On 1/31/2022 12:34 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 12:41:04 UTC+1:
>> måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:
>
>>>> That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
>>>>> For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
>>>
>>>> It exists for ALL
>>> Indexes exist only as long as infinitely many fractions remain not indexed.
>
>> You are still arguing "after finite steps"
>
> After finite steps = definable transpositions the matrix has provably no empty position
>
> 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, ...
> ...

I see..... So how are you going to fix that ?

>
> After all indices have been applied, the matrix, except the first column, is empty.

Your indices come as stickers. Just by a set of new ones off eBay.

And leave the matrix alone, that is Anti Cantor.

>
> This shows undefinable steps.

you took the wrong steps by messing with the first column, when you should have got a brand new set of counting numbers instead.

>
> Regards, WM

On 1/31/2022 1:08 PM, WM wrote:
> horand....@gmail.com schrieb am Montag, 31. Januar 2022 um 13:56:21 UTC+1:
>> On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
>>> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
>>>> On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
>>>
>>>>> Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
>>>> Again, so?? All you show continually is that you have no clue how infinity works.
>>> Infinity does not work.
>> You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it.
>
> Not at all. I accept all infinitely many terms of the sequence of transpositions 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, ...
> ...
> namely
>
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 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, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 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, ...
> ..., and so on.
>
> For *all* *infinitely* many of them I have proved that no matrix position gets empty. How can you accuse me of finitism?!
>
>> IF there are only finitely many integers, then the cardinality of N is finite.
>
> If there are more than all finite terms, then we get
>
> 1/1, O , O, O , ...
> 1/2, O , O, O , ...
> 2/1, O , O, O , ...
> 1/3, O , O, O , ...
> 2/2, O , O, O , ...
> ...,
>
> I agree. I would only mention that the exit of the fractions happens somewhat panic-stricken. We have no means of individually controlling them.(They have no Passierschein like the inhabitants imprisoned in the former GDR would have needed.)
>
> I think that you share my description of facts.
>
> Regards, WM
>

what facts ? you are too far off, lost in your own woods for any help

Best to study "Cantor's Enumeration" in detail.
Notice he does not mess around with the matrix of rationals at all, it is fixed and permanent.
All he does is count the elements.

but WM has to slide the elements around in a big tic-tac-toe game and stay confuseded, to make the magic dark ones not appear, only a "might be there",
but dont look, as you cannot see them anyway.

On 1/31/2022 3:31 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 30. Januar 2022 um 23:11:10 UTC+1:
>> On 1/29/2022 5:50 PM, WM wrote:

>>>>> [...] however no O covers a fraction because
>>>>> all fractions like 1/2 have settled in
>>>>> the first column.
>>>>
>>>> Yes, no O covers a fraction.
>>>
>>> In the limit!
>
> That is, when all indexes have been applied.

Okay,
I can translate "in the limit" here to
"when each index has been applied".

Yes,
when each index has been applied ("in the limit"),
no O covers a fraction.

We can learn about what happens or does not happen
when each index has been applied ("in the limit")
without "arriving at the limit" even in principle.

| Find reliable claims which describe one of the
| indexes, no matter which index.
| | Find more reliable claims about each index by
| taking only reliable statement-steps from what we
| already know are reliable claims.

Step one. Describe an index.
What is an index?

[1]
Certain collections either match in every order or
do not match in every order. We are familiar with
many examples of these certain collections:
a flocks of sheep, a pocket of pebbles, a glory of
unicorns. The technical term for one of these
collections is "finite".

For most finite collections, its members will have no
particular successors or predecessors. In one order,
it could be succeeded by one and, in another order,
it could be by another. What's special about these
collections is that the answer is the same, match or
don't match, in all the orders.

We get _indexes_ from assigning permanent successors
to possible members of these finite collections.
With a little more description, these are what we
call "natural numbers".

What's special about natural numbers, these
members-with-permanent-successors, is that there is
only one collection ⟨x,...,y⟩ starting with x, ending
with y, in which the actual successors are the
permanent successors.

| Let ⟨x,...,y⟩ and ⟪x,...,y⟫ both begin and end with
| x and y, and let the actual successors (in each
| collection) be the permanent successors ( j --> j+1 )
| | If there is any member k in ⟨x,...,y⟩ but not-in
| ⟪x,...,y⟫ there is a first such member k'
| The predecessor of k' in ⟨x,...,y⟩ is its permanent
| successor j, j+1 = k'
| j must be in ⟪x,...,y⟫ too, and its permanent
| successor k' = j+1 must be in ⟪x,...,y⟫ too.
| Contradiction.
| | Thus,
| there is no member in ⟨x,...,y⟩ but not-in ⟪x,...,y⟫
| ⟨x,...,y⟩ = ⟪x,...,y⟫

What's special about natural numbers is that we can
refer to a particular collection ⟨0,...,k⟩ ordered by
permanent successors by referring to the element k
which ends ⟨0,...,k⟩

| ⟨0,...,k⟩ has a total order such that...
| ⟨0,...,k⟩ begins with 0 and ends with k
| For each cut I,F of ⟨0,...,k⟩ the initial segment I
| ends and the final segment F begins, and
| the beginning j of F is the permanent successor of
| the end i of I, j = i+1

_The k that end these ⟨0,...,k⟩ are the indexes_

Some paragraphs back in my post, at [1], I refer to
"certain collections". Among these collections, there
are flocks of sheep and pockets of pebbles. These
certain collections either match in each of their orders
or don't match in each of their orders. The "finite"
collections.

What's special about the indexes k is that,
for each of these finite collections
(which only match or only not-match),
one of their ⟨0,...,k⟩ match it.

Recall the general architecture of the arguments we're
using. They start
"Find reliable claims which describe one of ... "
Here, we have the opportunity to learn about a flock of
sheep, a pocket of pebbles, a glory of unicorns,
all at once, by describing one of these ends of ⟨0,...,k⟩
There are people who find that prospect highly desirable.

However,
not all collections can be matched to one of these
⟨0,...,k⟩
For example,
the collection {⟨0,...,k⟩} of collections ⟨0,...,k⟩
cannot be matched to one of the ⟨0,...,k⟩

This is paradox to the same degree that
"not all plane figures are squares"
is a paradox: not at all.

