FromTheRafters schrieb am Freitag, 21. Januar 2022 um 11:10:22 UTC+1:
> WM used his keyboard to write :
> > FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 22:33:46 UTC+1:
> >
> >> So 'element by element' means the same as 'step by step' to you?
> >
> > What could be the difference?
> Does Sum_n=1^100(Sn) where Sn is the sequence of natural numbers mean
> to list the first one hundred triangular numbers?
>
> Or, does it just say to add "all" the 100 requisite elements together?

1) Wrong analogy. A sum is not a mapping.
2) The sum can be taken over every n, not only 100. There is no number prohibited.

Regards, WM

FromTheRafters schrieb am Freitag, 21. Januar 2022 um 11:52:10 UTC+1:
> WM explained :

> > Readers who read this but are unable to get it are victims of intellectual
> > abuse by matheologians.
> It is like you are saying that the infinitely many buses with
> infinitely many passengers on each bus is impossible because to can't
> itemize by the itemizing the passengers on the first bus first because
> you can never get to the start of the second bus.

Of course they are impossible. But whenever we accept their existence, then it is clear that also the matrix exists and its hOles will never decrease.
>
> You seem to think of this as depleting the index set.

The hOles are in no way depleted. Not in any step and not after all.

Regards, WM

On 1/21/2022 4:51 AM, FromTheRafters wrote:
> WM explained :
>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 22:38:39 UTC+1:
>>> WM presented the following explanation :
>>
>>>> That is our topic here. Going through the set 1, 2, 3, ... is called counting. Issuing natural numbers to fractions is counting them.
>>> I thought we were playing Tic-Tac-Toe with X's and O's
>>
>> The indeXes are in the first column. The hOles are in the rest.
>>
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> XOOOOOOOO...
>> ...
>>>
>>> Still, even if you just woke up from counting sheep, sets don't "become" anything.
>>
>> The indexes are claimed to cover the whole matrix when applied according to Cantor. That is wrong because no hOle leaves as long as indeXes block the
>> only possible drain, namely the first column. And afterwards, if there is an afterwards, nothing happens.
>>
>> Readers who read this but are unable to get it are victims of intellectual abuse by matheologians.
>
> It is like you are saying that the infinitely many buses with infinitely many passengers on each bus is impossible because to can't itemize by the
> itemizing the passengers on the first bus first because you can never get to the start of the second bus.
>
> You seem to think of this as depleting the index set.
>
> Equivalently, we can't itemize every passenger by itemizing the first passenger on each of infinitely many buses because we can never get to the second
> passenger on the first bus.
>
> Try first bus first passenger, then second bus first passenger, then first bus second passenger, first bus third passenger, second bus second passenger,
> third bus first passenger ...
>
>
> 1  3  4  10 11 21 ...
>
> 2  5  9  12 20 ...
>
> 6  8  13 19 ...
>
> 7  14 18 ...
>
> 15 17 ...
>
> 16 ...
>
> ...
>
>
> Here, the index set is neither horizontal nor vertical on the graphic, it is serpentine and only 'used' once.

that works

On 1/21/2022 8:21 AM, WM wrote:
> FromTheRafters schrieb am Freitag, 21. Januar 2022 um 11:10:22 UTC+1:
>> WM used his keyboard to write :
>>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 22:33:46 UTC+1:
>>>
>>>> So 'element by element' means the same as 'step by step' to you?
>>>
>>> What could be the difference?
>> Does Sum_n=1^100(Sn) where Sn is the sequence of natural numbers mean
>> to list the first one hundred triangular numbers?
>>
>> Or, does it just say to add "all" the 100 requisite elements together?
>
> 1) Wrong analogy. A sum is not a mapping.

wrong, a sum and mapping is an operation between two numbers.

> 2) The sum can be taken over every n, not only 100. There is no number prohibited.

as you know 100 is an example, no numbers were prohibited

>
> Regards, WM

how about some more red herrings ?

