sergio schrieb am Donnerstag, 20. Januar 2022 um 18:15:31 UTC+1:
> On 1/20/2022 9:12 AM, WM wrote:
> > zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:

> >> It proves nothing of the sort. You changing the bijection proves nothing about it.
> >
> > I did not change it. I used Cantor's enumeration
> > k = (m + n - 1)(m + n - 2)/2 + m
> Liar.

Where does my model deviate?

Regards, WM

WM wrote:

> sergio schrieb am Donnerstag, 20. Januar 2022 um 18:15:31 UTC+1:
>> On 1/20/2022 9:12 AM, WM wrote:
>> > zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31
>> > UTC+1:
>
>> >> It proves nothing of the sort. You changing the bijection proves
>> >> nothing about it.
>> >
>> > I did not change it. I used Cantor's enumeration k = (m + n - 1)(m +
>> > n - 2)/2 + m
>> Liar.
>
> Where does my model deviate?

model?? fucking stupid, you don't even know what a subdomain is.

On 1/20/2022 12:47 PM, WM wrote:
> sergio schrieb am Donnerstag, 20. Januar 2022 um 18:15:31 UTC+1:
>> On 1/20/2022 9:12 AM, WM wrote:
>>> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:
>
>>>> It proves nothing of the sort. You changing the bijection proves nothing about it.
>>>
>>> I did not change it. I used Cantor's enumeration
>>> k = (m + n - 1)(m + n - 2)/2 + m
>> Liar.
>
> Where does my model deviate?

you change the matrix as you go step by step, fail. Cantor does not do that.

>
> Regards, WM

On 1/20/2022 9:05 AM, FromTheRafters wrote:
> WM wrote on 1/20/2022 :
>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:28:18 UTC+1:
>>> WM presented the following explanation :
>>
>>>> Here is another picture of the same facts. Consider the infinite matrix
>>>> 1/1, _, _, _, ... 2/1, _, _, _, ... 3/1, _, _, _, ... 4/1, _, _, _, ... ...
>>>> and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.
>>> Indeed, so why do you insist upon doing it this wrong way. Start at the corner and serpentine. There is nothing to stop that 'process' so every
>>> element will 'eventually' be 'hit'.
>>
>> There is no drain for O's either. So they remain on the matrix.
>
> Poor maintenance on the drains is probably the reason for your leaky sets. Ensure the drains are properly seated before 'filling' your sets with elements.

you can plug leaks for a while with a mixture of aeather and dark numbers...

On Thursday, 20 January 2022 at 15:36:11 UTC-4, sergio wrote:
> On 1/20/2022 9:05 AM, FromTheRafters wrote:
> > WM wrote on 1/20/2022 :
> >> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:28:18 UTC+1:
> >>> WM presented the following explanation :
> >>
> >>>> Here is another picture of the same facts. Consider the infinite matrix
> >>>> 1/1, _, _, _, ... 2/1, _, _, _, ... 3/1, _, _, _, ... 4/1, _, _, _, ... ...
> >>>> and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.
> >>> Indeed, so why do you insist upon doing it this wrong way. Start at the corner and serpentine. There is nothing to stop that 'process' so every
> >>> element will 'eventually' be 'hit'.
> >>
> >> There is no drain for O's either. So they remain on the matrix.
> >
> > Poor maintenance on the drains is probably the reason for your leaky sets. Ensure the drains are properly seated before 'filling' your sets with elements.
> you can plug leaks for a while with a mixture of aeather and dark numbers...

The right consistency of shit should do the trick also. WM is an expert at this.

On Thursday, January 20, 2022 at 6:24:32 PM UTC+1, WM wrote:
> JVR schrieb am Donnerstag, 20. Januar 2022 um 16:41:52 UTC+1:
> > On Thursday, January 20, 2022 at 4:02:30 PM UTC+1, WM wrote:
> > > FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
> > > > WM presented the following explanation :
> > > > > Every actually infinite set either will become completed or not.
> > > > Sets don't "become" anything.
> > > Counting is a process which, according to matheology, can become completed. We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß, Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291, 293, 305f, 413]
> > >
> > Finally - that's your problem - you don't know what 'Zuordnung' means.
> > Wie can fix that easily enough:
> >
> > Literally it means 'assignment'.
> > In mathematics it means 'function' or 'relation' or 'mapping'.
> > It doesn't mean eenie-meenie-mynie-mo.
> Either you are stupid or you are a liar, but probably you are both. When Cantor said "element by element", then this means precisely one by one and one after the other. Of course the frauds of matheology have tried to pervert this simple Zuordnung into an "all at once" in order to prevent inspection because they must have known for a long time already that there is no complete mapping between infinite sets - unless all are as stupid as you.
>
> Regards, WM