>>> Each O covers something in every finite term of
>>> the sequence.
>>> In the limit each O does not cover anything.
>>> Therefore infinitely many fractions have passed
>>> between all finite terms and the limit.
>>> They are not indexed.
>>> They are dark.

There is Cantor's sequence of fractions f[k]
⟨ ⟨1,f[1]⟩, ⟨2,f[2]⟩, ⟨3,f[3]⟩, ... ⟩

Each term ⟨k,f[k]⟩ of Cantor's sequence is a
finite term.

By "finite term", I mean that,
for each of its terms, the sequence up to that term
⟨ ⟨1,f[1]⟩, ⟨2,f[2]⟩, ..., ⟨k,f[k]⟩ ⟩
has a two ends and a total order such that,
for each cut, the initial segment has an end and
the final segment has a beginning.

Each term is a finite term.
This is unavoidable,
given what it means for something to be a sequence.

More avoidable, but still true here:
each fraction n/d is in a term in Cantor's sequence,
⟨k,n/d⟩ such that k = (n+d-1)*(n+d-2)/2 + n
A _finite_ term.
There are no terms in Cantor's sequence which are
not finite terms.

>>> Each O covers something in every finite term of
>>> the sequence.

For each O,
there is a sub-sequence of Cantor's noting the changes
to the fraction covered by that O
⟨ ⟨k[1],f[k[1]]⟩, ⟨k[2],f[k[2]]⟩, ⟨k[3],f[k[3]]⟩, ... ⟩

For example Bob has the sequence
⟨ ⟨2,1/2⟩, ⟨3,2/1⟩, ⟨6,21/1⟩, ⟨21,231/1⟩, ... ⟩

For each of the O's sub-sequence terms,
the sequence up to that term
⟨ ⟨k[1],f[k[1]]⟩, ⟨k[2],f[k[2]]⟩, ..., ⟨k[j],f[k[j]]⟩ ⟩
has a two ends and a total order such that,
for each cut, the initial segment has an end and
the final segment has a beginning.

There are no terms in the O's sub-sequence which are
not finite terms.
This follows from it being a sub-sequence of a sequence
which has no terms which are not finite terms.

>>> They are not indexed.
>>> They are dark.

No.
We have described the fractions, the indexes, the
sequences.
Seen or unseen, those are what we're working with.

måndag 31 januari 2022 kl. 19:34:25 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 12:41:04 UTC+1:
> > måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:
>
> > > > That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
> > > > >For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
> > >
> > > > It exists for ALL
> > > Indexes exist only as long as infinitely many fractions remain not indexed.
> > You are still arguing "after finite steps"
> After finite steps = definable transpositions the matrix has provably no empty position
> 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, ...
> ...
>
> After all indices have been applied, the matrix, except the first column, is empty.
>
> This shows undefinable steps.
>
> Regards, WM
your "undefinable" is meaningless

You cannot argue with finite steps. Get over it already

zelos...@gmail.com schrieb am Dienstag, 1. Februar 2022 um 07:06:22 UTC+1:
> måndag 31 januari 2022 kl. 19:34:25 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 12:41:04 UTC+1:
> > > måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:
> >
> > > > > That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
> > > > > >For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
> > > >
> > > > > It exists for ALL
> > > > Indexes exist only as long as infinitely many fractions remain not indexed.
> > > You are still arguing "after finite steps"
> > After finite steps = definable transpositions the matrix has provably no empty position
> > 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, ...
> > ...
> >
> > After all indices have been applied, the matrix, except the first column, is empty.
> >
> > This shows undefinable steps.
> >
> You cannot argue with finite steps.

I can argue that whenever a finite index is issued to a fraction all positions of the matrix are occupied by fractions. Therefore the clearing of the matrix is not possible by attaching finite indexes to the remaining fractions.

Regards, WM

Jim Burns schrieb am Dienstag, 1. Februar 2022 um 01:50:31 UTC+1:
> On 1/31/2022 3:31 AM, WM wrote:

> I can translate "in the limit" here to
> "when each index has been applied".
>
> Yes,
> when each index has been applied ("in the limit"),
> no O covers a fraction.

But whenever a finite index is issued to a fraction all O's are occupied by fractions. Therefore the clearing of the O-columns is not possible by attaching finite indexes to the remaining fractions.

> We can learn about what happens or does not happen
> when each index has been applied ("in the limit")
> without "arriving at the limit" even in principle.

Whenever a finite index is issued to a fraction all O's are occupied by fractions
>
> Step one. Describe an index.
> What is an index?
>
1, 2, 3, ..., in short a number enumerating a step where the whole matrix is filled.

> >>> They are not indexed.
> >>> They are dark.
> No.

Whenever a finite index is issued to a fraction all O's are occupied by fractions. If all are indexed in the limit, then this cannot be seen.

> Seen or unseen, those are what we're working with.

Unseen are dark.

Regards, WM

On Monday, 31 January 2022 at 15:08:51 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Montag, 31. Januar 2022 um 13:56:21 UTC+1:
> > On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
> > > horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
> > > > On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
> > >
> > > > > Determine an instance where one of the O's gets empty. Impossible.. All fractions leaving O's move simultaneously.
> > > > Again, so?? All you show continually is that you have no clue how infinity works.
> > > Infinity does not work.
> > You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it.
> Not at all. I accept all infinitely many terms of the sequence of transpositions 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, ...
> ...
> namely
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 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, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 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, ...
> ..., and so on.
>
> For *all* *infinitely* many of them I have proved that no matrix position gets empty. How can you accuse me of finitism?!

Stings, does it?

But you are entirely missing the point, as usual. Infinite processes have limits, and as we both know, you do not know how to handle limits.

you do not even know what "all infinitely many transpositions" means.

What about the transpositions

> 1/1, 1/3, 1/4, 1/5, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 2/1, 3/2, 3/3, 3/4, ...
> 3/1, 4/2, 4/3, 4/4, ...
> ...

> 1/1, 1/4, 1/5, 1/6, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 2/1, 3/2, 3/3, 3/4, ...
> 1/3, 4/2, 4/3, 4/4, ...
> ...

etc?

> > IF there are only finitely many integers, then the cardinality of N is finite.
> If there are more than all finite terms, then we get
>
> 1/1, O , O, O , ...
> 1/2, O , O, O , ...
> 2/1, O , O, O , ...
> 1/3, O , O, O , ...
> 2/2, O , O, O , ...
> ...,
>
> I agree. I would only mention that the exit of the fractions happens somewhat panic-stricken.

