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Enumerating all real numbers
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WM
Dec 26, 2022, 5:21:04 AM12/26/22
to
Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality |ω^ω| = ℵo.
The virtue of this method is that it avoids to squander the natural numbers: The entries of the Cantor-list are not enumerated by all natural numbers but by powers of 2. The antidiagonals of this list (never more than a countable set) are enumerated by the powers of 3. If a new list is created including these antidiagonals then its entries are enumerated by the powers of 5 and its antidiagonals (never more than a countable set) are enumerated by the powers of 7. This is going on, in every step using a new prime number. According to Euclid the set of prime numbers will never get exhausted.
Regards, WM
The virtue of this method is that it avoids to squander the natural numbers: The entries of the Cantor-list are not enumerated by all natural numbers but by powers of 2. The antidiagonals of this list (never more than a countable set) are enumerated by the powers of 3. If a new list is created including these antidiagonals then its entries are enumerated by the powers of 5 and its antidiagonals (never more than a countable set) are enumerated by the powers of 7. This is going on, in every step using a new prime number. According to Euclid the set of prime numbers will never get exhausted.
Regards, WM
Fritz Feldhase
Dec 26, 2022, 6:48:19 AM12/26/22
to
On Monday, December 26, 2022 at 11:21:04 AM UTC+1, WM wrote:
> I show that all real numbers belong to a countable set of cardinality [...] ℵo.
You can't "show" that, idiot, since it is wrong.
> I show that all real numbers belong to a countable set of cardinality [...] ℵo.
You can't "show" that, idiot, since it is wrong.
Ben Bacarisse
Dec 26, 2022, 9:07:36 AM12/26/22
to
WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
This reply is intended only for students who might fall for this sort of
thing. Some do, apparently... WM is impervious to logic (after all, he
explicitly assumes his own conclusion here) so he's never going to
learn.
> Assuming that countability is a sensible notion, I show that all real
> numbers belong to a countable set of cardinality |ω^ω| = ℵo.
>
> The virtue of this method is that it avoids to squander the natural
> numbers: The entries of the Cantor-list are not enumerated by all
> natural numbers but by powers of 2. The antidiagonals of this list
> (never more than a countable set)
Here WM explicitly assumes a fact logically equivalent to the conclusion
he is supposedly proving. There are, in fact, uncountably many
anti-diagonals. If there were, as WM incorrectly assumes, only
countably many anti-diagonals there would be no need for the powers of
primes construction: number any initial list of with the even numbers
and then enumerate the supposedly countable anti-diagonals using the odd
numbers.
So how can we show that there are /always/ uncountably many
anti-diagonals for any enumeration of the reals? If you a student,
particularly one of WM's students, pause here and see if you can prove
this fact.
.
.
.
.
.
OK, is this how you did it...
The conventional way to construct the anti-diagonal is to change digits
by, say, incriminating each digit mod 10, with the technical detail of
avoiding using a 9, so we might map both 8 and 9 to 0.
Here's an example of list of reals in [0,1] indexed by WM's powers of
two:
2: 0.612087234...
4: 0.971023145...
8: 0.651825344...
16: 0.443511672...
...
The diagonal is 0.6715... so the "incriminating" anti-diagonal starts
0.7826... But we could also have a "decrementing" anti-diagonal that
starts 0.5604...
But we could also choose to either increment (-) or decrement (-) at
each digit and still have an anti-diagonal. I.e. there are at least as
many anti-diagonals as there are infinite sequences of +s and -s. Can
you finish off the proof now?
Of course you can! Map + to 1 and - to 0 and we have the binary digits
of the reals in [0,1], i.e. WM is assuming the result he is trying to
prove.
> are enumerated by the powers of
> 3. If a new list is created including these antidiagonals then its
> entries are enumerated by the powers of 5 and its antidiagonals (never
> more than a countable set) are enumerated by the powers of 7. This is
> going on, in every step using a new prime number. According to Euclid
> the set of prime numbers will never get exhausted.
So, students, if WM presents this nonsense to you, be brave and counter
it in all three ways:
1. The plain counter-argument: the result of this construction is,
supposedly, is a function wm from N to [0,1] yet the anti-diagonal
constructed from flipping digit n of wm(n) is provably not in the image
of wm.[1]
2. The complexity argument: if his assumption about there being only
countably many anti-diagonals was correct, why did he use such a complex
argument?
3. The why the argument is wrong argument: it assumes the conclusion.
There are uncountably many anti-diagonals and every new prime enumerates
only countably many of them. The resulting list has, yet again,
uncountably many anti-diagonals.
Be prepared for the resulting snow-storm of waffle.
--
Ben.
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
This reply is intended only for students who might fall for this sort of
thing. Some do, apparently... WM is impervious to logic (after all, he
explicitly assumes his own conclusion here) so he's never going to
learn.
> Assuming that countability is a sensible notion, I show that all real
> numbers belong to a countable set of cardinality |ω^ω| = ℵo.
>
> The virtue of this method is that it avoids to squander the natural
> numbers: The entries of the Cantor-list are not enumerated by all
> natural numbers but by powers of 2. The antidiagonals of this list
> (never more than a countable set)
he is supposedly proving. There are, in fact, uncountably many
anti-diagonals. If there were, as WM incorrectly assumes, only
countably many anti-diagonals there would be no need for the powers of
primes construction: number any initial list of with the even numbers
and then enumerate the supposedly countable anti-diagonals using the odd
numbers.
So how can we show that there are /always/ uncountably many
anti-diagonals for any enumeration of the reals? If you a student,
particularly one of WM's students, pause here and see if you can prove
this fact.
.
.
.
.
.
OK, is this how you did it...
The conventional way to construct the anti-diagonal is to change digits
by, say, incriminating each digit mod 10, with the technical detail of
avoiding using a 9, so we might map both 8 and 9 to 0.
Here's an example of list of reals in [0,1] indexed by WM's powers of
two:
2: 0.612087234...
4: 0.971023145...
8: 0.651825344...
16: 0.443511672...
...
The diagonal is 0.6715... so the "incriminating" anti-diagonal starts
0.7826... But we could also have a "decrementing" anti-diagonal that
starts 0.5604...
But we could also choose to either increment (-) or decrement (-) at
each digit and still have an anti-diagonal. I.e. there are at least as
many anti-diagonals as there are infinite sequences of +s and -s. Can
you finish off the proof now?
Of course you can! Map + to 1 and - to 0 and we have the binary digits
of the reals in [0,1], i.e. WM is assuming the result he is trying to
prove.
> are enumerated by the powers of
> 3. If a new list is created including these antidiagonals then its
> entries are enumerated by the powers of 5 and its antidiagonals (never
> more than a countable set) are enumerated by the powers of 7. This is
> going on, in every step using a new prime number. According to Euclid
> the set of prime numbers will never get exhausted.
it in all three ways:
1. The plain counter-argument: the result of this construction is,
supposedly, is a function wm from N to [0,1] yet the anti-diagonal
constructed from flipping digit n of wm(n) is provably not in the image
of wm.[1]
2. The complexity argument: if his assumption about there being only
countably many anti-diagonals was correct, why did he use such a complex
argument?
3. The why the argument is wrong argument: it assumes the conclusion.
There are uncountably many anti-diagonals and every new prime enumerates
only countably many of them. The resulting list has, yet again,
uncountably many anti-diagonals.
Be prepared for the resulting snow-storm of waffle.
--
Ben.
Sergi o
Dec 26, 2022, 10:11:09 AM12/26/22
to
On 12/26/2022 4:21 AM, WM wrote:
> Assuming that countability is a sensible notion,
Are you for real ?
> Assuming that countability is a sensible notion,
You assume you have a notion of sensible thinking, a mistake!
https://www.math.toronto.edu/ivan/mat327/docs/notes/04-countability.pdf
> I show that all real numbers belong to a countable set
>
> Regards, WM
WM
Dec 26, 2022, 12:12:49 PM12/26/22
to
Ben Bacarisse schrieb am Montag, 26. Dezember 2022 um 15:07:36 UTC+1:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
>
> This reply is intended only for students who might fall for this sort of
> thing. Some do, apparently...
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
>
> This reply is intended only for students who might fall for this sort of
> thing. Some do, apparently...
> > Assuming that countability is a sensible notion, I show that all real
> > numbers belong to a countable set of cardinality |ω^ω| = ℵo.
> >
> > The virtue of this method is that it avoids to squander the natural
> > numbers: The entries of the Cantor-list are not enumerated by all
> > natural numbers but by powers of 2. The antidiagonals of this list
> > (never more than a countable set)
> Here WM explicitly assumes a fact logically equivalent to the conclusion
> he is supposedly proving. There are, in fact, uncountably many
> anti-diagonals.
Your "in fact" is in fact wrong. The antidiagonal can be constructed by a set of instructions as is well known and need not be discussed here. How many sets of instructions can be given? Finitely many. That yields countably many antidiagonals.
> > numbers belong to a countable set of cardinality |ω^ω| = ℵo.
> >
> > The virtue of this method is that it avoids to squander the natural
> > numbers: The entries of the Cantor-list are not enumerated by all
> > natural numbers but by powers of 2. The antidiagonals of this list
> > (never more than a countable set)
> Here WM explicitly assumes a fact logically equivalent to the conclusion
> he is supposedly proving. There are, in fact, uncountably many
> anti-diagonals.
> If there were, as WM incorrectly assumes, only
> countably many anti-diagonals there would be no need for the powers of
> primes construction: number any initial list of with the even numbers
> and then enumerate the supposedly countable anti-diagonals using the odd
> numbers.
>
> So how can we show that there are /always/ uncountably many
> anti-diagonals for any enumeration of the reals?
Not by your above approach.
> So how can we show that there are /always/ uncountably many
> anti-diagonals for any enumeration of the reals?
> 1. The plain counter-argument: the result of this construction is,
> supposedly, is a function wm from N to [0,1] yet the anti-diagonal
> constructed from flipping digit n of wm(n) is provably not in the image
> of wm.[1]
>
> 2. The complexity argument: if his assumption about there being only
> countably many anti-diagonals was correct, why did he use such a complex
> argument?
That's because the constructed antidiagonals can be inserted into the first list.
> 2. The complexity argument: if his assumption about there being only
> countably many anti-diagonals was correct, why did he use such a complex
> argument?
>
> 3. The why the argument is wrong argument: it assumes the conclusion.
On the contrary. You assume your conclusion:
> 3. The why the argument is wrong argument: it assumes the conclusion.
> There are uncountably many anti-diagonals
>
> Be prepared for the resulting snow-storm of waffle.
Your simple bunch of mistakes does not raise any storm.
> Be prepared for the resulting snow-storm of waffle.
But I never have mentioned this argument because of finite time and because of the much better argument "Conquer the Binary Tree": Buy a path for one cent, get one cent for each of its nodes. Buy further paths and get cents for each node not belonging to a path owned already. Get bankrupt finally because there are more paths than nodes. I never have heard a protest - not even from students of the university, visiting accidentally.
