> > >
> > > 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, 2/4,
> > > 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1 ... will appear in the first
> > > column. So 6/1 has to occupy line 21. Therefore 21/1 sitting there will be
> > > exchanged by 6/1 and extend the sequence in (1, 2): 1/2, 2/1, 3/1, 3/1, 6/1,
> > > 21/1.
> > Why does 3/1 appear twice?
> An error. In fact it appears thrice, and then 6/1 several times until 21/1 appears.

Good! So you actually have thought about the sequence. In every position other than the first column, you will get an increasing sequence of integers, so the structure of M(oo) is

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

All fractions are in the first column!

On 2/25/2022 3:49 AM, FromTheRafters wrote:
> Chris M. Thomasson pretended :
>> On 2/24/2022 4:54 AM, WM wrote:
>>> Fritz Feldhase schrieb am Mittwoch, 23. Februar 2022 um 16:18:03 UTC+1:
>>>
>>>> Can you actually PROVE your claim? (I got the impression that
>>>> actually, say, (M_k)_1,2 --> oo. So far I see (M_k)_1,2 = (2/1, 3/1,
>>>> 6/1, ...) and a little consideration shows that (M_k)_1,2 diverges,
>>>> no?)
>>>
>>> Like the sequence 1, 2, 3, ... . Important is only that never an
>>> empty place can occur when two fractions are exchanged.
>>
>> 1, 2, 3, ... can be used to index every fraction.
>
> I disagree. We should be talking about rational numbers, not fractions.

I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
to index any other infinite set?

On 2/15/2022 1:03 AM, FromTheRafters wrote:
> Chris M. Thomasson submitted this idea :
>> On 2/13/2022 3:15 AM, WM wrote:
>>> Chris M. Thomasson schrieb am Samstag, 12. Februar 2022 um 10:15:42
>>> UTC+1:
>>>
>>>> think of Cantor Pairing creating an infinite number of single numbers
>>>> from two numbers, and vise versa.
>>>
>>> All positive fractions m/n are contained in the matrix
>>>
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>>
>>> Cantor claims that all can be indexed by natural numbers, i.e.,
>>> written in a sequence, when he uses the sequence
>> [...]
>>
>> The set of natural numbers is infinite. Therefore, it can be used to
>> index any other infinite set. Period.
>
> Countably infinite.

It sure seems like every infinite set can be counted wrt iteration, on a
per-step basis.

On 2/25/2022 3:24 AM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 24. Februar 2022 um 23:25:34 UTC+1:
>> I am not interested in the sequence of fractions; I am interested in the individual elements of the matrices M(k).
>>
>> 1/1, 1/2, 1/3, 1/4, ... 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, 6/1, 4/1, 1/4, ...
>> 2/1, 2/2, 2/3, 2/4, ... 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, ... 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, ... 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, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
>> 6/1, 6/2, 6/3, 6/4, ... 6/1, 6/2, 6/3, 6/4, ... 6/1, 6/2, 6/3, 6/4, ... 6/1, 6/2, 6/3, 6/4, ... 3/1, 6/2, 6/3, 6/4, ...
>> ... ... ... ...
>>
>> The elements in the first column will, of course, be the positive fractions in Cantor's order, and after step k, m_{k,1} will be constant, since the k'th fraction will have been swapped into place. But look at m_{1,2}(k), for instance. The sequence starts 1/2, 2/1, 3/1, 3/1, 6/1, .... How does it continue, and does it have a limit? If it does have a limit, do you think it is one of the fractions you swapped into the first column? (Which one?)
>
> It has no limit. It vanishes in darkness.

Why does it vanish? You are thinking to finite here. An infinite set
goes on forever.

[...]