Not true. At every step you can determine the position of every fraction, if you are so inclined. If you assume that every transposition takes the same amount of time as every other, you will see an orderly process, but you are going to have to spend a loooong time to observe the terminal position. You can, of course, speed up the process by compressing time nonlinearly, in which case you may arrive at the terminal position more quickly, but any "panic" is inferred by the observer. It is not part of the process itself.

horand....@gmail.com schrieb am Dienstag, 1. Februar 2022 um 13:20:47 UTC+1:
> On Monday, 31 January 2022 at 15:08:51 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Montag, 31. Januar 2022 um 13:56:21 UTC+1:
> > > On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
> > > > horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
> > > > > On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
> > > >
> > > > > > Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
> > > > > Again, so?? All you show continually is that you have no clue how infinity works.
> > > > Infinity does not work.
> > > You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it.
> > Not at all. I accept all infinitely many terms of the sequence of transpositions 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, ...
> > ...
> > namely
> > 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> > 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 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, ...
> > 4/1, 4/2, 4/3, 4/4, ... 4/1, 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, ...
> > ..., and so on.
> >
> > For *all* *infinitely* many of them I have proved that no matrix position gets empty. How can you accuse me of finitism?!

> Infinite processes have limits, and as we both know, you do not know how to handle limits.

I prove that the limit in the present case is not produced by definable indexes.
>
> What about the transpositions
>
> > 1/1, 1/3, 1/4, 1/5, ...
> > 1/2, 2/2, 2/3, 2/4, ...
> > 2/1, 3/2, 3/3, 3/4, ...
> > 3/1, 4/2, 4/3, 4/4, ...
> > ...
>
> > 1/1, 1/4, 1/5, 1/6, ...
> > 1/2, 2/2, 2/3, 2/4, ...
> > 2/1, 3/2, 3/3, 3/4, ...
> > 1/3, 4/2, 4/3, 4/4, ...
> > ...
>
> etc?

Every set of transpositions can be used. The matrix will remain filled. But I have in particular Cantor's sequence in mind in order to disarm the critcs.

> > I would only mention that the exit of the fractions happens somewhat panic-stricken.
> Not true.

True since all definable exists leave the matrix filled.

> At every step you can determine the position of every fraction, if you are so inclined.

At every step where this is possible, the matrix is completely fillded.

> If you assume that every transposition takes the same amount of time as every other, you will see an orderly process, but you are going to have to spend a loooong time to observe the terminal position.

I am not interested in the details because the only fact that counts is this: all definable exits leave the matrix completely filled.

> You can, of course, speed up the process by compressing time nonlinearly, in which case you may arrive at the terminal position more quickly, but any "panic" is inferred by the observer. It is not part of the process itself.

It describes a process where infinitely many elements exit whitout control. That is TND because every controlled exit leaves the matrix filled.

Regards, WM

On Tuesday, 1 February 2022 at 09:09:07 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 1. Februar 2022 um 13:20:47 UTC+1:
> > On Monday, 31 January 2022 at 15:08:51 UTC-4, WM wrote:
> > > horand....@gmail.com schrieb am Montag, 31. Januar 2022 um 13:56:21 UTC+1:
> > > > On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
> > > > > horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
> > > > > > On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
> > > > >
> > > > > > > Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
> > > > > > Again, so?? All you show continually is that you have no clue how infinity works.
> > > > > Infinity does not work.
> > > > You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it.
> > > Not at all. I accept all infinitely many terms of the sequence of transpositions 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, ...
> > > ...
> > > namely
> > > 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> > > 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 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, ...
> > > 4/1, 4/2, 4/3, 4/4, ... 4/1, 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, ...
> > > ..., and so on.
> > >
> > > For *all* *infinitely* many of them I have proved that no matrix position gets empty. How can you accuse me of finitism?!
> > Infinite processes have limits, and as we both know, you do not know how to handle limits.
> I prove that the limit in the present case is not produced by definable indexes.

You prove yet again that you have no idea about limits, nor mathematical proofs.

> > What about the transpositions
> >
> > > 1/1, 1/3, 1/4, 1/5, ...
> > > 1/2, 2/2, 2/3, 2/4, ...
> > > 2/1, 3/2, 3/3, 3/4, ...
> > > 3/1, 4/2, 4/3, 4/4, ...
> > > ...
> >
> > > 1/1, 1/4, 1/5, 1/6, ...
> > > 1/2, 2/2, 2/3, 2/4, ...
> > > 2/1, 3/2, 3/3, 3/4, ...
> > > 1/3, 4/2, 4/3, 4/4, ...
> > > ...
> >
> > etc?
> Every set of transpositions can be used. The matrix will remain filled. But I have in particular Cantor's sequence in mind

So do I. You seem too dense to realize that.

> > > I would only mention that the exit of the fractions happens somewhat panic-stricken.
> > Not true.
> True since all definable exists leave the matrix filled.
> > At every step you can determine the position of every fraction, if you are so inclined.
> At every step where this is possible, the matrix is completely fillded.
> > If you assume that every transposition takes the same amount of time as every other, you will see an orderly process, but you are going to have to spend a loooong time to observe the terminal position.
> I am not interested in the details because the only fact that counts is this: all definable exits leave the matrix completely filled.

Ah, yes! When all else fails, switch quantifiers. That seems to be the only tool in your proof kit.

> > You can, of course, speed up the process by compressing time nonlinearly, in which case you may arrive at the terminal position more quickly, but any "panic" is inferred by the observer. It is not part of the process itself.
> It describes a process where infinitely many elements exit whitout control. That is TND

What does Thursday Night Dinner have to do with anything?

On 2/1/2022 7:12 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 1. Februar 2022 um 01:50:31 UTC+1:
>> On 1/31/2022 3:31 AM, WM wrote:

>> I can translate "in the limit" here to
>> "when each index has been applied".
>>
>> Yes,
>> when each index has been applied ("in the limit"),
>> no O covers a fraction.
>
> But whenever a finite index is issued to a fraction

.... less than all indexes have been issued to fractions.

> But whenever a finite index is issued to a fraction
> all O's are occupied by fractions.
> Therefore the clearing of the O-columns is not possible
> by attaching

.... less than all indexes to fractions.

