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On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but
all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree. Where is XXXX? Ahh, conversing with you is a shot in
the dark.
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but
all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but
all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions can
be enumerated.
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but
all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions can
be enumerated.
On 8/3/2025 3:55 AM, WM wrote:
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but
all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions
can be enumerated.
When we say N is countable and complete, we still know that there is no largest natural number... See?
On 8/3/2025 3:55 AM, WM wrote:
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears but
all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions
can be enumerated.
What fraction cannot be realized? Give me an example? :)
On 03.08.2025 22:06, Chris M. Thomasson wrote:That's a "proof", not a proof. It depends on a misunderstanding of
On 8/3/2025 3:55 AM, WM wrote:The proof shows that no O can leave the matrix. All visible fractions
On 03.08.2025 05:20, Chris M. Thomasson wrote:What fraction cannot be realized? Give me an example? :)
On 8/2/2025 1:46 PM, WM wrote:Here is no mention of a tree but of the fact that not all fractions
On 02.08.2025 21:51, Chris M. Thomasson wrote:Map that to a tree.
0, 1, 2, 3, 4, 5, ...Can you understand that in thi re-ordering never an O disappears but >>>>> all visible matrix elements get free of O?
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
can be enumerated.
get X's. This proves the existence of dark fractions.
Or do you know how O's disappear from the matrix? (They only disappearYes, it's obvious.
from the visible part.)
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
Or do you know how O's disappear from the matrix? (They only disappear
from the visible part.)
Yes, it's obvious.
In your step by step process for reordering the X's and O's, the O's move steadily rightwards and downwards. (At least, as I remember it from a previous post - I can't be bothered to get into the details again.) At
each and every finite step, there are O's in the matrix.
In the limit as the number of steps tends to infinity, the O's have
"moved so far" that they are no longer "at any finite position". This is
not difficult to visualise for people who aren't WM.
There is no need for "dark fractions" or some ill-defined partition of
the matrix into a "visible" part and an "invisible" part.
On 03.08.2025 22:04, Chris M. Thomasson wrote:
On 8/3/2025 3:55 AM, WM wrote:
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears
but all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions
can be enumerated.
When we say N is countable and complete, we still know that there is
no largest natural number... See?
There is no mention of a largest natural number. All natural numbers (indices X) are distributed according to Cantor but most fractions keep
O's, proving that they are not indexed. Proof: No O can leave the matrix.
On 8/4/2025 3:37 AM, WM wrote:
On 03.08.2025 22:04, Chris M. Thomasson wrote:
On 8/3/2025 3:55 AM, WM wrote:
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark???
Can you understand that in thi re-ordering never an O disappears
but all visible matrix elements get free of O?
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions
can be enumerated.
When we say N is countable and complete, we still know that there is
no largest natural number... See?
There is no mention of a largest natural number. All natural numbers
(indices X) are distributed according to Cantor but most fractions
keep O's, proving that they are not indexed. Proof: No O can leave the
matrix.
Have you ever implemented Cantor Pairing? They are all indexed.
On 04.08.2025 21:22, Chris M. Thomasson wrote:
On 8/4/2025 3:37 AM, WM wrote:
On 03.08.2025 22:04, Chris M. Thomasson wrote:
On 8/3/2025 3:55 AM, WM wrote:
On 03.08.2025 05:20, Chris M. Thomasson wrote:
On 8/2/2025 1:46 PM, WM wrote:
On 02.08.2025 21:51, Chris M. Thomasson wrote:
Can you understand that in thi re-ordering never an O disappears >>>>>>> but all visible matrix elements get free of O?
0, 1, 2, 3, 4, 5, ...
WM thinks 6 is dark? Or 6 + 1 is dark? Not sure... WM is dark??? >>>>>>>
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...-a-a-a-a ...
Map that to a tree.
Here is no mention of a tree but of the fact that not all fractions >>>>> can be enumerated.
When we say N is countable and complete, we still know that there is
no largest natural number... See?
There is no mention of a largest natural number. All natural numbers
(indices X) are distributed according to Cantor but most fractions
keep O's, proving that they are not indexed. Proof: No O can leave
the matrix.
Have you ever implemented Cantor Pairing? They are all indexed.
What is all?
Any rational you can think of has an
index.
On 04.08.2025 21:41, Chris M. Thomasson wrote:
Any rational you can think of has an index.
Correct. Nevertheless almost all rationals are not indexed (proved by
the presence of O's). Conclusion? You cannot think of most.
On 04.08.2025 16:29, Alan Mackenzie wrote:You haven't proved that for the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:
Or do you know how O's disappear from the matrix? (They only disappear
from the visible part.)
Yes, it's obvious.
In the limit as the number of steps tends to infinity, the O's have
"moved so far" that they are no longer "at any finite position". This
is not difficult to visualise for people who aren't WM.
We do not visualize but prove that never an O leaves the matrix.
The limit of the sequence of the number of O's is infinite,There is no need for "dark fractions" or some ill-defined partition of
the matrix into a "visible" part and an "invisible" part.
According to analysis the function f(n) = C has limit C too.
On 8/4/2025 12:44 PM, WM wrote:
On 04.08.2025 21:41, Chris M. Thomasson wrote:
Any rational you can think of has an index.
Correct. Nevertheless almost all rationals are not indexed (proved by
the presence of O's). Conclusion? You cannot think of most.
any rational is indexed in the bidirectional cantor pairing. Therefore,
they all have a unique index.
On 04.08.2025 16:29, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Or do you know how O's disappear from the matrix? (They only disappear
from the visible part.)
Yes, it's obvious.
In your step by step process for reordering the X's and O's, the O's move
steadily rightwards and downwards. (At least, as I remember it from a
previous post - I can't be bothered to get into the details again.) At
each and every finite step, there are O's in the matrix.
Namely as many O's as at the beginning.
In the limit as the number of steps tends to infinity, the O's have
"moved so far" that they are no longer "at any finite position". This is
not difficult to visualise for people who aren't WM.
We do not visualize .....
.... but prove that never an O leaves the matrix.
There is no need for "dark fractions" or some ill-defined partition of
the matrix into a "visible" part and an "invisible" part.
According to analysis the function f(n) = C has limit C too. Either this
is wrong or all O's stay at dark cells of the matrix.
Regards, WM--
Am Mon, 04 Aug 2025 16:45:09 +0200 schrieb WM:
According to analysis the function f(n) = C has limit C too.The limit of the sequence of the number of O's is infinite,
but the number of O's in the limit of the matrices is zero.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 04.08.2025 16:29, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Or do you know how O's disappear from the matrix? (They only disappear >>>> from the visible part.)
Yes, it's obvious.
In your step by step process for reordering the X's and O's, the O's move >>> steadily rightwards and downwards. (At least, as I remember it from a
previous post - I can't be bothered to get into the details again.) At
each and every finite step, there are O's in the matrix.
Namely as many O's as at the beginning.
Well, it is a countably infinite number of O's, but yes. It stays
countably infinite for each finite step. But NOT in the limit as the
number of steps tends to infinity, as I explained in my next paragraph.
.... but prove that never an O leaves the matrix.
That word "leave" is mathematically meaningless there. You haven't
proved anything about the process.
The plain fact is that after a finite number of steps of the swapping
process you envisage, there are a countably infinite number of both X's
and O's.
In the infinite limit, by contrast, there are just a countably
infinite number of X's and no O's.
There is no contradiction here. The
limit of a thing is not necessarily the thing of the limit.
As I said, it would help you if you could envisage the process by which
the O's "move" steadily further from the origin of your infinite matrix.
There is no need for "dark fractions" or some ill-defined partition of
the matrix into a "visible" part and an "invisible" part.
According to analysis the function f(n) = C has limit C too. Either this
is wrong or all O's stay at dark cells of the matrix.
That's a complete non-sequitur.
The constant function has the limit C.
That has no bearing whatsoever on the non-existence of "dark cells" in
the matrix.
On 04.08.2025 22:19, joes wrote:No. This is not the same as below.
Am Mon, 04 Aug 2025 16:45:09 +0200 schrieb WM:
Therefore they all remain.According to analysis the function f(n) = C has limit C too.The limit of the sequence of the number of O's is infinite,
The limit matrix has only X's. The O's "go" to infinite indices,but the number of O's in the limit of the matrices is zero.That is in cotradiction with the limit C. But if assumed so, where do
they go?
On 04.08.2025 22:36, Alan Mackenzie wrote:Only for finite indices, not for the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:
.... but prove that never an O leaves the matrix.
That word "leave" is mathematically meaningless there. You haven't
proved anything about the process.
I have proved that only X and O are exchanged, never any is deleted.
Dude. Either you deny the limit or you accept that a matrix full ofThe plain fact is that after a finite number of steps of the swappingAn unsupported claim - nothing else.
process you envisage, there are a countably infinite number of both X's
and O's.
In the infinite limit, by contrast, there are just a countably infinite
number of X's and no O's.
Yes, but not the value of the limit.There is no contradiction here. TheThe limit of a constant function is this constant.
limit of a thing is not necessarily the thing of the limit.
Again, not in the limit, only for finite indices.As I said, it would help you if you could envisage the process by whichBut remaining within the matrix.
the O's "move" steadily further from the origin of your infinite
matrix.
Am Mon, 04 Aug 2025 22:37:16 +0200 schrieb WM:
On 04.08.2025 22:19, joes wrote:No. This is not the same as below.
Am Mon, 04 Aug 2025 16:45:09 +0200 schrieb WM:Therefore they all remain.
According to analysis the function f(n) = C has limit C too.The limit of the sequence of the number of O's is infinite,
The limit matrix has only X's.but the number of O's in the limit of the matrices is zero.That is in cotradiction with the limit C. But if assumed so, where do
they go?
The O's "go" to infinite indices,
which lie outside the matrix.
Am Mon, 04 Aug 2025 22:45:19 +0200 schrieb WM:
On 04.08.2025 22:36, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Only for finite indices, not for the limit..... but prove that never an O leaves the matrix.
That word "leave" is mathematically meaningless there. You haven't
proved anything about the process.
I have proved that only X and O are exchanged, never any is deleted.
Dude. Either you deny the limit or you accept that a matrix full ofThe plain fact is that after a finite number of steps of the swappingAn unsupported claim - nothing else.
process you envisage, there are a countably infinite number of both X's
and O's.
In the infinite limit, by contrast, there are just a countably infinite
number of X's and no O's.
only X's does not contain any O's.
Yes, but not the value of the limit.There is no contradiction here. TheThe limit of a constant function is this constant.
limit of a thing is not necessarily the thing of the limit.
Again, not in the limit, only for finite indices.As I said, it would help you if you could envisage the process by whichBut remaining within the matrix.
the O's "move" steadily further from the origin of your infinite
matrix.
WM <wolfgang.mueckenheim@tha.de> wrote:
According to analysis the function f(n) = C has limit C too. Either this
is wrong or all O's stay at dark cells of the matrix.
That's a complete non-sequitur. The constant function has the limit C.
That has no bearing whatsoever on the non-existence of "dark cells" in
the matrix.
Note that for M|+ckenheim
-a-a-a lim(card(A_n)) = card(lim(A_n)).
is a fact.
Which "proves" WM's "stance": "The 'O's can't leave the matrix; not even
in the limit!".
On 05.08.2025 15:39, Moebius wrote:I already told you in a post yesterday. The O's "move" steadily away
Note that for M|+ckenheimFact is that the number of O's does never change.
-a-a-a lim(card(A_n)) = card(lim(A_n)).
is a fact.
Which "proves" WM's "stance": "The 'O's can't leave the matrix; not evenOf course not. If you disagree tell me how this could happen by pure exchange with X's.
in the limit!".
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.08.2025 15:39, Moebius wrote:
Which "proves" WM's "stance": "The 'O's can't leave the matrix [or matrices]; not even
in the limit!".
[...] tell me how this could happen by pure exchange with X's.
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to infinity", every last one of them.
Please forgive me trying to explain it in terms you might understand.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.08.2025 15:39, Moebius wrote:
Which "proves" WM's "stance": "The 'O's can't leave the matrix [or matrices]; not even
in the limit!".
[...] tell me how this could happen by pure exchange with X's.
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to infinity", every last one of them.
Please forgive me trying to explain it in terms you might understand.
Regards, WM
Am 05.08.2025 um 16:13 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.08.2025 15:39, Moebius wrote:
Which "proves" WM's "stance": "The 'O's can't leave the matrix [or
matrices]; not even
in the limit!".
[...] tell me how this could happen by pure exchange with X's.
If I may add (using WM-lingo too):
1. The limit is not "reached" "by pure exchange with X's". "In the
limit" is only a figure of speech, not something we actually "arive at".
2. The limit of a sequence of objects may have different properties than
the objects (terms) in the sequence.
For example: For all n e IN: {1, ..., n} is FINITE. But lim {1, ..., n}infinity",
is INFINITE (namely = IN).
Hint: Each and every matrix in your sequence contains (infinitely many)
Os. The LIMIT of this sequence of matrices doesn't contain ANY O. (A
simple mathematical fact.)
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to
every last one of them.
<argl>
"My opponent's reasoning reminds me of the heathen, who, being asked on
what the world stood, replied, "On a tortoise." But on what does the tortoise stand? "On another tortoise." With Mr. Barker, too, there are tortoises all the way down."
rCorCe"Second Evening: Remarks of Rev. Dr. Berg"
Please forgive me trying to explain it in terms you might understand.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to infinity", every last one of them.
Please forgive me trying to explain it in terms you might understand.
On 05.08.2025 16:13, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to infinity",
every last one of them.
The O's are exchanged with X's, never deleted, never leaving the finite domain. All exchanges happen at finite places.
Please forgive me trying to explain it in terms you might understand.
You should try to understand that all happens at finite places. No O
will ever reach an infinite index.
How can you dare to propose such illogical nonsense!
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.08.2025 16:13, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to infinity", >>> every last one of them.
The O's are exchanged with X's, never deleted, never leaving the finite
domain. All exchanges happen at finite places.
Of course. But you don't understand the concept "limit as the steps
tend to infinity".
For every step the X's indeed never leave the finite
domain, but in the limit they have vanished.
You don't understand that, and you're not trying to understand it.
Every mathematics undergraduate understands it, but you don't.M
Please forgive me trying to explain it in terms you might understand.
You should try to understand that all happens at finite places. No O
will ever reach an infinite index.
As already said, you don't understand "limit ... tends to infinity".
How can you dare to propose such illogical nonsense!
It's basic mathematics.
On 05.08.2025 21:21, Alan Mackenzie wrote:In no *finite* case. Why do you accept the limit?
WM <wolfgang.mueckenheim@tha.de> wrote:I understand that never an O can be deleted. In no case!
On 05.08.2025 16:13, Alan Mackenzie wrote:Of course. But you don't understand the concept "limit as the steps
WM <wolfgang.mueckenheim@tha.de> wrote:
I already told you in a post yesterday. The O's "move" steadily awayThe O's are exchanged with X's, never deleted, never leaving the
from the origin. In the limit they have "moved all the way to
infinity", every last one of them.
finite domain. All exchanges happen at finite places.
tend to infinity".
What do you think the limit is?For every step the X's indeed never leave the finite domain, but in theYou mean the O's. So you wish to apply magic, I prefer mathematics.
limit they have vanished.
Any sufficiently advanced...You don't understand that, and you're not trying to understand it.I am refuting to apply magic.
No, no finite(ly indexed) matrix represents the bijection. You need thePlease forgive me trying to explain it in terms you might understand.You should try to understand that all happens at finite places. No O
will ever reach an infinite index.
Accept what limit?As already said, you don't understand "limit ... tends to infinity".Note that Cantor does not accept a limit.
When dealing with Cantor's mappings between infinite sets, it is argued usually that these mappings require a "limit" to be completed or thatWeren't you the one who complained that the process never finished?
they cannot be completed. Such arguing has to be rejected flatly.
For this reason some of Cantor's statements are quoted below.Off topic.
On 05.08.2025 10:28, joes wrote:An infinite number of times. Tell me, what is the limit of that?
Am Mon, 04 Aug 2025 22:45:19 +0200 schrieb WM:There is nothing else except X and O are exchanged.
On 04.08.2025 22:36, Alan Mackenzie wrote:Only for finite indices, not for the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:
I have proved that only X and O are exchanged, never any is deleted..... but prove that never an O leaves the matrix.That word "leave" is mathematically meaningless there. You haven't
proved anything about the process.
Ok, so you deny the limit. What is it then?That is wrong. No O can leave. The visible part however contains onlyDude. Either you deny the limit or you accept that a matrix full ofThe plain fact is that after a finite number of steps of the swappingAn unsupported claim - nothing else.
process you envisage, there are a countably infinite number of both
X's and O's.
In the infinite limit, by contrast, there are just a countably
infinite number of X's and no O's.
only X's does not contain any O's.
X's.
Proof pending.That is nonsense.Yes, but not the value of the limit.There is no contradiction here. TheThe limit of a constant function is this constant.
limit of a thing is not necessarily the thing of the limit.
Show me the finite column and row where an O stays.An unsupported and wrong claim.Again, not in the limit, only for finite indices.As I said, it would help you if you could envisage the process byBut remaining within the matrix.
which the O's "move" steadily further from the origin of your
infinite matrix.
Am Tue, 05 Aug 2025 12:16:59 +0200 schrieb WM:
On 05.08.2025 10:28, joes wrote:An infinite number of times. Tell me, what is the limit of that?
Am Mon, 04 Aug 2025 22:45:19 +0200 schrieb WM:There is nothing else except X and O are exchanged.
On 04.08.2025 22:36, Alan Mackenzie wrote:Only for finite indices, not for the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:I have proved that only X and O are exchanged, never any is deleted.
.... but prove that never an O leaves the matrix.That word "leave" is mathematically meaningless there. You haven't
proved anything about the process.
Ok, so you deny the limit. What is it then?That is wrong. No O can leave. The visible part however contains onlyDude. Either you deny the limit or you accept that a matrix full ofThe plain fact is that after a finite number of steps of the swapping >>>>> process you envisage, there are a countably infinite number of bothAn unsupported claim - nothing else.
X's and O's.
In the infinite limit, by contrast, there are just a countably
infinite number of X's and no O's.
only X's does not contain any O's.
X's.
Proof pending.That is nonsense.Yes, but not the value of the limit.There is no contradiction here. TheThe limit of a constant function is this constant.
limit of a thing is not necessarily the thing of the limit.
Show me the finite column and row where an O stays.An unsupported and wrong claim.Again, not in the limit, only for finite indices.As I said, it would help you if you could envisage the process byBut remaining within the matrix.
which the O's "move" steadily further from the origin of your
infinite matrix.
On 04.08.2025 21:52, Chris M. Thomasson wrote:
On 8/4/2025 12:44 PM, WM wrote:
On 04.08.2025 21:41, Chris M. Thomasson wrote:
Any rational you can think of has an index.