Regards, WM
WM
Dec 26, 2022, 1:01:43 PM12/26/22
to
Fritz Feldhase schrieb am Montag, 26. Dezember 2022 um 12:48:19 UTC+1:
> On Monday, December 26, 2022 at 11:21:04 AM UTC+1, WM wrote:
>
> > I show that all real numbers belong to a countable set of cardinality [...] ℵo.
>
> You can't "show" that,
I have shown it under the premise that countability is a sensible notion.
> On Monday, December 26, 2022 at 11:21:04 AM UTC+1, WM wrote:
>
> > I show that all real numbers belong to a countable set of cardinality [...] ℵo.
>
> You can't "show" that,
But the better proof is this: Buy a path for one cent, get one cent for each of its nodes. Buy further paths and get cents for each node not belonging to a path owned already. Get bankrupt finally because there are more paths than nodes. Of course this arguments ptroves uncountabiliyt is nonsense.I never have heard a protest - not even from students of the university, who were visiting accidentally. So much stupidity is reserved for readers of sci.logic
Regards, WM
Sergi o
Dec 26, 2022, 1:32:39 PM12/26/22
to
On 12/26/2022 12:01 PM, WM wrote:
> Fritz Feldhase schrieb am Montag, 26. Dezember 2022 um 12:48:19 UTC+1:
>> On Monday, December 26, 2022 at 11:21:04 AM UTC+1, WM wrote:
>>
>>> I show that all real numbers belong to a countable set of cardinality [...] ℵo.
>>
>> You can't "show" that,
>
> I have shown it under the premise that countability is a sensible notion.
you have yet to show you can count all real numbers...
> Fritz Feldhase schrieb am Montag, 26. Dezember 2022 um 12:48:19 UTC+1:
>> On Monday, December 26, 2022 at 11:21:04 AM UTC+1, WM wrote:
>>
>>> I show that all real numbers belong to a countable set of cardinality [...] ℵo.
>>
>> You can't "show" that,
>
> I have shown it under the premise that countability is a sensible notion.
we are waiting....
>
> But the better proof is this: Buy a path for one cent, get one cent for each of its nodes.
a node can have thousands of paths connected, or none at all.
you cannot get anywhere without a path.
Each node is numbered in black paint with a natural number.
Each path is identified by the two end nodes it attaches to.
Assuming I have all the nodes, I will sell them and make $$$.
obviously there are far more nodes than paths.
and this has nothing to do with your claim of Enumerating all real numbers, Where did that go ?
WM
Dec 26, 2022, 1:40:19 PM12/26/22
to
Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
> obviously there are far more nodes than paths.
Obviously. (Tomorrow you will claim the contrary.)
> obviously there are far more nodes than paths.
>
> and this has nothing to do with your claim of Enumerating all real numbers, Where did that go ?
The nodes are countable.
> and this has nothing to do with your claim of Enumerating all real numbers, Where did that go ?
Regards, WM
Python
Dec 26, 2022, 2:08:03 PM12/26/22
to
Crank Wolfgang Müeckenheim, aka WM wrote:
> [snip usual Mückenheim's nonsense]
> I never have heard a protest - not even from students of the university, visiting accidentally.
Interesting. So you had an "accidental" visit of your "course" by
students from a real University. Was it Augsburg University? Were
there students in maths? Are they organizing visit as if Hochschule
Augsburg were a kind of zoo? Or because they knew a crank there, YOU,
is pretending to teach math while his whole career, especially in
teaching, is a FRAUD?
> [snip usual Mückenheim's nonsense]
> I never have heard a protest - not even from students of the university, visiting accidentally.
students from a real University. Was it Augsburg University? Were
there students in maths? Are they organizing visit as if Hochschule
Augsburg were a kind of zoo? Or because they knew a crank there, YOU,
is pretending to teach math while his whole career, especially in
teaching, is a FRAUD?
WM
Dec 26, 2022, 5:01:06 PM12/26/22
to
Python schrieb am Montag, 26. Dezember 2022 um 20:08:03 UTC+1:
> > I never have heard a protest - not even from students of the university, visiting accidentally.
> Interesting. So you had an "accidental" visit of your "course" by
> students from a real University. Was it Augsburg University?
It happens that math-students of Augsburg University (possibly also from other universities, but I do not inspect my audience) visit my course. It also happens that students of Augsburg university switch to Augsburg University of Applied Sciences as well as vice versa that students of Augsburg University of Applied Sciences switch to Augsburg University.
> > I never have heard a protest - not even from students of the university, visiting accidentally.
> Interesting. So you had an "accidental" visit of your "course" by
> students from a real University. Was it Augsburg University?
Regards, WM
JVR
Dec 26, 2022, 5:47:42 PM12/26/22
to
Student's from Mücke's technical college have to satisfy the same requirements as everybody else in order to matriculate at a real university and, in general, they don't have the prerequisites, else they wouldn't be in McMuck-school.
Mücke's course on "The Infinite" is a joke, and not typical for schools of this kind. Mücke as a 'Professor of Mathematics' who has spent 20 years trying to understand the natural numbers is also a joke. But these things happen. Read 'Professor Unrat' ('The Blue Angel') by Heinrich Mann.
Sergi o
Dec 26, 2022, 7:53:27 PM12/26/22
to
On 12/26/2022 12:40 PM, WM wrote:
> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
>
>> obviously there are far more nodes than paths.
>
> Obviously. (Tomorrow you will claim the contrary.)
Obviously, since you provided no structure at all to what you are talking about.
> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
>
>> obviously there are far more nodes than paths.
>
> Obviously. (Tomorrow you will claim the contrary.)
>>
>> and this has nothing to do with your claim of Enumerating all real numbers, Where did that go ?
>
> The nodes are countable.
>
> Regards, WM
...and this has nothing to do with your claim of *Enumerating all real numbers*, Where did that go ?
did you make mistake ?
Sergi o
Dec 26, 2022, 8:57:38 PM12/26/22
to
WM
Dec 27, 2022, 3:28:29 AM12/27/22
to
Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
> On 12/26/2022 12:40 PM, WM wrote:
> > Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
> >
> >> obviously there are far more nodes than paths.
> >
> > Obviously. (Tomorrow you will claim the contrary.)
> Obviously, since you provided no structure at all to what you are talking about.
You don't know a Binary Tree?
> On 12/26/2022 12:40 PM, WM wrote:
> > Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
> >
> >> obviously there are far more nodes than paths.
> >
> > Obviously. (Tomorrow you will claim the contrary.)
> Obviously, since you provided no structure at all to what you are talking about.
> > The nodes are countable.
> because each node has its serial number stamped on it's bottom from the factory, they come that way.
> ...and this has nothing to do with your claim of *Enumerating all real numbers*, Where did that go ?
An infinite set which is smaller than a countable set is countable.
>
> did you make mistake ?
No. And never anybody listening to my proof has mentioned a counter argument:
> did you make mistake ?
Conquer the Binary Tree
You get one cent. Buy a path for that cent, get one cent for each of its nodes. Buy further paths for one cent each and get one cent for each node not belonging to a path owned already. If set theory is wrong, you will get infinitely rich. If set theory is true, you will get bankrupt finally because there are more paths than nodes. (Every transaction can be performed in half of the remaining time.)
I have never met anybody who supports the latter. Do you know or can you imagine of true fools who do?
Regards, WM
Python
Dec 27, 2022, 8:19:20 AM12/27/22
to
is "teaching" pseudo-mathematics and abuse students at Hochschule
Augsburg?
Sergi o
Dec 27, 2022, 9:23:49 AM12/27/22
to
On 12/27/2022 2:28 AM, WM wrote:
> Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
>> On 12/26/2022 12:40 PM, WM wrote:
>>> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
>>>
>>>> obviously there are far more nodes than paths.
>>>
>>> Obviously. (Tomorrow you will claim the contrary.)
>> Obviously, since you provided no structure at all to what you are talking about.
>
> You don't know a Binary Tree?
you did not state you were talking about a binary tree at all. your OP topic is your claim of *Enumerating all real numbers*, Where did that go ?
> Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
>> On 12/26/2022 12:40 PM, WM wrote:
>>> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
>>>
>>>> obviously there are far more nodes than paths.
>>>
>>> Obviously. (Tomorrow you will claim the contrary.)
>> Obviously, since you provided no structure at all to what you are talking about.
>
> You don't know a Binary Tree?
>
>>> The nodes are countable.
>> because each node has its serial number stamped on it's bottom from the factory, they come that way.
>
>> ...and this has nothing to do with your claim of *Enumerating all real numbers*, Where did that go ?
>
> An infinite set which is smaller than a countable set is countable.
cat got your tongue ?
>>
>> did you make mistake ?
>
> No. And never anybody listening to my proof has mentioned a counter argument:
>
you made a HUGE obvious mistake and now try to change subject.
>
> Regards, WM
>
WM
Dec 27, 2022, 11:40:58 AM12/27/22
to
Look my game "Conquer the Binary Tree" shows this in utmost clarity.
You get one cent. Buy a path for that cent, get one cent for each of its nodes. Buy further paths for one cent each and get one cent for each node not belonging to a path you own already. If set theory is wrong, you will get infinitely rich. If set theory is true, you will get bankrupt finally because there are more paths than nodes. (Every transaction can be performed in half of the remaining time.)
I have never received a counter argument, even from students of mathematics who occasionally have visited my lecture. Well it has been mentioned that receiving infinitely many cents per path makes things uncertain. I don't think so, but I can meet this argument: From every bought path you get only cents for the first 100 nodes not yet owned by you.
That is sufficient to prove my case. If the number of paths should be larger than the number of nodes, then you must some time start to buy paths which do not yield any node. But that is impossible.
If you have a valid counter argument, then let me know please. Otherwise recognize and admit that all your life you have adhered to a nonsense theory, which is not only wrong but also completely useless.
Regards, WM
WM
Dec 27, 2022, 11:44:43 AM12/27/22
to
Sergi o schrieb am Dienstag, 27. Dezember 2022 um 15:23:49 UTC+1:
> On 12/27/2022 2:28 AM, WM wrote:
> > Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
> >> On 12/26/2022 12:40 PM, WM wrote:
> >>> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
> >>>
> >>>> obviously there are far more nodes than paths.
> >>>
> >>> Obviously. (Tomorrow you will claim the contrary.)
> >> Obviously, since you provided no structure at all to what you are talking about.
> >
> > You don't know a Binary Tree?
> you did not state you were talking about a binary tree at all.
But you talked about it.
> On 12/27/2022 2:28 AM, WM wrote:
> > Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
> >> On 12/26/2022 12:40 PM, WM wrote:
> >>> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
> >>>
> >>>> obviously there are far more nodes than paths.
