On 12/21/23 6:05 AM, WM wrote:
> Le 20/12/2023 à 19:04, immibis a écrit :
>> On 12/20/23 14:17, Richard Damon wrote:
>>> Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
>>> rationals.
>>
>> I am missing the start of this conversation. Aren't integral rationals
>> exactly the integers?
>
> Of course. Therefore it is irrelevant whether we start with the first
> column of
>
> 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, ...
> ..
>
> or first biject the integer fractions with the natural numbers
>
> 1, 1/2, 1/3, 1/4, ...
> 2, 2/2, 2/3, 2/4, ...
> 3, 3/2, 3/3, 3/4, ...
> 4, 4/2, 4/3, 4/4, ...
> 5, 5/2, 5/3, 5/4, ...
> ..

So, what are you trying to do here? This isn't a matrix of any normal
domain.

>
> In no case it is possible to cover the whole matrix by indeXes
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ..
>
> Regards, WM
>
>
>
Your arguement doesn't make sense.

IF you are trying to say that there isn't a mapping for the values in
the first column to the whole set, you are just wrong and the mapping
has been given.

If you are arguing that the shape of the first column doesn't ever cover
the matrix, that is just an irrelevent stupidity, as no one says it should.

Cantor said there was a bijection between the integers and the
rationals, not that all rationals were integers.

Your inability to understand the difference shows your lack of
understanding and unsound reasoning.

We cover 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, ...
....

with values mappes as following:

1 2 4 7 11 16 22
3 5 8 12 17 23
6 9 13 18 24
10 14 19 25
15 20 26
21 26
27

Note, all of these *VALUES* can be found in the first column of the
rationals, but that is irrelevant, as we are showing a mapping of
indexes from the Natural Numbers to rational matrix.

On 12/21/23 5:57 AM, WM wrote:
> Le 21/12/2023 à 10:41, Mikko a écrit :
>> On 2023-12-20 16:28:09 +0000, WM said:
>
>>> The matrix remains, only the covering by X changes.
>>
>> Your first "covering" is just another matrix with the same idicex.
>
> It has the same (m, n), but not all positions have been counted by k
> k = (m + n - 1)(m + n - 2)/2 + m.
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ..
>
>> When it "changes" it does not really change, there just is another
>> "covering", and after successive changes a sequence of such "coverings".
>
> And none covers the whole matrix by indeXes.
>
> Regards, WM
>

So, you just don't understand what the bijection is doing.

You aren't indexing the values of k to the matrix, but the index
locations m,n

You are just doing it backwards, as seems to be your nature,

On 2023-12-21 11:56:10 +0000, WM said:

> Le 21/12/2023 à 12:48, Mikko a écrit :
>> On 2023-12-21 10:57:10 +0000, WM said:
>>
>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>
>>>>> The matrix remains, only the covering by X changes.
>>>>
>>>> Your first "covering" is just another matrix with the same idicex.
>>>
>>> It has the same (m, n), but not all positions have been counted by k
>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> Positions are not covered by. Some of them are covered by X.
>>
>> Each position as m and n so each position has k.
>
> That is an error. Each O remains in the matrix at a not indexed place.

By the definition of "matrix" there are only indexed places.
A structure that contains unindexec places is not a matrix.

Mikko

On 12/20/2023 5:51 AM, WM wrote:
> Le 19/12/2023 à 21:07, Richard Damon a écrit :
>> On 12/19/23 1:54 PM, WM wrote:

>>> What forbids
>>> to index the first column first?
>>> Or what forbids
>>> to use the integer fractions n/1 for indexing?
>>
>> Nothing "forbids" it,
>> but the rule is that
>> if ANY indexing works,
>> then we have established that
>> they are equal sizes.
>
> First, this is a stupid rule.

The rule is the same for finite and infinite sets.