Correct. Nevertheless almost all rationals are not indexed (proved by
the presence of O's). Conclusion? You cannot think of most.
any rational is indexed in the bidirectional cantor pairing.
Therefore, they all have a unique index.
What about the O's indicating not indexed fractions?
WM <wolfgang.mueckenheim@tha.de> wrote:
On 04.08.2025 16:29, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Or do you know how O's disappear from the matrix? (They only disappear >>>> from the visible part.)
Yes, it's obvious.
In your step by step process for reordering the X's and O's, the O's move >>> steadily rightwards and downwards. (At least, as I remember it from a
previous post - I can't be bothered to get into the details again.) At
each and every finite step, there are O's in the matrix.
Namely as many O's as at the beginning.
Well, it is a countably infinite number of O's, but yes. It stays
countably infinite for each finite step. But NOT in the limit as the
number of steps tends to infinity, as I explained in my next paragraph.[...]
joes explained :
Am Tue, 05 Aug 2025 12:16:59 +0200 schrieb WM:
On 05.08.2025 10:28, joes wrote:An infinite number of times. Tell me, what is the limit of that?
Am Mon, 04 Aug 2025 22:45:19 +0200 schrieb WM:There is nothing else except X and O are exchanged.
On 04.08.2025 22:36, Alan Mackenzie wrote:Only for finite indices, not for the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:I have proved that only X and O are exchanged, never any is deleted.
.... but prove that never an O leaves the matrix.That word "leave" is mathematically meaningless there.-a You haven't >>>>>> proved anything about the process.
Ok, so you deny the limit. What is it then?That is wrong. No O can leave. The visible part however contains onlyDude. Either you deny the limit or you accept that a matrix full ofThe plain fact is that after a finite number of steps of the swapping >>>>>> process you envisage, there are a countably infinite number of both >>>>>> X's and O's.An unsupported claim - nothing else.
In the infinite limit, by contrast, there are just a countably
infinite number of X's and no O's.
only X's does not contain any O's.
X's.
Proof pending.That is nonsense.Yes, but not the value of the limit.-a There is no contradiction here.-a TheThe limit of a constant function is this constant.
limit of a thing is not necessarily the thing of the limit.
Show me the finite column and row where an O stays.An unsupported and wrong claim.Again, not in the limit, only for finite indices.As I said, it would help you if you could envisage the process byBut remaining within the matrix.
which the O's "move" steadily further from the origin of your
infinite matrix.
There is overflow housing available at the Hilbert.
Am Tue, 05 Aug 2025 22:01:23 +0200 schrieb WM:
On 05.08.2025 21:21, Alan Mackenzie wrote:
I understand that never an O can be deleted. In no case!In no *finite* case.
Why do you accept the limit?
What do you think the limit is?For every step the X's indeed never leave the finite domain, but in theYou mean the O's. So you wish to apply magic, I prefer mathematics.
limit they have vanished.
No, no finite(ly indexed) matrix represents the bijection. You need the limit. No O can be at a finite index.
Accept what limit?As already said, you don't understand "limit ... tends to infinity".Note that Cantor does not accept a limit.
On 05.08.2025 21:21, Alan Mackenzie wrote:You don't understand at all.
WM <wolfgang.mueckenheim@tha.de> wrote:I understand ....
On 05.08.2025 16:13, Alan Mackenzie wrote:Of course. But you don't understand the concept "limit as the steps
WM <wolfgang.mueckenheim@tha.de> wrote:The O's are exchanged with X's, never deleted, never leaving the finite
I already told you in a post yesterday. The O's "move" steadily away
from the origin. In the limit they have "moved all the way to infinity", >>>> every last one of them.
domain. All exchanges happen at finite places.
tend to infinity".
.... that never an O can be deleted. In no case!In the current scenario, O's don't get deleted. They just move away to
Yes. Sorry.For every step the X's indeed never leave the finiteYou mean the O's.
domain, but in the limit they have vanished.
So you wish to apply magic, I prefer mathematics.Your preferences are beyond your abilities. What you call "magic" is established mathematics, as developed by minds far superior to either of
You are refusing to apply established mathematics.You don't understand that, and you're not trying to understand it.I am refuting to apply magic.
That statement confirms you as a crank, but we knew that anyway.Every mathematics undergraduate understands it, but you don't.Unfortunately they have been spoilt by stupid teachers.
As a pre-eminent mathematician, Cantor understood full well what limitsNote that Cantor does not accept a limit.As already said, you don't understand "limit ... tends to infinity".Please forgive me trying to explain it in terms you might understand.You should try to understand that all happens at finite places. No O
will ever reach an infinite index.
Don't know about "usually", I never heard any such silly arguments till I encountered this newsgroup.Cantor, rejecting the limit ideaHow can you dare to propose such illogical nonsense!It's basic mathematics.
When dealing with Cantor's mappings between infinite sets, it is argued usually that these mappings require a "limit" to be completed or that
they cannot be completed.
Such arguing has to be rejected flatly.Such arguing is non-sensical, based on misunderstandings of basic maths.
For this reason some of Cantor's statements are quoted below.None of the following cites (which I would normally snip as they are off
"If we think the numbers p/q in such an order [...] then every number--
p/q comes at an absolutely fixed position of a simple infinite sequence"
[E. Zermelo: "Georg Cantor rCo Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 126]
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
"thus we get the epitome (-e) of all real algebraic numbers [...] and
with respect to this order we can talk about the NU<th algebraic number where not a single one of this epitome (NU+) has been forgotten." [E. Zermelo: "Georg Cantor rCo Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]
"such that every element of the set stands at a definite position of
this sequence" [E. Zermelo: "Georg Cantor rCo Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152] The clarity of these expressions is noteworthy: all and every,
completely, at an absolutely fixed position, NU<th number, where not a single one has been forgotten.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
So you wish to apply magic, I prefer mathematics.
Your preferences are beyond your abilities. What you call "magic" is established mathematics, as developed by minds far superior to either of
ours over the last few centuries.
That you fail to understand this
"magic" should be your problem alone.
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
Which however is always finite. ....
.... So the O's remain in the matrix.
So you wish to apply magic, I prefer mathematics.
Your preferences are beyond your abilities. What you call "magic" is
established mathematics, as developed by minds far superior to either of
ours over the last few centuries.
No, that has not been the subject of analysis. It is only Cantor's invention. And he has not used infinite numbers to enumerate.
That you fail to understand this
"magic" should be your problem alone.
You claim that exchange with X's at finite places can remove O's into
the infinite. This is simply nonsense.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
As a pre-eminent mathematician, Cantor understood full well what limits
were and how to use them.
Cantor, rejecting the limit idea
When dealing with Cantor's mappings between infinite sets, it is argued
usually that these mappings require a "limit" to be completed or that
they cannot be completed.
Don't know about "usually", I never heard any such silly arguments till I encountered this newsgroup.
For this reason some of Cantor's statements are quoted below.
None of the following cites (which I would normally snip as they are off topic) comes close to "rejecting the limit idea".
Am Tue, 05 Aug 2025 22:01:23 +0200 schrieb WM:
When dealing with Cantor's mappings between infinite sets, it is arguedWeren't you the one who complained that the process never finished?
usually that these mappings require a "limit" to be completed or that
they cannot be completed. Such arguing has to be rejected flatly.
Am Tue, 05 Aug 2025 12:16:59 +0200 schrieb WM:
There is nothing else except X and O are exchanged.An infinite number of times. Tell me, what is the limit of that?
Proof pending.That is nonsense.Yes, but not the value of the limit.There is no contradiction here. TheThe limit of a constant function is this constant.
limit of a thing is not necessarily the thing of the limit.
Show me the finite column and row where an O stays.An unsupported and wrong claim.Again, not in the limit, only for finite indices.As I said, it would help you if you could envisage the process byBut remaining within the matrix.
which the O's "move" steadily further from the origin of your
infinite matrix.
WM seems to be ultra finite. He seems to think that Cantor pairing
cannot possibly index all positive rationals where the pair (x, y) can
be x/y. I keep trying to ask him to show me a (x, y) pair (aka x/y) that cannot be indexed... He fails to do so.
On 06.08.2025 00:47, Chris M. Thomasson wrote:
WM seems to be ultra finite. He seems to think that Cantor pairing
cannot possibly index all positive rationals where the pair (x, y) can
be x/y. I keep trying to ask him to show me a (x, y) pair (aka x/y) that
cannot be indexed... He fails to do so.
Show me the first (or any) O leaving the matrix.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
Which however is always finite. ....
Yes.
.... So the O's remain in the matrix.
Yes. For any number of steps. But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1. This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1 "remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
No, that has not been the subject of analysis. It is only Cantor's
invention. And he has not used infinite numbers to enumerate.
To what do your "that" and your "it" refer?
That you fail to understand this
"magic" should be your problem alone.
You claim that exchange with X's at finite places can remove O's into
the infinite. This is simply nonsense.
It may be nonsense, but it was a plausible argument to try to get you to understand the notion of limits.
Maybe the sequence I depict above
might do a better job.
On 06.08.2025 14:43, Alan Mackenzie wrote:That is untrue. Please don't lie about what I've written. That is
WM <wolfgang.mueckenheim@tha.de> wrote:Of course. He called -e a limit. But -e does not contribute to the enumeration of the fractions.
As a pre-eminent mathematician, Cantor understood full well what limits
were and how to use them.
You proposed it yourself.Cantor, rejecting the limit ideaDon't know about "usually", I never heard any such silly arguments till I
When dealing with Cantor's mappings between infinite sets, it is argued
usually that these mappings require a "limit" to be completed or that
they cannot be completed.
encountered this newsgroup.
Without limit never an O is lost.Yes.
The O's prove that most fractions are not indexed.That's untrue. You fail to understand what a mathematical proof is,
Nobody's arguing with that, and it says precisely nothing about limits.The quotes show that every fraction occupies a fixed and *finite* placeFor this reason some of Cantor's statements are quoted below.None of the following cites (which I would normally snip as they are off
topic) comes close to "rejecting the limit idea".
in the sequence.
Nothing happens in the infinite. In particular, never an O leaves the<sigh>
matrix.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
As a pre-eminent mathematician, Cantor understood full well what limits
were and how to use them.
Of course. He called -e a limit. But -e does not contribute to the
enumeration of the fractions.
Cantor, rejecting the limit idea
When dealing with Cantor's mappings between infinite sets, it is argued >>>> usually that these mappings require a "limit" to be completed or that
they cannot be completed.
Don't know about "usually", I never heard any such silly arguments till I >>> encountered this newsgroup.
You proposed it yourself.
That is untrue. Please don't lie about what I've written. That is
merely your misunderstanding (possibly deliberate) of what I've written.
Without limit never an O is lost.
Yes.
The O's prove that most fractions are not indexed.
That's untrue. You fail to understand what a mathematical proof is,
just as you fail to understand limits.
For this reason some of Cantor's statements are quoted below.
None of the following cites (which I would normally snip as they are off >>> topic) comes close to "rejecting the limit idea".
The quotes show that every fraction occupies a fixed and *finite* place
in the sequence.
Nobody's arguing with that, and it says precisely nothing about limits.
Nothing happens in the infinite. In particular, never an O leaves the
matrix.
<sigh>
On 06.08.2025 18:59, Alan Mackenzie wrote:There is no index "where" they leave, it happens in the limit process,
WM <wolfgang.mueckenheim@tha.de> wrote:How and where do they leave?
On 06.08.2025 14:43, Alan Mackenzie wrote:Yes. For any number of steps. But NOT in the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move awayWhich however is always finite. So the O's remain in the matrix.
to an unbounded distance.
But don't try to stultify students. If all n fail to enumerate the
infinitely many fractions equipped with O's but you claim that
afterwards all fractions are indexed (by what), then every intelligent student recognizes that you don't tell the truth.
Of course. The limit is not a term.Using an analogous, but simpler example, consider the sequence of realAnd the limit is never reached by any f(n). That's the same as with
numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
Cantor's enumeration. Note that indexing is only possible by natural
numbers, not by -e.
They don't.The limit of the sequence (I hope you can agree to this) is 1. ThisNot by a long way. There are infinitely many O's. They cannot vanish immediately.
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1
"remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
Cantor invented completion?The completion of an enumeration.No, that has not been the subject of analysis. It is only Cantor'sTo what do your "that" and your "it" refer?
invention. And he has not used infinite numbers to enumerate.
So which term enumerates all positions?Maybe the sequence I depict above might do a better job.This is a nice example. It shows that you confuse the enumeration of all terms of the sequence (which in fact is done by the position of the
second 1) and the limit which has nothing to do with this enumeration.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
As a pre-eminent mathematician, Cantor understood full well what limits
were and how to use them.
Of course. He called -e a limit. But -e does not contribute to the
enumeration of the fractions.
Cantor, rejecting the limit idea
When dealing with Cantor's mappings between infinite sets, it is argued >>>> usually that these mappings require a "limit" to be completed or that
they cannot be completed.
Don't know about "usually", I never heard any such silly arguments till I >>> encountered this newsgroup.
You proposed it yourself.
That is untrue. Please don't lie about what I've written. That is
merely your misunderstanding (possibly deliberate) of what I've written.
Without limit never an O is lost.
Yes.
The O's prove that most fractions are not indexed.
That's untrue.
The quotes show that every fraction occupies a fixed and *finite* place
in the sequence.
Nobody's arguing with that,
Le 06/08/2025 |a 19:46, Alan Mackenzie a |-crit :
WM <wolfgang.mueckenheim@tha.de> wrote:
WM completely overlooks that however the mapping where there is only "X"Nothing happens in the infinite. In particular, never an O leaves the
matrix.
in the matrix is obtained doesn't matter as long as it exists
So he insists in obtaining it as the limit of a sequence of mappings
where there are always some "O" there, even an infinity of them i.e. it
is not a mapping from N to NxN. This is definitely not the most clever
way, it may help to visualize the final mapping though. But the "final" mapping can be built directly without using such a sequence.
He's again, assuming that if a property is true for all members of a sequence then it is a property of its limit.
WM completely overlooks that however the mapping where there is only "X"
in the matrix is obtained doesn't matter as long as it exists and
express a bijective mapping between N and NxN.
... he insists in obtaining it as the limit of a sequence of mappings
where there are always some "O" there, even an infinity of them i.e. it
is not a mapping from N to NxN. This is definitely not the most clever
way, it may help to visualize the final mapping though. But the "final" mapping can be built directly without using such a sequence.
He's again, assuming that if a property is true for all members of a sequence then it is a property of its limit.
Something that is so many times wrong even out of Set Theory... With such a kind of "argument" you
could also prove that any positive number is zero or that zero is positive...
Le 06/08/2025 |a 19:19, WM a |-crit :
Show me the first (or any) O leaving the matrix.
again and again the same sophistry from Wolfgang M|+ckenheim, debunked
for ages.
Le 06/08/2025 |a 19:19, WM a |-crit :
Show me the first (or any) O leaving the matrix.
again and again the same sophistry from Wolfgang M|+ckenheim, debunked
for ages.
On 05.08.2025 23:33, joes wrote:
Am Tue, 05 Aug 2025 22:01:23 +0200 schrieb WM:
When dealing with Cantor's mappings between infinite sets, it is arguedWeren't you the one who complained that the process never finished?
usually that these mappings require a "limit" to be completed or that
they cannot be completed. Such arguing has to be rejected flatly.
Either the enumeration of the rationals is never finished, or, if it is claimed to be finished, dark numbers are needed.
WM <wolfgang.mueckenheim@tha.de> wrote:[...]
On 05.08.2025 21:21, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.08.2025 16:13, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
I already told you in a post yesterday. The O's "move" steadily away >>>>> from the origin. In the limit they have "moved all the way to infinity", >>>>> every last one of them.
The O's are exchanged with X's, never deleted, never leaving the finite >>>> domain. All exchanges happen at finite places.
Of course. But you don't understand the concept "limit as the steps
tend to infinity".
I understand ....
You don't understand at all.
.... that never an O can be deleted. In no case!
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
For every step the X's indeed never leave the finite
domain, but in the limit they have vanished.
You mean the O's.
Yes. Sorry.
So you wish to apply magic, I prefer mathematics.
Your preferences are beyond your abilities. What you call "magic" is established mathematics, as developed by minds far superior to either of
ours over the last few centuries. That you fail to understand this
"magic" should be your problem alone.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
Which however is always finite. ....
Yes.
.... So the O's remain in the matrix.
Yes. For any number of steps. But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1. This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1 "remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
So you wish to apply magic, I prefer mathematics.
Your preferences are beyond your abilities. What you call "magic" is
established mathematics, as developed by minds far superior to either of >>> ours over the last few centuries.
No, that has not been the subject of analysis. It is only Cantor's
invention. And he has not used infinite numbers to enumerate.
To what do your "that" and your "it" refer?
That you fail to understand this
"magic" should be your problem alone.
You claim that exchange with X's at finite places can remove O's into
the infinite. This is simply nonsense.
It may be nonsense, but it was a plausible argument to try to get you to understand the notion of limits. Maybe the sequence I depict above
might do a better job.
Regards, WM
On 8/6/2025 10:11 AM, WM wrote:
On 05.08.2025 23:33, joes wrote:
Am Tue, 05 Aug 2025 22:01:23 +0200 schrieb WM:
When dealing with Cantor's mappings between infinite sets, it is argued >>>> usually that these mappings require a "limit" to be completed or thatWeren't you the one who complained that the process never finished?
they cannot be completed. Such arguing has to be rejected flatly.
Either the enumeration of the rationals is never finished, or, if it is
claimed to be finished, dark numbers are needed.
Do you think complete and/or all means finite?
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
As a pre-eminent mathematician, Cantor understood full well what limits
were and how to use them.
Of course. He called -e a limit. But -e does not contribute to the enumeration of the fractions.
Cantor, rejecting the limit idea
When dealing with Cantor's mappings between infinite sets, it is argued
usually that these mappings require a "limit" to be completed or that
they cannot be completed.
Don't know about "usually", I never heard any such silly arguments till I
encountered this newsgroup.
You proposed it yourself. Without limit never an O is lost. The O's
prove that most fractions are not indexed.
For this reason some of Cantor's statements are quoted below.
None of the following cites (which I would normally snip as they are off
topic) comes close to "rejecting the limit idea".
The quotes show that every fraction occupies a fixed and *finite* place
in the sequence. Nothing happens in the infinite. In particular, never
an O leaves the matrix.
Regards, WM
On 8/6/2025 10:11 AM, WM wrote:
On 05.08.2025 23:33, joes wrote:
Am Tue, 05 Aug 2025 22:01:23 +0200 schrieb WM:
When dealing with Cantor's mappings between infinite sets, it is argued >>>> usually that these mappings require a "limit" to be completed or thatWeren't you the one who complained that the process never finished?
they cannot be completed. Such arguing has to be rejected flatly.
Either the enumeration of the rationals is never finished, or, if it is
claimed to be finished, dark numbers are needed.
Do you think complete and/or all means finite?