> >>>
> >>> Obviously. (Tomorrow you will claim the contrary.)
> >> Obviously, since you provided no structure at all to what you are talking about.
> >
> > You don't know a Binary Tree?
> you did not state you were talking about a binary tree at all.
> > No. And never anybody listening to my proof has mentioned a counter argument:
> >
> your OP topic is your claim of *Enumerating all real numbers*, Where did that go ?
Regards, WM
JVR
Dec 27, 2022, 1:10:54 PM12/27/22
to
WM
Dec 27, 2022, 5:05:03 PM12/27/22
to
JVR schrieb am Dienstag, 27. Dezember 2022 um 19:10:54 UTC+1:
> On Tuesday, December 27, 2022 at 2:19:20 PM UTC+1, Python wrote:
> Why would math students at Augsburg University care what goes on at some Technical College?
You mean Augsburg University of Applied Sciences?
> On Tuesday, December 27, 2022 at 2:19:20 PM UTC+1, Python wrote:
> Why would math students at Augsburg University care what goes on at some Technical College?
I have not asked them. Only by accident I became aware of one or the other. Perhaps it has become known that I give a lecture which is unique in Germany and which contains this irrefutable proof:
You get one cent. Buy a path for that cent, get one cent for each of its nodes. Buy further paths for one cent each and get one cent for each node not belonging to a path you own already. If set theory is wrong, you will get infinitely rich. If set theory is true, you will get bankrupt finally because there are more paths than nodes. (Every transaction can be performed in half of the remaining time.)
Sergi o
Dec 27, 2022, 8:45:40 PM12/27/22
to
On 12/27/2022 10:44 AM, WM wrote:
> Sergi o schrieb am Dienstag, 27. Dezember 2022 um 15:23:49 UTC+1:
>> On 12/27/2022 2:28 AM, WM wrote:
>>> Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
>>>> On 12/26/2022 12:40 PM, WM wrote:
>>>>> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
>>>>>
>>>>>> obviously there are far more nodes than paths.
>>>>>
>>>>> Obviously. (Tomorrow you will claim the contrary.)
>>>> Obviously, since you provided no structure at all to what you are talking about.
>>>
>>> You don't know a Binary Tree?
>> you did not state you were talking about a binary tree at all.
>
> But you talked about it.
no. I talked about nodes and paths, without any structure, as you provided none, and proved there are more paths than nodes.
> Sergi o schrieb am Dienstag, 27. Dezember 2022 um 15:23:49 UTC+1:
>> On 12/27/2022 2:28 AM, WM wrote:
>>> Sergi o schrieb am Dienstag, 27. Dezember 2022 um 01:53:27 UTC+1:
>>>> On 12/26/2022 12:40 PM, WM wrote:
>>>>> Sergi o schrieb am Montag, 26. Dezember 2022 um 19:32:39 UTC+1:
>>>>>
>>>>>> obviously there are far more nodes than paths.
>>>>>
>>>>> Obviously. (Tomorrow you will claim the contrary.)
>>>> Obviously, since you provided no structure at all to what you are talking about.
>>>
>>> You don't know a Binary Tree?
>> you did not state you were talking about a binary tree at all.
>
> But you talked about it.
>
>>> No. And never anybody listening to my proof has mentioned a counter argument:
>>>
>> your OP topic is your claim of *Enumerating all real numbers*, Where did that go ?
>
> I show that there are less paths than nodes. That is sufficient.
>
> Regards, WM
Fritz Feldhase
Dec 28, 2022, 3:28:39 AM12/28/22
to
On Wednesday, December 28, 2022 at 2:45:40 AM UTC+1, Sergi o wrote:
> So you "enumerated all real number" by saying "there are less paths than nodes".
Nutjob WM:
> So you "enumerated all real number" by saying "there are less paths than nodes".
"All real numbers can be enumerated since there are less paths than nodes".
"All real numbers can be enumerated since there are less paths than nodes".
"All real numbers can be enumerated since there are less paths than nodes".
:
WM
Dec 28, 2022, 5:32:39 AM12/28/22
to
Sergi o schrieb am Mittwoch, 28. Dezember 2022 um 02:45:40 UTC+1:
> On 12/27/2022 10:44 AM, WM wrote:
> >>>>>> obviously there are far more nodes than paths.
> >>>>>
> >>>>> Obviously. (Tomorrow you will claim the contrary.)
> On 12/27/2022 10:44 AM, WM wrote:
> >>>>>> obviously there are far more nodes than paths.
> >>>>>
> >>>>> Obviously. (Tomorrow you will claim the contrary.)
> > I show that there are less paths than nodes. That is sufficient.
> So you "enumerated all real number" by saying "there are less paths than nodes".
Enumerating of infinity sets is impossible, as I have shown elsewhere: https://www.researchgate.net/publication/365605468_Proof_of_the_existence_of_dark_numbers_bilingual_version
> So you "enumerated all real number" by saying "there are less paths than nodes".
Under the *assumption* that enumerating infinite sets is possible, the nodes of the Binary Tree could be enumerated and every smaller set like the set of paths could be enumerated too.
Regards, WM
Julio Di Egidio
Dec 28, 2022, 9:30:06 AM12/28/22
to
Julio
Sergi o
Dec 28, 2022, 11:12:15 AM12/28/22
to
On 12/28/2022 4:32 AM, WM wrote:
> Sergi o schrieb am Mittwoch, 28. Dezember 2022 um 02:45:40 UTC+1:
>> On 12/27/2022 10:44 AM, WM wrote:
>
>>>>>>>> obviously there are far more nodes than paths.
>>>>>>>
>>>>>>> Obviously. (Tomorrow you will claim the contrary.)
>
>>> I show that there are less paths than nodes. That is sufficient.
>> So you "enumerated all real number" by saying "there are less paths than nodes".
>
>
> Sergi o schrieb am Mittwoch, 28. Dezember 2022 um 02:45:40 UTC+1:
>> On 12/27/2022 10:44 AM, WM wrote:
>
>>>>>>>> obviously there are far more nodes than paths.
>>>>>>>
>>>>>>> Obviously. (Tomorrow you will claim the contrary.)
>
>>> I show that there are less paths than nodes. That is sufficient.
>> So you "enumerated all real number" by saying "there are less paths than nodes".
>
>
> Under the *assumption* that enumerating infinite sets is possible, the nodes of the Binary Tree could be enumerated and every smaller set like the set of paths could be enumerated too.
>
> Regards, WM
QUAcK!!
>
> Regards, WM
WM
Dec 28, 2022, 11:43:20 AM12/28/22
to
Sergi o schrieb am Mittwoch, 28. Dezember 2022 um 17:12:15 UTC+1:
> >>>>>>>> obviously there are far more nodes than paths.
> >>>>>>>
> >>>>>>> Obviously. (Tomorrow you will claim the contrary.)
Regards, WM
> >>>>>>>> obviously there are far more nodes than paths.
> >>>>>>>
> >>>>>>> Obviously. (Tomorrow you will claim the contrary.)
WM
Dec 28, 2022, 11:52:18 AM12/28/22
to
Fritz Feldhase schrieb am Mittwoch, 28. Dezember 2022 um 09:28:39 UTC+1:
> On Wednesday, December 28, 2022 at 2:45:40 AM UTC+1, Sergi o wrote:
>
> > So you "enumerated all real number" by saying "there are less paths than nodes".
>
> On Wednesday, December 28, 2022 at 2:45:40 AM UTC+1, Sergi o wrote:
>
> > So you "enumerated all real number" by saying "there are less paths than nodes".
>
> "All real numbers can be enumerated since there are less paths than nodes".
Sergi o
Dec 28, 2022, 11:53:32 AM12/28/22
to
Sergi o
Dec 28, 2022, 12:40:51 PM12/28/22
to
meaningless: "Under the *assumption* that enumerating infinite sets is possible" [no assumption needed]
meaningless: "every smaller set like the set of paths could be enumerated too." [every? how small?]
meaningless: binary tree and enumerating the real numbers.
>
> Regards, WM
Sergi o
Dec 30, 2022, 4:29:35 PM12/30/22
to
On 12/26/2022 4:21 AM, WM wrote:
> Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality |ω^ω| = ℵo.
so WHY are the real numbers countable ?
>
> The virtue of this method is that it avoids to squander the natural numbers: The entries of the Cantor-list are not enumerated by all natural numbers but by powers of 2. The antidiagonals of this list (never more than a countable set) are enumerated by the powers of 3. If a new list is created including these antidiagonals then its entries are enumerated by the powers of 5 and its antidiagonals (never more than a countable set) are enumerated by the powers of 7. This is going on, in every step using a new prime number. According to Euclid the set of prime numbers will never get exhausted.
>
> Regards, WM
> Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality |ω^ω| = ℵo.
so WHY are the real numbers countable ?
>
> The virtue of this method is that it avoids to squander the natural numbers: The entries of the Cantor-list are not enumerated by all natural numbers but by powers of 2. The antidiagonals of this list (never more than a countable set) are enumerated by the powers of 3. If a new list is created including these antidiagonals then its entries are enumerated by the powers of 5 and its antidiagonals (never more than a countable set) are enumerated by the powers of 7. This is going on, in every step using a new prime number. According to Euclid the set of prime numbers will never get exhausted.
>
> Regards, WM
Sergi o
Dec 30, 2022, 4:30:19 PM12/30/22
to
On 12/28/2022 4:32 AM, WM wrote:
Sergi o
Dec 31, 2022, 5:57:58 PM12/31/22
to
On 12/28/2022 10:43 AM, WM wrote:
Jim Burns
Dec 31, 2022, 6:15:43 PM12/31/22
to
Happy New Year.
olcott
Dec 31, 2022, 7:14:48 PM12/31/22
to
On 12/26/2022 4:21 AM, WM wrote:
> Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality |ω^ω| = ℵo.
>
> Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality |ω^ω| = ℵo.
>
> The virtue of this method is that it avoids to squander the natural numbers: The entries of the Cantor-list are not enumerated by all natural numbers but by powers of 2. The antidiagonals of this list (never more than a countable set) are enumerated by the powers of 3. If a new list is created including these antidiagonals then its entries are enumerated by the powers of 5 and its antidiagonals (never more than a countable set) are enumerated by the powers of 7. This is going on, in every step using a new prime number. According to Euclid the set of prime numbers will never get exhausted.
>
> Regards, WM
Real numbers can be counted by simply counting all of the immediately
>
> Regards, WM
adjacent points on a number line. The immediately adjacent points on a
number line can be identified using interval notation.
The line segment [0,1] is exactly one geometric point larger then the
line segment [0,1). I call this the infinitesimal number system.
Points immediately adjacent to a point on the number line can also be
reference by geometric point offsets.
The right endpoint of the line segment [0,1) can be referenced by an
offset from the right endpoint of the line segment [0,1] -
Geometric_Points(1)
Prior to my [Infinitesimal Number System] people did not seem to be
aware that immediately adjacent points on a number line existed.