If,
for each b ∈ B
b has its own aᵇ ∈ A
then
A holds at least as many as B
|B| ≤ |A|

What _isn't_ a rule, and
which would be a rule for only sets like
flocks of sheep or bags of pebbles,
is that
if
1.to.1 f: B ↣ A
1.to.1 g: B ↣ A
then
f(B) = A and g(B) = A
or
f(B) ≠ A and g(B) ≠ A

What you (WM) are probably calling stupid
isn't so much a _rule_ rule
as it is the _observation_ that
the flock-of-sheep rule isn't correct
for all sets.

For example.

The flock-of-sheep rule is correct for
each ⟨0,…,i⟩, each ⟨1,…,j+1⟩

Define 1.to.1 f(k) = k-1

if i=j
f⟨1,…,j+1⟩ = ⟨0,…,i⟩
and
for each 1.to.1 g: ⟨1,…,j+1⟩ ↣ ⟨0,…,i⟩
g⟨1,…,j+1⟩ = ⟨0,…,i⟩

if i≠j
f⟨1,…,j+1⟩ ≠ ⟨0,…,i⟩
and
for each 1.to.1 g: ⟨1,…,j+1⟩ ↣ ⟨0,…,i⟩
g⟨1,…,j+1⟩ ≠ ⟨0,…,i⟩

However,
we observe that
the flock-of-sheep rule does not work
for ᴸᵁᴮ⟨⟨0,…,i⟩⟩ ᴸᵁᴮ⟨⟨1,…,j+1⟩⟩
least upper bounds
of all ⟨0,…,i⟩ and of all ⟨1,…,j+1⟩

for 1.to.1 f(k) = k-1
fᴸᵁᴮ⟨⟨1,…,j+1⟩⟩ = ᴸᵁᴮ⟨⟨0,…,i⟩⟩

for 1.to.1 g(k) = k
gᴸᵁᴮ⟨⟨1,…,j+1⟩⟩ = ᴸᵁᴮ⟨⟨1,…,j+1⟩⟩
ᴸᵁᴮ⟨⟨1,…,j+1⟩⟩ ≠ ᴸᵁᴮ⟨⟨0,…,i⟩⟩

ᴸᵁᴮ⟨⟨1,…,j+1⟩⟩ is
a proper subset of ᴸᵁᴮ⟨⟨0,…,i⟩⟩
and
ᴸᵁᴮ⟨⟨1,…,j+1⟩⟩ is
the same size as ᴸᵁᴮ⟨⟨0,…,i⟩⟩

We observe that
not all sets are flocks of sheep.

Le 21/12/2023 à 16:16, Richard Damon a écrit :
> On 12/21/23 6:05 AM, WM wrote:
>> Le 20/12/2023 à 19:04, immibis a écrit :
>>> On 12/20/23 14:17, Richard Damon wrote:
>>>> Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
>>>> rationals.
>>>
>>> I am missing the start of this conversation. Aren't integral rationals
>>> exactly the integers?
>>
>> Of course. Therefore it is irrelevant whether we start with the first
>> column of
>>
>> 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, ...
>> ..
>>
>> or first biject the integer fractions with the natural numbers
>>
>> 1, 1/2, 1/3, 1/4, ...
>> 2, 2/2, 2/3, 2/4, ...
>> 3, 3/2, 3/3, 3/4, ...
>> 4, 4/2, 4/3, 4/4, ...
>> 5, 5/2, 5/3, 5/4, ...
>> ..
>>
>> In no case it is possible to cover the whole matrix by indeXes
>>
>> XOOO...
>> XOOO...
>> XOOO...
>> XOOO...
>> ..
>
> IF you are trying to say that there isn't a mapping for the values in
> the first column to the whole set, you are just wrong and the mapping
> has been given.

It appears so, but the O indicate not indeXed fractions. This has to be
taken into account.
>
> If you are arguing that the shape of the first column doesn't ever cover
> the matrix, that is just an irrelevent stupidity, as no one says it should.

That is Cantor's claim. All indexes have to be appßlied. They can be
taken from the first column.