On 2/8/2022 2:27 PM, Chris M. Thomasson wrote:
> On 2/8/2022 1:01 AM, FromTheRafters wrote:
>> Chris M. Thomasson expressed precisely :
>>> On 2/7/2022 6:25 AM, WM wrote:
>>>> zelos...@gmail.com schrieb am Montag, 7. Februar 2022 um 14:25:08
>>>> UTC+1:
>>>>
>>>>> your "definable" is meaningless
>>>>
>>>> Definable enumeration of the matrix of fractions would distinguish
>>>> which column was emptied first, which second and so on in the matrix
>>>> of all positive fractions m/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, ...
>>>> ...,
>>>>
>>>> when indexing them. You believe in the result
>>>>
>>>> 1/1, __, __, __, ...
>>>> 1/2, __, __, __, ...
>>>> 2/1, __, __, __, ...
>>>> 1/3, __, __, __, ...
>>>> 2/2, __, __, __, ...
>>>> ...
>>>>
>>>> It should also be possible to model this configuration by exchanging
>>>> pairs of fractions
>>>>
>>>> 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, ...
>>>> ... ... ... ...
>>>>
>>>> but it is not. The first column contains as many places as before
>>>> but every exchange leaves all positions of the matrix occupied.
>>>> There is no definable term of the sequence of configurations which
>>>> shows an empty place for the first time. If clearing happens, then
>>>> not be definable terms.
>>>>
>>>> What causes this failure? It is impossible to empty the matrix. This
>>>> holds also in the first case, but it is not as obvious there as in
>>>> the second case. If the matrix really became empty by well-ordered,
>>>> indexed terms, then one of its columns would be the first to get
>>>> empty. Of course this is not possible, but this necessity is often
>>>> denied by switching to the potentially infinite which has no last
>>>> elements. But by exchanges of matrix-elements it becomes obvious
>>>> that not even any single position can be cleared in a definable way,
>>>> let alone the whole matrix. That means the matrix cannot be cleared
>>>> in a definable way.
>>>
>>> Think of:
>>>
>>>
>>> [0...n]
>>> ________
>>>
>>> [1...n]
>>>
>>>
>>> All of the rationals? ;^)
>
> The iteration would be:
>
> [0] = 0/1, 0/2, 0/...n
> [1] = 1/1, 1/2, 1/...n
> [2] = 2/1, 2/2, 2/...n
> [3] = 3/1, 3/2, 3/...n
>
> [...]
>
>
>> Numerators can be negative.
>
> Oops! Negative numbers were not on my mind. ;^o

To get at the negative numbers, take the mirror image.

[0] = -0/1, -0/2, -0/...n
[1] = -1/1, -1/2, -1/...n
[2] = -2/1, -2/2, -2/...n
[3] = -3/1, -3/2, -3/...n
[...]

The other side, so to speak... ;^)