Copyright 2018 Olcott
--
Copyright 2022 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
Sergi o
Dec 31, 2022, 9:09:13 PM12/31/22
to
Julio Di Egidio
Jan 1, 2023, 10:37:14 AM1/1/23
to
Julio
Julio Di Egidio
Jan 1, 2023, 10:40:49 AM1/1/23
to
Julio
Message has been deleted
Sergi o
Jan 1, 2023, 12:35:21 PM1/1/23
to
On 1/1/2023 10:47 AM, Fritz Feldhase wrote:
>
> Since for every node there is at least one path that "runs through" this node, there are at least as many paths as nodes.
>
> Hence the set of paths is at least "as large as" the set of nodes.
binary tree nodes can be sequentially numbered, as # of nodes is 1, 2, 4, 8, 16, ... to levels (or layers from the top) 1, 2, 3, 4, 5, ... (2^(level#))
all the paths can also be numbered sequentially as well, as you can draw out the binary tree as a flat map, and each node has 3 paths, one entering and
2 leaving. So number the paths between layers, between layers 1 and 2 there are 2 paths, between layers 2 and 3 there are 4 paths... (2,4,8,16,...)
so the entire binary tree can be mapped one to one to the rationals, and so can the paths.
If we stop at some level, k, it has 2^k nodes and 2^k paths in front of the nodes, and 2^(k+1) paths after the nodes.
so the answer to more or less nodes to paths, it depends upon before or after the node when measured, or it alternates as you go further out on the
tree, from a factor of 1 to a factor of 2, then factor of 1, then 2 etc.
or nodes/paths = 1 or 1/2
[this has nothing at all to do with enumerating the real numbers as WM claims he has]
> On Wednesday, December 28, 2022 at 5:52:18 PM UTC+1, WM wrote:
>> Fritz Feldhase schrieb am Mittwoch, 28. Dezember 2022 um 09:28:39 UTC+1:
>>> On Wednesday, December 28, 2022 at 2:45:40 AM UTC+1, Sergi o wrote:
>>>>
>>>> So you "enumerated all real number" by saying "there are less paths than nodes".
>>>>
>>> Nutjob WM: "All real numbers can be enumerated since there are less paths than nodes".
>> Fritz Feldhase schrieb am Mittwoch, 28. Dezember 2022 um 09:28:39 UTC+1:
>>> On Wednesday, December 28, 2022 at 2:45:40 AM UTC+1, Sergi o wrote:
>>>>
>>>> So you "enumerated all real number" by saying "there are less paths than nodes".
>>>>
>>>
>> Under the *assumption* that enumerating infinite sets is possible, the nodes of the Binary Tree could be enumerated and every smaller set like the set of paths could be enumerated too.
>
> Except that the set of paths is not smaller than the set of nodes.
>> Under the *assumption* that enumerating infinite sets is possible, the nodes of the Binary Tree could be enumerated and every smaller set like the set of paths could be enumerated too.
>
>
> Since for every node there is at least one path that "runs through" this node, there are at least as many paths as nodes.
>
> Hence the set of paths is at least "as large as" the set of nodes.
binary tree nodes can be sequentially numbered, as # of nodes is 1, 2, 4, 8, 16, ... to levels (or layers from the top) 1, 2, 3, 4, 5, ... (2^(level#))
all the paths can also be numbered sequentially as well, as you can draw out the binary tree as a flat map, and each node has 3 paths, one entering and
2 leaving. So number the paths between layers, between layers 1 and 2 there are 2 paths, between layers 2 and 3 there are 4 paths... (2,4,8,16,...)
so the entire binary tree can be mapped one to one to the rationals, and so can the paths.
If we stop at some level, k, it has 2^k nodes and 2^k paths in front of the nodes, and 2^(k+1) paths after the nodes.
so the answer to more or less nodes to paths, it depends upon before or after the node when measured, or it alternates as you go further out on the
tree, from a factor of 1 to a factor of 2, then factor of 1, then 2 etc.
or nodes/paths = 1 or 1/2
[this has nothing at all to do with enumerating the real numbers as WM claims he has]
Mostowski Collapse
Jan 1, 2023, 12:49:25 PM1/1/23
to
What about each real number?
LoL
WM schrieb:
> Assuming that countability is a sensible notion, I show that
> all real numbers belong to a countable set of cardinality |ω^ω| = ℵo.
Sergi o
Jan 1, 2023, 2:21:17 PM1/1/23
to
WM is still counting, he is stuck roughly in the interval between 0.0000000000000000000000000 and 0.000000000000000000000000001
Ben Bacarisse
Jan 1, 2023, 4:02:25 PM1/1/23
to
Sergi o <inv...@invalid.com> writes:
> On 1/1/2023 10:47 AM, Fritz Feldhase wrote:
>> Except that the set of paths is not smaller than the set of nodes.
>> Since for every node there is at least one path that "runs through" this node, there are at least as many paths as nodes.
>> Hence the set of paths is at least "as large as" the set of nodes.
>
>
> binary tree nodes can be sequentially numbered, as # of nodes is 1, 2,
> 4, 8, 16, ... to levels (or layers from the top) 1, 2, 3, 4, 5, ...
> (2^(level#))
>
> all the paths can also be numbered sequentially as well, as you can
> draw out the binary tree as a flat map, and each node has 3 paths, one
> entering and 2 leaving. So number the paths between layers, between
> layers 1 and 2 there are 2 paths, between layers 2 and 3 there are 4
> paths... (2,4,8,16,...)
>
> so the entire binary tree can be mapped one to one to the rationals,
> and so can the paths.
Only the finite paths. Every string in {L,R}* describes a path that
> On 1/1/2023 10:47 AM, Fritz Feldhase wrote:
>> Except that the set of paths is not smaller than the set of nodes.
>> Since for every node there is at least one path that "runs through" this node, there are at least as many paths as nodes.
>> Hence the set of paths is at least "as large as" the set of nodes.
>
>
> binary tree nodes can be sequentially numbered, as # of nodes is 1, 2,
> 4, 8, 16, ... to levels (or layers from the top) 1, 2, 3, 4, 5, ...
> (2^(level#))
>
> all the paths can also be numbered sequentially as well, as you can
> draw out the binary tree as a flat map, and each node has 3 paths, one
> entering and 2 leaving. So number the paths between layers, between
> layers 1 and 2 there are 2 paths, between layers 2 and 3 there are 4
> paths... (2,4,8,16,...)
>
> so the entire binary tree can be mapped one to one to the rationals,
> and so can the paths.
ends in a node and both the set of nodes and the strings in {L,R}* are
"countable".
But the infinite paths do not end. The infinite strings of L and R
(functions from N to {L,R}) are not "countable".
> [this has nothing at all to do with enumerating the real numbers as WM
> claims he has]
map to rational numbers as do those paths that have a repeating tail.
But the rest of the infinite paths map to irrational numbers.
--
Ben.
WM
Jan 1, 2023, 4:34:44 PM1/1/23
to
Mostowski Collapse schrieb am Sonntag, 1. Januar 2023 um 18:49:25 UTC+1:
> What about each real number?
>
Conquer the Binary Tree
> What about each real number?
>
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
WM
Jan 1, 2023, 4:36:19 PM1/1/23
to
Ben Bacarisse schrieb am Sonntag, 1. Januar 2023 um 22:02:25 UTC+1:
> The paths in the tree trivially map to reals in [0,1]. The finite paths
> map to rational numbers as do those paths that have a repeating tail.
> But the rest of the infinite paths map to irrational numbers.
There is no rest. There are limits - but not in the Binary Tree.
> The paths in the tree trivially map to reals in [0,1]. The finite paths
> map to rational numbers as do those paths that have a repeating tail.
> But the rest of the infinite paths map to irrational numbers.
Jim Burns
Jan 1, 2023, 5:08:41 PM1/1/23
to
I agree, very cool.
One day, my sister's phone started labeling
menu options in ????????
My first guess was Klingon.
It turned out to be Amharic.
Yes, we turned the English back on,
though I've been tempted to experiment
with my own phone.
----
Google says
(እያንዳንዱ ልውውጥ ተከስቷል) ⇒ ¬(ቦብ ከሴሎች በአንዱ ውስጥ አለ።)
==
(Each exchange has occurred) ⇒ ¬(Bob is in one of the cells.)
Sergi o
Jan 1, 2023, 10:55:29 PM1/1/23
to
your last paragraph is interesting, Im guessing that a path in the binary tree is a sequence of 1s,0s depending upon choice at a node, so you get
strings of binary numbers, put a dot in front, and you have all numbers in the interval [0,1], and the rest as you say.
Mostowski Collapse
Jan 2, 2023, 3:31:49 AM1/2/23
to
If set theory, such as ZFC, is right then
there is something wrong.
Its only an axiom system. Every theorem is
stated relative to it, like for example:
Theorem:
ZFC |- |ω^ω| > ℵo
So why would one claim |ω^ω| = ℵo in ZFC?
Thats a contradiction.
LoL
P.S.: Its an easy consequence of Cantors Theorem,
and |ω| = ℵo. By Cantors Theorem we have:
|2^ω| > |ω|
And for set exponentiation its also easy to see that
we have at least |ω^ω| >= |2^ω|, which gives by transitivity:
|ω^ω| > ℵo
there is something wrong.
Its only an axiom system. Every theorem is
stated relative to it, like for example:
Theorem:
ZFC |- |ω^ω| > ℵo
So why would one claim |ω^ω| = ℵo in ZFC?
Thats a contradiction.
LoL
P.S.: Its an easy consequence of Cantors Theorem,
and |ω| = ℵo. By Cantors Theorem we have:
|2^ω| > |ω|
And for set exponentiation its also easy to see that
we have at least |ω^ω| >= |2^ω|, which gives by transitivity:
|ω^ω| > ℵo
Mostowski Collapse
Jan 2, 2023, 3:46:29 AM1/2/23
to
The problem are words like "most". So when WM uses phrases
like "most paths", he is fuzzying set theory concepts, and his
cardinality |..| is not the same cardinality as set theory cardinality
|..|. We have basically two cardinalities, WMs claim based on fuzzying:
/* WMs Claim */
|ω^ω|_2 = ℵo
and then the usual set theory claim, based on cardinality:
/* ZFCs Claim */
|ω^ω|_1 > ℵo
Mostlikely you can establish some fuzzying set theory and make
WMs Claim true. There is something similar with Fermats Last
Theorem, we have this here, it has no non-trivial solution:
/* Fermats Last Theorem */
x^3 + y^3 = z^3
On the other hand this here has infinitely many solutions
/* Almost Fermat Last Theorem */
x^3 + y^3 = z^3 +/- 1
See also:
An amazing identity of Ramanujan
https://web.maths.unsw.edu.au/~mikeh/webpapers/paper39.pdf
like "most paths", he is fuzzying set theory concepts, and his
cardinality |..| is not the same cardinality as set theory cardinality
|..|. We have basically two cardinalities, WMs claim based on fuzzying:
/* WMs Claim */
|ω^ω|_2 = ℵo
and then the usual set theory claim, based on cardinality:
/* ZFCs Claim */
|ω^ω|_1 > ℵo
Mostlikely you can establish some fuzzying set theory and make
WMs Claim true. There is something similar with Fermats Last
Theorem, we have this here, it has no non-trivial solution:
/* Fermats Last Theorem */
x^3 + y^3 = z^3
On the other hand this here has infinitely many solutions
/* Almost Fermat Last Theorem */
x^3 + y^3 = z^3 +/- 1
See also:
An amazing identity of Ramanujan
https://web.maths.unsw.edu.au/~mikeh/webpapers/paper39.pdf
Ben Bacarisse
Jan 2, 2023, 6:40:52 AM1/2/23
to
graph with node set N⁺ and edge set { (i, 2i), (1, 2i+1) | i in N⁺ }.
