zelos...@gmail.com schrieb am Dienstag, 8. Februar 2022 um 06:35:55 UTC+1:
> måndag 7 februari 2022 kl. 15:25:47 UTC+1 skrev WM:
> > 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, ...
> > ...,

> your "definable" is still meaningless, it has no mathematical definition

The index of 2/2 is definable. It is 5. The index of any fraction remaining in the matrix after the third column has been emptied is not definable.

Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?

Regards, WM

sergio schrieb am Montag, 7. Februar 2022 um 19:01:25 UTC+1:
> On 2/7/2022 11:18 AM, WM wrote:

> > And yet all of them provable fail to clear any position, let alone all positions, because exchanging two elements will not reduce the elements..
> wrong, all elements are already enumerated,

but they are not yet in the first column.

> their index can be specifically calculated.

What is the index of the first fraction following the last indexed fraction of the seventh column? You do believe that the fractions of the seventh column can be indexed completely, don't you?

Regards, WM

horand....@gmail.com schrieb am Montag, 7. Februar 2022 um 19:58:54 UTC+1:
> On Monday, 7 February 2022 at 13:18:36 UTC-4, WM wrote:

> Again. You have no clue about limits.

Every exchange leaves all positions of the matrix occupied. Never a place is emptied, let alone all places. And “never” does not mean “in the infinite”!

> Until you acquire at least a minimal working knowledge, there is no point continuing this fruitless exchange.

"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" dann ist kein Grenzwert beteiligt.

Gruß, WM

WM expressed precisely :
> sergio schrieb am Montag, 7. Februar 2022 um 19:01:25 UTC+1:
>> On 2/7/2022 11:18 AM, WM wrote:
>
>>> And yet all of them provable fail to clear any position, let alone all
>>> positions, because exchanging two elements will not reduce the elements..
>> wrong, all elements are already enumerated,
>
> but they are not yet in the first column.

Nobody cares. All your shuffling only shows that there are ways to
obfuscate the otherwise obvious bijection.

To paraphrase Mitch, stop moving your Q's around.

Jim Burns schrieb am Montag, 7. Februar 2022 um 22:04:23 UTC+1:
> On 2/6/2022 8:07 AM, WM wrote:
> > Jim Burns schrieb

> We know some things about these semi-dark collections,
> by starting from a description and taking only
> reliable statement-steps.

You don't know anything about dark indexes. But you know a lot about the transpositions:
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, ...
.... ... ... ...
Never a place is emptied, let alone all places. And “never” does not mean “in the infinite”!

Would you like to exchange these two aspects of knowing and not knowing? It is impossible.
> One thing we know is that no column runs empty
> before any other column.

Sorry. All fractions are indexed one after the other, and the columns are disjubct. Therefore yoour claim is provably wrong.

> If a whole column came before any fraction,
> it would be a semi-dark collection inside
> a fully-lit collection. And that cannot happen.
> So, it doesn't.

That is true. But it shows that Cantor's claim about the trinity of completeness, infinity, and well-order has burst.

> > If so, you could know which column of the matrix
> > is the first running empty.

> We know that no first-running-empty exists.

If an infinite set of disjunct subsets is emptied in a sequential way, then different subsets have to be emptied in a sequential way too.

> Surely, that's better than "knowing" a falsehood?

I am discussing this here in order to exorcize Cantor's falsehood.

Regards, WM

FromTheRafters schrieb am Dienstag, 8. Februar 2022 um 10:26:57 UTC+1:
> WM expressed precisely :
> > sergio schrieb am Montag, 7. Februar 2022 um 19:01:25 UTC+1:
> >> On 2/7/2022 11:18 AM, WM wrote:
> >
> >>> And yet all of them provable fail to clear any position, let alone all
> >>> positions, because exchanging two elements will not reduce the elements..
> >> wrong, all elements are already enumerated,
> >
> > but they are not yet in the first column.
> Nobody cares.