> But whenever a finite index is issued to a fraction
> all O's are occupied by fractions.
> Therefore the clearing of the O-columns is not possible
> by attaching finite indexes to the remaining fractions.

Each ⟨1,...,k⟩ has a _counting-order_ which begins at 1
and ends somewhere. Where it ends is an _index_

( A counting-order is a stepping-order in which
( each step is a count.

Some collections have a two-ended _stepping-order_

( A stepping-order is a total-order in which
( each cut is a step.

Each collection which has a two-ended stepping-order
matches one and only one of the ⟨1,...,k⟩
(Each ⟨1,...,k⟩ ends somewhere.)

Not all collections have a two-ended stepping-order.

For example, even though
_each_ ⟨1,...,k⟩ has a two-ended stepping-order,
the collection {⟨1,...,k⟩} of _all_ ⟨1,...,k⟩
does not have a two-ended stepping-order.

( Because...
( The successor k+1 of the second end of ⟨1,...,k⟩
( cannot be in ⟨1,...,k⟩
( But ⟨1,...,k+1⟩ also has
( a two-ended counting-order beginning at 1.
( And ⟨1,...,k+1⟩ is after ⟨1,...,k⟩
( ⟨1,...,k⟩ does not end the collection {⟨1,...,k⟩}
( Generalizing, no ⟨1,...,k⟩ ends {⟨1,...,k⟩}

For each end k, there is exactly one ⟨1,...,k⟩
which it ends.

The collection ⋃{⟨1,...,k⟩} of the ends of ⟨1,...,k⟩
can be assigned the same order as {⟨1,...,k⟩}
| j =< k iff ⟨1,...,j⟩ ⊆ ⟨1,...,k⟩

ℕ⁺ = ⋃{⟨1,...,k⟩}

ℕ⁺ also does not have a two-ended stepping-order.
This is in contrast to its subset _up to_ any k ∈ ℕ⁺
⟨1,...,k⟩ has a two-ended stepping-order.

{ Both are true because...
( Only ⟨1,...,k⟩ with two-ended stepping orders
( are included in {⟨1,...,k⟩}
( k+1 cannot be in ⟨1,...,k⟩

----
A property which
collections with two-ended stepping orders have
and which
collections without two-ended stepping-orders don't have
is,
if two collections match in one total order,
then they match in every total order.

Therefore,
swaps among the elements of ℕ⁺ _up to_ any k ∈ ℕ⁺
are within a two-ended-stepping-order-able ⟨1,...,k⟩
and cannot lose any O's or leave any O's fraction-less.
But
swaps among _all_ the elements of ℕ⁺ do not have
that property.

>> We can learn about what happens or does not happen
>> when each index has been applied ("in the limit")
>> without "arriving at the limit" even in principle.
>
> Whenever a finite index is issued to a fraction

....less than all indexes have been issued.

> Whenever a finite index is issued to a fraction
> all O's are occupied by fractions
>
>> Step one. Describe an index.
>> What is an index?
>
> 1, 2, 3, ..., in short a number enumerating a step
> where the whole matrix is filled.

You're being much too vague.

Each ⟨1,...,k⟩ has a counting-order which begins at 1
and ends somewhere. Where it ends is an _index_

A counting-order is a stepping-order in which
each step is a count.

A stepping-order is a total-order in which
each cut is a step.

>>>>> They are not indexed.
>>>>> They are dark.
>>
>> No.
>
> Whenever a finite index is issued to a fraction

....less than all indexes have been issued.

> Whenever a finite index is issued to a fraction
> all O's are occupied by fractions.
> If all are indexed in the limit,
> then this cannot be seen.

>> I can translate "in the limit" here to
>> "when each index has been applied".

All indexes have been issued when all indexes have
been issued.

This news must come as a shock to you.
Take deep breaths.
You can get through this.

>> Seen or unseen, those are what we're working with.
>
> Unseen are dark.

The unseen are reasoned about the same way that
the seen are reasoned about. Describe, and take
only reliable statement-steps from the description.

I propose that this ability to reason about the unseen
is Eugene Wigner's "unreasonable effectiveness of
mathematics in the natural sciences".

tisdag 1 februari 2022 kl. 13:06:45 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 1. Februar 2022 um 07:06:22 UTC+1:
> > måndag 31 januari 2022 kl. 19:34:25 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 12:41:04 UTC+1:
> > > > måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:
> > >
> > > > > > That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
> > > > > > >For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
> > > > >
> > > > > > It exists for ALL
> > > > > Indexes exist only as long as infinitely many fractions remain not indexed.
> > > > You are still arguing "after finite steps"
> > > After finite steps = definable transpositions the matrix has provably no empty position
> > > 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, ...
> > > ...
> > >
> > > After all indices have been applied, the matrix, except the first column, is empty.
> > >
> > > This shows undefinable steps.
> > >
> > You cannot argue with finite steps.
> I can argue that whenever a finite index is issued to a fraction all positions of the matrix are occupied by fractions. Therefore the clearing of the matrix is not possible by attaching finite indexes to the remaining fractions.
>
> Regards, WM

This just says you cannot do it in finite steps which is wholy irrelevant!

horand....@gmail.com schrieb am Dienstag, 1. Februar 2022 um 17:11:32 UTC+1:
> On Tuesday, 1 February 2022 at 09:09:07 UTC-4, WM wrote:

> > I am not interested in the details because the only fact that counts is this: all definable exits leave the matrix completely filled.
> Ah, yes! When all else fails, switch quantifiers.

Why should I?

Regards, WM

Jim Burns schrieb am Dienstag, 1. Februar 2022 um 17:44:31 UTC+1:
> On 2/1/2022 7:12 AM, WM wrote:

> swaps among the elements of ℕ⁺ _up to_ any k ∈ ℕ⁺
> are within a two-ended-stepping-order-able ⟨1,...,k⟩
> and cannot lose any O's or leave any O's fraction-less.

That are the definable swaps.

> But
> swaps among _all_ the elements of ℕ⁺ do not have
> that property.

That are the undefinable or dark swaps. Would you prefer another name? "All-swaps" for instance.

> > Whenever a finite index is issued to a fraction
> ...less than all indexes have been issued.

Yes. Whenever a definable index has been issued, most indexes have not yet been issued.
They are called "all indexes".