(Using N⁺ to be clear that the nodes start at 1 not 0).
> your last paragraph is interesting, Im guessing that a path in the
> binary tree is a sequence of 1s,0s depending upon choice at a node, so
> you get strings of binary numbers, put a dot in front, and you have
> all numbers in the interval [0,1], and the rest as you say.
nonsense about binary trees for years...
--
Ben.
JVR
Jan 2, 2023, 7:35:45 AM1/2/23
to
Sometimes he even decorates it with tinsel (Lametta in German) and paints
paths in various colors.
I think it is literally true that this crackpot has learned nothing in the 20+ years
that he has been spreading the news that all mathematics formulated after
the year 1642 is nonsense.
When did he first show up in the usenet?
JVR
Jan 2, 2023, 8:46:21 AM1/2/23
to
but are not as unbelievable and are not as persistently held as delusions."
--- APA dictionary of psychology, 2nd ed
WM
Jan 2, 2023, 9:29:04 AM1/2/23
to
==> There is no limit matrix of only X.
==> There is no enumeration of all fractions.
Regards, WM
WM
Jan 2, 2023, 9:32:36 AM1/2/23
to
Sergi o schrieb am Montag, 2. Januar 2023 um 04:55:29 UTC+1:
>
> I had "links", confused with "paths", where a "link" is the path between only 2 nodes, however a "path" is an arbitrary path of nodes and links.
>
Conquer the Binary Tree
>
> I had "links", confused with "paths", where a "link" is the path between only 2 nodes, however a "path" is an arbitrary path of nodes and links.
>
.
/ \
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 ∈ ℕ_0} of nodes a_k is countable as shown by the indices of the nodes.
WM
Jan 2, 2023, 9:34:11 AM1/2/23
to
Ben Bacarisse schrieb am Montag, 2. Januar 2023 um 12:40:52 UTC+1:
> Yes, sorry, I thought that did not need saying. WM has been
convincing all his students with this story for years:
> Yes, sorry, I thought that did not need saying. WM has been
Sergi o
Jan 2, 2023, 9:34:39 AM1/2/23
to
Ramanujan was a true Genius
here is lost notebook;
https://archive.org/details/lostnotebookothe0000rama
here is lost notebook;
https://archive.org/details/lostnotebookothe0000rama
WM
Jan 2, 2023, 9:35:00 AM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 09:46:29 UTC+1:
> The problem are words like "most". So when WM uses phrases
> like "most paths", he is fuzzying set theory concepts, and his
> cardinality |..| is not the same cardinality as set theory cardinality
>
> The problem are words like "most". So when WM uses phrases
> like "most paths", he is fuzzying set theory concepts, and his
> cardinality |..| is not the same cardinality as set theory cardinality
>
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 ∈ ℕ_0} of nodes a_k is countable as shown by the indices of the nodes.
.
/ \
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 ∈ ℕ_0} of nodes a_k is countable as shown by the indices of the nodes.
Fritz Feldhase
Jan 2, 2023, 9:59:17 AM1/2/23
to
On Monday, January 2, 2023 at 3:29:04 PM UTC+1, WM wrote:
> There is no enumeration of all fractions.
Sure there is. See: https://mathworld.wolfram.com/PairingFunction.html
> There is no enumeration of all fractions.
Mostowski Collapse
Jan 2, 2023, 10:33:55 AM1/2/23
to
If he doesn't use fuzzy notions like "most paths", he just
falls back to this relatively simple paradox:
- The binary tree has countably infinite many nodes.
- The binary tree has uncountably infinite many infinite paths.
And he tries to confuse node and path. A node can be
adressed by a finite bit path which can be easily encoded
as natural numbers, showing that there are countably
infinite many nodes. That there are uncountably infinite
many paths, can be shown by Cantor's Theorem:
Cantor's theorem
https://en.wikipedia.org/wiki/Cantor's_theorem
falls back to this relatively simple paradox:
- The binary tree has countably infinite many nodes.
- The binary tree has uncountably infinite many infinite paths.
And he tries to confuse node and path. A node can be
adressed by a finite bit path which can be easily encoded
as natural numbers, showing that there are countably
infinite many nodes. That there are uncountably infinite
many paths, can be shown by Cantor's Theorem:
Cantor's theorem
https://en.wikipedia.org/wiki/Cantor's_theorem
WM
Jan 2, 2023, 11:04:57 AM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 16:33:55 UTC+1:
> A node can be
> adressed by a finite bit path which can be easily encoded
> as natural numbers, showing that there are countably
> infinite many nodes. That there are uncountably infinite
> many paths, can be shown by Cantor's Theorem:
This shows that there is a contradiction in set theory. Or do you believe that the game Conquer the Binary Tree could be lost? Even if only 2 nodes per selected path could be collected, the player would win.
> A node can be
> adressed by a finite bit path which can be easily encoded
> as natural numbers, showing that there are countably
> infinite many nodes. That there are uncountably infinite
> many paths, can be shown by Cantor's Theorem:
Regards, WM
WM
Jan 2, 2023, 11:08:54 AM1/2/23
to
Regards, WM
Mostowski Collapse
Jan 2, 2023, 11:16:44 AM1/2/23
to
You would need to first show that your game
is even related to Cantors Theorem.
Usually a game has at least two players ∃ and
∀. But your game has only one player.
Such things do not really deserve the name
game. Whats the opponent move?
WM schrieb:
is even related to Cantors Theorem.
Usually a game has at least two players ∃ and
∀. But your game has only one player.
Such things do not really deserve the name
game. Whats the opponent move?
WM schrieb:
Mostowski Collapse
Jan 2, 2023, 11:20:51 AM1/2/23
to
If you want games in set theory, check this out:
In mathematics, the axiom of determinacy
(abbreviated as AD) is a possible axiom
for set theory introduced by Jan Mycielski
and Hugo Steinhaus in 1962. It refers to
certain two-person topological games of length ω.
https://en.wikipedia.org/wiki/Axiom_of_determinacy
But the axiom of determinancy is not part of
ZFC. You need to add it separately, and you
will then get another set theory than ZFC itself.
You get something a little stronger, that
might even have certain cardinal numbers.
So the question would then not be whether
ZFC is consistent, but whether this extended
ZFC is consistent. Such games are also part
of so called descriptive set theory.
Mostowski Collapse schrieb:
In mathematics, the axiom of determinacy
(abbreviated as AD) is a possible axiom
for set theory introduced by Jan Mycielski
and Hugo Steinhaus in 1962. It refers to
certain two-person topological games of length ω.
https://en.wikipedia.org/wiki/Axiom_of_determinacy
But the axiom of determinancy is not part of
ZFC. You need to add it separately, and you
will then get another set theory than ZFC itself.
You get something a little stronger, that
might even have certain cardinal numbers.
So the question would then not be whether
ZFC is consistent, but whether this extended
ZFC is consistent. Such games are also part
of so called descriptive set theory.
Mostowski Collapse schrieb:
WM
Jan 2, 2023, 11:26:59 AM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 17:16:44 UTC+1:
> You would need to first show that your game
> is even related to Cantors Theorem.
Everybody, except very uneducated ones, knows that the paths represent the real numbers in the unit interval and that Cantor's result is that they re uncountable, i.e., more than nodes..
> You would need to first show that your game
> is even related to Cantors Theorem.
> Usually a game has at least two players ∃ and
> ∀. But your game has only one player.
>
> Such things do not really deserve the name
> game. Whats the opponent move?
The opponent's move has been done in 1891. Then the mine had been laid.
> Such things do not really deserve the name
> game. Whats the opponent move?
Regards, WM
Mostowski Collapse
Jan 2, 2023, 11:29:05 AM1/2/23
to
Unfortunately the full axiom of determinacy (AD),
contradicts the axiom of choice, so your other
set theory would be ZF+AD I guess, and not ZFC+AD.
See also here:
"The axiom of determinacy is a proposed axiom of
set theory that is consistent with Zermelo-Fraenkel
set theory (ZF) but is inconsistent with the axiom
of choice (and hence ZFC)."
https://brilliant.org/wiki/axiom-of-determinacy/
BTW: About descriptive set theory, i.e. where
determinacy already fails in ZFC:
The properties of the projective sets are not
completely determined by ZFC. This is related to
the fact that ZFC proves Borel determinacy, but
not projective determinacy.
https://en.wikipedia.org/wiki/Descriptive_set_theory#Projective_sets_and_Wadge_degrees
But I doubt that you can challenge Cantors
Theorem by stealing the idea of games. Since
Cantors Theorem doesn't need much to be
provable. Its only a diagonal argument, based
on the axiom of comprehension. So it even doesn't
need the axiom of choice to be proved.
Mostowski Collapse schrieb:
contradicts the axiom of choice, so your other
set theory would be ZF+AD I guess, and not ZFC+AD.
See also here:
"The axiom of determinacy is a proposed axiom of
set theory that is consistent with Zermelo-Fraenkel
set theory (ZF) but is inconsistent with the axiom
of choice (and hence ZFC)."
https://brilliant.org/wiki/axiom-of-determinacy/
BTW: About descriptive set theory, i.e. where
determinacy already fails in ZFC:
The properties of the projective sets are not
completely determined by ZFC. This is related to
the fact that ZFC proves Borel determinacy, but
not projective determinacy.
https://en.wikipedia.org/wiki/Descriptive_set_theory#Projective_sets_and_Wadge_degrees
But I doubt that you can challenge Cantors
Theorem by stealing the idea of games. Since
Cantors Theorem doesn't need much to be
provable. Its only a diagonal argument, based
on the axiom of comprehension. So it even doesn't
need the axiom of choice to be proved.
Mostowski Collapse schrieb:
Mostowski Collapse
Jan 2, 2023, 11:33:13 AM1/2/23
to
Solitär isn't a game, its only a puzzle.