I think we are in luck - it is, without any doubt, true that not everybody is as stupid as I am. And, obviously,
very few people are dumber than you.
Cantor writes formally when he formulates definitions, theorems and proofs. He also writes informally,
discursively, appealing to the reader's intuition, using heuristics, when he discusses and describes the consequences of the
formal theory. He does both extremely well.
Your innumerable quotes, from Cantor and many other people, almost always come from private correspondence,
pop science presentations, introductions to books. They almost never come from formal mathematical writings, which
you are obviously not able to understand.
Cantor never - und I mean literally never - uses an argument of the kind that leads to many of your errors,
namely "eenie, meenie, minee, mo until we reach the end ... ".

On 1/20/2022 1:44 PM, JVR wrote:
> On Thursday, January 20, 2022 at 6:24:32 PM UTC+1, WM wrote:
>> JVR schrieb am Donnerstag, 20. Januar 2022 um 16:41:52 UTC+1:
>>> On Thursday, January 20, 2022 at 4:02:30 PM UTC+1, WM wrote:
>>>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
>>>>> WM presented the following explanation :
>>>>>> Every actually infinite set either will become completed or not.
>>>>> Sets don't "become" anything.
>>>> Counting is a process which, according to matheology, can become completed. We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß, Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f, 291, 293, 305f, 413]
>>>>
>>> Finally - that's your problem - you don't know what 'Zuordnung' means.
>>> Wie can fix that easily enough:
>>>
>>> Literally it means 'assignment'.
>>> In mathematics it means 'function' or 'relation' or 'mapping'.
>>> It doesn't mean eenie-meenie-mynie-mo.
>> Either you are stupid or you are a liar, but probably you are both. When Cantor said "element by element", then this means precisely one by one and one after the other. Of course the frauds of matheology have tried to pervert this simple Zuordnung into an "all at once" in order to prevent inspection because they must have known for a long time already that there is no complete mapping between infinite sets - unless all are as stupid as you.
>>
>> Regards, WM
>
> I think we are in luck - it is, without any doubt, true that not everybody is as stupid as I am. And, obviously,
> very few people are dumber than you.
> Cantor writes formally when he formulates definitions, theorems and proofs. He also writes informally,
> discursively, appealing to the reader's intuition, using heuristics, when he discusses and describes the consequences of the
> formal theory. He does both extremely well.
> Your innumerable quotes, from Cantor and many other people, almost always come from private correspondence,
> pop science presentations, introductions to books. They almost never come from formal mathematical writings, which
> you are obviously not able to understand.
> Cantor never - und I mean literally never - uses an argument of the kind that leads to many of your errors,
> namely "eenie, meenie, minee, mo until we reach the end ... ".

the name Cantor should never appear in a post with WM of any kind. WM uses it to gloss up his image.

WM does not know any math at all, and always, Always makes mistakes in every post.

sergio schrieb am Donnerstag, 20. Januar 2022 um 20:34:05 UTC+1:
> On 1/20/2022 12:47 PM, WM wrote:
> > sergio schrieb am Donnerstag, 20. Januar 2022 um 18:15:31 UTC+1:
> >> On 1/20/2022 9:12 AM, WM wrote:
> >>> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:
> >
> >>>> It proves nothing of the sort. You changing the bijection proves nothing about it.
> >>>
> >>> I did not change it. I used Cantor's enumeration
> >>> k = (m + n - 1)(m + n - 2)/2 + m
> >> Liar.
> >
> > Where does my model deviate?
> you change the matrix as you go step by step, fail. Cantor does not do that.

He does.

If two well-defined manifolds, M and N, can be related completely, element by element, to each other [...],
two sets are ascribed the same cardinality if they mutually uniquely, element by element, can be mapped onto each other."

"Two sets are called 'equivalent' if they mutually uniquely, element by element, can be mapped onto each other."

Can you read this?

Regards, WM

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 to
its own i/1 of its first column.

You take "finished" to mean that there is a last entry
in the matrix, in the first column. That's not true
and it's not what I said.

"All" is not determined by reaching some last individual.
Not with our describe-and-take-reliable-steps method.

"All" is determined by falling within the description.
If the description is of some collection without a
last individual, "all" works the same as elsewhere.