> >> I can translate "in the limit" here to
> >> "when each index has been applied".
> All indexes have been issued when all indexes have
> been issued.

That is when also all all-indexes have been issued.
>
> This news must come as a shock to you.
> Take deep breaths.
> You can get through this.

Not by definable indexes. You just denied it few lines above: "Whenever a finite index is issued to a fraction ... less than all indexes have been issued." Short memory?

> > Unseen are dark.
> The unseen are reasoned about the same way that
> the seen are reasoned about. Describe, and take
> only reliable statement-steps from the description.

The unseen numbers are numbers. But they cannot be taken individually. Otherwise we would take all of them such that none remained. This is only possible collectively

|ℕ \ {1, 2, 3, ...}| = 0

not individually:

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> I propose that this ability to reason about the unseen

So you have understood that unseen numbers must be existing?
> is Eugene Wigner's "unreasonable effectiveness of
> mathematics in the natural sciences".

I don't think that mathematical effectiveness is unreasonable. Mathematics was constructed for this purpose by abstracting from nature.

Regards, WM

On Wednesday, 2 February 2022 at 05:06:15 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 1. Februar 2022 um 17:11:32 UTC+1:
> > On Tuesday, 1 February 2022 at 09:09:07 UTC-4, WM wrote:
>
> > > I am not interested in the details because the only fact that counts is this: all definable exits leave the matrix completely filled.
> > Ah, yes! When all else fails, switch quantifiers.
> Why should I?

You did! Whether your refusal to acknowledge this is malice or incompetence is irrelevant. I will go as far as saying that whenever you use the words "all" or "every", you have already switched quantifiers in your heart.

On 2/2/2022 3:22 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 1. Februar 2022 um 17:44:31 UTC+1:
>> On 2/1/2022 7:12 AM, WM wrote:
>
>> swaps among the elements of ℕ⁺ _up to_ any k ∈ ℕ⁺
>> are within a two-ended-stepping-order-able ⟨1,...,k⟩
>> and cannot lose any O's or leave any O's fraction-less.
>
> That are the definable swaps.

so you use the term "definable" when you want to obfuscate.

On 2/1/2022 6:06 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 1. Februar 2022 um 07:06:22 UTC+1:
>> måndag 31 januari 2022 kl. 19:34:25 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 12:41:04 UTC+1:
>>>> måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:
>>>
>>>>>> That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
>>>>>>> For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
>>>>>
>>>>>> It exists for ALL
>>>>> Indexes exist only as long as infinitely many fractions remain not indexed.
>>>> You are still arguing "after finite steps"
>>> After finite steps = definable transpositions the matrix has provably no empty position
>>> 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, ...
>>> ...
>>>
>>> After all indices have been applied, the matrix, except the first column, is empty.
>>>
>>> This shows undefinable steps.
>>>
>> You cannot argue with finite steps.
>
> I can argue

you can argue, but you are ALWAYS wrong, ALWAYS.

On 2/1/2022 7:08 AM, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 1. Februar 2022 um 13:20:47 UTC+1:
>> On Monday, 31 January 2022 at 15:08:51 UTC-4, WM wrote:
>>> horand....@gmail.com schrieb am Montag, 31. Januar 2022 um 13:56:21 UTC+1:
>>>> On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
>>>>> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
>>>>>> On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
>>>>>
>>>>>>> Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
>>>>>> Again, so?? All you show continually is that you have no clue how infinity works.
>>>>> Infinity does not work.
>>>> You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it.
>>> Not at all. I accept all infinitely many terms of the sequence of transpositions 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, ...
>>> ...
>>> namely
>>> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
>>> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 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, ...
>>> 4/1, 4/2, 4/3, 4/4, ... 4/1, 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, ...
>>> ..., and so on.
>>>
>>> For *all* *infinitely* many of them I have proved that no matrix position gets empty. How can you accuse me of finitism?!
>
>> Infinite processes have limits, and as we both know, you do not know how to handle limits.
>
> I prove that the limit in the present case is not produced by definable indexes.

liar. You do not know how to prove anything.

"definable" is a word you use to obscure your fakery

the rest of your post is not math at all.

You have failed to disprove Cantor, you are flailing at it.

On 2/2/2022 4:22 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 1. Februar 2022 um 17:44:31 UTC+1:

>> swaps among the elements of ℕ⁺ _up to_ any k ∈ ℕ⁺
>> are within a two-ended-stepping-order-able ⟨1,...,k⟩
>> and cannot lose any O's or leave any O's fraction-less.
>
> That are the definable swaps.
>
>> But
>> swaps among _all_ the elements of ℕ⁺ do not have
>> that property.
>
> That are the undefinable or dark swaps.

Your "definable" swaps and your "undefinable" swaps
are the same swaps, but in different collections,
two-ended-stepping-order-able collections and
un-two-ended-stepping-order-able collections.

A property which
collections with two-ended stepping orders have
and which
collections without two-ended stepping-orders don't have
is,
if two collections match in one total order,
then they match in every total order.

> Would you prefer another name?
> "All-swaps" for instance.

Your dark swaps are the
un-two-ended-stepping-order-ably-collected swaps.

We might need a shorter name for them, though.
"TESOAC" and "unTESOAC"? It looks like Gaelic.

The thing about using "all-swaps" is that the first column
ℕ⁺⨯{1} is not all of the matrix ℕ⁺⨯ℕ⁺
but it does have "dark" fractions.
On the other hand, the first column is unTESOAC.
I think "TESOAC" and "unTESOAC" capture the distinction
you are talking about.

The same indexes that are TESOAC in ⟨1,...,k⟩ are
unTESOAC in ℕ⁺ = ⋃{⟨1,...,k⟩}
What changes is the collection.

If someone were to refuse to believe in sets,
this would be difficult distinction to grasp.

----
For some collections, its members cannot be matched with
fewer than all its members.
These are the two-ended-stepping-order-able collections.
Its members are two-ended-stepping-order-ably-collected.
TESOAC.

For some collections, its members CAN be matched with
fewer than all its members.
These are the UN-two-ended-stepping-order-able collections.
Its members are un-two-ended-stepping-order-ably-collected.
unTESOAC.

It takes a proof to show that the matchability is paired
with the kind of order. This matchability seems to sit at
the core of your posts, but you always attribute its
presence or absence to the members, not to the collection.