On 1/21/2022 2:56 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 06:52:19 UTC+1:
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...

your X on the bottom is in the wrong place

>
>>> The sequence of configurations is such that by transpositions the number of X's cannot increase and the number of O's cannot decrease. That is basic to all transpositions.
>>>
>>> You are wrong. I am using no entirely different function but the well known
>>>
>>> k = (m + n - 1)(m + n - 2)/2 + m.
>
> Here for k = 8.
>
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...

your X on the bottom is in the wrong place

> ...
>
>>> Dark numbers cannot be shown. But the indirect existence proof is done by the invariable number of O's.
>
>>> You must believe:
>>> 1) For all k ∈ ℕ the number of X's and O's does not change by more than plus minus zero.
>>> Or do you deny this?
>>> 2) The enumeration becomes ready. When this happens the number of O's is zero although there is no way where they could have gone.
>
>>> I did not change it. I used Cantor's enumeration
>> You are not when you change the function.
>
> See above.
>
>>> The sequence of configurations is such that by transpositions the number of X's cannot increase and the number of O's cannot decrease. That is basic to all transpositions.
>> Still changing the function and trying to obfuscate it. The original function works, get over it.
>>> Dark numbers cannot be shown. But the indirect existence proof is done by the invariable number of O's
>> Empty assertion, I can show for ALL numbers.
>
> Then show me the first O that leaves the board.
>>
> Regards, WM

On 1/21/2022 8:17 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 10:25:13 UTC+1:
>> fredag 21 januari 2022 kl. 09:56:41 UTC+1 skrev WM:
>
>>>> Empty assertion, I can show for ALL numbers.
>>> Then show me the first O that leaves the board.
>>>>
>> I do not care for your "board"
>
> And the board does not care for you and whether you care. It is existing and shows that matheologians who claim that all hOles disappear are stupid.

you modify the board on the fly,

it is simple 2D to 1D conversion

>>
>> for any rational number n/m I can find a natural number to map to it.
>
> But mathematical conclusion finds no hOle hone, all hOles present before after.

it is your board, you need to fix its problem

>That proves you wrong.

no, it proves you board is wrong

>
> Regards, WM

WM presented the following explanation :
> FromTheRafters schrieb am Freitag, 21. Januar 2022 um 11:10:22 UTC+1:
>> WM used his keyboard to write :
>>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 22:33:46 UTC+1:
>>>
>>>> So 'element by element' means the same as 'step by step' to you?
>>>
>>> What could be the difference?
>> Does Sum_n=1^100(Sn) where Sn is the sequence of natural numbers mean
>> to list the first one hundred triangular numbers?
>>
>> Or, does it just say to add "all" the 100 requisite elements together?
>
> 1) Wrong analogy. A sum is not a mapping.

It is not necessarily a 100 count step-by-step process which is the
point I'm making and you are missing.

sergio expressed precisely :
> On 1/21/2022 4:51 AM, FromTheRafters wrote:
>> WM explained :
>>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 22:38:39 UTC+1:
>>>> WM presented the following explanation :
>>>
>>>>> That is our topic here. Going through the set 1, 2, 3, ... is called
>>>>> counting. Issuing natural numbers to fractions is counting them.
>>>> I thought we were playing Tic-Tac-Toe with X's and O's
>>>
>>> The indeXes are in the first column. The hOles are in the rest.
>>>
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> XOOOOOOOO...
>>> ...
>>>>
>>>> Still, even if you just woke up from counting sheep, sets don't "become"
>>>> anything.
>>>
>>> The indexes are claimed to cover the whole matrix when applied according
>>> to Cantor. That is wrong because no hOle leaves as long as indeXes block
>>> the only possible drain, namely the first column. And afterwards, if there
>>> is an afterwards, nothing happens.
>>>
>>> Readers who read this but are unable to get it are victims of intellectual
>>> abuse by matheologians.
>>
>> It is like you are saying that the infinitely many buses with infinitely
>> many passengers on each bus is impossible because to can't itemize by the
>> itemizing the passengers on the first bus first because you can never get
>> to the start of the second bus.
>>
>> You seem to think of this as depleting the index set.
>>
>> Equivalently, we can't itemize every passenger by itemizing the first
>> passenger on each of infinitely many buses because we can never get to the
>> second passenger on the first bus.
>>
>> Try first bus first passenger, then second bus first passenger, then first
>> bus second passenger, first bus third passenger, second bus second
>> passenger, third bus first passenger ...
>>
>>
>> 1  3  4  10 11 21 ...
>>
>> 2  5  9  12 20 ...
>>
>> 6  8  13 19 ...
>>
>> 7  14 18 ...
>>
>> 15 17 ...
>>
>> 16 ...
>>
>> ...
>>
>>
>> Here, the index set is neither horizontal nor vertical on the graphic, it
>> is serpentine and only 'used' once.
>
>
> that works