Le 06/08/2025 |a 19:46, Alan Mackenzie a |-crit :
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
As a pre-eminent mathematician, Cantor understood full well what limits >>>> were and how to use them.
Of course. He called -e a limit. But -e does not contribute to the
enumeration of the fractions.
Cantor, rejecting the limit idea
When dealing with Cantor's mappings between infinite sets, it is
argued
usually that these mappings require a "limit" to be completed or that >>>>> they cannot be completed.
Don't know about "usually", I never heard any such silly arguments
till I
encountered this newsgroup.
You proposed it yourself.
That is untrue.-a Please don't lie about what I've written.-a That is
merely your misunderstanding (possibly deliberate) of what I've written.
Without limit never an O is lost.
Yes.
The O's prove that most fractions are not indexed.
That's untrue.-a You fail to understand what a mathematical proof is,
just as you fail to understand limits.
For this reason some of Cantor's statements are quoted below.
None of the following cites (which I would normally snip as they are
off
topic) comes close to "rejecting the limit idea".
The quotes show that every fraction occupies a fixed and *finite*
place in the sequence.
Nobody's arguing with that, and it says precisely nothing about limits.
Nothing happens in the infinite. In particular, never an O leaves the
matrix.
<sigh>
WM completely overlooks that however the mapping where there is only "X"
in the matrix is obtained doesn't matter as long as it exists and
express a bijective mapping between N and NxN.
So he insists in obtaining it as the limit of a sequence of mappings
where there are always some "O" there, even an infinity of them i.e. it
is not a mapping from N to NxN. This is definitely not the most clever
way, it may help to visualize the final mapping though. But the "final" mapping can be built directly without using such a sequence.
He's again, assuming that if a property is true for all members of a sequence then it is a property of its limit. Something that is so many
times wrong even out of Set Theory... With such a kind of "argument" you could also prove that any positive number is zero or that zero is positive...
On 06.08.2025 19:46, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
As a pre-eminent mathematician, Cantor understood full well what limits >>>> were and how to use them.
Of course. He called -e a limit. But -e does not contribute to the
enumeration of the fractions.
Cantor, rejecting the limit idea
When dealing with Cantor's mappings between infinite sets, it is
argued
usually that these mappings require a "limit" to be completed or that >>>>> they cannot be completed.
Don't know about "usually", I never heard any such silly arguments
till I
encountered this newsgroup.
You proposed it yourself.
That is untrue.-a Please don't lie about what I've written.-a That is
merely your misunderstanding (possibly deliberate) of what I've written.
Without limit never an O is lost.
Yes.
You claim that an O is lost. Therefore you lied above.
The O's prove that most fractions are not indexed.
That's untrue.
That is their definition. The O's denote not indexed fractions.
The quotes show that every fraction occupies a fixed and *finite* place
in the sequence.
Nobody's arguing with that,
Cantor is. I am.
On 06.08.2025 20:00, Python wrote:
Le 06/08/2025 |a 19:46, Alan Mackenzie a |-crit :
WM <wolfgang.mueckenheim@tha.de> wrote:
WM completely overlooks that however the mapping where there is onlyNothing happens in the infinite. In particular, never an O leaves the
matrix.
"X" in the matrix is obtained doesn't matter as long as it exists
Every matrix contains O's, i.e. not indexed fractions.
So he insists in obtaining it as the limit of a sequence of mappings
where there are always some "O" there, even an infinity of them i.e.
it is not a mapping from N to NxN. This is definitely not the most
clever way, it may help to visualize the final mapping though. But the
"final" mapping can be built directly without using such a sequence.
I have applied Cantor's mapping. Only this is discussed in "Proof of the existence of dark numbers (bilingual version)", OSFPREPRINTS (Nov 2022)
He's again, assuming that if a property is true for all members of a
sequence then it is a property of its limit.
There is no limit of the enumeration other than the complete application
of all indices. In all matrices the O's exist with not a single loss.
The sequence 0.1, 0.01. 0.001, ... for example has the limit 0, but the complete enumeration is the complete sequence without this limit. That
is Cantor's "limit".
Obviously you confuse the analytical limit which has no digit 1 with the complete enumeration where all terms have one digit 1. This property is
true for all terms like the O's remaining in all matrices.
Regards, WM
Chris M. Thomasson formulated on Wednesday :
On 8/6/2025 10:11 AM, WM wrote:
On 05.08.2025 23:33, joes wrote:
Am Tue, 05 Aug 2025 22:01:23 +0200 schrieb WM:
When dealing with Cantor's mappings between infinite sets, it isWeren't you the one who complained that the process never finished?
argued
usually that these mappings require a "limit" to be completed or that >>>>> they cannot be completed. Such arguing has to be rejected flatly.
Either the enumeration of the rationals is never finished, or, if it
is claimed to be finished, dark numbers are needed.
Do you think complete and/or all means finite?
He seems to mean that finite means no dark numbers are needed, therefore infinite means they are needed. :D
He is insane.
On 8/6/2025 1:35 PM, WM wrote:
Every matrix contains O's, i.e. not indexed fractions.
Name a fraction, aka a Cantor Pair in the form of (x, y) as (x/y) that
is not indexed?
Am Wed, 06 Aug 2025 19:34:27 +0200 schrieb WM:
On 06.08.2025 18:59, Alan Mackenzie wrote:There is no index "where" they leave, it happens in the limit process,
WM <wolfgang.mueckenheim@tha.de> wrote:How and where do they leave?
On 06.08.2025 14:43, Alan Mackenzie wrote:Yes. For any number of steps. But NOT in the limit.
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away >>>>> to an unbounded distance.Which however is always finite. So the O's remain in the matrix.
But don't try to stultify students. If all n fail to enumerate the
infinitely many fractions equipped with O's but you claim that
afterwards all fractions are indexed (by what), then every intelligent
student recognizes that you don't tell the truth.
the total of all the steps.
Of course. The limit is not a term.Using an analogous, but simpler example, consider the sequence of realAnd the limit is never reached by any f(n). That's the same as with
numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
Cantor's enumeration. Note that indexing is only possible by natural
numbers, not by -e.
They don't.The limit of the sequence (I hope you can agree to this) is 1. ThisNot by a long way. There are infinitely many O's. They cannot vanish
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1
"remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
immediately.
Do you agree that the limit is 1?
Cantor invented completion?The completion of an enumeration.No, that has not been the subject of analysis. It is only Cantor'sTo what do your "that" and your "it" refer?
invention. And he has not used infinite numbers to enumerate.
So which term enumerates all positions?Maybe the sequence I depict above might do a better job.This is a nice example. It shows that you confuse the enumeration of all
terms of the sequence (which in fact is done by the position of the
second 1) and the limit which has nothing to do with this enumeration.
The limit is a handy compression of a sequence.
On 07.08.2025 01:38, Chris M. Thomasson wrote:
On 8/6/2025 1:35 PM, WM wrote:
Every matrix contains O's, i.e. not indexed fractions.
Name a fraction, aka a Cantor Pair in the form of (x, y) as (x/y) that
is not indexed?
All fractions that can be named get indexed.
Nevertheless most fractions remain unindexed.
It is impossible to shuffle one X per line over the matrix such that the whole matrix is covered.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.08.2025 01:38, Chris M. Thomasson wrote:
On 8/6/2025 1:35 PM, WM wrote:
Every matrix contains O's, i.e. not indexed fractions.
Name a fraction, aka a Cantor Pair in the form of (x, y) as (x/y) that
is not indexed?
All fractions that can be named get indexed.
All fractions can be named, and all get indexed. That's what Cantor demonstrated. If you _really_ believe this isn't the case, meet Chris's challenge and name a fraction which cannot be indexed.
Nevertheless most fractions remain unindexed.
Quatsch! Again, name a single fraction which will not be indexed.
It is impossible to shuffle one X per line over the matrix such that the
whole matrix is covered.
We've already discussed that to death. The plain fact is you are wrong
here, too.
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions that can be named get indexed.
All fractions can be named,
and all get indexed. That's what Cantor demonstrated.
Nevertheless most fractions remain unindexed.
Quatsch!
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions can be named, and all get indexed.
That's what Cantor
demonstrated.
If you _really_ believe this isn't the case, meet Chris's.
challenge and name a fraction which cannot be indexed.
Nevertheless most fractions remain unindexed.
Quatsch! Again, name a single fraction which will not be indexed.
It is impossible to shuffle one X per line over the matrix such that the
whole matrix is covered.
We've already discussed that to death.
here, too.
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions that can be named get indexed.
All fractions can be named,
Indeed.
On 07.08.2025 20:20, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions can be named, and all get indexed.
Naming is done by exchange of X and O.
That's what Cantor
demonstrated.
He did so for definable numbers not knowing that most numbers are undefinable. This is proved by the O's.
-aIf you _really_ believe this isn't the case, meet Chris's.
challenge and name a fraction which cannot be indexed.
I do not believe but have proved. But most dark numbers cannot be defined.
Nevertheless most fractions remain unindexed.
Quatsch!-a Again, name a single fraction which will not be indexed.
Have you understood that your example with the analytical limit of the sequence is nonsense?
It is impossible to shuffle one X per line over the matrix such that the >>> whole matrix is covered.
We've already discussed that to death.
You have discussed the analytical limit which has nothing to do with enumerating terms. Have you understood my explanation?
The terms 10^-n of the sequence (10^-n) are enumerated by n. The limit 0
is not a term and is not enumerated. It has nothing to do with Cantor's theory.
-a The plain fact is you are wrong
here, too.
The plain fact is that you have no arguments but your belief.
Regards, WM
On 07.08.2025 22:03, Moebius wrote:Get and deliver? I don't understand what you want to say.
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:Indeed counting requires according to Cantor and me *) to get an X and
WM <wolfgang.mueckenheim@tha.de> wrote:Indeed.
All fractions that can be named get indexed.All fractions can be named,
to deliver an O. Therefore only few fractions can be counted. But all definable fractios can be counted. That proves undefinable fractions.
*) WM: Die Cantorsche Formel ergibt meine Matrizen. FF: Das hatte ichWrong group, but Cantor's formula produces none of your matrices but
mir schon gedacht. Und das stimmt auch.
Merke: Die Cantorsche Formel hat nichts mit dem analytischen Grenzwert
der Matrizen zu tun.
On 06.08.2025 22:12, joes wrote:Yes yes, the limit is the result of that. No single step finishes it.
Am Wed, 06 Aug 2025 19:34:27 +0200 schrieb WM:
On 06.08.2025 18:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
There is no limit process. There is only the process of enumerating. ItThere is no index "where" they leave, it happens in the limit process,How and where do they leave? If all n fail to enumerate theWhich however is always finite. So the O's remain in the matrix.Yes. For any number of steps. But NOT in the limit.
infinitely many fractions equipped with O's but you claim that
afterwards all fractions are indexed (by what), then every intelligent
student recognizes that you don't tell the truth.
the total of all the steps.
is defined by exchange of O and X
What's your point?But indexing is done in the terms only.Of course. The limit is not a term.Using an analogous, but simpler example, consider the sequence ofAnd the limit is never reached by any f(n). That's the same as with
real numbers in decimal: 1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
Cantor's enumeration. Note that indexing is only possible by natural
numbers, not by -e.
How can the 1 disappear?Of course. But it has no bearing on the indexing.They don't. Do you agree that the limit is 1?The limit of the sequence (I hope you can agree to this) is 1. ThisNot by a long way. There are infinitely many O's. They cannot vanish
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That
1 "remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
immediately.
That refers to the sequence given by his formula being a bijection.So he said. "Die so definirte unendliche Reihe hat nun das merkw|+rdigeCantor invented completion?The completion of an enumeration.No, that has not been the subject of analysis. It is only Cantor'sTo what do your "that" and your "it" refer?
invention. And he has not used infinite numbers to enumerate.
an sich, s|nmtliche positiven rationalen Zahlen und jede von ihnen nur
einmal an einer bestimmten Stelle zu enthalten." Note: s|nmtliche.
Answer the question.Alas it has nothing to do with counting of the terms.This is a nice example. It shows that you confuse the enumeration ofSo which term enumerates all positions?
all terms of the sequence (which in fact is done by the position of
the second 1) and the limit which has nothing to do with this
enumeration.
The limit is a handy compression of a sequence.
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions that can be named get indexed.All fractions can be named,
all get indexed. That's what Cantor demonstrated.
Nevertheless most fractions remain unindexed.
Quatsch!
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions that can be named get indexed.All fractions can be named,
Indeed. If n/m is a fraction, the string consisting of n "|"s followed
by an "/" followed by m "|"s may be considered a name for n/m.
You see, M|+ckenheim:
1/1 is referred to by "|/|". In other words, "|/|" is a name for 1/1.
1/2 is referred to by "|/||". In other words, "|/||" is a name for 1/2.
2/1 is referred to by "||/|". In other words, "||/|" is a name for 2/1.
and so on.
[Hint @ M|+ckenheim: The mathematical "reality" is not
"bound"/"restricted" by the physical "reality". Mathematical objects do
not "exist" ("reside") in the physical reality. That's why mathematical theories do NOT refer to the "physical universe". Except in your delusion.]
On the other hand,
all get indexed. That's what Cantor demonstrated.
Indeed! Actually, this does not depend on M|+ckenheim's condition "can be named". [And even if it were, your claim would still be true.]
Nevertheless most fractions remain unindexed.
Quatsch!
Right. Complete nonsense.
Hint @ M|+ckenheim: If n/m is a fraction then m + ((m + n reA 1) (m + n reA 2))/2 is its index. Too complicated for you? <facepalm>
So there is no fraction which "remains unindexed".
.
.
.
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions that can be named get indexed.All fractions can be named,
Indeed. If n/m is a fraction, the string consisting of n "|"s followed
by an "/" followed by m "|"s may be considered a name for n/m.
You see, M|+ckenheim:
1/1 is referred to by "|/|". In other words, "|/|" is a name for 1/1.
1/2 is referred to by "|/||". In other words, "|/||" is a name for 1/2.
2/1 is referred to by "||/|". In other words, "||/|" is a name for 2/1.
and so on.
[Hint @ M|+ckenheim: The mathematical "reality" is not
"bound"/"restricted" by the physical "reality". Mathematical objects do
not "exist" ("reside") in the physical reality. That's why mathematical theories do NOT refer to the "physical universe". Except in your delusion.]
On the other hand,
all get indexed. That's what Cantor demonstrated.
Indeed! Actually, this does not depend on M|+ckenheim's condition "can be named". [And even if it were, your claim would still be true.]
Nevertheless most fractions remain unindexed.
Quatsch!
Right. Complete nonsense.
Hint @ M|+ckenheim: If n/m is a fraction then m + ((m + n reA 1) (m + n reA 2))/2 is its index. Too complicated for you? <facepalm>
So there is no fraction which "remains unindexed".
Am Thu, 07 Aug 2025 22:43:15 +0200 schrieb WM:
On 07.08.2025 22:03, Moebius wrote:Get and deliver?
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:Indeed counting requires according to Cantor and me *) to get an X and
WM <wolfgang.mueckenheim@tha.de> wrote:Indeed.
All fractions that can be named get indexed.All fractions can be named,
to deliver an O. Therefore only few fractions can be counted. But all
definable fractios can be counted. That proves undefinable fractions.
Fortunately Cantor was only concerned with "definable" fractinos.
Am Thu, 07 Aug 2025 19:06:58 +0200 schrieb WM:
On 06.08.2025 22:12, joes wrote:
Am Wed, 06 Aug 2025 19:34:27 +0200 schrieb WM:
On 06.08.2025 18:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Yes yes, the limit is the result of that. No single step finishes it.There is no limit process. There is only the process of enumerating. ItThere is no index "where" they leave, it happens in the limit process,How and where do they leave? If all n fail to enumerate theWhich however is always finite. So the O's remain in the matrix.Yes. For any number of steps. But NOT in the limit.
infinitely many fractions equipped with O's but you claim that
afterwards all fractions are indexed (by what), then every intelligent >>>> student recognizes that you don't tell the truth.
the total of all the steps.
is defined by exchange of O and X
What's your point?But indexing is done in the terms only.Of course. The limit is not a term.Using an analogous, but simpler example, consider the sequence ofAnd the limit is never reached by any f(n). That's the same as with
real numbers in decimal: 1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
Cantor's enumeration. Note that indexing is only possible by natural
numbers, not by -e.
How can the 1 disappear?Of course. But it has no bearing on the indexing.They don't. Do you agree that the limit is 1?The limit of the sequence (I hope you can agree to this) is 1. ThisNot by a long way. There are infinitely many O's. They cannot vanish
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That >>>>> 1 "remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and >>>>> O's.
immediately.
That refers to the sequence given by his formula being a bijection.So he said. "Die so definirte unendliche Reihe hat nun das merkw|+rdigeCantor invented completion?The completion of an enumeration.No, that has not been the subject of analysis. It is only Cantor's >>>>>> invention. And he has not used infinite numbers to enumerate.To what do your "that" and your "it" refer?
an sich, s|nmtliche positiven rationalen Zahlen und jede von ihnen nur
einmal an einer bestimmten Stelle zu enthalten." Note: s|nmtliche.
Answer the question.Alas it has nothing to do with counting of the terms.This is a nice example. It shows that you confuse the enumeration ofSo which term enumerates all positions?
all terms of the sequence (which in fact is done by the position of
the second 1) and the limit which has nothing to do with this
enumeration.
The limit is a handy compression of a sequence.
On 07.08.2025 23:01, joes wrote:
Am Thu, 07 Aug 2025 22:43:15 +0200 schrieb WM:
On 07.08.2025 22:03, Moebius wrote:Get and deliver?
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:Indeed counting requires according to Cantor and me *) to get an X and
WM <wolfgang.mueckenheim@tha.de> wrote:Indeed.
All fractions that can be named get indexed.All fractions can be named,
to deliver an O. Therefore only few fractions can be counted. But all
definable fractios can be counted. That proves undefinable fractions.
A fraction marked by an O takes an X and return an O.
Fortunately Cantor was only concerned with "definable" fractinos.
He thought so. But he was wrong. Proof: All exchanges between X and O
can only happen in terms of the sequence, not in the analytical limit. However many O's remain.
Regards, WM
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot be uniquely indexed?
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot be
uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/tyvnk_v1 or "New proof of dark numbers by means of the thinned out harmonic series", OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/osf/53qg2_v1?view_only=. If you have questions, then ask.
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot be
uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/tyvnk_v1 or "New proof of dark numbers by means of the thinned out harmonic series", OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/osf/53qg2_v1?view_only=.
If you have questions, then ask.
Regards, WM--
Le 08/08/2025 |a 22:41, WM a |-crit :
On 08.08.2025 22:00, Chris M. Thomasson wrote:
[M|+ckenheim, n]ame an [ordered] pair (x, y), [...] that cannot be uniquely indexed.