On 2/26/2022 9:47 PM, Chris M. Thomasson wrote:
> On 2/8/2022 2:27 PM, Chris M. Thomasson wrote:
>> On 2/8/2022 1:01 AM, FromTheRafters wrote:
>>> Chris M. Thomasson expressed precisely :
>>>> On 2/7/2022 6:25 AM, WM wrote:
>>>>> zelos...@gmail.com schrieb am Montag, 7. Februar 2022 um 14:25:08 UTC+1:
>>>>>
>>>>>> your "definable" is meaningless
>>>>>
>>>>> Definable enumeration of the matrix of fractions would distinguish which column was emptied first, which second and so on in the matrix of all
>>>>> positive fractions m/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, ...
>>>>> ...,
>>>>>
>>>>> when indexing them. You believe in the result
>>>>>
>>>>> 1/1, __, __, __, ...
>>>>> 1/2, __, __, __, ...
>>>>> 2/1, __, __, __, ...
>>>>> 1/3, __, __, __, ...
>>>>> 2/2, __, __, __, ...
>>>>> ...
>>>>>
>>>>> It should also be possible to model this configuration by exchanging pairs of fractions
>>>>>
>>>>> 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, ...
>>>>> ... ... ... ...
>>>>>
>>>>> but it is not. The first column contains as many places as before but every exchange leaves all positions of the matrix occupied. There is no
>>>>> definable term of the sequence of configurations which shows an empty place for the first time. If clearing happens, then not be definable terms.
>>>>>
>>>>> What causes this failure? It is impossible to empty the matrix. This holds also in the first case, but it is not as obvious there as in the second
>>>>> case. If the matrix really became empty by well-ordered, indexed terms, then one of its columns would be the first to get empty. Of course this is
>>>>> not possible, but this necessity is often denied by switching to the potentially infinite which has no last elements. But by exchanges of
>>>>> matrix-elements it becomes obvious that not even any single position can be cleared in a definable way, let alone the whole matrix. That means the
>>>>> matrix cannot be cleared in a definable way.
>>>>
>>>> Think of:
>>>>
>>>>
>>>> [0...n]
>>>> ________
>>>>
>>>> [1...n]
>>>>
>>>>
>>>> All of the rationals? ;^)
>>
>> The iteration would be:
>>
>> [0] = 0/1, 0/2, 0/...n
>> [1] = 1/1, 1/2, 1/...n
>> [2] = 2/1, 2/2, 2/...n
>> [3] = 3/1, 3/2, 3/...n
>>
>> [...]
>>
>>
>>> Numerators can be negative.
>>
>> Oops! Negative numbers were not on my mind. ;^o
>
> To get at the negative numbers, take the mirror image.
>
> [0] = -0/1, -0/2, -0/...n
> [1] = -1/1, -1/2, -1/...n
> [2] = -2/1, -2/2, -2/...n
> [3] = -3/1, -3/2, -3/...n
> [...]
>
> The other side, so to speak... ;^)

that's only a quadrant, lets do the whole 2D space, all 4 quads

map that from n,m to x, y (note each rational goes to only coordinates of natural numbers)

origin x axis (y=0)

....-0/3,-0/2,-0/1,0/0,+0/1, 0/2, 0/3 ...

Y axis (x=0)

....+3/0,2/0,1/0,0/0,-1/0,-2/0,-3/0...

so the y axis blows up (no problem), Pesky 0/0 is in the middle, and each fraction shows up in 2 different quadrants

where is 3/5 ?

well at (3,5) AND (-3,-5) (?)

On 2/26/2022 7:47 PM, Chris M. Thomasson wrote:
> On 2/8/2022 2:27 PM, Chris M. Thomasson wrote:
>> On 2/8/2022 1:01 AM, FromTheRafters wrote:
>>> Chris M. Thomasson expressed precisely :
>>>> On 2/7/2022 6:25 AM, WM wrote:
>>>>> zelos...@gmail.com schrieb am Montag, 7. Februar 2022 um 14:25:08
>>>>> UTC+1:
>>>>>
>>>>>> your "definable" is meaningless
>>>>>
>>>>> Definable enumeration of the matrix of fractions would distinguish
>>>>> which column was emptied first, which second and so on in the
>>>>> matrix of all positive fractions m/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, ...
>>>>> ...,
>>>>>
>>>>> when indexing them. You believe in the result
>>>>>
>>>>> 1/1, __, __, __, ...
>>>>> 1/2, __, __, __, ...
>>>>> 2/1, __, __, __, ...
>>>>> 1/3, __, __, __, ...
>>>>> 2/2, __, __, __, ...
>>>>> ...
>>>>>
>>>>> It should also be possible to model this configuration by
>>>>> exchanging pairs of fractions
>>>>>
>>>>> 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, ...
>>>>> ... ... ... ...
>>>>>
>>>>> but it is not. The first column contains as many places as before
>>>>> but every exchange leaves all positions of the matrix occupied.
>>>>> There is no definable term of the sequence of configurations which
>>>>> shows an empty place for the first time. If clearing happens, then
>>>>> not be definable terms.
>>>>>
>>>>> What causes this failure? It is impossible to empty the matrix.
>>>>> This holds also in the first case, but it is not as obvious there
>>>>> as in the second case. If the matrix really became empty by
>>>>> well-ordered, indexed terms, then one of its columns would be the
>>>>> first to get empty. Of course this is not possible, but this
>>>>> necessity is often denied by switching to the potentially infinite
>>>>> which has no last elements. But by exchanges of matrix-elements it
>>>>> becomes obvious that not even any single position can be cleared in
>>>>> a definable way, let alone the whole matrix. That means the matrix
>>>>> cannot be cleared in a definable way.
>>>>
>>>> Think of:
>>>>
>>>>
>>>> [0...n]
>>>> ________
>>>>
>>>> [1...n]
>>>>
>>>>
>>>> All of the rationals? ;^)
>>
>> The iteration would be:
>>
>> [0] = 0/1, 0/2, 0/...n
>> [1] = 1/1, 1/2, 1/...n
>> [2] = 2/1, 2/2, 2/...n
>> [3] = 3/1, 3/2, 3/...n
>>
>> [...]
>>
>>
>>> Numerators can be negative.
>>
>> Oops! Negative numbers were not on my mind. ;^o
>
> To get at the negative numbers, take the mirror image.
>
> [0] = -0/1, -0/2, -0/...n
> [1] = -1/1, -1/2, -1/...n
> [2] = -2/1, -2/2, -2/...n
> [3] = -3/1, -3/2, -3/...n
> [...]
>
> The other side, so to speak... ;^)