Define ℕ⁺ = ⋃{{1,...,k}}

Describe k ∈ ℕ⁺ as ending {1,...,k}
Describe {1,...,k} as beginning at 1 and having
a counting-order
(each cut is a step, each step is a count)

Step reliably from these claims about k to another
claim about k. It will be as true of k as the first
claims were. And the first claims are true of each
element of ℕ⁺ This is how we "complete" ℕ⁺ by
starting the argument with and staying with claims
about the _complete_ ℕ⁺ Ends not required.

>> in order to know each location formerly holding n/d
>> has i/1 now.
>
> That is irrelevant in geometry.

"Geometry"?
You've stolen another term with which to obfuscate.

In geometry, there are collections which match
point-to-point with a proper subset. For example,
a line segment with half of that line segment.
This is well-known.

It's also probably not relevant to whatever you're
hiding with your use of "geometry". Still, it's
interesting.

>> I think that the best way to think about the O's
>> not-landing and ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1} is that
>> there are these special collections
>> ( ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1} are not among them),
>> the ones with two-ended stepping-orders, for which
>> the O's not-landing would need explaining.
>
> Sorry, I have not got what you mean,
> but it appears interesting.

Collections with two-ended stepping-orders are special.
Two-ended stepping-orders are useful to those
responsible for keeping track of sheep and pebbles,
useful primarily because, if they match in one order,
they match in all orders. It's a property that can
send shepherds out into the darkening fields to look
for a lost sheep.

Not all collections have two-ended stepping-orders.
Not all collections, if they match in one order,
match in all orders. Not all collections are special.

Our describe-and-take-reliable-steps method works for
both special and non-special collections. It works for
any correctly described collection, from the description.

Your argument for the presence of O's in the matrix,
despite i = (n+d-1)*(n+d-2)/2 + n, calls upon the special
property linked to having a two-ended stepping-order.
Not all collections have a two-ended stepping order.

Each {0,...,k} has a two-ended stepping-order and,
matched element-to-element, cannot lose any sheep.
Each {0,...,k} is special.

The {0,...,k} might not seem special, because, everywhere
we look, there they are. That does not take into account
just how excellently useful the special collections are.
Everywhere I drive, there's pavement under my tires! Amazing!
But, no, not really. Driving where there is pavement is
useful. It doesn't mean the whole world is paved.

In summary and in conclusion,
not all collections are special.

JVR schrieb am Donnerstag, 20. Januar 2022 um 20:44:57 UTC+1:

> Your innumerable quotes, from Cantor and many other people, almost always come from private correspondence,
> pop science presentations, introductions to books. They almost never come from formal mathematical writings, which
> you are obviously not able to understand.

You are wrong. This is the basis of set theory:

"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" [Cantor, Collected Works, p. 119, Crelles Journal f. Mathematik Bd. 84, S. 242 - 258 (1878)]

"wobei zwei Mengen dieselbe Mächtigkeit zugeschrieben wird, wenn sie sich gegenseitig eindeutig, Element für Element einander zuordnen lassen." [Cantor, Collected Works, p. 167, Grundlagen einer allgemeinen Mannigfaltigkeitslehre (Leipzig 1883)]

"Zwei Mengen werden hierbei 'äquivalent' genannt, wenn sie sich gegenseitig eindeutig, Element für Element, einander zuordnen lassen." [Cantor, Collected Works, pp. 380 & 441]

Cantor never - und I mean literally never - uses an argument of the kind that leads to many of your errors,
namely "eenie, meenie, minee, mo until we reach the end ... ".

Above I have proved you are wrong: Ene mene mu und raus bist Du. But of course you will never admit it.

Regards, WM

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.

> to
> its own i/1 of its first column.

That means no n/d remains. Hence you get finished.
>
> You take "finished" to mean that there is a last entry
> in the matrix, in the first column. That's not true
> and it's not what I said.

I take it only as : None remains.
>
> "All" is not determined by reaching some last individual.

That means all is not all in linear order. That requires dark elements.

> "All" is determined by falling within the description.

But the description is not comprehensive. Otherwise all would imply a last one.
> If the description is of some collection without a
> last individual, "all" works the same as elsewhere.

All O's will remain works fine.

> a counting-order
> (each cut is a step, each step is a count)

And every count opens a gap where the i/1 is removed.
>
> Step reliably from these claims about k to another
> claim about k.

Reliably every count opens a gap where the i/1 is removed.

> >> in order to know each location formerly holding n/d
> >> has i/1 now.
> >
> > That is irrelevant in geometry.
> "Geometry"?
> You've stolen another term with which to obfuscate.