Finding some which don't work does not change anything no matter how
much WM wishes it did. But he keeps trying.

FromTheRafters schrieb am Freitag, 21. Januar 2022 um 17:23:50 UTC+1:
> WM presented the following explanation :

> >> Or, does it just say to add "all" the 100 requisite elements together?
> >
> > 1) Wrong analogy. A sum is not a mapping.
> It is not necessarily a 100 count step-by-step process which is the
> point I'm making and you are missing.

But it is possible to check every step. That is the important point, at least in mathematics.

Regards, WM

On Friday, January 21, 2022 at 6:54:54 PM UTC+1, WM wrote:

> it is possible to check every step. That is the important point, at least in mathematics.

Indeed!

Chuck Norris counted to infinity - twice.

On 1/21/2022 11:54 AM, WM wrote:
> FromTheRafters schrieb am Freitag, 21. Januar 2022 um 17:23:50 UTC+1:
>> WM presented the following explanation :
>
>>>> Or, does it just say to add "all" the 100 requisite elements together?
>>>
>>> 1) Wrong analogy. A sum is not a mapping.
>> It is not necessarily a 100 count step-by-step process which is the
>> point I'm making and you are missing.
>
> But it is possible to check every step. That is the important point, at least in mathematics.
>
> Regards, WM

another red herring...

On 1/20/2022 4:24 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 20. Januar 2022 um 22:00:33 UTC+1:
>> On 1/20/2022 1:46 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Donnerstag, 20. Januar 2022 um 19:16:18 UTC+1:
>>>> On 1/20/2022 5:15 AM, WM wrote:

>>>>> then you will succeed? No, it is only owed to your
>>>>> short-sightedness. Because you cannot see the end of
>>>>> the first line and of any line or column.
>>>>
>>>> No, your argument fails.
>>>> We do not need to see the ends (which do not exist)
>>>
>>> But you claim that you get finished.
>>
>> I claim that, with i = (n+d-1)*(n+d-2)/2 + n,
>> each n/d of the whole matrix is matched
>
> That cannot happen because the matching i/1 are taken
> away from their places. And every i/1 replacing them
> is also taken away from its place.
>
> The gaps cannot be filled.

....if the replacing has a last replacement.

However, the replacing does not have a last replacement.

The O at n/d goes to i/1, the X at i/1 goes to n/d,
where i = (n+d-1)*(n+d-2)/2 + n

The O (which was an X) at i/1 goes to i'/1,
the X at i'/1 goes to i/1,
where i' = (i+1-1)*(i+1-2)/2 + i

_Up to_ any stage in the replacing, the X's and O's
and their locations are special collections, like
those of sheep and pebbles. _Up to_ any stage, we have
Conservation of O's.

_All_ the replacements is not _up to_ any stage in
the replacing. The collection of all does not have
Conservation of O's. It does not have a two-ended
stepping-order.

The Conservation of O's and the two-ended stepping-order
are provably linked.

Jim Burns schrieb am Samstag, 22. Januar 2022 um 00:54:09 UTC+1:
> On 1/20/2022 4:24 PM, WM wrote:
> > Jim Burns schrieb
> > am Donnerstag, 20. Januar 2022 um 22:00:33 UTC+1:
> >> On 1/20/2022 1:46 PM, WM wrote:
> >>> Jim Burns schrieb

> > That cannot happen because the matching i/1 are taken
> > away from their places. And every i/1 replacing them
> > is also taken away from its place.
> >
> > The gaps cannot be filled.
> ...if the replacing has a last replacement.