Negative result from -2/1 is : -2
Negative result from 2/-1 is : -2

On 2/26/2022 10:09 PM, Chris M. Thomasson wrote:
> On 2/26/2022 7:47 PM, Chris M. Thomasson wrote:
>> On 2/8/2022 2:27 PM, Chris M. Thomasson wrote:
>>> On 2/8/2022 1:01 AM, FromTheRafters wrote:
>>>> Chris M. Thomasson expressed precisely :
>>>>> On 2/7/2022 6:25 AM, WM wrote:
>>>>>> zelos...@gmail.com schrieb am Montag, 7. Februar 2022 um 14:25:08
>>>>>> UTC+1:
>>>>>>
>>>>>>> your "definable" is meaningless
>>>>>>
>>>>>> Definable enumeration of the matrix of fractions would distinguish
>>>>>> which column was emptied first, which second and so on in the
>>>>>> matrix of all positive fractions m/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, ...
>>>>>> ...,
>>>>>>
>>>>>> when indexing them. You believe in the result
>>>>>>
>>>>>> 1/1, __, __, __, ...
>>>>>> 1/2, __, __, __, ...
>>>>>> 2/1, __, __, __, ...
>>>>>> 1/3, __, __, __, ...
>>>>>> 2/2, __, __, __, ...
>>>>>> ...
>>>>>>
>>>>>> It should also be possible to model this configuration by
>>>>>> exchanging pairs of fractions
>>>>>>
>>>>>> 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, ...
>>>>>> ... ... ... ...
>>>>>>
>>>>>> but it is not. The first column contains as many places as before
>>>>>> but every exchange leaves all positions of the matrix occupied.
>>>>>> There is no definable term of the sequence of configurations which
>>>>>> shows an empty place for the first time. If clearing happens, then
>>>>>> not be definable terms.
>>>>>>
>>>>>> What causes this failure? It is impossible to empty the matrix.
>>>>>> This holds also in the first case, but it is not as obvious there
>>>>>> as in the second case. If the matrix really became empty by
>>>>>> well-ordered, indexed terms, then one of its columns would be the
>>>>>> first to get empty. Of course this is not possible, but this
>>>>>> necessity is often denied by switching to the potentially infinite
>>>>>> which has no last elements. But by exchanges of matrix-elements it
>>>>>> becomes obvious that not even any single position can be cleared
>>>>>> in a definable way, let alone the whole matrix. That means the
>>>>>> matrix cannot be cleared in a definable way.
>>>>>
>>>>> Think of:
>>>>>
>>>>>
>>>>> [0...n]
>>>>> ________
>>>>>
>>>>> [1...n]
>>>>>
>>>>>
>>>>> All of the rationals? ;^)
>>>
>>> The iteration would be:
>>>
>>> [0] = 0/1, 0/2, 0/...n
>>> [1] = 1/1, 1/2, 1/...n
>>> [2] = 2/1, 2/2, 2/...n
>>> [3] = 3/1, 3/2, 3/...n
>>>
>>> [...]
>>>
>>>
>>>> Numerators can be negative.
>>>
>>> Oops! Negative numbers were not on my mind. ;^o
>>
>> To get at the negative numbers, take the mirror image.
>>
>> [0] = -0/1, -0/2, -0/...n
>> [1] = -1/1, -1/2, -1/...n
>> [2] = -2/1, -2/2, -2/...n
>> [3] = -3/1, -3/2, -3/...n
>> [...]
>>
>> The other side, so to speak... ;^)
>
> Negative result from -2/1 is : -2
> Negative result from 2/-1 is : -2
>