>>> Whenever a finite index is issued to a fraction
>>
>> ...less than all indexes have been issued.
>
> Yes. Whenever a definable index has been issued,
> most indexes have not yet been issued.
> They are called "all indexes".

For each index k,
k is TESOAC in ⟨1,...,k⟩
and k is unTESOAC in ℕ⁺ = ⋃{⟨1,...,k⟩}

On 2/2/2022 12:21 PM, Jim Burns wrote:
> On 2/2/2022 4:22 AM, WM wrote:
>> Jim Burns schrieb
>> am Dienstag, 1. Februar 2022 um 17:44:31 UTC+1:

>>> swaps among the elements of ℕ⁺ _up to_ any k ∈ ℕ⁺
>>> are within a two-ended-stepping-order-able ⟨1,...,k⟩
>>> and cannot lose any O's or leave any O's fraction-less.
>>
>> That are the definable swaps.
>>
>>> But
>>> swaps among _all_ the elements of ℕ⁺ do not have
>>> that property.
>>
>> That are the undefinable or dark swaps.
>
> Your "definable" swaps and your "undefinable" swaps
> are the same swaps, but in different collections,
> two-ended-stepping-order-able collections and
> un-two-ended-stepping-order-able collections.
>
> A property which
> collections with two-ended stepping orders have
> and which
> collections without two-ended stepping-orders don't have
> is,
> if two collections match in one total order,
> then they match in every total order.
>
>> Would you prefer another name?
>> "All-swaps" for instance.
>
> Your dark swaps are the
> un-two-ended-stepping-order-ably-collected swaps.
>
> We might need a shorter name for them, though.
> "TESOAC" and "unTESOAC"? It looks like Gaelic.
>
> The thing about using "all-swaps" is that the first column
> ℕ⁺⨯{1} is not all of the matrix ℕ⁺⨯ℕ⁺
> but it does have "dark" fractions.
> On the other hand, the first column is unTESOAC.
> I think "TESOAC" and "unTESOAC" capture the distinction
> you are talking about.
>
> The same indexes that are TESOAC in ⟨1,...,k⟩ are
> unTESOAC in ℕ⁺ = ⋃{⟨1,...,k⟩}
> What changes is the collection.
>
> If someone were to refuse to believe in sets,
> this would be difficult distinction to grasp.
>
> ----
> For some collections, its members cannot be matched with
> fewer than all its members.
> These are the two-ended-stepping-order-able collections.
> Its members are two-ended-stepping-order-ably-collected.
> TESOAC.
>
> For some collections, its members CAN be matched with
> fewer than all its members.
> These are the UN-two-ended-stepping-order-able collections.
> Its members are un-two-ended-stepping-order-ably-collected.
> unTESOAC.
>
> It takes a proof to show that the matchability is paired
> with the kind of order. This matchability seems to sit at
> the core of your posts, but you always attribute its
> presence or absence to the members, not to the collection.
>
>>>> Whenever a finite index is issued to a fraction
>>>
>>> ...less than all indexes have been issued.
>>
>> Yes. Whenever a definable index has been issued,
>> most indexes have not yet been issued.
>> They are called "all indexes".
>
> For each index k,
> k is TESOAC in ⟨1,...,k⟩
> and k is unTESOAC in ℕ⁺ = ⋃{⟨1,...,k⟩}

For each index j,
j is TESOAC in each ⟨1,...,k⟩ which it is in
and j is unTESOAC in ℕ⁺ = ⋃{⟨1,...,k⟩}

onsdag 2 februari 2022 kl. 10:22:05 UTC+1 skrev WM:
> Jim Burns schrieb am Dienstag, 1. Februar 2022 um 17:44:31 UTC+1:
> > On 2/1/2022 7:12 AM, WM wrote:
>
> > swaps among the elements of ℕ⁺ _up to_ any k ∈ ℕ⁺
> > are within a two-ended-stepping-order-able ⟨1,...,k⟩
> > and cannot lose any O's or leave any O's fraction-less.
> That are the definable swaps.
> > But
> > swaps among _all_ the elements of ℕ⁺ do not have
> > that property.
> That are the undefinable or dark swaps. Would you prefer another name? "All-swaps" for instance.
> > > Whenever a finite index is issued to a fraction
> > ...less than all indexes have been issued.
> Yes. Whenever a definable index has been issued, most indexes have not yet been issued.
> They are called "all indexes".
> > >> I can translate "in the limit" here to
> > >> "when each index has been applied".
> > All indexes have been issued when all indexes have
> > been issued.
> That is when also all all-indexes have been issued.
> >
> > This news must come as a shock to you.
> > Take deep breaths.
> > You can get through this.
> Not by definable indexes. You just denied it few lines above: "Whenever a finite index is issued to a fraction ... less than all indexes have been issued." Short memory?
> > > Unseen are dark.
> > The unseen are reasoned about the same way that
> > the seen are reasoned about. Describe, and take
> > only reliable statement-steps from the description.
> The unseen numbers are numbers. But they cannot be taken individually. Otherwise we would take all of them such that none remained. This is only possible collectively
>
> |ℕ \ {1, 2, 3, ...}| = 0
>
> not individually:
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
> >
> > I propose that this ability to reason about the unseen
> So you have understood that unseen numbers must be existing?
> > is Eugene Wigner's "unreasonable effectiveness of
> > mathematics in the natural sciences".
> I don't think that mathematical effectiveness is unreasonable. Mathematics was constructed for this purpose by abstracting from nature.
>
> Regards, WM
you sitll do not understand that you cannto go from finite to infinite

Jim Burns schrieb am Mittwoch, 2. Februar 2022 um 18:21:36 UTC+1:
> On 2/2/2022 4:22 AM, WM wrote:

> > Would you prefer another name?
> > "All-swaps" for instance.
> Your dark swaps are the
> un-two-ended-stepping-order-ably-collected swaps.
>
> We might need a shorter name for them, though.
> "TESOAC" and "unTESOAC"? It looks like Gaelic.

If you like.
>
> The thing about using "all-swaps" is that the first column
> ℕ⁺⨯{1} is not all of the matrix ℕ⁺⨯ℕ⁺
> but it does have "dark" fractions.
> On the other hand, the first column is unTESOAC.
> I think "TESOAC" and "unTESOAC" capture the distinction
> you are talking about.