Try to <bla bla bla>
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot be
uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/tyvnk_v1 or "New proof of dark numbers by means of the thinned out harmonic series", OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/osf/53qg2_v1? view_only=.
If you have questions, then ask.
Le 08/08/2025 |a 22:41, WM a |-crit :
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot
be uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual
version)", OSFPREPRINTS (Nov 2022)
https://osf.io/preprints/osf/tyvnk_v1 or
"New proof of dark numbers by means of the thinned out harmonic
series", OSFPREPRINTS (10 Mar 2025)
https://osf.io/preprints/osf/53qg2_v1?view_only=.
If you have questions, then ask.
Evading the question
by mentioning complete idiotic "articles" of yours
is not an acceptable answer crank Wolfgang M|+ckenheim, from Hochscule Augsburg.
On 8/8/2025 1:41 PM, WM wrote:
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot
be uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual
version)", OSFPREPRINTS (Nov 2022)
https://osf.io/preprints/osf/tyvnk_v1 or
"New proof of dark numbers by means of the thinned out harmonic
series", OSFPREPRINTS (10 Mar 2025)
https://osf.io/preprints/osf/53qg2_v1? view_only=.
If you have questions, then ask.
Sigh. Name a Cantor pair (x, y), where (x/y) is the fraction, that
cannot be uniquely indexed?
If (n, m) (n/m) is an ordered pair (a fraction) then
m + ((m + n reA 1) (m + n reA 2))/2 is its index.
WM <wolfgang.mueckenheim@tha.de> wrote:
Try to understand "Proof of the existence of dark numbers (bilingual
version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/tyvnk_v1 or >> "New proof of dark numbers by means of the thinned out harmonic series",
OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/osf/53qg2_v1?view_only=.
We do understand the contents of these web pages. They're not rigorous mathematics
having been discussed at length here and found wanting.
If you have questions, then ask.
Chris did ask a question, which you failed to answer, namely: Name a
Cantor pair (x, y), where (x/y) is the fraction, that cannot be uniquely indexed?
Please answer it now.
On 08.08.2025 22:53, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Try to understand "Proof of the existence of dark numbers (bilingual
version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/tyvnk_v1 or >>> "New proof of dark numbers by means of the thinned out harmonic series", >>> OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/osf/53qg2_v1?view_only=.
We do understand the contents of these web pages. They're not rigorous
mathematics
They are more rigorous than all your set-theoretical knowledge.
having been discussed at length here and found wanting.
In principle there is nothing to be discussed. Unless you were blinded
by Cantor's nonsense, you would recognize immediately (as most of my students)
Conquer the Binary Tree
.
/ \
0 1
/\ /\
0 1 0 1
/\ /\ /\ /\
...
The complete infinite Binary Tree consists of nodes representing bits
(binary digits 0 and 1) which are indexed by non-negative integers and connected by edges such that every node has two and only two child
nodes. Node number 2n + 1 is called the left child of node number n,
node number 2n + 2 is called the right child of node number n. The set
{a_k | k ree rao_0} of nodes a_k is countable as shown by the indices of the nodes.
To play the game Conquer the Binary Tree you start with one cent. For
one cent you can buy an infinite path of your choice in the Binary Tree.
For every node covered by this path you will get a cent. For every cent
you can buy another path of your choice. For every node covered by this
path (and not yet covered by previously chosen paths) you will get a
cent. For every cent you can buy another path. And so on. Since there
are only countably many nodes yielding as many cents but uncountably
many paths requiring as many cents, the player will get bankrupt before
all paths are conquered. If no player gets bankrupt, the number of paths cannot surpass the number of nodes.
Note: If set theory is right, then most paths that you can buy do not
contain new nodes.
Regards, WM
On 07/30/2025 10:29 AM, WM wrote:
Conquer the Binary Tree
.
/ \
0 1
/\ /\
0 1 0 1
/\ /\ /\ /\
...
The complete infinite Binary Tree consists of nodes representing bits
(binary digits 0 and 1) which are indexed by non-negative integers and
connected by edges such that every node has two and only two child
nodes. Node number 2n + 1 is called the left child of node number n,
node number 2n + 2 is called the right child of node number n. The set
{a_k | k ree rao_0} of nodes a_k is countable as shown by the indices of the >> nodes.
To play the game Conquer the Binary Tree you start with one cent. For
one cent you can buy an infinite path of your choice in the Binary Tree.
For every node covered by this path you will get a cent. For every cent
you can buy another path of your choice. For every node covered by this
path (and not yet covered by previously chosen paths) you will get a
cent. For every cent you can buy another path. And so on. Since there
are only countably many nodes yielding as many cents but uncountably
many paths requiring as many cents, the player will get bankrupt before
all paths are conquered. If no player gets bankrupt, the number of paths
cannot surpass the number of nodes.
Note: If set theory is right, then most paths that you can buy do not
contain new nodes.
Regards, WM
With the natural/unit equivalency function, it results a
Square Cantor Space, whose iteration is both a depth- and
breadth-first traversal, of this infinite structure.
There are at least three definitions of continuity,
at least three laws of large numbers, at least three
kinds of Cantor space, and at least three probabilistic
limit theorems.
Le 09/08/2025 |a 16:11, Ross Finlayson a |-crit :
On 07/30/2025 10:29 AM, WM wrote:
Conquer the Binary Tree
.
/ \
0 1
/\ /\
0 1 0 1
/\ /\ /\ /\
...
The complete infinite Binary Tree consists of nodes representing bits
(binary digits 0 and 1) which are indexed by non-negative integers and
connected by edges such that every node has two and only two child
nodes. Node number 2n + 1 is called the left child of node number n,
node number 2n + 2 is called the right child of node number n. The set
{a_k | k ree rao_0} of nodes a_k is countable as shown by the indices of the
nodes.
To play the game Conquer the Binary Tree you start with one cent. For
one cent you can buy an infinite path of your choice in the Binary Tree. >>> For every node covered by this path you will get a cent. For every cent
you can buy another path of your choice. For every node covered by this
path (and not yet covered by previously chosen paths) you will get a
cent. For every cent you can buy another path. And so on. Since there
are only countably many nodes yielding as many cents but uncountably
many paths requiring as many cents, the player will get bankrupt before
all paths are conquered. If no player gets bankrupt, the number of paths >>> cannot surpass the number of nodes.
Note: If set theory is right, then most paths that you can buy do not
contain new nodes.
Regards, WM
With the natural/unit equivalency function, it results a
Square Cantor Space, whose iteration is both a depth- and
breadth-first traversal, of this infinite structure.
There are at least three definitions of continuity,
at least three laws of large numbers, at least three
kinds of Cantor space, and at least three probabilistic
limit theorems.
And at least three kind of people : those who can count and those who
can't.
Seriously Ross, you are a joke, right?
In principle there is nothing to be discussed. Unless you were blinded
by Cantor's nonsense, you would recognize immediately (as most of my
students)
"most" ? So you encountered some resistance.
On 08.08.2025 23:32, Chris M. Thomasson wrote:
On 8/8/2025 1:41 PM, WM wrote:
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot
be uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual
version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/
tyvnk_v1 or
"New proof of dark numbers by means of the thinned out harmonic
series", OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/
osf/53qg2_v1? view_only=.
If you have questions, then ask.
Sigh. Name a Cantor pair (x, y), where (x/y) is the fraction, that
cannot be uniquely indexed?
Look, if there are really ra|reC natural numbers, then not all can be smaller than ra|reC/2. In fact half of them have to be larger. But none of them can be identified.
On 09.08.2025 15:37, Python wrote:
In principle there is nothing to be discussed. Unless you were
blinded by Cantor's nonsense, you would recognize immediately (as
most of my students)
"most" ? So you encountered some resistance.
No, but I could not ask each one.
On 07.08.2025 01:38, Chris M. Thomasson wrote:
On 8/6/2025 1:35 PM, WM wrote:
All fractions that can be named get indexed.Every matrix contains O's, i.e. not indexed fractions.
Name a fraction, aka a Cantor Pair in the form of (x, y) as (x/y) that
is not indexed?
Nevertheless most fractions remain unindexed.
It is impossible to shuffle one X per line over the matrix such that the whole matrix is covered.
Regards, WM
On 8/9/2025 6:13 AM, WM wrote:
On 08.08.2025 23:32, Chris M. Thomasson wrote:
On 8/8/2025 1:41 PM, WM wrote:
On 08.08.2025 22:00, Chris M. Thomasson wrote:
On 8/8/2025 5:23 AM, WM wrote:
A fraction marked by an O takes an X and return an O.
Name a Cantor pair (x, y), where (x/y) is the fraction, that cannot
be uniquely indexed?
Try to understand "Proof of the existence of dark numbers (bilingual
version)", OSFPREPRINTS (Nov 2022) https://osf.io/preprints/osf/
tyvnk_v1 or
"New proof of dark numbers by means of the thinned out harmonic
series", OSFPREPRINTS (10 Mar 2025) https://osf.io/preprints/
osf/53qg2_v1? view_only=.
If you have questions, then ask.
Sigh. Name a Cantor pair (x, y), where (x/y) is the fraction, that
cannot be uniquely indexed?
Look, if there are really ra|reC natural numbers, then not all can be
smaller than ra|reC/2. In fact half of them have to be larger. But none of >> them can be identified.
Huh?
On 8/7/2025 8:37 AM, WM wrote:
All fractions that can be named get indexed.
Nevertheless most fractions remain unindexed.
Wow! Any cantor pair (x, y) can be indexed.
The fraction (x/y) is just a
way to show a fraction from any cantor pair.
It is impossible to shuffle one X per line over the matrix such that
the whole matrix is covered.
On 09.08.2025 20:11, Chris M. Thomasson wrote:
On 8/7/2025 8:37 AM, WM wrote:
All fractions that can be named get indexed.
Nevertheless most fractions remain unindexed.
Wow! Any cantor pair (x, y) can be indexed.
Yes.
The fraction (x/y) is just a way to show a fraction from any cantor pair.
Alas there are, according to Cantor, |rao| natural numbers. Can all be smaller than |rao|/2? Hardly.
It is impossible to shuffle one X per line over the matrix such that
the whole matrix is covered.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element.
Regards, WM
On 08/10/2025 06:08 AM, WM wrote:
On 09.08.2025 20:11, Chris M. Thomasson wrote:
On 8/7/2025 8:37 AM, WM wrote:
All fractions that can be named get indexed.
Nevertheless most fractions remain unindexed.
Wow! Any cantor pair (x, y) can be indexed.
Yes.
The fraction (x/y) is just a way to show a fraction from any cantor
pair.
Alas there are, according to Cantor, |rao| natural numbers. Can all be
smaller than |rao|/2? Hardly.
It is impossible to shuffle one X per line over the matrix such that
the whole matrix is covered.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element.
Regards, WM
For an echo chamber, 'tis pretty big.
Whether writing "'tis" for "it is" emphasizes the verb rather than
subject, goes to show language has its meanings.
Yeah, a lot of time "that" goes a long way to establish meaning,
yet these days people can't even be bothered to include their commas,
each omission of which is a little loss of meaning.
Don't mean much.
Another usual example of an inductive impasse readily dispatched
with analytical bridges in the overall deductive, the wider deductive
and ab-ductive if you will yet that's a kind of deductive, inference,
once again we see there's a bridge of Zeno an invincible going-forwarder
yet may not cross.
That, ....
On 09.08.2025 20:11, Chris M. Thomasson wrote:
On 8/7/2025 8:37 AM, WM wrote:
All fractions that can be named get indexed.
Nevertheless most fractions remain unindexed.
Wow! Any cantor pair (x, y) can be indexed.
Yes.
The fraction (x/y) is just a way to show a fraction from any cantor pair.
Alas there are, according to Cantor, |rao| natural numbers. Can all be smaller than |rao|/2? Hardly.
It is impossible to shuffle one X per line over the matrix such that
the whole matrix is covered.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
Which however is always finite. ....
Yes.
.... So the O's remain in the matrix.
Yes. For any number of steps. But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1. This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1 "remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
Years ago I used a very specific simpler example, using 0 and 1 rather
than X an 0 and a one-dimensional "grid". One can use (the Cantor index
of) fractions or, even simpler, start with an alternating sequence and,
step by step, just swap the first 1 with the first following 0:
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s go?" he might ask his students.
Am 11.08.2025 um 00:31 schrieb Ben Bacarisse:
Years ago I used a very specific simpler example, using 0 and 1 rather than X an 0 and a one-dimensional "grid".-a One can use (the Cantor index of) fractions or, even simpler, start with an alternating sequence and, step by step, just swap the first 1 with the first following 0:
-a-a-a s_0-a =-a 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
-a-a-a s_1-a =-a 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
-a-a-a s_2-a =-a 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
-a-a-a s_3-a =-a 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s go?" he might ask his students.
Recently, I posted a similar example in de.sci.mathematic:
Wir betrachten eine Folge von Folgen und deren (punktweisen) Grenzwert.
Die Folge sei (f_0, f_1, f_2, f_3, ...) mit
f_0 = (0, 1, 0, 1, 0, 1, 0, 1, ...)
f_1 = (1, 0, 0, 1, 0, 1, 0, 1, ...)
f_2 = (1, 1, 0, 0, 0, 1, 0, 1, ...)
f_3 = (1, 1, 1, 0, 0, 0, 0, 1, ...)
usw.
Es ist dann lim f_n = (1, 1, 1, 1, ...) = (a_n)_(n e IN) mit a_n = 1 f|+r alle n e IN.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Of course, no meaningful answer from crank Wolfgang M|+ckenheim.
.
.
.
.
Am 11.08.2025 um 00:31 schrieb Ben Bacarisse:
Years ago I used a very specific simpler example, using 0 and 1 rather than X an 0 and a one-dimensional "grid". One can use (the Cantor index of) fractions or, even simpler, start with an alternating sequence and, step by step, just swap the first 1 with the first following 0:
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s go?" he might ask his students.
Recently, I posted a similar example in de.sci.mathematic:
Wir betrachten eine Folge von Folgen und deren (punktweisen) Grenzwert.
Die Folge sei (f_0, f_1, f_2, f_3, ...) mit
f_0 = (0, 1, 0, 1, 0, 1, 0, 1, ...)
f_1 = (1, 0, 0, 1, 0, 1, 0, 1, ...)
f_2 = (1, 1, 0, 0, 0, 1, 0, 1, ...)
f_3 = (1, 1, 1, 0, 0, 0, 0, 1, ...)
usw.
Es ist dann lim f_n = (1, 1, 1, 1, ...) = (a_n)_(n e IN) mit a_n = 1 f|+r alle n e IN.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Of course, no meaningful answer from crank Wolfgang M|+ckenheim.
.
.
.
.
On 08/10/2025 04:38 PM, Moebius wrote:
Am 11.08.2025 um 00:31 schrieb Ben Bacarisse:
Years ago I used a very specific simpler example, using 0 and 1 ratherindex
than X an 0 and a one-dimensional "grid". One can use (the Cantor
of) fractions or, even simpler, start with an alternating sequenceand,
step by step, just swap the first 1 with the first following 0:go?" he
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s
might ask his students.
Recently, I posted a similar example in de.sci.mathematic:
Wir betrachten eine Folge von Folgen und deren (punktweisen) Grenzwert.
Die Folge sei (f_0, f_1, f_2, f_3, ...) mit
f_0 = (0, 1, 0, 1, 0, 1, 0, 1, ...)
f_1 = (1, 0, 0, 1, 0, 1, 0, 1, ...)
f_2 = (1, 1, 0, 0, 0, 1, 0, 1, ...)
f_3 = (1, 1, 1, 0, 0, 0, 0, 1, ...)
usw.
Es ist dann lim f_n = (1, 1, 1, 1, ...) = (a_n)_(n e IN) mit a_n = 1 f|+r
alle n e IN.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Of course, no meaningful answer from crank Wolfgang M|+ckenheim.
.
.
.
.
Maybe think of it as being a Hilbert's Hotel franchisee. So, you
have not one, but three Hilbert's Hotels, and they're full. So,
the bellboy, has that each Hilbert Hotel is only one long corridor,
so, the bellboy can not reach any room unless first passing all
the preceding rooms, where each has a natural number.
Then, it's to make it more like a balls-and-vase problem where
you're not allowed to break the rules by claiming some capriciously
arbitrary construction exists, instead that here these sort of
things have to be done in an order.
So, imagine Hotels 2 and 3 don't have any towels, while Hotel 1
does, so, due their clamoring complaints, you send the bellboy
to take towels from Hotel 1 and back-and-forth provide towels
to Hotels 2 and 3. Yet, the bellboy's lazy, and will only serve
the first room respectively with or without a towel, depending
on whether he is without or with a towel.
So, you can provide towels to Hotels 2 and 3, yet now Hotel 1 has none.
Then, in this case the guy only had two hotels in his franchise to
begin with, and when you come up with a third hotel and this "Cantor Pairing", there's nothing he can do about it, because he would have
to start all over with a brand new hotel with infinitely many towels,
and another bellboy.
Or, you know, you could start right away, yet maybe one of the
reasons the bellboy is so lazy is because he's constantly doing
busywork with no recognition.
So, in mathematics, given that sort of contrivance, it's pretty simple
that you can't wish it away.
I.e., you're starting all over.
Alan Mackenzie <acm@muc.de> writes:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to >>>> an unbounded distance.
Which however is always finite. ....
Yes.
.... So the O's remain in the matrix.
Yes. For any number of steps. But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real
numbers in decimal:
One of the things I used to think was odd was the complexity a WM's
examples. But then I decided this was deliberate.
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1. This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1
"remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
Years ago I used a very specific simpler example, using 0 and 1 rather
than X an 0 and a one-dimensional "grid". One can use (the Cantor index
of) fractions or, even simpler, start with an alternating sequence and,
step by step, just swap the first 1 with the first following 0:
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s go?" he might ask his students.
One day he might get a student who (a) points out that such sequences
are just functions from N to {0,1}. (b) The sequence of functions s_n
has a well-defined limit. (c) WM's own textbook defines this limit and
shows how to calculate it!
[Also, he used to vehemently deny that any non-constant set sequences
have limits. But his textbook defines functions as sets (sets of pairs)
and defines limits for certain sequences of such sets.]
----
Ben.
On 09.08.2025 20:11, Chris M. Thomasson wrote:Maybe, just maybe, |rao|/2 isn't even defined.
On 8/7/2025 8:37 AM, WM wrote:Yes.
All fractions that can be named get indexed.Wow! Any cantor pair (x, y) can be indexed.
Nevertheless most fractions remain unindexed.
The fraction (x/y) is just aAlas there are, according to Cantor, |rao| natural numbers. Can all be smaller than |rao|/2? Hardly.
way to show a fraction from any cantor pair.
You are wrong there, and you know it. Your utterance of such blatantThe first column containing all natural numbers is infinite. But allIt is impossible to shuffle one X per line over the matrix such that
the whole matrix is covered.
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element.