And also if there is no last replacement. Remember that the drain is blocked.
>
> However, the replacing does not have a last replacement.

I know, but this excludes completeness.
>
> _Up to_ any stage in the replacing, the X's and O's
> and their locations are special collections, like
> those of sheep and pebbles. _Up to_ any stage, we have
> Conservation of O's.

So it is.
>
> _All_ the replacements is not _up to_ any stage in
> the replacing. The collection of all does not have
> Conservation of O's. It does not have a two-ended
> stepping-order.

Before a last step, there is no all steps.
>
> The Conservation of O's and the two-ended stepping-order
> are provably linked.

Nice to see at least one reader who understands this. Now there is only a small step for man but a rather big and healthy leap for mankind: The conservation of O's will never be violated. It cannot be violated, because a column has less area than all columns. Note that all are infinite and therefore of same length.

Regards, WM

On 1/22/2022 3:57 AM, WM wrote:
> Jim Burns schrieb am Samstag, 22. Januar 2022 um 00:54:09 UTC+1:
>> On 1/20/2022 4:24 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Donnerstag, 20. Januar 2022 um 22:00:33 UTC+1:
>>>> On 1/20/2022 1:46 PM, WM wrote:
>>>>> Jim Burns schrieb
>
>>> That cannot happen because the matching i/1 are taken
>>> away from their places. And every i/1 replacing them
>>> is also taken away from its place.
>>>
>>> The gaps cannot be filled.
>> ...if the replacing has a last replacement.
>
> And also if there is no last replacement. Remember that the drain is blocked.
>>
>> However, the replacing does not have a last replacement.
>
> I know, but this excludes completeness.
>>
>> _Up to_ any stage in the replacing, the X's and O's
>> and their locations are special collections, like
>> those of sheep and pebbles. _Up to_ any stage, we have
>> Conservation of O's.
>
> So it is.
>>
>> _All_ the replacements is not _up to_ any stage in
>> the replacing. The collection of all does not have
>> Conservation of O's. It does not have a two-ended
>> stepping-order.
>
> Before a last step, there is no all steps.
>>
>> The Conservation of O's and the two-ended stepping-order
>> are provably linked.
>
> Nice to see at least one reader who understands this. Now there is only a small step for man but a rather big and healthy leap for mankind: The conservation of O's will never be violated. It cannot be violated, because a column has less area than all columns. Note that all are infinite and therefore of same length.
>
> Regards, WM

why do you ignore the rows ?

Fritz Feldhase schrieb am Freitag, 21. Januar 2022 um 19:36:06 UTC+1:
> On Friday, January 21, 2022 at 6:54:54 PM UTC+1, WM wrote:
>
> > it is possible to check every step. That is the important point, at least in mathematics.
> Indeed!
>
> Chuck Norris counted to infinity - twice.

He even found all permutations of the natnumbers., implying the existence of all permutations of the fractions. Can you imagine that one by order of magnitude?

Regards, WM

On 1/22/2022 11:11 AM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 21. Januar 2022 um 19:36:06 UTC+1:
>> On Friday, January 21, 2022 at 6:54:54 PM UTC+1, WM wrote:
>>
>>> it is possible to check every step. That is the important point, at least in mathematics.
>> Indeed!
>>
>> Chuck Norris counted to infinity - twice.
>
> He even found all permutations of the natnumbers., implying the existence of all permutations of the fractions. Can you imagine that one by order of magnitude?
>
> Regards, WM

did Chuck find the Last One ?

WM wrote:

>> > it is possible to check every step. That is the important point, at
>> > least in mathematics.
>> Indeed! Chuck Norris counted to infinity - twice.
>
> He even found all permutations of the natnumbers., implying the
> existence of all permutations of the fractions. Can you imagine that one
> by order of magnitude?

good post. Here we have capitalist america.