Only dishonest or stupid bodies don't care.

Regards, WM

WM submitted this idea :
> FromTheRafters schrieb am Dienstag, 8. Februar 2022 um 10:26:57 UTC+1:
>> WM expressed precisely :
>>> sergio schrieb am Montag, 7. Februar 2022 um 19:01:25 UTC+1:
>>>> On 2/7/2022 11:18 AM, WM wrote:
>>>>> And yet all of them provable fail to clear any position, let alone all
>>>>> positions, because exchanging two elements will not reduce the elements..
>>>> wrong, all elements are already enumerated,
>>>
>>> but they are not yet in the first column.
>> Nobody cares.
>
> Only dishonest or stupid bodies don't care.

Also, those whom see the irrelevancy of your blather to the fact that
there is a bijection, don't care about your blather about several ways
to *not* show a bijection.

tisdag 8 februari 2022 kl. 10:12:43 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 8. Februar 2022 um 06:35:55 UTC+1:
> > måndag 7 februari 2022 kl. 15:25:47 UTC+1 skrev WM:
> > > 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, ...
> > > ...,
> > your "definable" is still meaningless, it has no mathematical definition
> The index of 2/2 is definable. It is 5. The index of any fraction remaining in the matrix after the third column has been emptied is not definable.
>
> Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

>The index of 2/2 is definable. It is 5

That is the what f(5) is for that surjective function, that still doesn't tell what "definable" means.

>The index of any fraction remaining in the matrix after the third column has been emptied is not definable.

This is meaningless

>Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?

Your "empty" is meaningless.

On 2/8/2022 3:16 AM, WM wrote:
> sergio schrieb am Montag, 7. Februar 2022 um 19:01:25 UTC+1:
>> On 2/7/2022 11:18 AM, WM wrote:
>
>>> And yet all of them provable fail to clear any position, let alone all positions, because exchanging two elements will not reduce the elements..
>> wrong, all elements are already enumerated,
>
> but they are not yet in the first column.

no, the first column is already occupied by rationals that have been enumerated.

>
>> their index can be specifically calculated.
>
> What is the index of the first fraction following the last indexed fraction of the seventh column?

there is no "last" as all columns are infinite.

You do believe that the fractions of the seventh column can be indexed completely, don't you?

Simply prove the converse, prove that one is not. Pick n/d in your seventh column, is it indexed ?

if you do not agree, then you are out of Math, and into imagination. Lack of knowing Algebra does that.

>
> Regards, WM

On 2/8/2022 3:20 AM, WM wrote:
> horand....@gmail.com schrieb am Montag, 7. Februar 2022 um 19:58:54 UTC+1:
>> On Monday, 7 February 2022 at 13:18:36 UTC-4, WM wrote:
>
>> Again. You have no clue about limits.
>
> Every exchange leaves all positions of the matrix occupied. Never a place is emptied, let alone all places. And “never” does not mean “in the infinite”!

none of that is relevant, you monkeyed with the matrix of rationals.

>
>> Until you acquire at least a minimal working knowledge, there is no point continuing this fruitless exchange.
>

On 2/8/2022 3:12 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 8. Februar 2022 um 06:35:55 UTC+1:
>> måndag 7 februari 2022 kl. 15:25:47 UTC+1 skrev WM:
>>> 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, ...
>>> ...,
>
>> your "definable" is still meaningless, it has no mathematical definition
>
> The index of 2/2 is definable.

No. The index of 2/2 is 5. Your "definable" is meaningless, you use it to obscure.

> It is 5. The index of any fraction remaining in the matrix after the third column has been emptied is not definable.

wrong, no column are "emptied" as
1. it is a set, a fixed matrix of the rationals
2. One cannot "empty" an infinite set, nor empty a column which is an infinite subset

>
> Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?

all rationals in the third column have been indexed already by k = (m + n - 1)(m + n - 2)/2 + m where m = 3

k = (3 + n - 1)(3 + n -2) + 3 (extra credit: simplify the equation, can you do it?)

now which number in the third column is not enumerated ? call that n.