>
> We cover 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, ...
> ...
>
> with values mappes as following:
>
> 1 2 4 7 11 16 22
> 3 5 8 12 17 23
> 6 9 13 18 24
> 10 14 19 25
> 15 20 26
> 21 26
> 27
>
>
> Note, all of these *VALUES* can be found in the first column of the
> rationals, but that is irrelevant, as we are showing a mapping of
> indexes from the Natural Numbers to rational matrix.

You are mistaken. All your indeXes up to every indeX applied in your
mapping belong to a finite set {1, 2, 3, ..., X} which has ℵo
successors. My proof shows that the O, indicating not indexed fractions,
will remain forever within the matrix, being even the majority.

Regards, WM

Le 21/12/2023 à 16:18, Richard Damon a écrit :
> On 12/21/23 5:57 AM, WM wrote:
>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>> On 2023-12-20 16:28:09 +0000, WM said:
>>
>>>> The matrix remains, only the covering by X changes.
>>>
>>> Your first "covering" is just another matrix with the same idicex.
>>
>> It has the same (m, n), but not all positions have been counted by k
>> k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> XOOO...
>> XOOO...
>> XOOO...
>> XOOO...
>> ..
>>
>>> When it "changes" it does not really change, there just is another
>>> "covering", and after successive changes a sequence of such "coverings".
>>
>> And none covers the whole matrix by indeXes.
>>
> So, you just don't understand what the bijection is doing.

A bijection should index every element of the matrix, but it does not,
because the O remain. Therefore it is not a bijection.

Regards, WM

Le 21/12/2023 à 17:45, Mikko a écrit :
> On 2023-12-21 11:56:10 +0000, WM said:
>
>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>
>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>
>>>>>> The matrix remains, only the covering by X changes.
>>>>>
>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>
>>>> It has the same (m, n), but not all positions have been counted by k
>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>
>>> Positions are not covered by. Some of them are covered by X.
>>>
>>> Each position as m and n so each position has k.
>>
>> That is an error. Each O remains in the matrix at a not indexed place.
>
> By the definition of "matrix" there are only indexed places.

But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.

Regards, WM

Le 21/12/2023 à 19:18, Jim Burns a écrit :
> On 12/20/2023 5:51 AM, WM wrote:
>> Le 19/12/2023 à 21:07, Richard Damon a écrit :
>>> On 12/19/23 1:54 PM, WM wrote:
>
>>>> What forbids
>>>> to index the first column first?
>>>> Or what forbids
>>>> to use the integer fractions n/1 for indexing?
>>>
>>> Nothing "forbids" it,
>>> but the rule is that
>>> if ANY indexing works,
>>> then we have established that
>>> they are equal sizes.
>>
>> First, this is a stupid rule.
>
> The rule is the same for finite and infinite sets.

No. For finite sets of equal numerosity any injection is a surjection and
any surjection is an injection.
>
> What you (WM) are probably calling stupid
> isn't so much a _rule_ rule
> as it is the _observation_ that
> the flock-of-sheep rule isn't correct
> for all sets.

That is not an observation but an assumption of people who deny or are
unable to understand actual infinity. Bob will never leave the matrix,
Therefore your arguments are invalid and worthless.

Regards, WM

On 12/22/23 3:04 AM, WM wrote:
> Le 21/12/2023 à 16:16, Richard Damon a écrit :
>> On 12/21/23 6:05 AM, WM wrote:
>>> Le 20/12/2023 à 19:04, immibis a écrit :
>>>> On 12/20/23 14:17, Richard Damon wrote:
>>>>> Cantor maps the NATURAL NUMBERS (not the intergral rationals) to
>>>>> the rationals.
>>>>
>>>> I am missing the start of this conversation. Aren't integral
>>>> rationals exactly the integers?
>>>
>>> Of course. Therefore it is irrelevant whether we start with the first
>>> column of
>>>
>>> 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, ...
>>> ..
>>>
>>> or first biject the integer fractions with the natural numbers
>>>
>>> 1, 1/2, 1/3, 1/4, ...
>>> 2, 2/2, 2/3, 2/4, ...
>>> 3, 3/2, 3/3, 3/4, ...
>>> 4, 4/2, 4/3, 4/4, ...
>>> 5, 5/2, 5/3, 5/4, ...
>>> ..
>>>
>>> In no case it is possible to cover the whole matrix by indeXes
>>>
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ..
>>
>> IF you are trying to say that there isn't a mapping for the values in
>> the first column to the whole set, you are just wrong and the mapping
>> has been given.
>
> It appears so, but the O indicate not indeXed fractions. This has to be
> taken into account.