If you don't have an opponennt, then
its only a puzzle and not a game.
A puzzle is "solved" by finding a path
to a solution. On the other hand to
always win a game, the effort is much
higher. A simple path doesn't suffice,
you need to develop a winning strategy.
Which is multiple paths depending
on the opponents move.
WM schrieb:
If you don't have an opponennt, then
its only a puzzle and not a game.
A puzzle is "solved" by finding a path
to a solution. On the other hand to
always win a game, the effort is much
higher. A simple path doesn't suffice,
you need to develop a winning strategy.
Which is multiple paths depending
on the opponents move.
WM schrieb:
Mostowski Collapse
Jan 2, 2023, 11:36:56 AM1/2/23
to
The difference between puzzle and game,
is the same difference as between a turing
machine, and an alternating turing machine.
Different complexity classes.
"The definition of NP uses the existential
mode of computation: if any choice leads
to an accepting state, then the whole
computation accepts.
The definition of co-NP uses the universal
mode of computation: only if all choices
lead to an accepting state does the whole
computation accept.
An alternating Turing machine (or to be more
precise, the definition of acceptance for
such a machine) alternates between these modes.
https://en.wikipedia.org/wiki/Alternating_Turing_machine
Mostowski Collapse schrieb:
is the same difference as between a turing
machine, and an alternating turing machine.
Different complexity classes.
"The definition of NP uses the existential
mode of computation: if any choice leads
to an accepting state, then the whole
computation accepts.
The definition of co-NP uses the universal
mode of computation: only if all choices
lead to an accepting state does the whole
computation accept.
An alternating Turing machine (or to be more
precise, the definition of acceptance for
such a machine) alternates between these modes.
https://en.wikipedia.org/wiki/Alternating_Turing_machine
Mostowski Collapse schrieb:
WM
Jan 2, 2023, 11:37:45 AM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 17:33:13 UTC+1:
> Solitär isn't a game, its only a puzzle.
> If you don't have an opponennt, then
> its only a puzzle and not a game.
Solitaire is any tabletop game which one can play by oneself, or a single-player game of concentration and skill using a set layout of tiles, pegs or stones. (Wiki)
> Solitär isn't a game, its only a puzzle.
> If you don't have an opponennt, then
> its only a puzzle and not a game.
Regards, WM
Mostowski Collapse
Jan 2, 2023, 11:45:24 AM1/2/23
to
Ignorance is bliss?
The vulgar term "game" is not the same term as the
mathematical term "game". There is even a game theory.
Modern game theory began with the idea of mixed-
strategy equilibria in two-person zero-sum game and
its proof by John von Neumann.
https://en.wikipedia.org/wiki/Game_theory
The vulgar term "game" is not the same term as the
mathematical term "game". There is even a game theory.
Modern game theory began with the idea of mixed-
strategy equilibria in two-person zero-sum game and
its proof by John von Neumann.
https://en.wikipedia.org/wiki/Game_theory
FredJeffries
Jan 2, 2023, 11:53:40 AM1/2/23
to
On Monday, January 2, 2023 at 12:31:49 AM UTC-8, Mostowski Collapse wrote:
> If set theory, such as ZFC, is right then
> there is something wrong.
>
> Its only an axiom system. Every theorem is
> stated relative to it, like for example:
>
> Theorem:
> ZFC |- |ω^ω| > ℵo
>
> So why would one claim |ω^ω| = ℵo in ZFC?
> Thats a contradiction.
Incorrect
> If set theory, such as ZFC, is right then
> there is something wrong.
>
> Its only an axiom system. Every theorem is
> stated relative to it, like for example:
>
> Theorem:
> ZFC |- |ω^ω| > ℵo
>
> So why would one claim |ω^ω| = ℵo in ZFC?
> Thats a contradiction.
https://mathworld.wolfram.com/OrdinalExponentiation.html
https://en.wikipedia.org/wiki/Ordinal_arithmetic#Exponentiation
>
> LoL
The lol is on you
Mostowski Collapse
Jan 2, 2023, 11:56:34 AM1/2/23
to
What do you think is incorrect?
To the best of my knowlegde we have |ω^ω| > ℵo and hence
|ω^ω| =/= ℵo and not |ω^ω| = ℵo as WM claims.
And |ω^ω| =/= ℵo & |ω^ω| = ℵo is a contradiction of
the form ~A & A, do you deny that also?
To the best of my knowlegde we have |ω^ω| > ℵo and hence
|ω^ω| =/= ℵo and not |ω^ω| = ℵo as WM claims.
And |ω^ω| =/= ℵo & |ω^ω| = ℵo is a contradiction of
the form ~A & A, do you deny that also?
Mostowski Collapse
Jan 2, 2023, 12:04:53 PM1/2/23
to
Please note also I didn't refer to ordinal exponentiation, only
to set exponentiation. Where did you see me mentioning ordinal
exponentiation. You were citing ordinal exponentiation, why?
I only was refering to set exponentiation in my post. Ordinal
exponentiation and set exponentiation are not the same.
Set exponentiation:
ω^ω = { f : ω -> ω }, also known as Baire space ω^ω
See also:
Note that ω^ω is not the ordinal exponentiation,
but rather the set of all functions from ω to itself.
It is easy, and I leave it to you, to verify that this is
in fact a set of cardinality 2^ℵ0 (if you give up, it
has been asked on this site infinitely many times
before), so definitely not ℵ0.
https://math.stackexchange.com/a/598226
to set exponentiation. Where did you see me mentioning ordinal
exponentiation. You were citing ordinal exponentiation, why?
I only was refering to set exponentiation in my post. Ordinal
exponentiation and set exponentiation are not the same.
Set exponentiation:
ω^ω = { f : ω -> ω }, also known as Baire space ω^ω
See also:
Note that ω^ω is not the ordinal exponentiation,
but rather the set of all functions from ω to itself.
It is easy, and I leave it to you, to verify that this is
in fact a set of cardinality 2^ℵ0 (if you give up, it
has been asked on this site infinitely many times
before), so definitely not ℵ0.
https://math.stackexchange.com/a/598226
Mostowski Collapse
Jan 2, 2023, 12:28:03 PM1/2/23
to
You need to ask WM what he means by ω^ω. But
because he ultimately puts a cardinal bracket around it,
and has |ω^ω| = ℵo as some formal statement
of his thread, it is most likely not ordinal nor cardinal
exponentiation, it would rather make sense that it
is set exponentiation. Which is what I was assuming.
because he ultimately puts a cardinal bracket around it,
and has |ω^ω| = ℵo as some formal statement
of his thread, it is most likely not ordinal nor cardinal
exponentiation, it would rather make sense that it
is set exponentiation. Which is what I was assuming.
FredJeffries
Jan 2, 2023, 1:29:22 PM1/2/23
to
On Monday, January 2, 2023 at 9:28:03 AM UTC-8, Mostowski Collapse wrote:
> You need to ask WM what he means by ω^ω. But
> because he ultimately puts a cardinal bracket around it,
> and has |ω^ω| = ℵo as some formal statement
Our beloved and esteemed professor merely copied and pasted the statement
> You need to ask WM what he means by ω^ω. But
> because he ultimately puts a cardinal bracket around it,
> and has |ω^ω| = ℵo as some formal statement
|ω^ω| = ℵo
from somewhere. He has no idea what it means, so it would be pointless to 'ask him'. But the context from where he most likely would have copied it is ordinal exponentiation, where it is a correct and familiar statement.
> of his thread, it is most likely not ordinal nor cardinal
> exponentiation, it would rather make sense that it
> is set exponentiation. Which is what I was assuming.
> Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 18:04:53 UTC+1:
> > Please note also I didn't refer to ordinal exponentiation, only
> > to set exponentiation. Where did you see me mentioning ordinal
> > exponentiation.
> > You were citing ordinal exponentiation, why?
> >
> > I only was refering to set exponentiation in my post. Ordinal
> > exponentiation and set exponentiation are not the same.
> >
> > Set exponentiation:
> > ω^ω = { f : ω -> ω }, also known as Baire space ω^ω
> >
> > See also:
> >
> > Note that ω^ω is not the ordinal exponentiation,
> > but rather the set of all functions from ω to itself.
> > It is easy, and I leave it to you, to verify that this is
> > in fact a set of cardinality 2^ℵ0 (if you give up, it
> > has been asked on this site infinitely many times
> > before), so definitely not ℵ0.
> > https://math.stackexchange.com/a/598226
https://math.stackexchange.com/questions/598224
Enough. The reason I posted in the first place was only because you seem to me to be an unmitigated pompous ass and it gave me a small, fleeting bit of pleasure to express that opinion. And your pathetic attempts at coverup only reinforces that opinion.
EOD
Mostowski Collapse
Jan 2, 2023, 2:03:03 PM1/2/23
to
Congratulations, you are getting rude and make
the false accusation that I tried to cover up something.
But you can re-read this thread, the mistake is on
clearly on your side, you didn't read my proof.
I literally wrote, you even cited it:
FredJeffries schrieb am Montag, 2. Januar 2023 um 17:53:40 UTC+1:
> P.S.: Its an easy consequence of Cantors Theorem,
> and |ω| = ℵo. By Cantors Theorem we have:
>
> |2^ω| > |ω|
>
> we have at least |ω^ω| >= |2^ω|, which gives by transitivity:
>
> |ω^ω| > ℵo
https://groups.google.com/g/sci.logic/c/FAWjFlV5b9I/m/P7TULIgOAgAJ
>
> |ω^ω| > ℵo
Such bull shit of yours doesn't need to be covered up.
It needs to be layed out in plain sight to show what moron you are.
Mostowski Collapse
Jan 2, 2023, 2:16:23 PM1/2/23
to
The SET EXPONENTIATION law that I used is this
monotonicity, which then translates to cardinality:
A ⊆ B => A^C ⊆ B^C
Anyway if you find an error in my proof, which
is a little weaker claim than what MSE claims,
I am happy if you post it. Otherwise please forward
your posts to smalldi...@getalife.com .
https://twitter.com/GretaThunberg/status/1608056944501178368
monotonicity, which then translates to cardinality:
A ⊆ B => A^C ⊆ B^C
Anyway if you find an error in my proof, which
is a little weaker claim than what MSE claims,
I am happy if you post it. Otherwise please forward
your posts to smalldi...@getalife.com .
https://twitter.com/GretaThunberg/status/1608056944501178368
Mostowski Collapse
Jan 2, 2023, 2:37:51 PM1/2/23
to
Since FredJeffries wasted so much energy in
being rude, its possibly a useless idea to ask him
for a review of my proof. So I retract that.
The ideal playground for his energy would be possibly
to put him some gun at hand and send him into
some Detroit Ghetto, so that he can let off some steam.
But be careful with looting, don't trap yourself.