Okay, you know that there are different swaps.
But meanwhile I have an idea how my swapping could be remedied. There is no control that the matrix remains completely occupied. You need only TESOAC except for the whole process.

Clear the first column of the well-known matrix

__, 1/2, 1/3, 1/4, ...
__, 2/2, 2/3, 2/4, ...
__, 3/2, 3/3, 3/4, ...
__, 4/2, 4/3, 4/4, ...
....

Deposit the remaining fractions there according to Cantor's prescription.

___, ___, ___, 1/4, ...
1/2, ___, 2/3, 2/4, ...
___, 3/2, 3/3, 3/4, ...
1/3, 4/2, 4/3, 4/4, ...
2/2, 5/2, 5/3, 5/4, ...
___ , 6/2, 6/3, 6/4, ...
....

There are no returns since the first column is empty. But if desired the integer fractions could be deposited at the free positions.

Do you believe that the matrix can be cleared in this way?
>
> For some collections, its members CAN be matched with
> fewer than all its members.

This would be the above procedure.

> These are the UN-two-ended-stepping-order-able collections.
> Its members are un-two-ended-stepping-order-ably-collected.
> unTESOAC.
>
> It takes a proof to show that the matchability is paired
> with the kind of order. This matchability seems to sit at
> the core of your posts, but you always attribute its
> presence or absence to the members, not to the collection.

The members are what is moved and indexed.

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 2. Februar 2022 um 12:37:21 UTC+1:
> On Wednesday, 2 February 2022 at 05:06:15 UTC-4, WM wrote:

> > > Ah, yes! When all else fails, switch quantifiers.
> > Why should I?
> You did! Whether your refusal to acknowledge this is malice or incompetence is irrelevant.

What about this version? Do I switch quantifiers there too?

Clear the first column of the well-known matrix

__, 1/2, 1/3, 1/4, ...
__, 2/2, 2/3, 2/4, ...
__, 3/2, 3/3, 3/4, ...
__, 4/2, 4/3, 4/4, ...
....

Deposit the remaining fractions there according to Cantor's prescription.

___, ___, ___, 1/4, ...
1/2, ___, 2/3, 2/4, ...
___, 3/2, 3/3, 3/4, ...
1/3, 4/2, 4/3, 4/4, ...
2/2, 5/2, 5/3, 5/4, ...
___ , 6/2, 6/3, 6/4, ...
....

There are no returns since the first column is empty. But if desired the integer fractions could be deposited at the free positions.

Is that possible? Is it quantifier switching?

Regards, WM

On Thursday, 3 February 2022 at 06:41:32 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 2. Februar 2022 um 12:37:21 UTC+1:
> > On Wednesday, 2 February 2022 at 05:06:15 UTC-4, WM wrote:
>
> > > > Ah, yes! When all else fails, switch quantifiers.
> > > Why should I?
> > You did! Whether your refusal to acknowledge this is malice or incompetence is irrelevant.
> What about this version? Do I switch quantifiers there too?
> Clear the first column of the well-known matrix
>
> __, 1/2, 1/3, 1/4, ...
> __, 2/2, 2/3, 2/4, ...
> __, 3/2, 3/3, 3/4, ...
> __, 4/2, 4/3, 4/4, ...
> ...
> Deposit the remaining fractions there according to Cantor's prescription.
>
> ___, ___, ___, 1/4, ...
> 1/2, ___, 2/3, 2/4, ...
> ___, 3/2, 3/3, 3/4, ...
> 1/3, 4/2, 4/3, 4/4, ...
> 2/2, 5/2, 5/3, 5/4, ...
> ___ , 6/2, 6/3, 6/4, ...
> ...
>
> There are no returns since the first column is empty. But if desired the integer fractions could be deposited at the free positions.
> Is that possible? Is it quantifier switching?

This is not quantifier switching. Yet. But you could simplify your approach by using two extra columns:

1, ___, 1/2, 1/3, 1/4, ...
2, ___, 2/1, 2/2, 2/3, 2/4, ...
3, ___, 3/1, 3/2, 3/3, 3/4, ...
4, ___, 4/1, 4/2, 4/3, 4/4, ...
5, ___, 5/1, 5/2, 5/3, 5/4, ...
6, ___, 6/1, 6/2, 6/3, 6/4, ...
....

and then shovel the fractions into the empty second column:

1, 1/1, ___, ___, ___, 1/4, ...
2, 1/2, ___, ___, 2/3, 2/4, ...
3, 2/1, ___, 3/2, 3/3, 3/4, ...
4, 1/3, 4/1, 4/2, 4/3, 4/4, ...
5, 2/2, 5/1, 5/2, 5/3, 5/4, ...
6, 3/1, 6/1, 6/2, 6/3, 6/4, ...
7, ___, 7/1, 7/2, 7/3, 7/4, ...
....

Now what? At every finite step there are aleph_0 (yes, aleph_0!) fractions remaining in the right part of the matrix (from column 3 onward). This cardinality tells you absolutely nothing about the limit of this process. We had that multiple times before, and your stupidity in this matter is astounding. You are brain-dead and have evidently been so for many years.

And just for the record, in the limit the right part of the matrix will be cleared, and the fractions neatly fill the second column. *EVERY* fraction has a place in the second column that can be computed using the formula you have been given. That you are too stupid to understand this formula is not my fault. I blame your defective wiring, and possibly your math teachers along the way.