No, you are confusing the index with the indexed.

the second x down (2) indexes the second element of the first row (1/2)
which you have marked with an O.

>>
>> If you are arguing that the shape of the first column doesn't ever
>> cover the matrix, that is just an irrelevent stupidity, as no one says
>> it should.
>
> That is Cantor's claim. All indexes have to be appßlied. They can be
> taken from the first column.

Right, and by the transformation of the index, they select all the
element of the matrix.

>
>>
>> We cover 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, ...
>> ...
>>
>> with values mappes as following:
>>
>> 1    2     4    7   11  16  22
>> 3    5     8    12  17  23
>> 6    9     13   18  24
>> 10   14    19   25
>> 15   20    26
>> 21   26
>> 27
>>
>>
>> Note, all of these *VALUES* can be found in the first column of the
>> rationals, but that is irrelevant, as we are showing a mapping of
>> indexes from the Natural Numbers to rational matrix.
>
> You are mistaken. All your indeXes up to every indeX applied in your
> mapping belong to a finite set {1, 2, 3, ..., X} which has ℵo
> successors. My proof shows that the O, indicating not indexed fractions,
> will remain forever within the matrix, being even the majority.

No, you are showing you can create a non-covering mapping, but the rule
says that if at least ONE covering mapping exists, we meet the requirement.

You can't prove non-existence by showing an example of something that isn't.

In other words, you just don't understand what Cantor is talking about,
because your understanding of the infinite is incomplete.

>
>
> Regards, WM

On 12/22/23 3:07 AM, WM wrote:
> Le 21/12/2023 à 16:18, Richard Damon a écrit :
>> On 12/21/23 5:57 AM, WM wrote:
>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>
>>>>> The matrix remains, only the covering by X changes.
>>>>
>>>> Your first "covering" is just another matrix with the same idicex.
>>>
>>> It has the same (m, n), but not all positions have been counted by k
>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ..
>>>
>>>> When it "changes" it does not really change, there just is another
>>>> "covering", and after successive changes a sequence of such
>>>> "coverings".
>>>
>>> And none covers the whole matrix by indeXes.
>>>
>> So, you just don't understand what the bijection is doing.
>
> A bijection should index every element of the matrix, but it does not,
> because the O remain. Therefore it is not a bijection.
>
> Regards, WM

So, you are showning that the bijecton for k to k/1 doesn't cover all of
m/n, but that doesn't show that the bijecton for k to m/n where
k = (m + n - 1)(m + n - 2)/2 + m doesn't work.

Since the requirement is only that there eists *A* bijection that works,
you haven't proven your point.

You are trying to prove non-existence by example, which just doesn't work.

On 2023-12-22 08:09:14 +0000, WM said:

> Le 21/12/2023 à 17:45, Mikko a écrit :
>> On 2023-12-21 11:56:10 +0000, WM said:
>>
>>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>>
>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>
>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>
>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>
>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>
>>>> Positions are not covered by. Some of them are covered by X.
>>>>
>>>> Each position as m and n so each position has k.
>>>
>>> That is an error. Each O remains in the matrix at a not indexed place.
>>
>> By the definition of "matrix" there are only indexed places.
>
> But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.

It is indexed by two natural numbers because the matrices meintioned
above and earlier in this discussion are matrices that are indexed by
two natural numbers. You have called these two numbers m and n.
For each natural number k there is only one pair of natural numbers
m and n so that k = (m + n - 1)(m + n - 2)/2 + m. Therefore k is
sufficient to identify a position in any of the matrices. But it does
not identify which matrix one is talking about so that must be told
separately.