>
> Regards, WM

You proved Cantor right again.

you did better with poor framed Achilles losing races to slow turtles.

you have no idea about infinity, you are finite.

On 2/8/2022 3:40 AM, WM wrote:
> FromTheRafters schrieb am Dienstag, 8. Februar 2022 um 10:26:57 UTC+1:
>> WM expressed precisely :
>>> sergio schrieb am Montag, 7. Februar 2022 um 19:01:25 UTC+1:
>>>> On 2/7/2022 11:18 AM, WM wrote:
>>>
>>>>> And yet all of them provable fail to clear any position, let alone all
>>>>> positions, because exchanging two elements will not reduce the elements..
>>>> wrong, all elements are already enumerated,
>>>
>>> but they are not yet in the first column.
>> Nobody cares.
>
> Only dishonest or stupid bodies don't care.
>
> Regards, WM

wrong again!

People "don't care" about your math, because it is full of errors, intentional misdirection, and it does not follow correct math at all.

You have been trying to disprove Cantor's Enumeration over and over, but you are using intentional mistakes to do it.

that is dishonest.

On 2/8/2022 4:39 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 7. Februar 2022 um 22:04:23 UTC+1:

>> We know some things about these semi-dark collections,
>> by starting from a description and taking only
>> reliable statement-steps.
>
> You don't know anything about dark indexes.

What is a dark index? Should I care?

I know enough about each second end k of a ⟨1,...,k⟩
( which has a counting-order beginning at 1 and
( ending somewhere
to take only reliable statement-steps from that
description to further, sometimes-surprising claims.

I know that not all collections of those k are fully-lit.
Each ⟨1,...,k⟩ is fully-lit.
ℕ⁺ = ⋃{⟨1,...,k⟩} is semi-dark.

If a dark index is in ℕ⁺ I know that it is
the second end of a collection with a counting-order
beginning at 1. That's a lot to know.

If a dark index is not in ℕ⁺ I am not making any
claims about it. It's irrelevant what I know or don't.

>> If a whole column came before any fraction,
>> it would be a semi-dark collection inside
>> a fully-lit collection. And that cannot happen.
>> So, it doesn't.
>
> That is true.
> But it shows that Cantor's claim about the trinity of
> completeness,

Define ℕ⁺ = ⋃{⟨1,...,k⟩}

We can reason _reliably_ about the elements of ℕ⁺
by starting with a claim that j ∈ ℕ⁺ is the second end
of a collection with a counting-order beginning at 1

For that purpose, _our_ purpose, ℕ⁺ is _complete_
Anything that should be included is included,
anything that should be excluded is excluded.

> infinity,

Fully-lit ("finite") collections either
(i)
match another collection in all total orders, or
(ii)
do not match that other collection in any total
order.

Semi-dark ("infinite") collections are not like that.

The same things can be in both fully-lit and semi-dark
collections. In fully-lit collections, injections
conserve. In semi-dark collections, they might not.

The things don't change. Their collections are different.

> and well-order has burst.

A fully-lit ("finite") collection is double-well-order-able.
There is an order in which each non-empty sub-collection
contains a beginning and an end.
This is what I'm saying with "two-ended-stepping-order-
-ability".

For some of the fully-lit collections, each second end k
is second end to one and only one collection ⟨1,...,k⟩
with a specific order, the counting-order, beginning at 1.
Because of this correspondence, we can discuss one of
these ⟨1,...,k⟩ by discussing its second end k. This
can be a very useful thing to do.

Each ⟨1,...,k⟩ has a counting-order with two ends, 1 and k.

All the second ends k of the ⟨1,...,k⟩ also have a
counting-order, inherited from the ⟨1,...,k⟩,
because one of ⟨1,...,j⟩ and ⟨1,...,k⟩ must be a
subset of the other.