Regards, WM--
On 8/10/2025 6:08 AM, WM wrote:
On 09.08.2025 20:11, Chris M. Thomasson wrote:
On 8/7/2025 8:37 AM, WM wrote:
All fractions that can be named get indexed.
Nevertheless most fractions remain unindexed.
Wow! Any cantor pair (x, y) can be indexed.
Yes.
So, any fraction wrt (x/y) are indexed
I don't think you
have ever implemented Cantor Pairing wrt going back and forth in the
sense of mapping an index into a unique pair and back again? Am I right?
Years ago I used a very specific simpler example, using 0 and 1 rather
than X an 0 and a one-dimensional "grid". One can use (the Cantor index
of) fractions or, even simpler, start with an alternating sequence and,
step by step, just swap the first 1 with the first following 0:
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s go?" he might ask his students.
One day he might get a student who (a) points out that such sequences
are just functions from N to {0,1}. (b) The sequence of functions s_n
has a well-defined limit. (c) MW's own textbook defines this limit and
shows how to calculate it!
[Also, he used to vehemently deny that any non-constant set sequences
have limits.
Ben Bacarisse <ben@bsb.me.uk> wrote:
One of the things I used to think was odd was the complexity a WM's
examples. But then I decided this was deliberate.
Possibly. On the other hand, it takes understanding to reduce
complicated things to their essentials.
1.1, 1.01, 1.001, 1.0001, ......
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
That's a neat example!
WM is lacking basic abstract understanding. He doesn't understand what a limit actually is.
WM <wolfgang.mueckenheim@tha.de> wrote:
Alas there are, according to Cantor, |rao| natural numbers. Can all be
smaller than |rao|/2? Hardly.
Maybe, just maybe, |rao|/2 isn't even defined.
It is impossible to shuffle one X per line over the matrix such that
the whole matrix is covered.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element.
You are wrong there, and you know it.
Your utterance of such blatant
nonsense explains the contempt in which you are held here.
Recently, I posted a similar example in de.sci.mathematic:
Of course, no meaningful answer
WM <wolfgang.mueckenheim@tha.de> wrote:
there are, according to Cantor, |rao| natural numbers. (WM)
Can all be smaller than |rao|/2?
[...] |rao|/2 isn't even defined.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element. (WM)
You [WM] are wrong there, and you know it. Your utterance of such blatant nonsense explains the contempt in which you are held here.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted. They just move away to
an unbounded distance.
Which however is always finite. ....
Yes.
.... So the O's remain in the matrix.
Yes. For any number of steps. But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1. This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted". That 1 "remains in the number". But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
Am 11.08.2025 um 16:44 schrieb Moebius: [...]
Let's consider a simpler "example".
In the context of classical mathematics we may define (and consider) the following two IN x 2 "matrices":
M =
1 0
2 0
3 0
: :
and
M' =
1 2
3 4
5 6
: :
Now M|+ckenheim's "argument" reads:
"[In M] The first column containing all natural numbers is infinite.
But the second column is just as long as the first. Therefore it is
impossible to attach a natural number to every matrix element [like
in M']."
In other words, M' does not (can't!) exist in M|+ckenheim's world.
Well...
.
.
.
But since even WM will accept the existence of all these sequences (I
hope), this proves (I guess) that actual infinite sequences are
impossible in math! [After all, the assumption of their existence just
leads to paradoxes, as we have seen!]
Am 11.08.2025 um 14:37 schrieb Alan Mackenzie:|IN|, |rao|/2 would be a natural number,
WM <wolfgang.mueckenheim@tha.de> wrote:
And if it were, assuming that |rao|/2 is meant to be "smaller" than
since |rao| is the smallest INFINITE cardinal number;
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element. (WM)
<Utter bullshit>
Am 06.08.2025 um 18:59 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted.-a They just move away to >>>> an unbounded distance.
Which however is always finite.-a ....
Yes.
.... So the O's remain in the matrix.
Yes.-a For any number of steps.-a But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real
numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1.-a This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted".-a That 1
"remains in the number".-a But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
You might as well consider the sequence:
1.10101010..., 1.01101010..., 1.00111010..., 1.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 1. This limit has only one non-zero digit.
WM <wolfgang.mueckenheim@tha.de> wrote:
[...] according to Cantor, [there are] |rao| natural numbers. Can all be smaller than |rao|/2?
Maybe, just maybe, |rao|/2 isn't even defined.
WM <wolfgang.mueckenheim@tha.de> wrote:
there are, according to Cantor, |rao| natural numbers. (WM)
Can all be smaller than |rao|/2?
[...] |rao|/2 isn't even defined.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element. (WM)
You [WM] are wrong there, and you know it. Your utterance of such blatant nonsense explains the contempt in which you are held here.
Am 11.08.2025 um 17:02 schrieb Moebius:
Am 06.08.2025 um 18:59 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted.-a They just move
away to
an unbounded distance.
Which however is always finite.-a ....
Yes.
.... So the O's remain in the matrix.
Yes.-a For any number of steps.-a But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real
numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1.-a This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted".-a That 1
"remains in the number".-a But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
You might as well consider the sequence:
1.10101010..., 1.01101010..., 1.00111010..., 1.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 1. This limit has only one non-zero digit.
You might as well consider the sequence:
0.10101010..., 0.01101010..., 0.00111010..., 0.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 0. This limit has no non-zero digit.
A wonder!
Where did the 1s go to?!
.
.
.
Am 11.08.2025 um 17:02 schrieb Moebius:
Am 06.08.2025 um 18:59 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 06.08.2025 14:43, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
In the current scenario, O's don't get deleted.-a They just move
away to
an unbounded distance.
Which however is always finite.-a ....
Yes.
.... So the O's remain in the matrix.
Yes.-a For any number of steps.-a But NOT in the limit.
Using an analogous, but simpler example, consider the sequence of real
numbers in decimal:
1.1, 1.01, 1.001, 1.0001, ......
Every element of that sequence has two non-zero digits.
The limit of the sequence (I hope you can agree to this) is 1.-a This
limit has only one non-zero digit.
At no element of the sequence does the second 1 get "deleted".-a That 1
"remains in the number".-a But in the limit, it has gone.
This is essentially the same thing which is happening to your X's and
O's.
You might as well consider the sequence:
1.10101010..., 1.01101010..., 1.00111010..., 1.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 1. This limit has only one non-zero digit.
You might as well consider the sequence:
0.10101010..., 0.01101010..., 0.00111010..., 0.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 0. This limit has no non-zero digit.
A wonder!
Where did the 1s go to?!
.
.
.
"Bijections [...] of actually infinite sets and IN are impossible." (WM,
in one of his 'papers')
So IN can't be an "actually infinite" set. After all, the identity
function id: IN --> IN defined with id(n) = n (for all n in IN) is
clearly a bijection between IN and IN (even WM had to admit THAT MUCH).
Hence sequences with index set IN (usually called /infinite sequneces/) aren't actually infinite!
.
.
.
"Bijections [...] of actually infinite sets and IN are impossible." (WM,
in one of his 'papers')
So IN can't be an "actually infinite" set. After all, the identity
function id: IN --> IN defined with id(n) = n (for all n in IN) is
clearly a bijection between IN and IN (even WM had to admit THAT MUCH).
Hence sequences with index set IN (usually called /infinite sequneces/) aren't actually infinite!
.
.
.
WM <wolfgang.mueckenheim@tha.de> wrote:
there are, according to Cantor, |rao| natural numbers. (WM)
Can all be smaller than |rao|/2?
[...] |rao|/2 isn't even defined.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element. (WM)
You [WM] are wrong there, and you know it. Your utterance of such blatant nonsense explains the contempt in which you are held here.
WM <wolfgang.mueckenheim@tha.de> wrote:
there are, according to Cantor, |rao| natural numbers. (WM)
Can all be smaller than |rao|/2?
[...] |rao|/2 isn't even defined.
The first column containing all natural numbers is infinite. But all
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element. (WM)
You [WM] are wrong there, and you know it. Your utterance of such blatant nonsense explains the contempt in which you are held here.
You might as well consider the sequence:
0.10101010..., 0.01101010..., 0.00111010..., 0.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 0. This limit has no non-zero digit.
A wonder!
Where did the 1s go to?!
Then |rao|/2 just would be |rao|, since |rao| * 2 = |rao| (and there is no OTHER
cardinal number k such that k * 2 = |rao|).
And right, in this case, all natural numbers WOULD BE smaller than |rao|/2.
On 11.08.2025 22:24, Moebius wrote:
You might as well consider the sequence:
0.10101010..., 0.01101010..., 0.00111010..., 0.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 0. This limit has no non-zero digit.
A wonder!
Where did the 1s go to?!
That's not a wonder. The limit is not created by the steps of the
sequence, namely exchange of 0 and 1. The limit of my matrices is not created by exchange of X and O. The limit is not a member of the counted set.
Regards, WM
On other words, in each and every term (matrix) in the sequence of
matrices (WM considers) there are infinitely many O's.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hint: It seems to me that WM has a tendency to consider his "matrix" as
a "variable" mathematical entity.
On 11.08.2025 14:37, Alan Mackenzie wrote:It's not.
WM <wolfgang.mueckenheim@tha.de> wrote:If |rao| is defined as an integer or whole number, ....
Alas there are, according to Cantor, |rao| natural numbers. Can all beMaybe, just maybe, |rao|/2 isn't even defined.
smaller than |rao|/2? Hardly.
.... then there must be as many natural numbers. Otherwise |rao| and rao would be a lie only.I'll accept your expertise on lies. But as a hint, "as many" doesn't
Not the definition is lacking. But the numbers between |rao|/2 and |rao| are dark.There are no "dark numbers". Their non-existence has been proven on this newsgroup at least twice. And, as already implied, |rao|/2 is not
Not from you, I wouldn't.Do you accept analysis?You are wrong there, and you know it.The first column containing all natural numbers is infinite. But allIt is impossible to shuffle one X per line over the matrix such that >>>>> the whole matrix is covered.
other columns are just as long as the first. Therefore it is impossible
to attach a natural number to every matrix element.
We don't accept false pseudo-mathematics, as propounded by mathematically ill-educated cranks.Your utterance of such blatantThat is based on the stupidity of the readers here. They claim that
nonsense explains the contempt in which you are held here.
infinite set theory is the basis of mathematics but don't accept the
results of mathematics, ....
.... for instance the share of indices n/1 within the infinite matrix.Well, so what? The proportion of these numbers counted in that way may
The number of indices n/1 in the first column of an n*n-matrix is n.
Its share in the matrix is n/n^2. Here the limit tells us about the
share of enumerated fractions in the infinite matrix: lim(n-->oo) n/n^2
= 0.
Regards, WM--
Maybe you have a clue to what a limit is not. As I've said already,
you're lacking any understanding of what a limit is. You can perhaps
recite a definition from somewhere, but you can't actually grasp and use
that definition.
WM <wolfgang.mueckenheim@tha.de> wrote:
If |rao| is defined as an integer or whole number, ....
It's not.
Not the definition is lacking. But the numbers between |rao|/2 and |rao| are >> dark.
There are no "dark numbers". Their non-existence has been proven on
this newsgroup at least twice.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Alas there are, according to Cantor, |rao| natural numbers. Can all be >>>> smaller than |rao|/2? Hardly.
Maybe, just maybe, |rao|/2 isn't even defined.
If |rao| is defined as an integer or whole number, ....
It's not.
Not the definition is lacking. But the numbers between |rao|/2 and |rao| are >> dark.
There are no "dark numbers".
Their non-existence has been proven on this
newsgroup at least twice.
.... for instance the share of indices n/1 within the infinite matrix.
The number of indices n/1 in the first column of an n*n-matrix is n.
Its share in the matrix is n/n^2. Here the limit tells us about the
share of enumerated fractions in the infinite matrix: lim(n-->oo) n/n^2
= 0.
Well, so what? The proportion of these numbers counted in that way may
tend to zero,
their absolute number in the limit is countably infinite.
As is the count of all these numbers. They can be put into a 1-1 correspondence, hence they are "the same" size.
On 12.08.2025 15:16, Alan Mackenzie wrote:None of your matrices is an enumeration.
Maybe you have a clue to what a limit is not. As I've said already,
you're lacking any understanding of what a limit is. You can perhaps
recite a definition from somewhere, but you can't actually grasp and
use that definition.
But I can grap the fact that enumerating does not use a limit.
You seem too stupid to understand that. But it is fact. If you don't
believe me believe Cantor:
"If we think the numbers p/q in such an order [...] then every number
p/q comes at an absolutely fixed position of a simple infinite sequence"
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place."
Any limit in sight?
On 12.08.2025 16:04, Alan Mackenzie wrote:What's that got to do with it? |rao| is not an integer.
WM <wolfgang.mueckenheim@tha.de> wrote:Cantor: "ich nenne deren Ordnungstypen allgemein reale ganze Zahlen."
On 11.08.2025 14:37, Alan Mackenzie wrote:It's not.
WM <wolfgang.mueckenheim@tha.de> wrote:If |rao| is defined as an integer or whole number, ....
Alas there are, according to Cantor, |rao| natural numbers. Can all be >>>>> smaller than |rao|/2? Hardly.Maybe, just maybe, |rao|/2 isn't even defined.
In as much as you have defined them, yes I have. The two pertinentYou have not yet grasped them.Not the definition is lacking. But the numbers between |rao|/2 and |rao| areThere are no "dark numbers".
dark.
Please, I don't lie on Usenet, ever.Their non-existence has been proven on this newsgroup at least twice.Liar.
Of course they can. There are an uncountably infinite number of them,They will never cover the matrix..... for instance the share of indices n/1 within the infinite matrix.Well, so what? The proportion of these numbers counted in that way may
The number of indices n/1 in the first column of an n*n-matrix is n.
Its share in the matrix is n/n^2. Here the limit tells us about the
share of enumerated fractions in the infinite matrix: lim(n-->oo) n/n^2
= 0.
tend to zero,
Moebius has demonstrated such a covering explicitly.their absolute number in the limit is countably infinite.That nonsense notion may fit, but they will never cover the matrix.
You mean, not according to cranky pseudo-mathematical "analysis".As is the count of all these numbers. They can be put into a 1-1Not according to mathematical analysis.
correspondence, hence they are "the same" size.
Regards, WM--
On 11.08.2025 22:24, Moebius wrote:Same goes for your sequence.
You might as well consider the sequence:
0.10101010..., 0.01101010..., 0.00111010..., 0.00011110..., ......
Every element of that sequence has infinitely many non-zero digits.
The limit of the sequence is 0. This limit has no non-zero digit.
A wonder!
Where did the 1s go to?!
That's not a wonder.
The limit is not created by the steps of theOf course the limit is not a term. But how is it "created"?
sequence, namely exchange of 0 and 1. The limit of my matrices is not
created by exchange of X and O. The limit is not a member of the counted
set.
They will never cover the matrix. (WM)
Of course they can. There are an uncountably << countably?
infinite number of them,just as there are an uncountably << countably?
infinite number of cells in the matrix.
Thus there is a 1-1 correspondence between them, i.e. a covering.
That nonsense notion may fit, but they will never cover the matrix.
Moebius has demonstrated such a covering explicitly.
Am Tue, 12 Aug 2025 16:24:46 +0200 schrieb WM:
On 12.08.2025 15:16, Alan Mackenzie wrote:None of your matrices is an enumeration.
Maybe you have a clue to what a limit is not. As I've said already,
you're lacking any understanding of what a limit is. You can perhaps
recite a definition from somewhere, but you can't actually grasp and
use that definition.
But I can grap the fact that enumerating does not use a limit.
You seem too stupid to understand that. But it is fact. If you don't
believe me believe Cantor:
"If we think the numbers p/q in such an order [...] then every number
p/q comes at an absolutely fixed position of a simple infinite sequence"
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place."
Any limit in sight?
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Alas there are, according to Cantor, |rao| natural numbers. Can all be >>>>>> smaller than |rao|/2? Hardly.
Maybe, just maybe, |rao|/2 isn't even defined.
If |rao| is defined as an integer or whole number, ....
It's not.
Cantor: "ich nenne deren Ordnungstypen allgemein reale ganze Zahlen."
What's that got to do with it? |rao| is not an integer.
Not the definition is lacking. But the numbers between |rao|/2 and |rao| are
dark.
There are no "dark numbers".
You have not yet grasped them.
In as much as you have defined them, yes I have. The two pertinent
things about a "dark number" are (i) it is an integer; (ii) its value
cannot be pinned down in any way.
Their non-existence has been proven on this newsgroup at least twice.
Liar.
Please, I don't lie on Usenet, ever.
The proof I gave runs as follows. Suppose the "dark numbers" are a
non-empty subset of the integers, from (i) above. Then this subset, as
any non-empty subset of the integers, has a least member. This least
member is now defined, pinned down.
Therefore, by (ii) above, it can't
be a "dark number". This is a contradiction. Thus there cannot be such "dark numbers".
.... for instance the share of indices n/1 within the infinite matrix. >>>> The number of indices n/1 in the first column of an n*n-matrix is n.
Its share in the matrix is n/n^2. Here the limit tells us about the
share of enumerated fractions in the infinite matrix: lim(n-->oo) n/n^2 >>>> = 0.
Well, so what? The proportion of these numbers counted in that way may
tend to zero,
They will never cover the matrix.
Of course they can.
There are an uncountably infinite number of them,
just as there are an uncountably infinite number of cells in the matrix.
Thus there is a 1-1 correspondence between them, i.e. a covering.
their absolute number in the limit is countably infinite.
That nonsense notion may fit, but they will never cover the matrix.
Moebius has demonstrated such a covering explicitly.
As is the count of all these numbers. They can be put into a 1-1
correspondence, hence they are "the same" size.
Not according to mathematical analysis.
You mean, not according to cranky pseudo-mathematical "analysis".
Am 12.08.2025 um 17:28 schrieb Alan Mackenzie:
They will never cover the matrix. (WM)
Of course they can.-a There are an uncountably << countably?
infinite number of them,just as there are an uncountably << countably?
infinite number of cells in the matrix.
Thus there is a 1-1 correspondence between them, i.e. a covering.
That nonsense notion may fit, but they will never cover the matrix.
Moebius has demonstrated such a covering explicitly.
Even a PROPER SUBSET of the set of natural numbers suffice to "cover"
the "matrix".
We just may consider the "matrix" (a_n,m)_(n,m e IN) defind with
-a-a-a-a-a-a-a-a a_n,m = 2^n * 3^m-a-a-a (for all n,m e IN).
It's easy to show that for any n,m,n',m' with (n,m) =/= (n',m'): a_n,m
=/= a_n',m'. (And it's clear that {2^n * 3^m e IN : n,m e IN} is a
proper subset of IN.)
It seems to me that M|+ckenheim must reject most of basic modern maths
stuff (as well as logical and/or coherent thinking, of course) in his crusade against "set theory".