Mikko

On 12/22/23 09:09, WM wrote:
> Le 21/12/2023 à 17:45, Mikko a écrit :
>> On 2023-12-21 11:56:10 +0000, WM said:
>>
>>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>>
>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>
>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>
>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>
>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>
>>>> Positions are not covered by. Some of them are covered by X.
>>>>
>>>> Each position as m and n so each position has k.
>>>
>>> That is an error. Each O remains in the matrix at a not indexed place.
>>
>> By the definition of "matrix" there are only indexed places.
>
> But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
>
> Regards, WM
>
All of them are indexed by Cantor

On 12/22/23 09:13, WM wrote:
> Le 21/12/2023 à 19:18, Jim Burns a écrit :
>> On 12/20/2023 5:51 AM, WM wrote:
>>> Le 19/12/2023 à 21:07, Richard Damon a écrit :
>>>> On 12/19/23 1:54 PM, WM wrote:
>>
>>>>> What forbids
>>>>> to index the first column first?
>>>>> Or what forbids
>>>>> to use the integer fractions n/1 for indexing?
>>>>
>>>> Nothing "forbids" it,
>>>> but the rule is that
>>>> if ANY indexing works,
>>>> then we have established that
>>>> they are equal sizes.
>>>
>>> First, this is a stupid rule.
>>
>> The rule is the same for finite and infinite sets.
>
> No. For finite sets of equal numerosity any injection is a surjection
> and any surjection is an injection.

Wrong rule. If any injection between two sets exists, they have equal sizes.

Le 22/12/2023 à 16:15, immibis a écrit :
> On 12/22/23 09:13, WM wrote:
>> Le 21/12/2023 à 19:18, Jim Burns a écrit :
>>> On 12/20/2023 5:51 AM, WM wrote:
>>>> Le 19/12/2023 à 21:07, Richard Damon a écrit :
>>>>> On 12/19/23 1:54 PM, WM wrote:
>>>
>>>>>> What forbids
>>>>>> to index the first column first?
>>>>>> Or what forbids
>>>>>> to use the integer fractions n/1 for indexing?
>>>>>
>>>>> Nothing "forbids" it,
>>>>> but the rule is that
>>>>> if ANY indexing works,
>>>>> then we have established that
>>>>> they are equal sizes.
>>>>
>>>> First, this is a stupid rule.
>>>
>>> The rule is the same for finite and infinite sets.
>>
>> No. For finite sets of equal numerosity any injection is a surjection
>> and any surjection is an injection.
>
> Wrong rule. If any injection between two sets exists, they have equal sizes.

Yes, that is awrong rule. See the injection of natnumbers into the reals.

Regards, WM

Le 22/12/2023 à 16:15, immibis a écrit :
> On 12/22/23 09:09, WM wrote:
>> Le 21/12/2023 à 17:45, Mikko a écrit :
>>> On 2023-12-21 11:56:10 +0000, WM said:
>>>
>>>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>>>
>>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>>
>>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>>
>>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>>
>>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>>
>>>>> Positions are not covered by. Some of them are covered by X.
>>>>>
>>>>> Each position as m and n so each position has k.
>>>>
>>>> That is an error. Each O remains in the matrix at a not indexed place.
>>>
>>> By the definition of "matrix" there are only indexed places.
>>
>> But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
>>
> All of them are indexed by Cantor

The remaining not indexed fractions, covered by O, prove the contrary.

Regards, WM

Le 22/12/2023 à 15:19, Mikko a écrit :
> On 2023-12-22 08:09:14 +0000, WM said:
>
>> Le 21/12/2023 à 17:45, Mikko a écrit :
>>> On 2023-12-21 11:56:10 +0000, WM said:
>>>
>>>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>>>
>>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>>
>>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>>
>>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>>
>>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>>
>>>>> Positions are not covered by. Some of them are covered by X.
>>>>>
>>>>> Each position as m and n so each position has k.
>>>>
>>>> That is an error. Each O remains in the matrix at a not indexed place.
>>>
>>> By the definition of "matrix" there are only indexed places.
>>
>> But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
>
> It is indexed by two natural numbers because the matrices meintioned
> above and earlier in this discussion are matrices that are indexed by
> two natural numbers.