Don't you like geometry?
>
> In geometry, there are collections which match
> point-to-point with a proper subset. For example,
> a line segment with half of that line segment.
> This is well-known.

He we have quares wich cover the squares of matrix elements.
>
> It's also probably not relevant to whatever you're
> hiding with your use of "geometry". Still, it's
> interesting.

The number of O's is constant.

> > Sorry, I have not got what you mean,
> > but it appears interesting.
> Collections with two-ended stepping-orders are special.
> Two-ended stepping-orders are useful to those
> responsible for keeping track of sheep and pebbles,
> useful primarily because, if they match in one order,
> they match in all orders.

And don't forget the O's! Every transposition is of first order.

It's a property that can
> send shepherds out into the darkening fields to look
> for a lost sheep.

Fortunately there cannot be a lost O.
>
> Not all collections have two-ended stepping-orders.
> Not all collections, if they match in one order,
> match in all orders. Not all collections are special.
>
> Our describe-and-take-reliable-steps method works for
> both special and non-special collections. It works for
> any correctly described collection, from the description.

Like my configurations at blocked drain.
>
> Your argument for the presence of O's in the matrix,
> despite i = (n+d-1)*(n+d-2)/2 + n, calls upon the special
> property linked to having a two-ended stepping-order.
> Not all collections have a two-ended stepping order.

But when the drain is blocked, the O's will stay. They will stay as long as the drain is blocked.

> In summary and in conclusion,
> not all collections are special.

But shuffling the squares covering less than 1 % of the whole area will never cover the full area.

Regards, WM

WM formulated the question :
> sergio schrieb am Donnerstag, 20. Januar 2022 um 20:34:05 UTC+1:
>> On 1/20/2022 12:47 PM, WM wrote:
>>> sergio schrieb am Donnerstag, 20. Januar 2022 um 18:15:31 UTC+1:
>>>> On 1/20/2022 9:12 AM, WM wrote:
>>>>> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31
>>>>> UTC+1:
>>>>>> It proves nothing of the sort. You changing the bijection proves nothing
>>>>>> about it.
>>>>>
>>>>> I did not change it. I used Cantor's enumeration
>>>>> k = (m + n - 1)(m + n - 2)/2 + m
>>>> Liar.
>>>
>>> Where does my model deviate?
>> you change the matrix as you go step by step, fail. Cantor does not do that.
>
> He does.
>
> If two well-defined manifolds, M and N, can be related completely, element by
> element, to each other [...],
>
> two sets are ascribed the same cardinality if they mutually uniquely, element
> by element, can be mapped onto each other."
>
> "Two sets are called 'equivalent' if they mutually uniquely, element by
> element, can be mapped onto each other."
>
> Can you read this?

So 'element by element' means the same as 'step by step' to you?

WM presented the following explanation :
> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 16:09:02 UTC+1:
>> WM has brought this to us :
>>> FromTheRafters schrieb am Donnerstag, 20. Januar 2022 um 12:29:53 UTC+1:
>>>> WM presented the following explanation :
>>>>> Every actually infinite set either will become completed or not.
>>>> Sets don't "become" anything.
>>>
>>> Counting is a process which, according to matheology, can become completed.
>>> We find the terms Zuordnungsgesetz, Zuordnungsmodus, Zuordnungsprozeß,
>>> Zuordnungsverfahren, Zuordnungsverhältnis. [Cantor, pp. 239, 283f, 286f,
>>> 291, 293, 305f, 413]
>> Who said anything about counting?
>
> 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

Still, even if you just woke up from counting sheep, sets don't
"become" anything.

On 1/20/2022 2:56 PM, WM wrote:
> sergio schrieb am Donnerstag, 20. Januar 2022 um 20:34:05 UTC+1:
>> On 1/20/2022 12:47 PM, WM wrote:
>>> sergio schrieb am Donnerstag, 20. Januar 2022 um 18:15:31 UTC+1:
>>>> On 1/20/2022 9:12 AM, WM wrote:
>>>>> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:
>>>
>>>>>> It proves nothing of the sort. You changing the bijection proves nothing about it.
>>>>>
>>>>> I did not change it. I used Cantor's enumeration
>>>>> k = (m + n - 1)(m + n - 2)/2 + m
>>>> Liar.
>>>
>>> Where does my model deviate?
>> you change the matrix as you go step by step, fail. Cantor does not do that.
>
> He does.

No. He does not. He does not change the matrix.