LoL
Mostowski Collapse
Jan 2, 2023, 2:49:11 PM1/2/23
to
Since you are a little clumsy, I also retract that
with the gun, you might shoot yourself in the foot.
Fritz Feldhase
Jan 2, 2023, 3:13:11 PM1/2/23
to
On Monday, January 2, 2023 at 7:29:22 PM UTC+1, FredJeffries wrote:
> Our beloved and esteemed professor merely copied and pasted the statement
>
> |ω^ω| = ℵo
>
> from somewhere. He has no idea what it means, so it would be pointless to 'ask him'.
Well, actually he stated:
> Our beloved and esteemed professor merely copied and pasted the statement
>
> |ω^ω| = ℵo
>
> from somewhere. He has no idea what it means, so it would be pointless to 'ask him'.
WM: "Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality |ω^ω| = ℵo."
So we may just ignore that "questionable" detail, and focus on the core of the statement:
WM: "Assuming that countability is a sensible notion, I show that all real numbers belong to a countable set of cardinality [...] ℵo."
Mostowski Collapse
Jan 2, 2023, 3:37:54 PM1/2/23
to
Its no secret that a positive real number r e R+ can be mapped
to an element of the Bair Space ω^ω. For each positive real
number r e R+ there exists a sequence a0,a1,a2,...
natural numbers. If r is rational we will end with zeros eventually,
otherwise if r is irrational the sequences goes on and on:
r0 = r
aj = floor(rj)
rj+1 = 1/(rj-aj) if rj-aj =\= 0
= 0 if rj-aj = 0
Example π:
Simple continued fraction expansion of Pi.
https://oeis.org/A001203
Here is a slightly different suggestion, only covering the irrational numbers:
https://en.wikipedia.org/wiki/Baire_space_%28set_theory%29#Relation_to_the_real_line
Another nifty coding would be to use a0=floor(r). And
a1,a2,.. alternating run-length encoding of the bits
r-a0, when looking at the binary expansion of this value.
Sergi o
Jan 2, 2023, 3:43:35 PM1/2/23
to
Mostowski Collapse
Jan 2, 2023, 3:47:39 PM1/2/23
to
In this here:
> aj = floor(rj)
> rj+1 = 1/(rj-aj)
One might want to do the reciprocal first and then
the flooring, eh voila one ends with this monster.
Its at least a monster namewise:
https://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin%E2%80%93Wirsing_operator
But its basically from the Greeks:
geometric anthyphairesis (Euclid’s Elements X.2)
https://webusers.imj-prg.fr/~salomon.ofman/Anthyphairesis_and_Theory_of_Proportions-I.pdf
Sorry for flushing sci.logic with so much stuff, but
it happens that I am already posting for weeks about
this stuff, in connection with SWI-Prologs rationalize/1
evaluable function, which gets currently a face lift.
So I am a little sensitive now and picking up every
new bit I see, this monster is new to me.
> aj = floor(rj)
> rj+1 = 1/(rj-aj)
the flooring, eh voila one ends with this monster.
Its at least a monster namewise:
https://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin%E2%80%93Wirsing_operator
But its basically from the Greeks:
geometric anthyphairesis (Euclid’s Elements X.2)
https://webusers.imj-prg.fr/~salomon.ofman/Anthyphairesis_and_Theory_of_Proportions-I.pdf
Sorry for flushing sci.logic with so much stuff, but
it happens that I am already posting for weeks about
this stuff, in connection with SWI-Prologs rationalize/1
evaluable function, which gets currently a face lift.
So I am a little sensitive now and picking up every
new bit I see, this monster is new to me.
WM
Jan 2, 2023, 5:08:34 PM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 17:45:24 UTC+1:
> The vulgar term "game" is not the same term as the
> mathematical term "game".
Conquer the Binary Tree is a vulgar game. It can be understood and played without game theory by all people except those with brains damaged by set theory.
> The vulgar term "game" is not the same term as the
> mathematical term "game".
Regards, WM
WM
Jan 2, 2023, 5:16:51 PM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 09:46:29 UTC+1:
> The problem are words like "most". So when WM uses phrases
> like "most paths", he is fuzzying set theory concepts, and his
> cardinality |..| is not the same cardinality as set theory cardinality
>
> |..|. We have basically two cardinalities, WMs claim based on fuzzying:
> The problem are words like "most". So when WM uses phrases
> like "most paths", he is fuzzying set theory concepts, and his
> cardinality |..| is not the same cardinality as set theory cardinality
>
>
> /* WMs Claim */
> |ω^ω|_2 = ℵo
>
> and then the usual set theory claim, based on cardinality:
>
> /* ZFCs Claim */
> |ω^ω|_1 > ℵo
>
> Mostlikely you can establish some fuzzying set theory and make
> WMs Claim true.
A model of ω^ω is the set of all finite sequences of natural numbers or, according to Hessenberg [G. Hessenberg: "Grundbegriffe der Mengenlehre", offprint from Abhandlungen der Fries'schen Schule, Vol. I, no. 4, Vandenhoeck & Ruprecht, Göttingen (1906) § 20], the ordering of the natural numbers by the number of prime factors and then by sizes of the factors.
Regards, WM
WM
Jan 2, 2023, 5:21:50 PM1/2/23
to
FredJeffries schrieb am Montag, 2. Januar 2023 um 19:29:22 UTC+1:
> On Monday, January 2, 2023 at 9:28:03 AM UTC-8, Mostowski Collapse wrote:
> > You need to ask WM what he means by ω^ω. But
> > because he ultimately puts a cardinal bracket around it,
> > and has |ω^ω| = ℵo as some formal statement
> Our beloved and esteemed professor merely copied and pasted the statement
>
> |ω^ω| = ℵo
>
> from somewhere.
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 42.
> On Monday, January 2, 2023 at 9:28:03 AM UTC-8, Mostowski Collapse wrote:
> > You need to ask WM what he means by ω^ω. But
> > because he ultimately puts a cardinal bracket around it,
> > and has |ω^ω| = ℵo as some formal statement
> Our beloved and esteemed professor merely copied and pasted the statement
>
> |ω^ω| = ℵo
>
> from somewhere.
A model of ω^ω is the set of all finite sequences of natural numbers or, according to Hessenberg [G. Hessenberg: "Grundbegriffe der Mengenlehre", offprint from Abhandlungen der Fries'schen Schule, Vol. I, no. 4, Vandenhoeck & Ruprecht, Göttingen (1906) § 20], the ordering of the natural numbers by the number of prime factors and then by sizes of the factors.
Regards, WM
WM
Jan 2, 2023, 5:24:02 PM1/2/23
to
Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 20:16:23 UTC+1:
> The SET EXPONENTIATION law that I used is this
> monotonicity, which then translates to cardinality:
>
> A ⊆ B => A^C ⊆ B^C
>
> Anyway if you find an error in my proof,
For education look into https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 42
> The SET EXPONENTIATION law that I used is this
> monotonicity, which then translates to cardinality:
>
> A ⊆ B => A^C ⊆ B^C
>
> Anyway if you find an error in my proof,
You will find then your error yourself.
Regards, WM
FredJeffries
Jan 2, 2023, 5:48:13 PM1/2/23
to
FredJeffries
Jan 2, 2023, 5:53:10 PM1/2/23
to
Sergi o
Jan 2, 2023, 7:00:20 PM1/2/23
to
On 1/2/2023 4:24 PM, WM wrote:
> Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 20:16:23 UTC+1:
>> The SET EXPONENTIATION law that I used is this
>> monotonicity, which then translates to cardinality:
>>
>> A ⊆ B => A^C ⊆ B^C
>>
>> Anyway if you find an error in my proof,
>
> For education look into https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 42
that is a troll book, mostly posts from this newsgroup!
> Mostowski Collapse schrieb am Montag, 2. Januar 2023 um 20:16:23 UTC+1:
>> The SET EXPONENTIATION law that I used is this
>> monotonicity, which then translates to cardinality:
>>
>> A ⊆ B => A^C ⊆ B^C
>>
>> Anyway if you find an error in my proof,
>
> For education look into https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 42
>
> Regards, WM
>
>
Sergi o
Jan 2, 2023, 7:01:33 PM1/2/23
to
On 1/2/2023 4:21 PM, WM wrote:
> FredJeffries schrieb am Montag, 2. Januar 2023 um 19:29:22 UTC+1:
>> On Monday, January 2, 2023 at 9:28:03 AM UTC-8, Mostowski Collapse wrote:
>>> You need to ask WM what he means by ω^ω. But
>>> because he ultimately puts a cardinal bracket around it,
>>> and has |ω^ω| = ℵo as some formal statement
>> Our beloved and esteemed professor merely copied and pasted the statement
>>
>> |ω^ω| = ℵo
>>
>> from somewhere.
>
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 42.
that is a troll book of postings in newsgroups
> FredJeffries schrieb am Montag, 2. Januar 2023 um 19:29:22 UTC+1:
>> On Monday, January 2, 2023 at 9:28:03 AM UTC-8, Mostowski Collapse wrote:
>>> You need to ask WM what he means by ω^ω. But
>>> because he ultimately puts a cardinal bracket around it,
>>> and has |ω^ω| = ℵo as some formal statement
>> Our beloved and esteemed professor merely copied and pasted the statement
>>
>> |ω^ω| = ℵo
>>
>> from somewhere.
>
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 42.
>
>
> Regards, WM
Ben Bacarisse
Jan 2, 2023, 8:24:23 PM1/2/23
to
WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
> Ben Bacarisse schrieb am Montag, 2. Januar 2023 um 12:40:52 UTC+1:
>
>> Yes, sorry, I thought that did not need saying. WM has been
>
> convincing all his students with this story for years:
How sad. Fortunately they are never going to need to learn the correct
answers as your college does not offer a degree in mathematics.
> 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 ∈ ℕ_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.
The player can't become bankrupt. Their wealth is a non-decreasing
function of the (numbered) steps in the game, though if they always buy
a sub-path of the previously chosen path then they will not hold on to
any particular cent coin indefinitely (since they will never earn any
more coins). This is true in WMaths as well as in proper mathematics.
> If no player gets bankrupt, the number of paths cannot surpass the
> number of nodes.
Has a student never asked you to define (i.e. not waffle about) what you
mean by one infinite number "surpassing" another? Maybe they all just
keep silent and write this incantation down ready for the exam. But if
your were ever asked to define it, what would you say?
> Note: If set theory is right, then most paths that you can buy do not
> contain new nodes.
At every step it is possible to buy a new path containing infinitely
many new nodes. It is also possible to buy a path containing no new
nodes at every step. Both are also true in WMaths as well.
Do your students ever ask you to define (i.e. no waffle about) what
"most" means in the context of an infinite collection? If someone were
brave enough to ask you to say what you mean, what would you tell them?
--
Ben.