OVER 40 MILLION FRAUDULENT DEEDS! Estimated 67 MILLION HOMELESS AMERICANS
- CRIMINAL CORRUPTION
https://www.bitchute.com/video/MsQtqc1EQKr1/

sergio schrieb am Samstag, 22. Januar 2022 um 18:26:13 UTC+1:
> On 1/22/2022 11:11 AM, WM wrote:
> > Fritz Feldhase schrieb am Freitag, 21. Januar 2022 um 19:36:06 UTC+1:
> >> On Friday, January 21, 2022 at 6:54:54 PM UTC+1, WM wrote:
> >>
> >>> it is possible to check every step. That is the important point, at least in mathematics.
> >> Indeed!
> >>
> >> Chuck Norris counted to infinity - twice.
> >
> > He even found all permutations of the natnumbers., implying the existence of all permutations of the fractions. Can you imagine that one by order of magnitude?
> >
> did Chuck find the Last One ?

I don't know. He claimed "all". But there need not be a smallest or largest fraction if only all existing fractions are ordered by size.

Regards, WM

On 1/22/2022 4:57 AM, WM wrote:
> Jim Burns schrieb
> am Samstag, 22. Januar 2022 um 00:54:09 UTC+1:

>> The Conservation of O's and the two-ended stepping-order
>> are provably linked.
>
> Nice to see at least one reader who understands this.

I take your approval to indicate I don't have to prove the
linking for you, that we're on the same page at least this much.

| What I mean by the terms I'm using:
| | A collection B with a _two-ended stepping-order_ '<'
| has a total order '<' of B with two ends in B
| in which each _cut_ of B is a _step_ of B
| | A _cut_ C of B with total order '<' is
| a non-empty downward-closed subset C of B with
| a non-empty upward-closed complement B\C
| | A _step_ C of B with total order '<' is
| a cut C of B for which
| C contains a last element before B\C, and
| B\C contains a first element after C

> Now there is only a small step for man but
> a rather big and healthy leap for mankind:
> The conservation of O's will never be violated.

No, i = (n+d-1)*(n+d-2)/2 + n violates O-conservation
for _all_ the transpositions
(NOT two-ended-stepping-orderable)
from _all_ the matrix
(NOT two-ended-stepping-orderable)
to _all_ the first column
(NOT two-ended-stepping-orderable).

Again,
the Conservation of O's and the two-ended stepping-order
are provably linked.

We don't have O-conservation and we don't have
two-ended stepping orders. No contradiction.

> It cannot be violated, because a column has less area
> than all columns. Note that all are infinite and
> therefore of same length.

"Finite" is a more broadly-used term for
"two-ended-stepping-orderable".
Finite collections obey O-conservation.
Infinite collections might violate O-conservation.

It is the O-conservation of finite, two-ended-stepping-
-orderable collections which make them useful to those
responsible for sheep and pebbles.
Not all collections are two-ended-stepping-orderable.

"Less area" and "same length" don't make sense
for collections NOT two-ended-stepping-orderable
because O's are not conserved for those collections.

----
>>> That cannot happen because the matching i/1 are taken
>>> away from their places. And every i/1 replacing them
>>> is also taken away from its place.
>>>
>>> The gaps cannot be filled.
>>
>> ...if the replacing has a last replacement.
>
> And also if there is no last replacement.
> Remember that the drain is blocked.

If there is no last replacement, then the replacements are
NOT two-ended-stepping-orderable.
Whatever the status of the drain, we might not have
O-conservation.

>> However, the replacing does not have a last replacement.
>
> I know, but this excludes completeness.

When we reason by description-and-reliable-steps,
the relevant sense of "completeness" is that in which
the claims we make are _completely_ true: true of
each individual about which we make the claims.

A description, because it is a description,
is true of each individual it describes.
_Complete? Check_

A reliable statement-step which starts on a _completely_
true statement ends on a _completely_ true statement,
always, in every circumstance, because that is what
makes a reliable statement-step reliable.
_Complete? Check_

Completeness cannot be excluded from description-and-
-reliable-steps reasoning. It is mixed into the concrete
that becomes its foundation.

> Before a last step, there is no all steps.

When we reason by description-and-reliable-steps,
the relevant sense of "all" is that in which
the claims we make are true of each individual
about which we make the claims.

Make a claim that k ends {1,..,k} and
{1,..,k} begins at 1 and has a counting-order
(a stepping-order in which each step is a count)
That's a claim true of _all_ individuals which end a
collection which starts at 1 and has a counting-order.