"One wonders by what [M|+ckenheim] would like to replace the mathematics created in the last 2500 years; if one takes Prof. M|+ckenheim seriously, then a fitting picture for the last page of this book ["The mathematics
of infinity"] would be the Ishango bone." (Franz Lemmermeyer)
https://en.wikipedia.org/wiki/Ishango_bone
.
.
.
Am 12.08.2025 um 17:28 schrieb Alan Mackenzie:
They will never cover the matrix. (WM)
Of course they can.-a There are an uncountably << countably?
infinite number of them,just as there are an uncountably << countably?
infinite number of cells in the matrix.
Thus there is a 1-1 correspondence between them, i.e. a covering.
That nonsense notion may fit, but they will never cover the matrix.
Moebius has demonstrated such a covering explicitly.
Even a PROPER SUBSET of the set of natural numbers suffice to "cover"
the "matrix".
We just may consider the "matrix" (a_n,m)_(n,m e IN) defind with
-a-a-a-a-a-a-a-a a_n,m = 2^n * 3^m-a-a-a (for all n,m e IN).
It's easy to show that for any n,m,n',m' with (n,m) =/= (n',m'): a_n,m
=/= a_n',m'. (And it's clear that {2^n * 3^m e IN : n,m e IN} is a
proper subset of IN.)
It seems to me that M|+ckenheim must reject most of basic modern maths
stuff (as well as logical and/or coherent thinking, of course) in his crusade against "set theory".
"One wonders by what [M|+ckenheim] would like to replace the mathematics created in the last 2500 years; if one takes Prof. M|+ckenheim seriously, then a fitting picture for the last page of this book ["The mathematics
of infinity"] would be the Ishango bone." (Franz Lemmermeyer)
https://en.wikipedia.org/wiki/Ishango_bone
.
.
.
On 12.08.2025 17:28, Alan Mackenzie wrote:I've been doing that for several decades. It's a damned difficult
WM <wolfgang.mueckenheim@tha.de> wrote:Try to learn German.
On 12.08.2025 16:04, Alan Mackenzie wrote:What's that got to do with it? |rao| is not an integer.
WM <wolfgang.mueckenheim@tha.de> wrote:Cantor: "ich nenne deren Ordnungstypen allgemein reale ganze Zahlen."
On 11.08.2025 14:37, Alan Mackenzie wrote:It's not.
WM <wolfgang.mueckenheim@tha.de> wrote:If |rao| is defined as an integer or whole number, ....
Alas there are, according to Cantor, |rao| natural numbers. Can all be >>>>>>> smaller than |rao|/2? Hardly.Maybe, just maybe, |rao|/2 isn't even defined.
You have never satisfactorally defined "dark numbers". All you've everFor most dark numbers, but not for all.In as much as you have defined them, yes I have. The two pertinentYou have not yet grasped them.Not the definition is lacking. But the numbers between |rao|/2 and |rao| areThere are no "dark numbers".
dark.
things about a "dark number" are (i) it is an integer; (ii) its value
cannot be pinned down in any way.
In otherwords, "dark numbers" are undefined. You're saying they're an undefinable subset of the integers, I think.Some dark numbers can get visible. The reason is the the visible numbersPlease, I don't lie on Usenet, ever.Their non-existence has been proven on this newsgroup at least twice.Liar.
The proof I gave runs as follows. Suppose the "dark numbers" are a
non-empty subset of the integers, from (i) above. Then this subset, as
any non-empty subset of the integers, has a least member. This least
member is now defined, pinned down.
are potentially infinite. There is no strict border.
Quatsch! I assumed well known properties of the integers, took your "definition", such as I can discern it, and followed through to theTherefore, by (ii) above, it can't be a "dark number". This is aYou assumed a strict border. That would contradict the potentially
contradiction. Thus there cannot be such "dark numbers".
infinite character of the collection of visible numbers.
Regards, WM--
Am 12.08.2025 um 17:28 schrieb Alan Mackenzie:
They will never cover the matrix. (WM)
Of course they can.-a There are an uncountably << countably?
infinite number of them,just as there are an uncountably << countably?
infinite number of cells in the matrix.
Thus there is a 1-1 correspondence between them, i.e. a covering.
That nonsense notion may fit, but they will never cover the matrix.
Moebius has demonstrated such a covering explicitly.
Even a PROPER SUBSET of the set of natural numbers suffice to "cover"
the "matrix".
We just may consider the "matrix" (a_n,m)_(n,m e IN) defind with
-a-a-a-a-a-a-a-a a_n,m = 2^n * 3^m-a-a-a (for all n,m e IN).
It's easy to show that for any n,m,n',m' with (n,m) =/= (n',m'): a_n,m
=/= a_n',m'. (And it's clear that {2^n * 3^m e IN : n,m e IN} is a
proper subset of IN.)
It seems to me that M|+ckenheim must reject most of basic modern maths
stuff (as well as logical and/or coherent thinking, of course) in his crusade against "set theory".
"One wonders by what [M|+ckenheim] would like to replace the mathematics created in the last 2500 years; if one takes Prof. M|+ckenheim seriously, then a fitting picture for the last page of this book ["The mathematics
of infinity"] would be the Ishango bone." (Franz Lemmermeyer)
https://en.wikipedia.org/wiki/Ishango_bone
.
.
.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Cantor: "ich nenne deren Ordnungstypen allgemein reale ganze Zahlen."
What's that got to do with it? |rao| is not an integer.
Try to learn German. [WM]
I've been doing that for several decades. It's a damned difficult
language. ;-)
Am 12.08.2025 um 19:15 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Cantor: "ich nenne deren Ordnungstypen allgemein reale ganze Zahlen."
Cantor: "*ich* nenne ..." ["*I* call ..."]
This not "German", but Cantor's TERMINOLOGY concerning some of the
notions in his "Transfinite Mengenlehre". <holy shit!>
Hint: "Modern set theory" did not adopt all of Cantor's TERMINOLOGY.
(See the works of Hessenberg (1906), Hausdorff (1914), Kamke (1928), etc.)
ORDINALs (ordinal numbers) are NOT referred to as "allgemein reale ganze Zahlen" these days, as the ASSHOLE M|+ckenheim knows very well.
What's that got to do with it?-a |rao| is not an integer.
Right. Using modern terminology: |rao| !e Z.
M|+ckenheim loves to play sleight of hand.
Actually, WM's "argument" would as well concern the following TRIVIAL case:
The prime numbers will never "cover" a 1 x IN "matrix" (i.e. a
sequence). So there can't be an infinite sequence (2, 3, 5, 7, ...) of
prime numbers.
After all there are as many terms in a sequence as natural numbers: a_1, a_2, a_3, ... (in other words, the index set of an infinite sequence is
IN), but there are far less prime numbers (than natural numbers), so HOW
CAN THEY cover ALL "places" a_1, a_2, a_3, ... in the sequence?!
Some dark numbers can get visible.
On 11.08.2025 00:31, Ben Bacarisse wrote:
Years ago I used a very specific simpler example, using 0 and 1 rather
than X an 0 and a one-dimensional "grid". One can use (the Cantor index
of) fractions or, even simpler, start with an alternating sequence and,
step by step, just swap the first 1 with the first following 0:
s_0 = 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
s_1 = 0, 0, 1, 1, 0, 1, 0, 1, 0, ...
s_2 = 0, 0, 0, 1, 1, 1, 0, 1, 0, ...
s_3 = 0, 0, 0, 0, 1, 1, 1, 1, 0, ...
In the limit, this sequence is all zeros. "Where did all the 1s go?" he
might ask his students.
Cantor's enumeration has nothing to do with the analytic limit. Only the terms of a sequence can index.
One day he might get a student who (a) points out that such sequences
are just functions from N to {0,1}. (b) The sequence of functions s_n
has a well-defined limit. (c) MW's own textbook defines this limit and
shows how to calculate it!
But it has no bearing on indexing. If Cantor had claimed that in the
limit all fractions were indexed, nobody would have paid attention.
[Also, he used to vehemently deny that any non-constant set sequences
have limits.
I use limits where they are appropriate. For instance the number of
indices in the first column of an n*n-matrix is n. Its share is n/n^2.
Here the limit tells us about the share of enumerated fractions in the infinite matrix: lim(n-->oo) n/n^2 = 0.
Result: Every mathematician accepting the analytical limit must deny
Cantor's claims.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Alas there are, according to Cantor, |rao| natural numbers. Can all be >>>>>>>> smaller than |rao|/2? Hardly.
Maybe, just maybe, |rao|/2 isn't even defined.
If |rao| is defined as an integer or whole number, ....
It's not.
Cantor: "ich nenne deren Ordnungstypen allgemein reale ganze Zahlen."
What's that got to do with it? |rao| is not an integer.
Try to learn German.
I've been doing that for several decades. It's a damned difficult
language. ;-)
For most dark numbers, but not for all.
You have never satisfactorally defined "dark numbers". All you've ever
done is written that some integers are dark numbers, some aren't, and
there's no criterion to separate those two cases.
Some dark numbers can get visible. The reason is the the visible numbers
are potentially infinite. There is no strict border.
In otherwords, "dark numbers" are undefined. You're saying they're an undefinable subset of the integers, I think.
Therefore, by (ii) above, it can't be a "dark number". This is a
contradiction. Thus there cannot be such "dark numbers".
You assumed a strict border. That would contradict the potentially
infinite character of the collection of visible numbers.
Quatsch! I assumed well known properties of the integers, took your "definition", such as I can discern it, and followed through to the
logical consequences. I assumed nothing about "borders".
Of course, maybe "dark numbers" aren't integers at all.
Actually, WM's "argument" would as well concern the following TRIVIAL case:
The prime numbers will never "cover" a 1 x IN "matrix" (i.e. a
sequence). So there can't be an infinite sequence (2, 3, 5, 7, ...) of
prime numbers.
After all there are as many terms in a sequence as natural numbers: a_1, a_2, a_3, ... (in other words, the index set of an infinite sequence is
IN), but there are far less prime numbers (than natural numbers), so HOW
CAN THEY cover ALL "places" a_1, a_2, a_3, ... in the sequence?!
|-a-a-a-a Die Menge der Kardinalzahlen ist IN_0 = {0, 1, 2, 3, ...}
Die /Menge der Kardinalzahlen/ ist sicherlich NICHT gleich IN u {0}.
Am 12.08.2025 um 19:15 schrieb Alan Mackenzie:
ORDINALs (ordinal numbers) are NOT referred to as "allgemein reale ganze Zahlen" these days, as M|+ckenheim knows very well.
Try to learn German. [WM]
I've been doing that for several decades.-a It's a damned difficult
language.-a ;-)
Try harder. :-P
Or read /The Awful German Language/ by Mark Twain.
-+Meine philologischen Studien haben mich davon |+berzeugt, dass ein begabter Mensch Englisch (au|fer Rechtschreibung und Aussprache) in 30 Stunden, Franz||sisch in 30 Tagen und Deutsch in 30 Jahren lernen kann.-2
On 12.08.2025 22:19, Moebius wrote:
Am 12.08.2025 um 19:15 schrieb Alan Mackenzie:
ORDINALs (ordinal numbers) are NOT referred to as "allgemein reale ganze
Zahlen" these days, as M|+ckenheim knows very well.
I even know why this is not liked.
Try to learn German. [WM]
How lucky to be born in Germany.I've been doing that for several decades.-a It's a damned difficult
language.-a ;-)
Try harder. :-P
Or read /The Awful German Language/ by Mark Twain.
-+Meine philologischen Studien haben mich davon |+berzeugt, dass ein
begabter Mensch Englisch (au|fer Rechtschreibung und Aussprache) in 30
Stunden, Franz||sisch in 30 Tagen und Deutsch in 30 Jahren lernen kann.-2
On 12.08.2025 19:15, Alan Mackenzie wrote:[ .... ]
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
These numbers are "whole numbers" because they can be counted or can be continued to be counted. Further fractions can be added: Die erweiterte ganze Zahlenreihe kann, wenn es die Zwecke fordern, ohne weiteres zuMore garbage from you. The term "whole number" is well understood by mathematicians. You're trying to pretend it has a different definition,
einer kontinuierlichen Zahlenmenge vervollst|nndigt werden, indem man zu jeder ganzen Zahl +# alle reellen Zahlen x, die gr|||fer als Null und kleiner als Eins sind, hinzuf|+gt.
From this you see that Cantor's transfinite numbers are whole numbers.
The reason why this is denied nowadays is their leading to contradictions.
That's not a mathematical definition, and can't be used in mathematics.You have never satisfactorally defined "dark numbers". All you've everHere is my definition: Definition: A natural number is "identified" or (individually) "defined" or "instantiated" if it can be communicated
done is written that some integers are dark numbers, some aren't, and
there's no criterion to separate those two cases.
such that sender and receiver understand the same and can link it by a finite initial segment to the origin 0. All other natural numbers are
called dark natural numbers.
The existence of dark numbers has recently been proved without any doubt: "New proof of dark numbers by means of the thinned out harmonic series", OSFPREPRINTS (10 Mar 2025)Presumably, that's your work. "Without any doubt" is a lie. There is
https://osf.io/preprints/osf/53qg2_v1?view_only=
All ways of saying that "dark numbers" are not a mathematical concept.Some dark numbers can get visible. The reason is the the visible numbers >>> are potentially infinite. There is no strict border.
Which are likewise undefined mathematically.In otherwords, "dark numbers" are undefined. You're saying they're anThey are defined as given above as the complement of visible or
undefinable subset of the integers, I think.
definable numbers.
But this property depends on time and system.Yes, it's nothing to do with mathematics, in other words.
The pocket calculator is a good example. An even better example is theThe "set of known prime numbers" is not a well defined set.
the potentially infinite set of known prime numbers. It has no upper
bound although almost all prime numbers will remain dark forever.
Maybe. But I didn't assume any such number. Rather I proved its existence. Any non-empty subset of the natural numbers has a least member. IfThe smallest dark number would be a border.Quatsch! I assumed well known properties of the integers, took yourTherefore, by (ii) above, it can't be a "dark number". This is aYou assumed a strict border. That would contradict the potentially
contradiction. Thus there cannot be such "dark numbers".
infinite character of the collection of visible numbers.
"definition", such as I can discern it, and followed through to the
logical consequences. I assumed nothing about "borders".
Try to understand the simple example of prime numbers. There is noIf I were to accept that "unknown-forever prime numbers" somehow defined
smallest unknown-forever prime number. Just the same true for the dark numbers,
I prefer to consider mathematics.Of course, maybe "dark numbers" aren't integers at all.It would be nice if you considered the unknown prime numbers.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 19:15, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
[ .... ]
These numbers are "whole numbers" because they can be counted or can be
continued to be counted. Further fractions can be added: Die erweiterte
ganze Zahlenreihe kann, wenn es die Zwecke fordern, ohne weiteres zu
einer kontinuierlichen Zahlenmenge vervollst|nndigt werden, indem man zu
jeder ganzen Zahl +# alle reellen Zahlen x, die gr|||fer als Null und
kleiner als Eins sind, hinzuf|+gt.
From this you see that Cantor's transfinite numbers are whole numbers.
The reason why this is denied nowadays is their leading to contradictions.
More garbage from you. The term "whole number" is well understood by mathematicians. You're trying to pretend it has a different definition,
and hence derail the main point of discussion here. That's a typical
crank move.
Le 13/08/2025 |a 18:13, Alan Mackenzie a |-crit :
More garbage from you.-a The term "whole number" is well understood by
mathematicians.-a You're trying to pretend it has a different definition,
and hence derail the main point of discussion here.-a That's a typical
crank move.
The typical crank move that consists in considering that the choice of
words (like "whole" and "numbers", or whatever) has a meaning in itself independent of the actual definition of these words (i.e. the d|-finition and properties associated with them, that may change given the context). John Gabriel, M|+ckenheim are always playing this silly sophistic game.
On 12.08.2025 19:15, Alan Mackenzie wrote:How else do you misunderstand Cantor, you fucker?
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Try to learn German.What's that got to do with it? |rao| is not an integer.Cantor: "ich nenne deren Ordnungstypen allgemein reale ganzeIt's not.If |rao| is defined as an integer or whole number, ....Alas there are, according to Cantor, |rao| natural numbers. Can >>>>>>>>> all be smaller than |rao|/2? Hardly.Maybe, just maybe, |rao|/2 isn't even defined.
Zahlen."
You do us the courtesy.I've been doing that for several decades. It's a damned difficultThen let Google translate. It has become a fairly good translator:
language. ;-)
Vielleicht wundern Sie sich |+ber meine K|+hnheit, die Dinge -e, -e + 1,Aha, "of the second class", i.e. not the integers Z.
..., +#,... auch ganze Zahlen, und zwar die ganzen, realen Zahlen der
zweiten Classe zu nennen,
dass -e die erste nach allen ++ geschaffene ganze Zahl sei, d. h. die...after all of them, i.e. not one of them.
erste, welche gr||sser genannt werden soll, als alle ++.
These numbers are "whole numbers" because they can be counted or can be continued to be counted. Further fractions can be added: Die erweiterteThey are called ordinals today. I think Cantor is talking about adding
ganze Zahlenreihe kann, wenn es die Zwecke fordern, ohne weiteres zu
einer kontinuierlichen Zahlenmenge vervollst|nndigt werden, indem man zu jeder ganzen Zahl +# alle reellen Zahlen x, die gr|||fer als Null und
kleiner als Eins sind, hinzuf|+gt.
From this you see that Cantor's transfinite numbers are whole numbers.Sensible.
The reason why this is denied nowadays is their leading to
contradictions.
Yeah, that describes an empty set.Here is my definition: Definition: A natural number is "identified" or (individually) "defined" or "instantiated" if it can be communicatedFor most dark numbers, but not for all.You have never satisfactorally defined "dark numbers". All you've ever
done is written that some integers are dark numbers, some aren't, and
there's no criterion to separate those two cases.
such that sender and receiver understand the same and can link it by a
finite initial segment to the origin 0. All other natural numbers are
called dark natural numbers.
Doesn't apply to the set of all primes, period.Some dark numbers can get visible. The reason is the the visible
numbers are potentially infinite. There is no strict border.
In otherwords, "dark numbers" are undefined. You're saying they're an
undefinable subset of the integers, I think.
They are defined as given above as the complement of visible or
definable numbers. But this property depends on time and system. The
pocket calculator is a good example. An even better example is the the potentially infinite set of known prime numbers. It has no upper bound although almost all prime numbers will remain dark forever.
By analogy, there are no "dark numbers", but you won't understand thatThe smallest dark number would be a border. Try to understand the simple example of prime numbers. There is no smallest unknown-forever primeQuatsch! I assumed well known properties of the integers, took yourTherefore, by (ii) above, it can't be a "dark number". This is aYou assumed a strict border. That would contradict the potentially
contradiction. Thus there cannot be such "dark numbers".
infinite character of the collection of visible numbers.
"definition", such as I can discern it, and followed through to the
logical consequences. I assumed nothing about "borders".
number. Just the same true for the dark numbers,
They are still primes.Of course, maybe "dark numbers" aren't integers at all.It would be nice if you considered the unknown prime numbers.