As the O, not indexed by Cantor prove, most of the matrix' (m, n) are
dark.
Only the upper left corner contains visible m and n.

> You have called these two numbers m and n.
> For each natural number k there is only one pair of natural numbers
> m and n so that k = (m + n - 1)(m + n - 2)/2 + m. Therefore k is
> sufficient to identify a position in any of the matrices.

The remaining O prove the contrary.

Regards, WM

Le 22/12/2023 à 15:05, Richard Damon a écrit :

> No, you are showing you can create a non-covering mapping,

I show that Cantor's mapping is not covering the matrix.

> but the rule
> says that if at least ONE covering mapping exists, we meet the requirement.

Who made that rule?
>
Regards, WM

Le 22/12/2023 à 15:08, Richard Damon a écrit :
> On 12/22/23 3:07 AM, WM wrote:
>> Le 21/12/2023 à 16:18, Richard Damon a écrit :
>>> On 12/21/23 5:57 AM, WM wrote:
>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>
>>>>>> The matrix remains, only the covering by X changes.
>>>>>
>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>
>>>> It has the same (m, n), but not all positions have been counted by k
>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>
>>>> XOOO...
>>>> XOOO...
>>>> XOOO...
>>>> XOOO...
>>>> ..
>>>>
>>>>> When it "changes" it does not really change, there just is another
>>>>> "covering", and after successive changes a sequence of such
>>>>> "coverings".
>>>>
>>>> And none covers the whole matrix by indeXes.
>>>>
>>> So, you just don't understand what the bijection is doing.
>>
>> A bijection should index every element of the matrix, but it does not,
>> because the O remain. Therefore it is not a bijection.
>>
> So, you are showning that the bijecton for k to k/1 doesn't cover all of
> m/n, but that doesn't show that the bijecton for k to m/n where
> k = (m + n - 1)(m + n - 2)/2 + m doesn't work.

If the first bijection k to k/1 is a bijection, then I apply in the second
stepCantor's procedure. If not, then there are no infinite bijections at
all.

Regards, WM

On 12/22/23 11:23 AM, WM wrote:
> Le 22/12/2023 à 15:05, Richard Damon a écrit :
>
>> No, you are showing you can create a non-covering mapping,
>
> I show that Cantor's mapping is not covering the matrix.

No, you didn't, because you didn't do Cantor's mapping.

>
>> but the rule says that if at least ONE covering mapping exists, we
>> meet the requirement.
>
> Who made that rule?

It wasn't "Made" but "Discovered". It is a natural consequence of the
need for a consistent set of logical rules to describe infinite set.

If you are willing for your logic to be in a system that creates
inconsistent results, and thus worthless for establishing the actual
truth of statement, go ahead and ignore it.

Of course, the fact that you even ask the question says you do not
understand what Cantor does, and thus your arguments are based on your
ignorance.

>>
> Regards, WM

On 12/22/23 11:15 AM, WM wrote:
> Le 22/12/2023 à 16:15, immibis a écrit :
>> On 12/22/23 09:13, WM wrote:
>>> Le 21/12/2023 à 19:18, Jim Burns a écrit :
>>>> On 12/20/2023 5:51 AM, WM wrote:
>>>>> Le 19/12/2023 à 21:07, Richard Damon a écrit :
>>>>>> On 12/19/23 1:54 PM, WM wrote:
>>>>
>>>>>>> What forbids
>>>>>>> to index the first column first?
>>>>>>> Or what forbids
>>>>>>> to use the integer fractions n/1 for indexing?
>>>>>>
>>>>>> Nothing "forbids" it,
>>>>>> but the rule is that
>>>>>> if ANY indexing works,
>>>>>> then we have established that
>>>>>> they are equal sizes.
>>>>>
>>>>> First, this is a stupid rule.
>>>>
>>>> The rule is the same for finite and infinite sets.
>>>
>>> No. For finite sets of equal numerosity any injection is a surjection
>>> and any surjection is an injection.
>>
>> Wrong rule. If any injection between two sets exists, they have equal
>> sizes.
>
> Yes, that is awrong rule. See the injection of natnumbers into the reals.
>
> Regards, WM
>
>

But the actual rule is either have a bijection, or an injection both ways.