So, all the second ends of the ⟨1,...,k⟩ also have a
counting-order.
And they have a first end, 1.
However, as a collection, they do not have a
second end. Each k is followed by k+1.

The lack of a second end breaks the _second_
well-order, the reversed one.
( The second well-order is still there in any
( subset which has a second end.

The first well-order, beginning at 1, is not broken.

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

FromTheRafters schrieb am Dienstag, 8. Februar 2022 um 14:18:28 UTC+1:

> your blather about several ways
> to *not* show a bijection.

My way
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, ...
.... ... ... ...
is ***precisely*** Cantor's way with the only exception that I am looking where the waste is deposited, Cantor however
1/1, __, __, __, ...
1/2, __, __, __, ...
2/1, __, __, __, ...
1/3, __, __, __, ...
2/2, __, __, __, ...
....
doesn't.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 8. Februar 2022 um 14:23:40 UTC+1:
> tisdag 8 februari 2022 kl. 10:12:43 UTC+1 skrev WM:

> >The index of any fraction remaining in the matrix after the third column has been emptied is not definable.

> This is meaningless

It is not meaningless if all columns can be emptied in a well-ordered way, i.e., not simultaneously.

> >Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?
> Your "empty" is meaningless.

My empty is the same as everyone's empty. Its meaning is: Nothing remains.

Regards, WM

Jim Burns schrieb am Dienstag, 8. Februar 2022 um 21:43:56 UTC+1:
> On 2/8/2022 4:39 AM, WM wrote:
> > Jim Burns schrieb
> > am Montag, 7. Februar 2022 um 22:04:23 UTC+1:
> >> We know some things about these semi-dark collections,
> >> by starting from a description and taking only
> >> reliable statement-steps.
> >
> > You don't know anything about dark indexes.
> What is a dark index? Should I care?

If you claim that all fractions are in one sequence, for instance in the first column, then all other columns must be empty. But since the fractions are well-ordered by their indexes, the columns cannot get empty simultaneously. That implies indexes that cannot be known. They are dark --- if the are at all.

> I know that not all collections of those k are fully-lit.

So you know more than 99.99 % of contemporary and all mathematicians.

> If a dark index is not in ℕ⁺ I am not making any
> claims about it. It's irrelevant what I know or don't.

It is possible that all dark numbers are nonsense. But then the real line must have gaps.

Regards, WM

WM brought next idea :
> FromTheRafters schrieb am Dienstag, 8. Februar 2022 um 14:18:28 UTC+1:
>
>> your blather about several ways
>> to *not* show a bijection.
>
> My way

Blah blah blah.

There is no need to change from what Cantor already did just to show
yet another way to *not* show the bijection.

FromTheRafters schrieb am Mittwoch, 9. Februar 2022 um 11:19:14 UTC+1:

> There is no need to change from what Cantor already did just to show
> yet another way to *not* show the bijection.

Why are you not interested to find out why my way fails and why Cantor's way would fail if he took into account the overburden? Defensor Dei attitude?

Regards, WM

onsdag 9 februari 2022 kl. 09:20:31 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 8. Februar 2022 um 14:23:40 UTC+1:
> > tisdag 8 februari 2022 kl. 10:12:43 UTC+1 skrev WM:
>
> > >The index of any fraction remaining in the matrix after the third column has been emptied is not definable.
>
> > This is meaningless
> It is not meaningless if all columns can be emptied in a well-ordered way, i.e., not simultaneously.
> > >Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?
> > Your "empty" is meaningless.
> My empty is the same as everyone's empty. Its meaning is: Nothing remains.
>
> Regards, WM

There is no "remains" or such to talk about.

You talk about finite changes never being enough, o fcourse not so just shut up already!