torsdag 3 februari 2022 kl. 11:37:59 UTC+1 skrev WM:
> Jim Burns schrieb am Mittwoch, 2. Februar 2022 um 18:21:36 UTC+1:
> > On 2/2/2022 4:22 AM, WM wrote:
>
> > > Would you prefer another name?
> > > "All-swaps" for instance.
> > Your dark swaps are the
> > un-two-ended-stepping-order-ably-collected swaps.
> >
> > We might need a shorter name for them, though.
> > "TESOAC" and "unTESOAC"? It looks like Gaelic.
> If you like.
> >
> > The thing about using "all-swaps" is that the first column
> > ℕ⁺⨯{1} is not all of the matrix ℕ⁺⨯ℕ⁺
> > but it does have "dark" fractions.
> > On the other hand, the first column is unTESOAC.
> > I think "TESOAC" and "unTESOAC" capture the distinction
> > you are talking about.
> Okay, you know that there are different swaps.
> But meanwhile I have an idea how my swapping could be remedied. There is no control that the matrix remains completely occupied. You need only TESOAC except for the whole process.
>
> Clear the first column of the well-known matrix
>
> __, 1/2, 1/3, 1/4, ...
> __, 2/2, 2/3, 2/4, ...
> __, 3/2, 3/3, 3/4, ...
> __, 4/2, 4/3, 4/4, ...
> ...
>
> Deposit the remaining fractions there according to Cantor's prescription.
>
> ___, ___, ___, 1/4, ...
> 1/2, ___, 2/3, 2/4, ...
> ___, 3/2, 3/3, 3/4, ...
> 1/3, 4/2, 4/3, 4/4, ...
> 2/2, 5/2, 5/3, 5/4, ...
> ___ , 6/2, 6/3, 6/4, ...
> ...
>
> There are no returns since the first column is empty. But if desired the integer fractions could be deposited at the free positions.
>
> Do you believe that the matrix can be cleared in this way?
> >
> > For some collections, its members CAN be matched with
> > fewer than all its members.
> This would be the above procedure.
> > These are the UN-two-ended-stepping-order-able collections.
> > Its members are un-two-ended-stepping-order-ably-collected.
> > unTESOAC.
> >
> > It takes a proof to show that the matchability is paired
> > with the kind of order. This matchability seems to sit at
> > the core of your posts, but you always attribute its
> > presence or absence to the members, not to the collection.
> The members are what is moved and indexed.
>
> Regards, WM
you finitely rearranging shit does nothing to it, just quit it already

On 2/3/2022 5:37 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 2. Februar 2022 um 18:21:36 UTC+1:
>> On 2/2/2022 4:22 AM, WM wrote:

>> These are the UN-two-ended-stepping-order-able collections.
>> Its members are un-two-ended-stepping-order-ably-collected.
>> unTESOAC.
>>
>> It takes a proof to show that the matchability is paired
>> with the kind of order. This matchability seems to sit at
>> the core of your posts, but you always attribute its
>> presence or absence to the members, not to the collection.
>
> The members are what is moved and indexed.

The collections of members are what are TESOA or unTESOA.
The collections conserve Bob or do not conserve Bob.
Bob himself, meaning no disrespect, is not special.

Consider a game of chess.
White has two bishops. So does Black.
One bishop can only land on white squares
the other can only land on black squares.
Q. What makes these bishops so different?

A. Nothing. They aren't different.
In the next game, their roles could be reversed.
it's the square each stands on that is different.

If Bob is collected in a TESOA collection,
then Bob is TESOAC.
If Bob is collected in an unTESOA collection,
then Bob is unTESOAC.
But Bob is Bob.

----
>>> Would you prefer another name?
>>> "All-swaps" for instance.
>>
>> Your dark swaps are the
>> un-two-ended-stepping-order-ably-collected swaps.
>>
>> We might need a shorter name for them, though.
>> "TESOAC" and "unTESOAC"? It looks like Gaelic.
>
> If you like.
>
>> The thing about using "all-swaps" is that the first column
>> ℕ⁺⨯{1} is not all of the matrix ℕ⁺⨯ℕ⁺
>> but it does have "dark" fractions.
>> On the other hand, the first column is unTESOAC.
>> I think "TESOAC" and "unTESOAC" capture the distinction
>> you are talking about.
>
> Okay, you know that there are different swaps.

There are different collections of the same swaps,
TESOA collections and unTESOA collections.

Only TESOA collections have Bob-conservation.

> But meanwhile I have an idea how my swapping could be
> remedied. There is no control that the matrix remains
> completely occupied. You need only TESOAC
> except for the whole process.

.... except for the whole process.

So each TESOA part has Bob-conservation,
but the unTESOA whole does not conserve Bob.

As I'm sure you know,
there can't be unTESOA parts of a TESOA whole,
but TESOA parts of an unTESOA whole is no problem.

> Clear the first column of the well-known matrix
>
> __, 1/2, 1/3, 1/4, ...
> __, 2/2, 2/3, 2/4, ...
> __, 3/2, 3/3, 3/4, ...
> __, 4/2, 4/3, 4/4, ...
> ...
>
> Deposit the remaining fractions there
> according to Cantor's prescription.
>
> ___, ___, ___, 1/4, ...
> 1/2, ___, 2/3, 2/4, ...
> ___, 3/2, 3/3, 3/4, ...
> 1/3, 4/2, 4/3, 4/4, ...
> 2/2, 5/2, 5/3, 5/4, ...
> ___ , 6/2, 6/3, 6/4, ...
> ...
>
> There are no returns since the first column is empty.
> But if desired the integer fractions could be
> deposited at the free positions.
>
> Do you believe that the matrix can be cleared in
> this way?

I very likely can prove that the matrix can be cleared
in this way.

I say no more than "very likely" because I'm having a
little trouble finding my enthusiasm for doing this
all over again. You very likely have described a
situation with TESOA parts but an unTESOA whole in which
Bob is not conserved or something like that.

Yes. Okay.
So each TESOA part has Bob-conservation,
but the unTESOA whole does not conserve Bob.
Et cetera, ceteris paribus, ad infinitum.

>> For some collections, its members CAN be matched with
>> fewer than all its members.
>
> This would be the above procedure.
>
>> These are the UN-two-ended-stepping-order-able collections.
>> Its members are un-two-ended-stepping-order-ably-collected.
>> unTESOAC.
>>
>> It takes a proof to show that the matchability is paired
>> with the kind of order. This matchability seems to sit at
>> the core of your posts, but you always attribute its
>> presence or absence to the members, not to the collection.
>
> The members are what is moved and indexed.

The collections of members are what are TESOA or unTESOA.
The collections conserve Bob or do not conserve Bob.
Bob himself, meaning no disrespect, is not special.

Consider a game of chess.
White has two bishops. So does Black.
One bishop can only land on white squares
the other can only land on black squares.
Q. What makes these bishops so different?

A. Nothing. They aren't different.
In the next game, their roles could be reversed.
it's the square each stands on that is different.

If Bob is collected in a TESOA collection,
then Bob is TESOAC.
If Bob is collected in an unTESOA collection,
then Bob is unTESOAC.
But Bob is Bob.