Negative negative -2/-1 is : 2
Positive Positive 2/1 is : 2

Chris M. Thomasson submitted this idea :
> On 2/25/2022 3:49 AM, FromTheRafters wrote:
>> Chris M. Thomasson pretended :
>>> On 2/24/2022 4:54 AM, WM wrote:
>>>> Fritz Feldhase schrieb am Mittwoch, 23. Februar 2022 um 16:18:03 UTC+1:
>>>>
>>>>> Can you actually PROVE your claim? (I got the impression that actually,
>>>>> say, (M_k)_1,2 --> oo. So far I see (M_k)_1,2 = (2/1, 3/1, 6/1, ...) and
>>>>> a little consideration shows that (M_k)_1,2 diverges, no?)
>>>>
>>>> Like the sequence 1, 2, 3, ... . Important is only that never an empty
>>>> place can occur when two fractions are exchanged.
>>>
>>> 1, 2, 3, ... can be used to index every fraction.
>>
>> I disagree. We should be talking about rational numbers, not fractions.
>
> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used to
> index any other infinite set?

It is only countably so. I retract my statement about fractions though.
It is just that I was taught long ago that fractions are part of a
whole, not just simply a whole number ratio as is evidently the case
now. In fact 'whole numbers' weren't used at all after about the third
grade. The naturals are 'positive integers' and the naturals with zero
are non-negative integers.

If you take a circle's diameter as the unit, and wrap that unit around
the circumference you get three 'whole units' as an 'integer part' and
a 'fractional part' of that whole which can be expressed in decimal
radix as <integer part>.<fractional part> which fractional part is not
even necessarily in Q.

The Continued Fractional Expansion of the value can sometimes tell us
more about it being irrational by revealing an unending pattern such as
with the noble numbers and other clear unending patterns. The rational
fractions will terminate.

So, what shall we call the 'parts of a whole' which are not terminating
in their CFEs if we cannot call them fractions?

Chris M. Thomasson laid this down on his screen :
> On 2/15/2022 1:03 AM, FromTheRafters wrote:
>> Chris M. Thomasson submitted this idea :
>>> On 2/13/2022 3:15 AM, WM wrote:
>>>> Chris M. Thomasson schrieb am Samstag, 12. Februar 2022 um 10:15:42
>>>> UTC+1:
>>>>
>>>>> think of Cantor Pairing creating an infinite number of single numbers
>>>>> from two numbers, and vise versa.
>>>>
>>>> All positive fractions m/n are contained in the matrix
>>>>
>>>> 1/1, 1/2, 1/3, 1/4, ...
>>>> 2/1, 2/2, 2/3, 2/4, ...
>>>> 3/1, 3/2, 3/3, 3/4, ...
>>>> 4/1, 4/2, 4/3, 4/4, ...
>>>> 5/1, 5/2, 5/3, 5/4, ...
>>>> ...
>>>>
>>>> Cantor claims that all can be indexed by natural numbers, i.e., written
>>>> in a sequence, when he uses the sequence
>>> [...]
>>>
>>> The set of natural numbers is infinite. Therefore, it can be used to index
>>> any other infinite set. Period.
>>
>> Countably infinite.
>
> It sure seems like every infinite set can be counted wrt iteration, on a
> per-step basis.