Take only reliable statement-steps from that claim and
you can only arrive at claims true of _all_ individuals
which end a collection which starts at 1 and has a
counting order.

What did I NOT mention here?
Anything about a last individual.

Reasoning by description-and-reliable-steps is _already_
complete in the relevant sense, from the get-go,
before anything can be said about a last individual.

On 1/21/2022 3:58 AM, WM wrote:
> FromTheRafters schrieb
> am Donnerstag, 20. Januar 2022 um 22:33:46 UTC+1:

>> So 'element by element' means the same as
>> 'step by step' to you?
>
> What could be the difference?

Let P be a collection of ordered pairs ⟨x,y⟩ with
x in collection X and y in collection Y.

This seems reasonable to me (JB):

P matches X and Y _element by element_ iff
for each x in X, there is one and only one ⟨x,y⟩ in P,
and
for each y in Y, there is one and only one ⟨x,y⟩ in P.

P matches X and Y _step by step_ iff
P matches X and Y element by element, and
P has a stepping-order.

A collection P with a _stepping-order_ '<'
has a total order '<' of P in which
each _cut_ of P is a _step_ of P

A _cut_ C of P with total order '<' is
a non-empty downward-closed subset C of P with
a non-empty upward-closed complement P\C

A _step_ C of P with total order '<' is
a cut C of P for which
C contains a last element before P\C, and
P\C contains a first element after C

Matching step-by-step has tighter requirements than
matching element-by-element (matching à la Cantor).

----
If X can be matched to any Y _step by step_
(the order of P is step by step)
then X and Y and P are _countable_

If that stepping-order of P also has two ends, then
X and Y and P are _finite_

If X can be matched to any Y with a two-ended stepping-
-order, then any other total order of X will be a
two-ended stepping-order.

However, if X can be matched to any Y with a total
order which is NOT two-ended-stepping, then any other
total order of X will be NOT two-ended-stepping.

Jim Burns schrieb am Samstag, 22. Januar 2022 um 20:35:22 UTC+1:
> On 1/22/2022 4:57 AM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 22. Januar 2022 um 00:54:09 UTC+1:
> >> The Conservation of O's and the two-ended stepping-order
> >> are provably linked.
> >
> > Nice to see at least one reader who understands this.
> I take your approval to indicate I don't have to prove the
> linking for you, that we're on the same page at least this much.

That is good to know. Most readres would deny.
>
> | What I mean by the terms I'm using:

I know your intention. Nevertheless we have the fact that the mapping from ℕ to the matrix M
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, ...
....
according to Cantor is believed to remove all O's whereas the mapping from ℕ to the first column and from the first columns to the whole matrix M: n --> n/1 & n/1 --> M will preserve all O's.

> No, i = (n+d-1)*(n+d-2)/2 + n violates O-conservation
> for _all_ the transpositions

That shows that the X in

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
....
are only changing places but will never multiply and never cover the whole M.

> Again,
> the Conservation of O's and the two-ended stepping-order
> are provably linked.

Every moving will preserve the moved objects - in eternity.

> > It cannot be violated, because a column has less area
> > than all columns. Note that all are infinite and
> > therefore of same length.
> "Finite" is a more broadly-used term for
> "two-ended-stepping-orderable".

Finite is a term covering all visible natural numbers and all visible indexes.

> Finite collections obey O-conservation.
> Infinite collections might violate O-conservation.

You are wrong. The drain is blocked forever.

> "Less area" and "same length" don't make sense

If a person believes that the X can cover the whole matrix although all columns have exctly the same area, then they are fools.

> >> However, the replacing does not have a last replacement.
> >
> > I know, but this excludes completeness.
> When we reason by description-and-reliable-steps,
> the relevant sense of "completeness" is that in which
> the claims we make are _completely_ true: true of
> each individual about which we make the claims.

That mans each individual transposition will not change the relative numbers.