You can not inject the Reals to the Natural Numbers as you run out of
Natural Numbers in all attempts to do so.

On 12/22/23 11:19 AM, WM wrote:
> Le 22/12/2023 à 15:19, Mikko a écrit :
>> On 2023-12-22 08:09:14 +0000, WM said:
>>
>>> Le 21/12/2023 à 17:45, Mikko a écrit :
>>>> On 2023-12-21 11:56:10 +0000, WM said:
>>>>
>>>>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>>>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>>>>
>>>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>>>
>>>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>>>
>>>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>>>
>>>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>>>
>>>>>> Positions are not covered by. Some of them are covered by X.
>>>>>>
>>>>>> Each position as m and n so each position has k.
>>>>>
>>>>> That is an error. Each O remains in the matrix at a not indexed place.
>>>>
>>>> By the definition of "matrix" there are only indexed places.
>>>
>>> But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> It is indexed by two natural numbers because the matrices meintioned
>> above and earlier in this discussion are matrices that are indexed by
>> two natural numbers.
>
> As the O, not indexed by Cantor prove, most of the matrix' (m, n) are dark.
> Only the upper left corner contains visible m and n.

Except the matrix with the O's isn't Cantor injection in the rationals,
so doesn't prove anything but that you like to play with Strawmen, or
you just don't understand Cantor.

>
>> You have called these two numbers m and n.
>> For each natural number k there is only one pair of natural numbers
>> m and n so that k = (m + n - 1)(m + n - 2)/2 + m. Therefore k is
>> sufficient to identify a position in any of the matrices.
>
> The remaining O prove the contrary.
>
> Regards, WM
>
>

On 12/22/23 11:25 AM, WM wrote:
> Le 22/12/2023 à 15:08, Richard Damon a écrit :
>> On 12/22/23 3:07 AM, WM wrote:
>>> Le 21/12/2023 à 16:18, Richard Damon a écrit :
>>>> On 12/21/23 5:57 AM, WM wrote:
>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>
>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>
>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>
>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>>
>>>>> XOOO...
>>>>> XOOO...
>>>>> XOOO...
>>>>> XOOO...
>>>>> ..
>>>>>
>>>>>> When it "changes" it does not really change, there just is another
>>>>>> "covering", and after successive changes a sequence of such
>>>>>> "coverings".
>>>>>
>>>>> And none covers the whole matrix by indeXes.
>>>>>
>>>> So, you just don't understand what the bijection is doing.
>>>
>>> A bijection should index every element of the matrix, but it does
>>> not, because the O remain. Therefore it is not a bijection.
>>>
>> So, you are showning that the bijecton for k to k/1 doesn't cover all
>> of m/n, but that doesn't show that the bijecton for k to m/n where
>> k = (m + n - 1)(m + n - 2)/2 + m doesn't work.
>
> If the first bijection k to k/1 is a bijection, then I apply in the
> second stepCantor's procedure. If not, then there are no infinite
> bijections at all.
>
> Regards, WM

Excpet that it CAN'T be a "bijection" to the set of m/n as it isn't an
injection when reversed.

Yes, it can be a bijection of natural to integral rationals, but that
isn't the set you are talking about, so irrelevent.

I don't think you understand anything that you are talking about.

On 12/22/2023 3:13 AM, WM wrote:
> Le 21/12/2023 à 19:18, Jim Burns a écrit :
>> On 12/20/2023 5:51 AM, WM wrote:

>>> First, this is a stupid rule.