>
> If two well-defined manifolds, M and N, can be related completely, element by element, to each other [...],
>
> two sets are ascribed the same cardinality if they mutually uniquely, element by element, can be mapped onto each other."
>
> "Two sets are called 'equivalent' if they mutually uniquely, element by element, can be mapped onto each other."
>
> Can you read this?

that is diversion.

Cantors enumeration is so very simple, and yet so hard for you to comprehend.

>
> Regards, WM

On 1/20/2022 3:24 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.

your intentional mistake. nothing is taken away. If you think so, you think wrongly.

> And every i/1 replacing them is also taken away from its place.

silly. there is no replacing them nore taken away either.

go take some pre algebra courses, and buy a few more sets of all the natural numbers from eBay.

> The gaps cannot be filled.

gaps in your noggin.

torsdag 20 januari 2022 kl. 16:12:11 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 20. Januar 2022 um 12:37:31 UTC+1:
> > torsdag 20 januari 2022 kl. 11:21:31 UTC+1 skrev WM:
>
> > > > Your re-arrangement to find a non-bijection is entirely irrelevant then.
> > > A contradiction is very relevant.
> > > Here is another picture of the same facts. Consider the infinite matrix
> > > 1/1, _, _, _, ...
> > > 2/1, _, _, _, ...
> > > 3/1, _, _, _, ...
> > > 4/1, _, _, _, ...
> > > ...
> > > and try to fill the whole matrix *including the first column* with the fractions. It is obviously foolish, only to consider this attempt.
> > >
> >
> > >But you have not. This is proved by the blocked drain for unmatched fractions O.
> > It proves nothing of the sort. You changing the bijection proves nothing about it.
> I did not change it. I used Cantor's enumeration
> k = (m + n - 1)(m + n - 2)/2 + m
> > The formula is such that it will take every rational to a natural number.
> 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.
>
> > >If there is a bijection, then no transposition can destroy it.
> > but yours isn't a transpoition, it is infinitely many and that can.
> No. Infinitely many black elements do not contain a white element.
> >
> > >A contradiction is very relevant.
> > Except you are doing nothing to show a contradiction
> > "This entirely different function doesn't work as a bijection" is not a contradiction to my function functioning.
> You are wrong. I am using no entirely different function but the well known
>
> k = (m + n - 1)(m + n - 2)/2 + m.
> >
> > To show it is not a bijection, you have to show that there is at least one I missed. I can show that the function hits all of them. You have failed to do this.
> Dark numbers cannot be shown. But the indirect existence proof is done by the invariable number of O's.
> >
> > Fortunately for us I wouldn't even need to do that, if I can show a surjection from A to B, and a surjection from B to A, then there exists a bijection between both sets. It is trivial to make these surjections here.
> Between the visible elements this is trivial. Nevertheless the invariable number of X's and O's shows the dark side.
>
> 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.
>
> Regards, WM

>I did not change it. I used Cantor's enumeration

You are not when you change the function.

>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.

>You are wrong. I am using no entirely different function but the well known

You are not when you "transpose" shit you lying sack of shit.

>Between the visible elements this is trivial. Nevertheless the invariable number of X's and O's shows the dark side.

Empty assertion

>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 say it is irrelevant because your argument there is nothing but obfuscation.

Show me a rational number the function does not cover, simple as that.

zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 06:52:19 UTC+1:
XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

> > 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...
....

> > 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

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?

Regards, WM

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.

Regards, WM

fredag 21 januari 2022 kl. 09:56:41 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 06:52:19 UTC+1:
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
> > > 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...
> ...
> > > 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

I do not care for your "board"

for any rational number n/m I can find a natural number to map to it. That is all there is. Your board is just to obfuscate.

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?

When young Gauss was told by his instructor to add the first one
hundred natural numbers together, the instructor was apparently
thinking, like you, that Gauss would be adding them step by step
creating a list of triangular numbers leading up to the final 5050
figure. Surely it would occupy Gauss for the duration of the recess.

Gauss didn't do it your way, and he was out at recess in very little
time. He didn't use your step-by-step process because he wasn't told
that he must do it the stupid way.

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.

fredag 21 januari 2022 kl. 11:52:10 UTC+1 skrev FromTheRafters:
> 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.
WM is mentally retarded

zelos...@gmail.com pretended :
> fredag 21 januari 2022 kl. 11:52:10 UTC+1 skrev FromTheRafters:
>> 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.
> WM is mentally retarded

He's too stupid to be so arrogant, but he is relentless in proving his
ineptness.

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.
>
> 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. That proves you wrong.

Regards, WM