How do you interate the reals? I give you a real number and you go look
for the successor to that number and label it with its natural index?

If I give you the eighth real number (or whichever I say it is) can you
give me the ninth or next real number?

Chris M. Thomasson expressed precisely :
> On 2/8/2022 2:27 PM, Chris M. Thomasson wrote:
>> On 2/8/2022 1:01 AM, FromTheRafters wrote:
>>> Chris M. Thomasson expressed precisely :
>>>> On 2/7/2022 6:25 AM, WM wrote:
>>>>> zelos...@gmail.com schrieb am Montag, 7. Februar 2022 um 14:25:08 UTC+1:
>>>>>
>>>>>> your "definable" is meaningless
>>>>>
>>>>> Definable enumeration of the matrix of fractions would distinguish which
>>>>> column was emptied first, which second and so on in the matrix of all
>>>>> positive fractions m/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, ...
>>>>> ...,
>>>>>
>>>>> when indexing them. You believe in the result
>>>>>
>>>>> 1/1, __, __, __, ...
>>>>> 1/2, __, __, __, ...
>>>>> 2/1, __, __, __, ...
>>>>> 1/3, __, __, __, ...
>>>>> 2/2, __, __, __, ...
>>>>> ...
>>>>>
>>>>> It should also be possible to model this configuration by exchanging
>>>>> pairs of fractions
>>>>>
>>>>> 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, ...
>>>>> ... ... ... ...
>>>>>
>>>>> but it is not. The first column contains as many places as before but
>>>>> every exchange leaves all positions of the matrix occupied. There is no
>>>>> definable term of the sequence of configurations which shows an empty
>>>>> place for the first time. If clearing happens, then not be definable
>>>>> terms.
>>>>>
>>>>> What causes this failure? It is impossible to empty the matrix. This
>>>>> holds also in the first case, but it is not as obvious there as in the
>>>>> second case. If the matrix really became empty by well-ordered, indexed
>>>>> terms, then one of its columns would be the first to get empty. Of
>>>>> course this is not possible, but this necessity is often denied by
>>>>> switching to the potentially infinite which has no last elements. But by
>>>>> exchanges of matrix-elements it becomes obvious that not even any single
>>>>> position can be cleared in a definable way, let alone the whole matrix.
>>>>> That means the matrix cannot be cleared in a definable way.
>>>>
>>>> Think of:
>>>>
>>>>
>>>> [0...n]
>>>> ________
>>>>
>>>> [1...n]
>>>>
>>>>
>>>> All of the rationals? ;^)
>>
>> The iteration would be:
>>
>> [0] = 0/1, 0/2, 0/...n
>> [1] = 1/1, 1/2, 1/...n
>> [2] = 2/1, 2/2, 2/...n
>> [3] = 3/1, 3/2, 3/...n
>>
>> [...]
>>
>>
>>> Numerators can be negative.
>>
>> Oops! Negative numbers were not on my mind. ;^o
>
> To get at the negative numbers, take the mirror image.
>
> [0] = -0/1, -0/2, -0/...n
> [1] = -1/1, -1/2, -1/...n
> [2] = -2/1, -2/2, -2/...n
> [3] = -3/1, -3/2, -3/...n
> [...]
>
> The other side, so to speak... ;^)

Fuggetaboutit -- As Fritz Feldhase pointed out, I missed that WM was
talking about a superset of positive rationals which includes improper
fractions and reducible fractions as separate elements of the set
rather than 'equal' elements chosen from an equivalence class.

horand....@gmail.com schrieb am Samstag, 26. Februar 2022 um 23:13:46 UTC+1:
> On Saturday, 26 February 2022 at 09:02:57 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Freitag, 25. Februar 2022 um 17:25:27 UTC+1:
> >
> > > The sequences {m_{i,j}(k):k \in IN} are sequences of rational numbers with well-defined formation rules. (The definitions are recursive, but they are finite.) Does the sequence {1/2, 2/1, 3/1, 3/1, 6/1, ...} have a limit or not?
> > The sequence does not have a limit like 1, 2, 3, ... does not have a limit.
> Prove it|!