>> What you (WM) are probably calling stupid
>> isn't so much a _rule_ rule
>> as it is the _observation_ that
>> the flock-of-sheep rule isn't correct
>> for all sets.
>
> That is not an observation

𝕍 is the least.upper.bound of FISONs ⟨1,…,n⟩
∀A: ∀⟨1,…,n⟩ ⊆ A ⟹ ∀⟨1,…,n⟩ ⊆ 𝕍 ⊆ A
Standardly, 𝕍 = ℕ, however,
whatever ℕ is, consider least.upper.bound 𝕍

𝕍 is all.visibleᵂᴹ

cantor: 𝕍×𝕍 ↣ 𝕍×{1}
cantor⟨i,j⟩ = ⟨kᵢⱼ,1⟩
kᵢⱼ = i+(i+j-1)(i+j-2)/2

cantor⁻¹⟨k,1⟩ = ⟨iₖ,jₖ⟩
sₖ = max{h| (h-1)(h-2)/2 < k }
iₖ+jₖ = sₖ
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ(sₖ-1)/2-k+1

We observe that
cantor: 𝕍×𝕍 ↣ 𝕍×{1}
is 1.to.1

We observe that
there are at least as many
in 𝕍×{1} as there are in 𝕍×𝕍

1.to.1 𝕍×𝕍 ↣ 𝕍×{1}
|𝕍×𝕍| ≤ |𝕍×{1}|
and
𝕍×𝕍 ⊃≠ 𝕍×{1}

We observe that
𝕍×𝕍 and 𝕍×{1} are not flocks of sheep.

> That is not an observation
> but an assumption

an assumption
of the correctness of arithmetic

> of people who deny or are unable to
> understand actual infinity.

Your (WM's) actually.infiniteᵂᴹ set A
does _not_ have internal de.Bob.ification
but has a _subset_ B ⊆ A
which _does_ have internal de.Bob.ification.

Your explanation is that A contains elements
which "cannot be said" to self-equal.

I don't see what I can add to that.

On 2023-12-22 16:19:08 +0000, WM said:

> Le 22/12/2023 à 15:19, Mikko a écrit :
>> On 2023-12-22 08:09:14 +0000, WM said:
>>
>>> Le 21/12/2023 à 17:45, Mikko a écrit :
>>>> On 2023-12-21 11:56:10 +0000, WM said:
>>>>
>>>>> Le 21/12/2023 à 12:48, Mikko a écrit :
>>>>>> On 2023-12-21 10:57:10 +0000, WM said:
>>>>>>
>>>>>>> Le 21/12/2023 à 10:41, Mikko a écrit :
>>>>>>>> On 2023-12-20 16:28:09 +0000, WM said:
>>>>>>>
>>>>>>>>> The matrix remains, only the covering by X changes.
>>>>>>>>
>>>>>>>> Your first "covering" is just another matrix with the same idicex.
>>>>>>>
>>>>>>> It has the same (m, n), but not all positions have been counted by k
>>>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
>>>>>>
>>>>>> Positions are not covered by. Some of them are covered by X.
>>>>>>
>>>>>> Each position as m and n so each position has k.
>>>>>
>>>>> That is an error. Each O remains in the matrix at a not indexed place.
>>>>
>>>> By the definition of "matrix" there are only indexed places.
>>>
>>> But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> It is indexed by two natural numbers because the matrices meintioned
>> above and earlier in this discussion are matrices that are indexed by
>> two natural numbers.
>
> As the O, not indexed by Cantor prove, most of the matrix' (m, n) are dark.
> Only the upper left corner contains visible m and n.

Every m and every n is "visible". There are no other rows and no other
columns in the matrix.

Mikko

Le 22/12/2023 à 17:59, Richard Damon a écrit :
> On 12/22/23 11:23 AM, WM wrote:

>>> but the rule says that if at least ONE covering mapping exists, we
>>> meet the requirement.
>>
>> Who made that rule?
>
> It wasn't "Made" but "Discovered".

It was "discovered" that there are as many fractions as prime numbers. No
it is believed by matheologians who cannot think from 1 to 3. (They would
discover that between every pair of natural numebers there are
infinitely,many fraction.

> It is a natural consequence of the
> need for a consistent set of logical rules to describe infinite set.

That is not consistent but foolish.

Regards, WM