On 2/9/2022 2:16 AM, WM wrote:
> FromTheRafters schrieb am Dienstag, 8. Februar 2022 um 14:18:28 UTC+1:
>
>> your blather about several ways
>> to *not* show a bijection.
>
> My way
> 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, ...
> ... ... ... ...
> is ***precisely*** Cantor's way

LIAR.

with the only exception that I am looking where the waste is deposited, Cantor however
> 1/1, __, __, __, ...
> 1/2, __, __, __, ...
> 2/1, __, __, __, ...
> 1/3, __, __, __, ...
> 2/2, __, __, __, ...
> ...
> doesn't.
>
> Regards, WM
>

WM wrote on 2/9/2022 :
> FromTheRafters schrieb am Mittwoch, 9. Februar 2022 um 11:19:14 UTC+1:
>
>> There is no need to change from what Cantor already did just to show
>> yet another way to *not* show the bijection.
>
> Why are you not interested to find out why my way fails and why Cantor's way
> would fail if he took into account the overburden? Defensor Dei attitude?

Because the cleverness of what he did lies in the ways he did it. He
came up with a way to show the bijection quite obviously. You keep
trying to change it so as to show a way which doesn't work -- aparently
convinced that showing a way which doesn't work affects the ones that
do work.

They do not!

Scenario:

Two classes of students. One of boys, one of girls. It just so happens
that each boy is dating exactly one girl in the other class and the
same for each girl dating the boy who is dating her. No 'two-timers' at
all.

The couples are a bijection and if you attempt to pair them
'willy-nilly' instead by ethnicity, hair color, height, weight or
whatever, you might not show a bijection -- this won't change the fact
that the two sets are the same size as shown by the former bijection.

The notion of 'size' here carries into the infinite sets whereas the
explicit features associated with "same number of elements" when
dealing with finite sets might not.

On 2/9/2022 5:38 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 9. Februar 2022 um 11:19:14 UTC+1:
>
>> There is no need to change from what Cantor already did just to show
>> yet another way to *not* show the bijection.
>
> Why are you not interested to find out why my way fails and why Cantor's way would fail if he took into account the overburden? Defensor Dei attitude?
>
> Regards, WM

there is no overburden. Your way fails because you construct it to fail.

Cantors Enumeration is so simple, clear and direct, yet you fail to acknowledge it is right.

sergio submitted this idea :
> On 2/9/2022 5:38 AM, WM wrote:
>> FromTheRafters schrieb am Mittwoch, 9. Februar 2022 um 11:19:14 UTC+1:
>>
>>> There is no need to change from what Cantor already did just to show
>>> yet another way to *not* show the bijection.
>>
>> Why are you not interested to find out why my way fails and why Cantor's
>> way would fail if he took into account the overburden? Defensor Dei
>> attitude?
>>
>> Regards, WM
>
> there is no overburden. Your way fails because you construct it to fail.
>
> Cantors Enumeration is so simple, clear and direct, yet you fail to
> acknowledge it is right.

The ordering of Cantor's list of rationals is simple and direct. The
'table' of rationals is simply a visual aid to show how the sequence
winds its way through *all* of the cells.

The idiot attacks the visual aid thinking it changes everything and
proves himself to be smarter than Cantor. Typical crank behavior, not
unlike 'attacking' nomenclature and notation thinking it will change
what numbers are.

On 2/9/2022 2:20 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 8. Februar 2022 um 14:23:40 UTC+1:
>> tisdag 8 februari 2022 kl. 10:12:43 UTC+1 skrev WM:
>
>>> The index of any fraction remaining in the matrix after the third column has been emptied is not definable.
>
>> This is meaningless
>
> It is not meaningless if all columns can be emptied in a well-ordered way, i.e., not simultaneously.

your premise is wrong.

>
>>> Do you believe that the third column can be emptied by k = (m + n - 1)(m + n - 2)/2 + m ?
>> Your "empty" is meaningless.
>
> My empty is the same as everyone's empty. Its meaning is: Nothing remains.

No. Each and every element of the matrix of rationals is indexed by the natural numbers.

Your leaky rubber transmuted sets have holes in their logic.

>
> Regards, WM