Le 13/08/2025 |a 17:01, WM a |-crit :
On 12.08.2025 22:19, Moebius wrote:
ORDINALs (ordinal numbers) are NOT referred to as "allgemein reale
ganze Zahlen" these days, as M|+ckenheim knows very well.
I even know why this is not liked.
WM <wolfgang.mueckenheim@tha.de> wrote:
Some dark numbers can get visible.
Am 12.08.2025 um 19:15 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
Some dark numbers can get visible.
In other words, there are numbers with "changing" properties. :-)
I DOUBT that there are numbers with "changing" MATHEMATICAL properties.
Sure. The number 9861396396139619386083028396836963701730173017309713071237107310730173073017301730173017301730173017301730173017301371
may have been "dark" until now (since no one has ever written down a term/name referring to it before). But writing down its decimal representation certainly didn't "change" its MATHEMATICAL properties. It either is an odd number or an even number, a prime number or not a prime number etc., no matter if someone has written down its decimal representation or not.
Hint@M|+ckenheim:
(1) Thinking of a certain number does not change the MATHEMATICAL
properties of this number.
(2) Thinking of a certain physical object does not change the PHYSICAL properties of this object.
.
.
.
Am 13.08.2025 um 17:06 schrieb Python:
Le 13/08/2025 |a 17:01, WM a |-crit :
On 12.08.2025 22:19, Moebius wrote:
ORDINALs (ordinal numbers) are NOT referred to as "allgemein reale
ganze Zahlen" these days, as M|+ckenheim knows very well.
I even know why this is not liked.
Actually, he knows shit, this idiot.
@M|+ckenheim: It's "not liked" because we want to be able to
differentiate between the set of all /whole numbers/ (=positive integers including zero), i. e. IN_0, and the class (hint: not a set!) of all
ordinal numbers. <facepalm>
A student is definitely not lucky if she or he had the misfortune to
have you,
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 19:15, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
[ .... ]
These numbers are "whole numbers" because they can be counted or can be
continued to be counted. Further fractions can be added: Die erweiterte
ganze Zahlenreihe kann, wenn es die Zwecke fordern, ohne weiteres zu
einer kontinuierlichen Zahlenmenge vervollst|nndigt werden, indem man zu
jeder ganzen Zahl +# alle reellen Zahlen x, die gr|||fer als Null und
kleiner als Eins sind, hinzuf|+gt.
From this you see that Cantor's transfinite numbers are whole numbers.
The reason why this is denied nowadays is their leading to contradictions.
More garbage from you.
mathematicians.
You're trying to pretend it has a different definition,
You have never satisfactorally defined "dark numbers". All you've ever
done is written that some integers are dark numbers, some aren't, and
there's no criterion to separate those two cases.
Here is my definition: Definition: A natural number is "identified" or
(individually) "defined" or "instantiated" if it can be communicated
such that sender and receiver understand the same and can link it by a
finite initial segment to the origin 0. All other natural numbers are
called dark natural numbers.
That's not a mathematical definition, and can't be used in mathematics.
It's off-topic on this newsgroup.
The existence of dark numbers has recently been proved without any doubt:
"New proof of dark numbers by means of the thinned out harmonic series",
OSFPREPRINTS (10 Mar 2025)
https://osf.io/preprints/osf/53qg2_v1?view_only=
Presumably, that's your work. "Without any doubt" is a lie. There is
doubt, to say the least, on this newsgroup. If it were to be reviewed by
a competent mathematician who declared it correct, it might be worth
looking at.
Some dark numbers can get visible. The reason is the the visible numbers >>>> are potentially infinite. There is no strict border.
All ways of saying that "dark numbers" are not a mathematical concept.
Even if they did exist (which they don't), they wouldn't be useful in a mathematical setting.
In otherwords, "dark numbers" are undefined. You're saying they're an
undefinable subset of the integers, I think.
They are defined as given above as the complement of visible or
definable numbers.
Which are likewise undefined mathematically.
But this property depends on time and system.
Yes, it's nothing to do with mathematics, in other words.
The "set of known prime numbers" is not a well defined set.
Therefore, by (ii) above, it can't be a "dark number". This is a >>>>> contradiction. Thus there cannot be such "dark numbers".
You assumed a strict border. That would contradict the potentially
infinite character of the collection of visible numbers.
Quatsch! I assumed well known properties of the integers, took your
"definition", such as I can discern it, and followed through to the
logical consequences. I assumed nothing about "borders".
The smallest dark number would be a border.
Maybe. But I didn't assume any such number. Rather I proved its existence.
Any non-empty subset of the natural numbers has a least member. If
you're arguing that a set of "dark numbers" has no least member, then you
are agreeing that it isn't a subset of the natural numbers.
Try to understand the simple example of prime numbers. There is no
smallest unknown-forever prime number. Just the same true for the dark
numbers,
If I were to accept that "unknown-forever prime numbers" somehow defined
a subset of N (which I don't), then it would indeed have a least member.
Of course, maybe "dark numbers" aren't integers at all.
It would be nice if you considered the unknown prime numbers.
I prefer to consider mathematics.
On 13.08.2025 17:06, Python wrote:[ ... as a teacher. ]
A student is definitely not lucky if she or he had the misfortune to
have you [Wolfgang M|+ckenheim],
You mean he or she can no longer believe in Cantor's nonsense after classical mathematics showsHe or she is definitely not lucky to be "taught" about set theory by a
lim(n-->oo) n/n^2 = 0 and not 1?
There are worse things than to get happy by belief in nonsense.You call mathematics "nonsense" simply because it's beyond your
Regards, WM--
So for WM it is clear that there are more elements in the set A = {0, 1,
2, 3, ...} than in the set B = {1, 2, 3, ...}. After all, B c A and 0 is missing in B!
MOREOVER, this means: |A| > |B|, where |X| denotes the number of
elements in X. (After all, "more" means more!)
Of course. Do never use Cantor's deceptive "bijections". If a bijection exists between two sets, then *every* injection is a bijection.
An even better example is the the
potentially infinite set of known prime numbers. It has no upper bound
although almost all prime numbers will remain dark forever.
Doesn't apply to the set of all primes, period.
There is no smallest unknown-forever prime
number. Just the same true for the dark numbers,
By analogy, there are no "dark numbers",
It would behoove you to find
out why you need to deny a "border".
They are still primes.Of course, maybe "dark numbers" aren't integers at all.It would be nice if you considered the unknown prime numbers.
(1) Thinking of a certain number does not change the MATHEMATICAL
properties of this number.
(2) Thinking of a certain physical object does not change the PHYSICAL properties of this object.
On 14.08.2025 00:43, Moebius wrote:
(1) Thinking of a certain number does not change the MATHEMATICAL
properties of this number.
(2) Thinking of a certain physical object does not change the PHYSICAL
properties of this object.
Thinking of thought objects changes their properties from not thought of
to thought of.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 00:43, Moebius wrote:
(1) Thinking of a certain number does not change the MATHEMATICAL
properties of this number.
(2) Thinking of a certain physical object does not change the PHYSICAL
properties of this object.
Thinking of thought objects changes their properties from not thought of
to thought of.
This is getting ludicrous. It's supposed to be a newsgroup about mathematics.
"Being thought of" is not a property of an object. It is a property of
the person doing the thinking of.
On 14.08.2025 14:33, Alan Mackenzie wrote:
A lot of angry words. [ .... ]
Mathematicians would discuss arguments.
Are there prime numbers which will never be known by humans?
Is the limit lim(n-->oo) n/n^2 = 0 or 1?
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
Thinking of thought objects changes their properties from not thought of
to thought of.
This is getting ludicrous. It's supposed to be a newsgroup about mathematics.
"Being thought of" is not a property of an object. It is a property of
the person doing the thinking of.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 14:33, Alan Mackenzie wrote:
A lot of angry words. [ .... ]
No, sadness, not anger.
Mathematicians would discuss arguments.
Are there prime numbers which will never be known by humans?
Is the limit lim(n-->oo) n/n^2 = 0 or 1?
Mathematicians don't spend time discussing silly non-mathematical things
like "will never be known by humans",
and certainly not things about
limits which would bore a 14 year old.
On 14.08.2025 15:40, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 14:33, Alan Mackenzie wrote:
A lot of angry words. [ .... ]
No, sadness, not anger.
Mathematicians would discuss arguments.
Are there prime numbers which will never be known by humans?
Is the limit lim(n-->oo) n/n^2 = 0 or 1?
Mathematicians don't spend time discussing silly non-mathematical things
like "will never be known by humans",
Will never be known by anyone even if time is infinite.
and certainly not things about
limits which would bore a 14 year old.
The limit 0 is not difficult to find. But that "modern mathematics" contradicts 0 and claims 1 is not so boring.
Regards, WM--
On 13.08.2025 19:37, joes wrote:
An even better example is the the
potentially infinite set of known prime numbers. It has no upper bound
although almost all prime numbers will remain dark forever.
Doesn't apply to the set of all primes, period.
Only to the visible primes.
There is no smallest unknown-forever prime
number. Just the same true for the dark numbers,
By analogy, there are no "dark numbers",
There are prime numbers which never will be addressed. They are dark.
On 13.08.2025 19:10, Moebius wrote:Do please prove that.
So for WM it is clear that there are more elements in the set A = {0,Of course I use mathematics,
1, 2, 3, ...} than in the set B = {1, 2, 3, ...}. After all, B c A
and 0 is missing in B!
MOREOVER, this means: |A| > |B|, where |X| denotes the number ofOf course. Do never use Cantor's deceptive "bijections". If a bijection exists between two sets, then *every* injection is a bijection.
elements in X. (After all, "more" means more!)
On 13.08.2025 18:13, Alan Mackenzie wrote:He didnrCOt mean what is today called the integers Z.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 19:15, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 17:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.08.2025 16:04, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 11.08.2025 14:37, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
These numbers are "whole numbers" because they can be counted or can
be continued to be counted. Further fractions can be added: Die
erweiterte ganze Zahlenreihe kann, wenn es die Zwecke fordern, ohne
weiteres zu einer kontinuierlichen Zahlenmenge vervollst|nndigt werden,
indem man zu jeder ganzen Zahl +# alle reellen Zahlen x, die gr|||fer als >>> Null und kleiner als Eins sind, hinzuf|+gt.
From this you see that Cantor's transfinite numbers are whole
numbers.
The reason why this is denied nowadays is their leading to
contradictions.
The term "whole number" is well understood by mathematicians.I gave you Cantor's definition and his reason why he defined them this
You're trying to pretend it has a different definition,
way.
I looked it up, and there is no list of known primes, since it is cheaperWhich are likewise undefined mathematically.In otherwords, "dark numbers" are undefined. You're saying they'reThey are defined as given above as the complement of visible or
an undefinable subset of the integers, I think.
definable numbers.
The "set of known prime numbers" is not a well defined set.In fact, the known prime numbers without fixed point of time are not a
set. They are a potentially infinite collection.
The set of presently known prime numbers is a set.
Yes, rCRdark numbersrCY are not a set.Wrong. You assumed sets where potentially infinite collections have toMaybe. But I didn't assume any such number. Rather I proved itsThe smallest dark number would be a border.Quatsch! I assumed well known properties of the integers, took yourTherefore, by (ii) above, it can't be a "dark number". This is >>>>>> a contradiction. Thus there cannot be such "dark numbers".You assumed a strict border. That would contradict the potentially
infinite character of the collection of visible numbers.
"definition", such as I can discern it, and followed through to the
logical consequences. I assumed nothing about "borders".
existence.
be applied.
Ok, but we are unfortunately only interested in naturals.Any non-empty subset of the natural numbers has a least member. IfYes.
you're arguing that a set of "dark numbers" has no least member, then
you are agreeing that it isn't a subset of the natural numbers.
You are way hung up on CantorrCOs phrasing of rCRtill the last onerCY.Do you believe that all prime numbers can be known? Including a lastTry to understand the simple example of prime numbers. There is noIf I were to accept that "unknown-forever prime numbers" somehow
smallest unknown-forever prime number. Just the same true for the dark
numbers,
defined a subset of N (which I don't), then it would indeed have a
least member.
one?
Only for an arbitrarily large, but bounded, time.Of course there are prime numbers which are unknown for ever.I prefer to consider mathematics.Of course, maybe "dark numbers" aren't integers at all.It would be nice if you considered the unknown prime numbers.
"Being thought of" is not a property of an object. It is a property of
the person doing the thinking of.
Remember that we are dealing with a crank so insane that he considers
that "being a member of a finite set" for a single mathematical object x
is a meaningful property of x...
"Being thought of" is not a property of an object. It is a property of
the person doing the thinking of.
Remember that we are dealing with a crank so insane that he considers
that "being a member of a finite set" for a single mathematical object x
is a meaningful property of x...
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a "Being thought of" is not an extrinisc property of an object.
I'd tend to agree.
Lit.: https://en.wikipedia.org/wiki/ Intrinsic_and_extrinsic_properties_(philosophy)
Le 14/08/2025 |a 15:09, Alan Mackenzie a |-crit :
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 00:43, Moebius wrote:
(1) Thinking of a certain number does not change the MATHEMATICAL
properties of this number.
Thinking of thought objects [bla bla bla]
Remember that we are dealing with a crank [...]
On 13.08.2025 19:37, joes wrote:
An even better example is the the
potentially infinite set of known prime numbers. It has no upper bound
although almost all prime numbers will remain dark forever.
Doesn't apply to the set of all primes, period.
Only to the visible primes.
There is no smallest unknown-forever prime
number. Just the same true for the dark numbers,
By analogy, there are no "dark numbers",
There are prime numbers which never will be addressed. They are dark.
It would behoove you to find
out why you need to deny a "border".
I explained it often enough: Potentially infinite collections have no
last element. Therefore the complement has no first element.
They are still primes.Of course, maybe "dark numbers" aren't integers at all.It would be nice if you considered the unknown prime numbers.
Like dark natnumbers are still natural numbers.
Am 14.08.2025 um 17:56 schrieb Moebius:
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a-a "Being thought of" is not an extrinisc property of an object.
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no MATHEMATICAL property of any mathematical object.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 15:40, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 14:33, Alan Mackenzie wrote:
A lot of angry words. [ .... ]
No, sadness, not anger.
Mathematicians would discuss arguments.
Are there prime numbers which will never be known by humans?
Is the limit lim(n-->oo) n/n^2 = 0 or 1?
Mathematicians don't spend time discussing silly non-mathematical things >>> like "will never be known by humans",
Will never be known by anyone even if time is infinite.
and certainly not things about
limits which would bore a 14 year old.
The limit 0 is not difficult to find. But that "modern mathematics"
contradicts 0 and claims 1 is not so boring.
Modern mathematics does no such thing.
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If a bijectionDo please prove that.
exists between two sets, then *every* injection is a bijection.
Am Thu, 14 Aug 2025 14:24:43 +0200 schrieb WM:
On 13.08.2025 18:13, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
He didnrCOt mean what is today called the integers Z.The term "whole number" is well understood by mathematicians.I gave you Cantor's definition and his reason why he defined them this
You're trying to pretend it has a different definition,
way.
The set of presently known prime numbers is a set.I looked it up, and there is no list of known primes, since it is cheaper
to compute them.
Yes, rCRdark numbersrCY are not a set.
Of course there are prime numbers which are unknown for ever.Only for an arbitrarily large, but bounded, time.
On 14.08.2025 17:33, joes wrote:Every injection between bijective sets is bijective?
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
It is a property of complete sets that no exchange can create or delete elements. Therefore every reordering maintains the injection. OnlyYes, infinite permutations donrCOt preserve injectivity.
potential infinity can lead to Cantor's "bijections".
On 14.08.2025 16:18, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The limit 0 is not difficult to find. But that "modern mathematics"
contradicts 0 and claims 1 is not so boring.
Modern mathematics does no such thing.
You are mistaken again. Either there is a bijection between first column
and matrix, then the first column has exactly as many element as the
matrix (that is the meaning of bijection) and the limit is 1. Or this is wrong and the limit is 0. It is impossible that both is true because 0
=/= 1.
All my students have a better brain than you.
They all with no exception have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the potentially infinite sequence).
Therefore the sum of the not individually definable numbers of the
complement (and the complement itself) must be infinite.
May a merciful God prevent that you ever recognize what a rubbish you
have learnt and continue to be advocating.
Regards, WM--
All my students have a better brain than you. They all with no exception have accepted and understood that the sum of all individually definable numbers is finite in every case (of the potentially infinite sequence). Therefore the sum of the not individually definable numbers of the complement (and the complement itself) must be infinite.
May a merciful God prevent that you ever recognize what a rubbish you
have learnt and continue to be advocating.
WM <wolfgang.mueckenheim@tha.de> wrote:
[My students] have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the
potentially infinite sequence).
Therefore the sum of the not individually definable numbers of the
complement (and the complement itself) must be infinite.
Total gibberish.
Le 15/08/2025 |a 14:59, WM a |-crit :
All my students have a better brain than you. They all with no
exception have accepted and understood that the sum of all
individually definable numbers is finite in every case (of the
potentially infinite sequence). Therefore the sum of the not
individually definable numbers of the complement (and the complement
itself) must be infinite.
May a merciful God prevent that you ever recognize what a rubbish you
have learnt and continue to be advocating.
Reading THIS is 100% equivalent to reading a pedophile bragging about
how he molested and raped children.
"Academic freedom" ? Puke. Puke. Puke.
Hochschule Augsburg, it is time to ACT.
WM <wolfgang.mueckenheim@tha.de> wrote:
[My students] have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the
potentially infinite sequence).
Therefore the sum of the not individually definable numbers of the
complement (and the complement itself) must be infinite.
Total gibberish.
Am 14.08.2025 um 17:56 schrieb Moebius:
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a-a "Being thought of" is not an extrinisc property of an object.
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no MATHEMATICAL property of any mathematical object.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 16:18, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
[ .... ]
The limit 0 is not difficult to find. But that "modern mathematics"
contradicts 0 and claims 1 is not so boring.
Modern mathematics does no such thing.
You are mistaken again. Either there is a bijection between first column
and matrix, then the first column has exactly as many element as the
matrix (that is the meaning of bijection) and the limit is 1. Or this is
wrong and the limit is 0. It is impossible that both is true because 0
=/= 1.
It is possible that the limit, as you perceive it, is not well defined.
That is in fact the case, as you have shown.
That first column does
indeed have as many elements as the entire matrix - there is a 1-1 correspondence between them.
All my students have a better brain than you.
Unlikely, though possible.
They all with no exception have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the
potentially infinite sequence).
In other words, they (having better brains than me) have been sullied by false crank mathatematics, and will never understand these matters any
better than you do.
Therefore the sum of the not individually definable numbers of the
complement (and the complement itself) must be infinite.
Total gibberish.
May a merciful God prevent that you ever recognize what a rubbish you
have learnt and continue to be advocating.