There is no ω in the matrix. But there are only transpositions of fractions possible.

> Your drivel below shows absolute ignorance of sequences and limits. For instance, if all the elements in a particular position of the matrices were to behave like 1/f(k), with f increasing to infinity, then the reciprocals (that is, the sequence of elements in that particular spot) would converge to 0.

Nevertheless 0 would not be possible at any position.

> > Every place of the matrix (except 1/1) is occupied by a fraction that before had occupied another place. If the matrix
> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, 3/2, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> >
> > contains only fractions before shuffling them, then all shuffled fractions are fractions too after all shuffling. Note that ω and ω/1 and ω/2 and ... and 1/ω and 2/ω and ... are not fractions
> Did I say that they were?

You drivel about limits.

> > and therefore are not in the matrix of fractions which are to be enumerated. Not before and not after. Enumerating has nothing, really, really nothing to do with limits.
> Where did I say anything about enumerating?

It is the topic here.

Regards, WM

horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 03:49:17 UTC+1:

> Good! So you actually have thought about the sequence. In every position other than the first column, you will get an increasing sequence of integers, so the structure of M(oo) is
>
> 1/1, oo, oo, oo, oo, oo, oo, ...
> 1/2, oo, oo, oo, oo, oo, oo, ...
> 2/1, oo, oo, oo, oo, oo, oo, ...
> 1/3, oo, oo, oo, oo, oo, oo, ...
> 2/2, oo, oo, oo, oo, oo, oo, ...
> 3/1, oo, oo, oo, oo, oo, oo, ...
> ..., ..., ..., ..., ..., ..., ...
>
> All fractions are in the first column!

Which position is the first to get the "fraction" oo? Which column is the first to be conquered by oo's completely? Note that counting is the topic here. Counting excludes that all positions become oo simultaneously.

Regards, WM

FromTheRafters schrieb am Sonntag, 27. Februar 2022 um 10:30:47 UTC+1:

> Fuggetaboutit -- As Fritz Feldhase pointed out, I missed that WM was
> talking about a superset of positive rationals

That is what Cantor did
k = (m + n - 1)(m + n - 2)/2 + m
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

I only mimic him. But instead of bringing the naturals to the serpent I bring the serpent to the naturals. This shows that never all fractions can be indexed. In appealing and easily understandable form you can read it here. https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability Contradicted. It is only one page, smashing 150 years of set theory.

Regards, WM

Chris M. Thomasson schrieb am Sonntag, 27. Februar 2022 um 04:43:25 UTC+1:
> On 2/25/2022 3:24 AM, WM wrote:

> > It has no limit. It vanishes in darkness.
> Why does it vanish? You are thinking to finite here. An infinite set
> goes on forever.
>
If it goes on forever, then you will never know whether two sets are in bijection or whether one of them is larger than the other. Cantor completes the indexing, so he believes at least. But that is impossible. You can read it here: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability Contradicted. One page.

Regards, WM

fredag 25 februari 2022 kl. 12:34:38 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 25. Februar 2022 um 06:31:03 UTC+1:
> > torsdag 24 februari 2022 kl. 13:59:32 UTC+1 skrev WM:
>
> > > My matrices have no empty slots by mathematical proof: When two positive fractions change their places they do not leave an empty place. Thus your statement is wrong.
> > >
> > The one you want as "final" is not a matrix.
> There is no final matrix visible. You can find every lit fraction in the first column. You will never find a lit fraction residing permanently outside of the first column. But you will never find an empty place in the matrix.