Regards, WM

Jim Burns schrieb am Samstag, 22. Januar 2022 um 22:18:36 UTC+1:

> P matches X and Y _element by element_ iff
> for each x in X, there is one and only one ⟨x,y⟩ in P,
> and
> for each y in Y, there is one and only one ⟨x,y⟩ in P.

Like for every transposition there is one X and one O at the beginnig and at the end.
> Matching step-by-step has tighter requirements than
> matching element-by-element (matching à la Cantor).

One of both is enough to be sure: No X will ever double and no O will disappear.

Regards, WM

Jim Burns wrote:

>> What could be the difference?
>
> Let P be a collection of ordered pairs ⟨x,y⟩ with x in collection X and
> y in collection Y.

Nigera destroys 1 Million vaccines. They are not stupid.
https://www.brighteon.com/f74b2cc5-7e37-4e05-9261-cdd2db7ef455

lech walensa kiss my ass.

On 1/22/2022 5:42 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 22. Januar 2022 um 20:35:22 UTC+1:
>> On 1/22/2022 4:57 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 22. Januar 2022 um 00:54:09 UTC+1:

>>>> The Conservation of O's and the two-ended stepping-order
>>>> are provably linked.
>>>
>>> Nice to see at least one reader who understands this.
>>
>> I take your approval to indicate I don't have to prove
>> the linking for you, that we're on the same page
>> at least this much.
>
> That is good to know. Most readres would deny.

I would bet that most readers who understand what
I'm saying would agree.

You, I would tell to go sit with the exceptions who,
if they understood what I'm saying, would disagree.

Yes, you agreed. But without understanding.
I get no joy from mentioning this.
I foolishly thought we had made progress.

https://www.youtube.com/watch?v=qajvC2LVrN4&t=75s

Will the circle be unbroken
By and by Lord, by and by
There's a better home awaiting
In the sky Lord, in the sky.

>> | What I mean by the terms I'm using:
>
> I know your intention.

Oh, you know, do you?
What is the reason for your change of heart
within the next few sentences?

>>> It cannot be violated, because a column has less area
>>> than all columns. Note that all are infinite and
>>> therefore of same length.
>>
>> "Finite" is a more broadly-used term for
>> "two-ended-stepping-orderable".
>
> Finite is a term covering all visible natural numbers
> and all visible indexes.
>
>> Finite collections obey O-conservation.
>> Infinite collections might violate O-conservation.
>
> You are wrong. The drain is blocked forever.

Again,
the Conservation of O's and the two-ended stepping-order
are provably linked.

Conserving <--> two-ended stepping
NOT conserving <--> NOT two-ended stepping

On 1/22/2022 4:42 PM, WM wrote:
> Jim Burns schrieb am Samstag, 22. Januar 2022 um 20:35:22 UTC+1:
>> On 1/22/2022 4:57 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 22. Januar 2022 um 00:54:09 UTC+1:
>>>> The Conservation of O's and the two-ended stepping-order
>>>> are provably linked.
>>>
>>> Nice to see at least one reader who understands this.
>> I take your approval to indicate I don't have to prove the
>> linking for you, that we're on the same page at least this much.
>
> That is good to know. Most readres would deny.
>>
>> | What I mean by the terms I'm using:
>
> I know your intention. Nevertheless we have the fact that the mapping from ℕ to the matrix M
> 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, ...
> ...
> according to Cantor is believed to remove all O's

with diagonal threading, not your modified matrix

> whereas the mapping from ℕ to the first column and from the first columns to the whole matrix M: n --> n/1 & n/1 --> M will preserve all O's.
>
>> No, i = (n+d-1)*(n+d-2)/2 + n violates O-conservation
>> for _all_ the transpositions
>
> That shows that the X in
>
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
> are only changing places but will never multiply and never cover the whole M.

that is WMs way, which fails. Not Cantors.

>
>> Again,
>> the Conservation of O's and the two-ended stepping-order
>> are provably linked.
>
> Every moving will preserve the moved objects - in eternity.

why are you modifying the matrix ?

>
>>> It cannot be violated, because a column has less area
>>> than all columns. Note that all are infinite and
>>> therefore of same length.
>> "Finite" is a more broadly-used term for
>> "two-ended-stepping-orderable".
>