What I have learnt was distilled by the best mathematicians of the last
2500 years
You are incapable of understanding this mathematics,
On 14.08.2025 18:03, Moebius wrote:
Am 14.08.2025 um 17:56 schrieb Moebius:
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a-a "Being thought of" is not an extrinisc property of an object. >>>
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no MATHEMATICAL
property of any mathematical object.
The dark numbers are an important part of mathematics, comparable to the irrational numbers. And they are a very difficult notion since most mathematicians are unable to comprehend them.
Le 15/08/2025 |a 17:55, WM a |-crit :
On 14.08.2025 18:03, Moebius wrote:
Am 14.08.2025 um 17:56 schrieb Moebius:The dark numbers are an important part of mathematics, comparable to
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a-a "Being thought of" is not an extrinisc property of an object. >>>>
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no
MATHEMATICAL property of any mathematical object.
the irrational numbers.
And they are a very difficult notion since most mathematicians are unable to comprehend them.
What looks more likely?
- They are a very difficult notion since most mathematicians are unable
to comprehend them nevertheless all students who followed Wolfgang M|+ckenheim "course", at undergraduate level, comprehend them (according
to himself)
- They are a idiotic meaningless notion proposed by a demented molester
and crank called Wolfgang M|+ckenheim
On 15.08.2025 15:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 14.08.2025 16:18, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
[ .... ]
The limit 0 is not difficult to find. But that "modern mathematics"
contradicts 0 and claims 1 is not so boring.
Modern mathematics does no such thing.
You are mistaken again. Either there is a bijection between first column >>> and matrix, then the first column has exactly as many element as the
matrix (that is the meaning of bijection) and the limit is 1. Or this is >>> wrong and the limit is 0. It is impossible that both is true because 0
=/= 1.
It is possible that the limit, as you perceive it, is not well defined.
That is in fact the case, as you have shown.
It is very well defined.
That first column does indeed have as many elements as the entire
matrix - there is a 1-1 correspondence between them.
You will never understand your mistake.
All my students have a better brain than you.
Unlikely, though possible.
Proven by your last remark. Why should the limit of n/n^2 not be well defined?>
They all with no exception have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the
potentially infinite sequence).
In other words, they (having better brains than me) have been sullied by
false crank mathatematics, and will never understand these matters any
better than you do.
Therefore the sum of the not individually definable numbers of the
complement (and the complement itself) must be infinite.
Total gibberish.
Why. No, it is fact.
May a merciful God prevent that you ever recognize what a rubbish you
have learnt and continue to be advocating.
What I have learnt was distilled by the best mathematicians of the last
2500 years
No, it has been created by an unfortunate accident. Every mathematician
who does not understand what I discuss here is either stupid or inflexible.
You are incapable of understanding this mathematics,
I have understood it and its mistakes.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
On 15.08.2025 15:59, Alan Mackenzie wrote:
It is possible that the limit, as you perceive it, is not well defined.
That is in fact the case, as you have shown.
All my students have a better brain than you.
Unlikely, though possible.
Proven by your last remark. Why should the limit of n/n^2 not be well
defined?>
How dare you insinuate I said that?
Of course lim n->infinity n/n^2 is
defined, and its value is 0.
That has nothing to do with two countably
infinite sets being in 1-1 correspondence with eachother.
They all with no exception have accepted and understood that the sum of >>>> all individually definable numbers is finite in every case (of the
potentially infinite sequence).
In other words, they (having better brains than me) have been sullied by >>> false crank mathatematics, and will never understand these matters any
better than you do.
Therefore the sum of the not individually definable numbers of the
complement (and the complement itself) must be infinite.
Total gibberish.
Why. No, it is fact.
What, when "not individually definable" is mathematically meaningless?
May a merciful God prevent that you ever recognize what a rubbish you
have learnt and continue to be advocating.
What I have learnt was distilled by the best mathematicians of the last
2500 years
No, it has been created by an unfortunate accident. Every mathematician
who does not understand what I discuss here is either stupid or inflexible.
Crank.
You are incapable of understanding this mathematics,
I have understood it and its mistakes.
Crank.
On 14.08.2025 18:03, Moebius wrote:
Am 14.08.2025 um 17:56 schrieb Moebius:
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a-a "Being thought of" is not an extrinisc property of an object. >>>
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no
MATHEMATICAL property of any mathematical object.
The dark numbers are an important part of mathematics, comparable to the irrational numbers. And they are a very difficult notion since most mathematicians are unable to comprehend them.
Am 15.08.2025 um 18:03 schrieb Python:
The dark numbers are an important part of mathematics, comparable to
the irrational numbers.
Oh, really?! In which MATHEMATICAL theory do the play ANY role?
Am Fri, 15 Aug 2025 15:04:45 +0200 schrieb WM:
On 14.08.2025 17:33, joes wrote:Every injection between bijective sets is bijective?
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
On 15.08.2025 15:38, joes wrote:
Am Fri, 15 Aug 2025 15:04:45 +0200 schrieb WM:
On 14.08.2025 17:33, joes wrote:Every injection between bijective sets is bijective?
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
If a bijection exists between two sets, then *every* injection is a bijection.
On 15.08.2025 15:38, joes wrote:Yes, I would like to see that proved.
Am Fri, 15 Aug 2025 15:04:45 +0200 schrieb WM:If a bijection exists between two sets, then *every* injection is a bijection.
On 14.08.2025 17:33, joes wrote:Every injection between bijective sets is bijective?
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
Am 15.08.2025 um 15:59 schrieb Alan Mackenzie:definable" with something MEANINGFUL, say, "finitely many", we get:
WM <wolfgang.mueckenheim@tha.de> wrote:
[My students] have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the
potentially infinite sequence).
In other words, replacing the undefined notion "[all] individually
/His students have accepted and understood that the sum of finitelymany numbers is finite in every case [...]./
Actually, if WM's notions "individually definable", "visible", etc.etc. are replaced by "finitely many",
On 15.08.2025 17:29, Moebius wrote:
Am 15.08.2025 um 15:59 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
[My students] have accepted and understood that the sum of
all individually definable numbers is finite in every case (of the
potentially infinite sequence).
In other words, replacing the undefined notion "[all] individuallydefinable" with something MEANINGFUL, say, "finitely many", we get:
/His students have accepted and understood that the sum of finitelymany numbers is finite in every case [...]./
Actually, if WM's notions "individually definable", "visible", etc.etc. are replaced by "finitely many",
Can you imagine to individually define more than finitely many numbers? Hardly. Therefore my statements are simple and correct.
Le 15/08/2025 |a 17:55, WM a |-crit :
On 14.08.2025 18:03, Moebius wrote:
Am 14.08.2025 um 17:56 schrieb Moebius:The dark numbers are an important part of mathematics, comparable to
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
-a-a-a-a-a "Being thought of" is not an extrinisc property of an object. >>>>
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no
MATHEMATICAL property of any mathematical object.
the irrational numbers.
[T]hey are a very difficult notion since most mathematicians are unable to comprehend them.
What looks more likely?
- They are a very difficult notion since most mathematicians are unable
to comprehend them nevertheless all students who followed Wolfgang M|+ckenheim "course", at undergraduate level, comprehend them (according
to himself).
- They are an idiotic meaningless notion proposed by a demented molester
and crank called Wolfgang M|+ckenheim.
WM <wolfgang.mueckenheim@tha.de> wrote:
You are mistaken again. Either there is a bijection between first column
and the matrix, then the first column has exactly as many element as the
matrix (that is the meaning of bijection) and the limit is 1. Or this is
wrong and the limit is 0. It is impossible that both is true because <bla>
It is possible that the limit, as you perceive it, is not well defined.
All my students have a better brain than you.
Am 15.08.2025 um 18:03 schrieb Python:
Le 15/08/2025 |a 17:55, WM a |-crit :
On 14.08.2025 18:03, Moebius wrote:
Am 14.08.2025 um 17:56 schrieb Moebius:The dark numbers are an important part of mathematics, comparable to
In philosophy there's the distinction between "intrinsic" and
"extrinisic" properties, which might be "helpful" here. Using this
approach one might claim:
"Being thought of" is not an extrinisc property of an object.
I'd tend to agree.
Moreover (and more relevant in this NG) it is CERTAINLY no
MATHEMATICAL property of any mathematical object.
the irrational numbers.
Oh, really?! In which MATHEMATICAL theory do they play ANY role? Can you
name some textbooks, M|+ckenheim?
Asking ChatGPT?
Question: "Is it true that M|+ckenheim's dark numbers are an important
part of mathematics?"
ChatGPT: "[...] no, it is not true that "M|+ckenheimrCOs dark numbers" are
an important part of mathematics".
Got it. It's called DELUSION. See: https://en.wikipedia.org/wiki/Delusion
[T]hey are a very difficult notion since most mathematicians are
unable to comprehend them.
What looks more likely?
- They are a very difficult notion since most mathematicians are
unable to comprehend them nevertheless all students who followed
Wolfgang M|+ckenheim "course", at undergraduate level, comprehend them
(according to himself).
- They are an idiotic meaningless notion proposed by a demented
molester and crank called Wolfgang M|+ckenheim.
Wait a second, let me think ...
:-)
.
.
.
On 07.08.2025 22:03, Moebius wrote:
Am 07.08.2025 um 20:20 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
All fractions that can be named get indexed.
All fractions can be named,
Indeed.
Indeed counting requires according to Cantor and me *) to get an X and
to deliver an O.
definable fractios can be counted. That proves undefinable fractions.
*) WM: Die Cantorsche Formel ergibt meine Matrizen. FF: Das hatte ich
mir schon gedacht. Und das stimmt auch.
Merke: Die Cantorsche Formel hat nichts mit dem analytischen Grenzwert
der Matrizen zu tun.
Regards, WM
Am 15.08.2025 um 15:59 schrieb Alan Mackenzie:[...]
WM <wolfgang.mueckenheim@tha.de> wrote:
You are mistaken again. Either there is a bijection between first column >>> and the matrix, then the first column has exactly as many element as the >>> matrix (that is the meaning of bijection) and the limit is 1. Or this is >>> wrong and the limit is 0. It is impossible that both is true because
<bla>
This fucking asshole is talking nonsense, and he knows it!
Am Fri, 15 Aug 2025 19:01:09 +0200 schrieb WM:
On 15.08.2025 15:38, joes wrote:Yes, I would like to see that proved.
Am Fri, 15 Aug 2025 15:04:45 +0200 schrieb WM:If a bijection exists between two sets, then *every* injection is a
On 14.08.2025 17:33, joes wrote:Every injection between bijective sets is bijective?
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
bijection.
Am 15.08.2025 um 15:59 schrieb Alan Mackenzie:
I'VE *defined* "the asymptotic density for infinite matrices"
(M|+ckenheim's "limit") in de.sci.mathematik. The asymptotic density of elements =/= 0 in
1-a 0-a 0-a 0 ...
2-a 0-a 0-a 0 ...
3-a 0-a 0-a 0 ...
4-a 0-a 0-a 0 ...
:-a :-a :-a :
of course is 0.
While on the other hand, in
-a1-a 2-a 4-a 7 ...
-a3-a 5-a 8 12 ...
-a6-a 9 13 18 ...
10 14 19 25 ...
-a:-a :-a :-a :
it is 1. (sigh)
On 8/15/2025 8:55 AM, WM wrote:
The dark numbers are an important part of mathematics, comparable to
the irrational numbers. And they are a very difficult notion since
most mathematicians are unable to comprehend them.
I see that a dark number is a number that somebody has not thought of
yet.
What is
the sum of all the dark numbers?
Therefore, four can be a dark number because an entity in the universe
has not thought of it yet, right?
On 15.08.2025 19:36, joes wrote:
Am Fri, 15 Aug 2025 19:01:09 +0200 schrieb WM:
On 15.08.2025 15:38, joes wrote:
Am Fri, 15 Aug 2025 15:04:45 +0200 schrieb WM:
On 14.08.2025 17:33, joes wrote:
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Yes, I would like to see that proved.If a bijection exists between two sets, then *every* injection is aEvery injection between bijective sets is bijective?Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
bijection.
It is a property of complete sets ....
.... that no exchange can create or delete elements.
Therefore every reordering maintains the injection.
Regards, WM--
WM <wolfgang.mueckenheim@tha.de> wrote:
"Complete set" isn't a mathematical notion.
No matter how many elements
a set has, it is possible to append a further element to form a new set.
There is no such thing as an "incomplete set".
.... that no exchange can create or delete elements.
That's too ill-defined to discern what you mean. "Exchange" of what, precisely? If you "exchange two elements in a set", you are left with
the same set.
On 15.08.2025 19:36, joes wrote:How does this apply to your attempt? A bijection is also injective.
Am Fri, 15 Aug 2025 19:01:09 +0200 schrieb WM:It is a property of complete sets that no exchange can create or delete elements. Therefore every reordering maintains the injection.
On 15.08.2025 15:38, joes wrote:Yes, I would like to see that proved.
Am Fri, 15 Aug 2025 15:04:45 +0200 schrieb WM:If a bijection exists between two sets, then *every* injection is a
On 14.08.2025 17:33, joes wrote:Every injection between bijective sets is bijective?
Am Thu, 14 Aug 2025 14:38:29 +0200 schrieb WM:
Of course. Do never use Cantor's deceptive "bijections". If aDo please prove that.
bijection exists between two sets, then *every* injection is a
bijection.
bijection.
On 16.08.2025 00:10, Moebius wrote:You still need to disprove that infinite permutations donrCOt preserve that.
Am 15.08.2025 um 15:59 schrieb Alan Mackenzie:
I'VE *defined* "the asymptotic density for infinite matrices"
(M|+ckenheim's "limit") in de.sci.mathematik. The asymptotic density of
elements =/= 0 in
1-a 0-a 0 ...
2-a 0-a 0 ...
3-a 0-a 0 ...
:-a :-a :-a
of course is 0.
Right. Therefore not all cells can carry natural numbers.
While on the other hand, in
-a1-a 2-a 4-a...
-a3-a 5-a 8 ...
6-a 9 13 ...
-a:-a :-a :
it is 1. (sigh)
Then mathematics would be inconsistent.
Only one of the two assertions can be right.
That is not remedied by your following examples which I therefore have deleted.Thanks for the notice.
On 16.08.2025 14:11, Alan Mackenzie wrote:Noted. YourCOre not working in ZF.
WM <wolfgang.mueckenheim@tha.de> wrote:
"Complete set" isn't a mathematical notion.It is never emphasized because in ZF all sets are complete. But it is a mathematical notion in ZF, contrary to potentially infinite sets as
appear in my books
There is no such thing as an "incomplete set".Not in ZF.
Your starting function m,n -> m,1 (?) is not a bijectionrCa.... that no exchange can create or delete elements.
That's too ill-defined to discern what you mean. "Exchange" of what,
precisely? If you "exchange two elements in a set", you are left with
the same set.
If we exchange two elements of the pairs of a bijection, then the
injection remains too. Cantor's "bijections" are not of that kind.
WM <wolfgang.mueckenheim@tha.de> wrote:
"If a bijection exists between two sets, then *every* injection is a bijection." (WM)
[...]--- Synchronet 3.21a-Linux NewsLink 1.2
Am 16.08.2025 um 14:11 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
"If a bijection exists between two sets, then *every* injection is a
bijection." (WM)
Sure.
The identity function id: IN --> IN, id(n) = n, is a bijektion from IN
onto IN. Hence (WM) the injection f: IN --> IN, f(n) = 2*n, is a
bijection from IN onto IN too.
Question: For which n in IN is 2*n = 1?
<facepalm>
[...]
Am Sat, 16 Aug 2025 14:29:31 +0200 schrieb WM:
On 16.08.2025 14:11, Alan Mackenzie wrote:
Noted. YourCOre not working in ZF.There is no such thing as an "incomplete set".Not in ZF.
If we exchange two elements of the pairs of a bijection, then theYour starting function m,n -> m,1 (?) is not a bijectionrCa
injection remains too. Cantor's "bijections" are not of that kind.
Am 16.08.2025 um 16:54 schrieb Moebius:
"If a bijection exists between two sets, then *every* injection is a
bijection." (WM)
Hint: If this were the case, it would imply that all sets are finite.
Lern some set theory, M|+ckenheim.
On 15.08.2025 18:57, Chris M. Thomasson wrote:
On 8/15/2025 8:55 AM, WM wrote:
The dark numbers are an important part of mathematics, comparable to
the irrational numbers. And they are a very difficult notion since
most mathematicians are unable to comprehend them.
I see that a dark number is a number that somebody has not thought of
yet.
Yes, it is dark in the system of that somebody.
What is the sum of all the dark numbers?
The sum of dark natural numbers is infinite because there are infinitely many dark natural numbers in every system and at every time.
Therefore, four can be a dark number because an entity in the universe
has not thought of it yet, right?
For an earthworm 4 is certainly dark.
On 8/15/2025 3:10 PM, Moebius wrote:
Am 15.08.2025 um 15:59 schrieb Alan Mackenzie:[...]
WM <wolfgang.mueckenheim@tha.de> wrote:
You are mistaken again. Either there is a bijection between first
column
and the matrix, then the first column has exactly as many element as
the
matrix (that is the meaning of bijection) and the limit is 1. Or
this is
wrong and the limit is 0. It is impossible that both is true because
<bla>
This fucking asshole is talking nonsense, and he knows it!
Oh yeah. To blow his mind, let him learn about a L-system.
A = AB
B = BA
ABA
ABBAAB
...
He wants D for dark or something? Shit.
On 16.08.2025 16:27, joes wrote:Listen, they can not be (finitely!) permuted into each other.
Am Sat, 16 Aug 2025 14:29:31 +0200 schrieb WM:
Neither is Cantor's enumeration of fractions.If we exchange two elements of the pairs of a bijection, then theYour starting function m,n -> m,1 (?) is not a bijectionrCa
injection remains too. Cantor's "bijections" are not of that kind.
Am 16.08.2025 um 17:01 schrieb Moebius:
Am 16.08.2025 um 16:54 schrieb Moebius:
"If a bijection exists between two sets, then *every* injection is a
-abijection." (WM)
Hint: If this were the case, it would imply that all sets are finite.
Am Sat, 16 Aug 2025 18:25:43 +0200 schrieb WM:
On 16.08.2025 16:27, joes wrote:
Am Sat, 16 Aug 2025 14:29:31 +0200 schrieb WM:
Listen, they can not be (finitely!) permuted into each other.Neither is Cantor's enumeration of fractions.If we exchange two elements of the pairs of a bijection, then theYour starting function m,n -> m,1 (?) is not a bijectionrCa
injection remains too. Cantor's "bijections" are not of that kind.
Le 06/08/2025 |a 19:19, WM a |-crit :
Show me the first (or any) O leaving the matrix.
again and again the same sophistry from Wolfgang M|+ckenheim, debunked
for ages.
http://bsb.me.uk/dd-wealth.pdf