That would be because by definition THERE IS NOTHING EMPTY IN A MATRIX!

>
> You can avoid the problem by gathering the fractions
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> somewhere else, in another place, outside of the matrix.

That works because that is a SEQUENCE, not a fucking matrix.

> Then you get the impression that nothing will remain in the matrix. Apparently.
> If you don't look too close, Cantor seems right. But is he right then? Are the remaining fractions, like electrons in quantum theory, not there if not being observed? That is not what a mathematician should believe.

They are all in the fucking sequence, what are you on about you fucking crank?

>
> Regards, WM

zelos...@gmail.com schrieb am Sonntag, 27. Februar 2022 um 11:23:31 UTC+1:
> fredag 25 februari 2022 kl. 12:34:38 UTC+1 skrev WM:

> > You can avoid the problem by gathering the fractions
> > 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> > somewhere else, in another place, outside of the matrix.
>
> That works because that is a SEQUENCE, not a fucking matrix.
>
> > Then you get the impression that nothing will remain in the matrix. Apparently.
> > If you don't look too close, Cantor seems right. But is he right then? Are the remaining fractions, like electrons in quantum theory, not there if not being observed? That is not what a mathematician should believe.
>
> They are all in the fucking sequence, what are you on about you fucking crank?

All fractions which you can define are in the first column. But every such fraction has ℵo successors which you will never have defined completely. Dark numbers are real if infinity is actual. Here you may find the most comprehensive essay about dark numbers presently available: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Dark Numbers.

Further discussion will be irrelevant. Either you can understand or you you cannot.

Regards, WM

zelos...@gmail.com wrote:

>> There is no final matrix visible. You can find every lit fraction in
>> the first column. You will never find a lit fraction residing
>> permanently outside of the first column. But you will never find an
>> empty place in the matrix.
>
> That would be because by definition THERE IS NOTHING EMPTY IN A MATRIX!

The CIA – 70 years in Ukraine
https://www.youtube.com/embed/jRicZc-cZ0I
https://theduran.com/the-cia-70-years-in-ukraine/

on 2/27/2022, WM supposed :
> horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 03:49:17 UTC+1:
>
>> Good! So you actually have thought about the sequence. In every position
>> other than the first column, you will get an increasing sequence of
>> integers, so the structure of M(oo) is
>>
>> 1/1, oo, oo, oo, oo, oo, oo, ...
>> 1/2, oo, oo, oo, oo, oo, oo, ...
>> 2/1, oo, oo, oo, oo, oo, oo, ...
>> 1/3, oo, oo, oo, oo, oo, oo, ...
>> 2/2, oo, oo, oo, oo, oo, oo, ...
>> 3/1, oo, oo, oo, oo, oo, oo, ...
>> ..., ..., ..., ..., ..., ..., ...
>>
>> All fractions are in the first column!
>
> Which position is the first to get the "fraction" oo? Which column is the
> first to be conquered by oo's completely? Note that counting is the topic
> here.

It is not about counting, it is about being "countable" where countable
uses the mathematical definition not the "tabletalk" definition that
you prefer.

WM submitted this idea :
> FromTheRafters schrieb am Sonntag, 27. Februar 2022 um 10:30:47 UTC+1:
>
>> Fuggetaboutit -- As Fritz Feldhase pointed out, I missed that WM was
>> talking about a superset of positive rationals
>
> That is what Cantor did
> k = (m + n - 1)(m + n - 2)/2 + m
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1,
> 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

Doesn't matter anyway, these are all the same size sets.

> I only mimic him. But instead of bringing the naturals to the serpent I bring
> the serpent to the naturals. This shows that never all fractions can be
> indexed. In appealing and easily understandable form you can read it here.
> https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability
> Contradicted. It is only one page, smashing 150 years of set theory.

Delusional.