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
> Ben Bacarisse schrieb am Montag, 2. Januar 2023 um 12:40:52 UTC+1:
>
>> Yes, sorry, I thought that did not need saying. WM has been
>
> convincing all his students with this story for years:
How sad. Fortunately they are never going to need to learn the correct
answers as your college does not offer a degree in mathematics.
> 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 ∈ ℕ_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.
The player can't become bankrupt. Their wealth is a non-decreasing
function of the (numbered) steps in the game, though if they always buy
a sub-path of the previously chosen path then they will not hold on to
any particular cent coin indefinitely (since they will never earn any
more coins). This is true in WMaths as well as in proper mathematics.
> If no player gets bankrupt, the number of paths cannot surpass the
> number of nodes.
Has a student never asked you to define (i.e. not waffle about) what you
mean by one infinite number "surpassing" another? Maybe they all just
keep silent and write this incantation down ready for the exam. But if
your were ever asked to define it, what would you say?
> Note: If set theory is right, then most paths that you can buy do not
> contain new nodes.
At every step it is possible to buy a new path containing infinitely
many new nodes. It is also possible to buy a path containing no new
nodes at every step. Both are also true in WMaths as well.
Do your students ever ask you to define (i.e. no waffle about) what
"most" means in the context of an infinite collection? If someone were
brave enough to ask you to say what you mean, what would you tell them?
--
Ben.
Ben Bacarisse
Jan 2, 2023, 8:33:50 PM1/2/23
to
WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
> Ben Bacarisse schrieb am Sonntag, 1. Januar 2023 um 22:02:25 UTC+1:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
>
>> The paths in the tree trivially map to reals in [0,1]. The finite paths
>> map to rational numbers as do those paths that have a repeating tail.
>> But the rest of the infinite paths map to irrational numbers.
>
> There is no rest.
Well in WMaths you have admitted you can't define set difference so I
can image that saying "rest" is a problem for you.
But what I mean is that I can define a one-to-one mapping between
irrational numbers in [0,1] and paths in tree. None of those mapped
paths have repeating tails.
Can't you define such a mapping? Have you tried asking one of your
students to do so?
--
Ben.
Ben Bacarisse
Jan 2, 2023, 8:40:24 PM1/2/23
to
WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
> Ben Bacarisse schrieb am Montag, 26. Dezember 2022 um 15:07:36 UTC+1:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>
>> This reply is intended only for students who might fall for this sort of
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>
>> thing. Some do, apparently...
>
>> > Assuming that countability is a sensible notion, I show that all real
>> > numbers belong to a countable set of cardinality |ω^ω| = ℵo.
>> >
>> > The virtue of this method is that it avoids to squander the natural
>> > Assuming that countability is a sensible notion, I show that all real
>> > numbers belong to a countable set of cardinality |ω^ω| = ℵo.
>> >
>> > numbers: The entries of the Cantor-list are not enumerated by all
>> > natural numbers but by powers of 2. The antidiagonals of this list
>> > (never more than a countable set)
>> Here WM explicitly assumes a fact logically equivalent to the conclusion
>> he is supposedly proving. There are, in fact, uncountably many
>> anti-diagonals.
>
> Your "in fact" is in fact wrong. The antidiagonal can be constructed
> by a set of instructions as is well known and need not be discussed
> here. How many sets of instructions can be given? Finitely many. That
> yields countably many antidiagonals.
If you assume your conclusion, you can't be wrong. You might as well
cut out all the rest of the argument as simply state that every real in
[0,1] must have from finite "instructions" (a formula, or whether else
you might choose to permit).
Why make it so complicated? If the only numbers not listed must derive
from finite instructions you can just stop there. No need for all the
guff about primes.
--
Ben.
Mostowski Collapse
Jan 2, 2023, 11:07:00 PM1/2/23
to
WM is a some obvious quack.
Thats some fucked up nonsense. Only Hessenberg did
nowhere write "A model of ω^ω is the set of all
https://archive.org/details/grundbegriffede00hessgoog/page/n7/mode/2up
In 1906 nobody used the term "model". On the
other hand what you find, is the continued fractions:
Section IX. Nichtabzählbare Mengen
Jede Irrationalzahl s zwischen 0 und 1
lässt sich in einen und nur einen unendlichen
Kettenbruch von gleicher Form entwickeln,
und jeder unendliche Kettenbruch mit ganzzahligen
Nennern (dessen Zähler alle gleich 1 sind),
definiert eine und nur eine Irrationalzahl.
He then uses the same in §52 to argue:
ℵo^ℵo > ℵo
Which is exactly what I also proved, using
Cantors Theorem. You only find in §20 some
juggling with set, for this decomposition
of a natural number:
n = p1^k1 * ... * pj^kj
Just its prime factors p1 and their multiplicity kj.
WM schrieb:
Thats some fucked up nonsense. Only Hessenberg did
nowhere write "A model of ω^ω is the set of all
finite sequences of natural numbers"
You can check yourself:
https://archive.org/details/grundbegriffede00hessgoog/page/n7/mode/2up
In 1906 nobody used the term "model". On the
other hand what you find, is the continued fractions:
Section IX. Nichtabzählbare Mengen
Jede Irrationalzahl s zwischen 0 und 1
lässt sich in einen und nur einen unendlichen
Kettenbruch von gleicher Form entwickeln,
und jeder unendliche Kettenbruch mit ganzzahligen
Nennern (dessen Zähler alle gleich 1 sind),
definiert eine und nur eine Irrationalzahl.
He then uses the same in §52 to argue:
ℵo^ℵo > ℵo
Which is exactly what I also proved, using
Cantors Theorem. You only find in §20 some
juggling with set, for this decomposition
of a natural number:
n = p1^k1 * ... * pj^kj
Just its prime factors p1 and their multiplicity kj.
WM schrieb:
Mostowski Collapse
Jan 2, 2023, 11:24:31 PM1/2/23
to
The theorem ℵo^ℵo > ℵo based on cardinal exponentiation is here:
https://archive.org/details/grundbegriffede00hessgoog/page/n95/mode/2up
He goes into ordinal exponentiation a little later,
and indeed remarks that for ordinal exponentiation
α^β, the cardinality does not increase, §80 to §84.
https://archive.org/details/grundbegriffede00hessgoog/page/n127/mode/2up
But he nowhere uses the cardinality braket |...|.
Basically he has typed variables, and when the type
is cardinal, the exponention is cardinal exponentiation.
And when the type is ordinal, the exponentiation
is ordinal exponentiation. He doesn't give much
room to set exponentiation. His set exponentiation
is first an finite/infinite product (Verbindung)
and then only in passing he mentions:
§ 53. Die Mächtigkeit m^m ist ganz allgemein die der Menge
aller vollständigen Funktionen, die zu einer Menge von
der Mächtigkeit m gehören.
So he does not really bootstrap cardinalty exponentiation
from set exponentiation. The above is also only the
special case of set exponentiation A^B where A = B.
You need to wait for Bourbaki to give more room
to function spaces. Dan Christensen is still struggling
today with function spaces.
LoL
Mostowski Collapse schrieb:
Mostowski Collapse
Jan 2, 2023, 11:44:19 PM1/2/23
to
In the context of Hessenberg my interpretation
of ω^ω as set exponentiation would be indeed
wrong, since ω is an expression of type ordinal,
so it should be ordinal exponentiation. On the other
hand MSE shows, and since Bourbaki we have A^B
for set exponentiation, it is nowadays also
customary to read ω^ω as set exponentiation.
This little booklet gives me this theorem for
ordinal exponentiation, they put a little dot to
make clear that its ordinal exponentiation, and
they use also the cardinality braket! (sic)
2.11 Theorem: Schönflies 1913
|a^.b| = max(|a|,|b|)
Basic Set Theory, Azriel Levy - Page 126
https://www.amazon.com/dp/0486420795
Pitty nobody suggests on MSE to use the little
dot like here a^.b, adopting Azriel Levy typography,
to distinguish ordinal exponentiation from
other exponentiation. Maybe WM could pioneer this
adoption, to make clear what he wants to say?
Mostowski Collapse
Jan 2, 2023, 11:51:53 PM1/2/23
to
One more proof that MSE, Math Stack Exchange, is
utter bit rot, just some nonsense like all Stack Exchanges.
WM
Jan 3, 2023, 11:16:20 AM1/3/23
to
Mostowski Collapse schrieb am Dienstag, 3. Januar 2023 um 05:51:53 UTC+1:
> One more proof that MSE, Math Stack Exchange, is
> utter bit rot, just some nonsense like all Stack Exchanges.
Like MathOverflow: https://www.hs-augsburg.de/~mueckenh/Transfinity/Material/big%20picture%20-%20Checking%20the%20intelligence%20of%20MO-users%20-%20MathOverflow.html
> One more proof that MSE, Math Stack Exchange, is
> utter bit rot, just some nonsense like all Stack Exchanges.
> > > > Thats some fucked up nonsense. Only Hessenberg did
> > > > nowhere write "A model of ω^ω is the set of all
> > > > finite sequences of natural numbers"
> > > >> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.
> > > >> 42.
> > > >>
> > > >> A model of ω^ω is the set of all finite sequences of natural numbers
> > > >> or, according to Hessenberg [G. Hessenberg: "Grundbegriffe der
> > > >> Mengenlehre", offprint from Abhandlungen der Fries'schen Schule, Vol.
> > > >> I, no. 4, Vandenhoeck & Ruprecht, Göttingen (1906) § 20], the ordering
> > > >> of the natural numbers by the number of prime factors and then by
> > > >> sizes of the factors.
Can you read "or, according"?
> > > >> 42.
> > > >>
> > > >> A model of ω^ω is the set of all finite sequences of natural numbers
> > > >> or, according to Hessenberg [G. Hessenberg: "Grundbegriffe der
> > > >> Mengenlehre", offprint from Abhandlungen der Fries'schen Schule, Vol.
> > > >> I, no. 4, Vandenhoeck & Ruprecht, Göttingen (1906) § 20], the ordering
> > > >> of the natural numbers by the number of prime factors and then by
> > > >> sizes of the factors.
Regards, WM
WM
Jan 3, 2023, 11:25:24 AM1/3/23
to
Mostowski Collapse schrieb am Dienstag, 3. Januar 2023 um 05:44:19 UTC+1:
> a^.b
Cardinality is a nonsense notion for infinite sets. Therefore exponantiation is nonsense too. Here is the simple proof:
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.281.
In a nutshell:
If in
lim 2^n =/= 2^lim n
the exponentiationis replaced by what it abbreviates, namely repeated multiplication, then we get
2*2*2*... =/= 2*2*2*...
Regards, WM
> a^.b
Cardinality is a nonsense notion for infinite sets. Therefore exponantiation is nonsense too. Here is the simple proof:
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.281.
In a nutshell:
If in
lim 2^n =/= 2^lim n
the exponentiationis replaced by what it abbreviates, namely repeated multiplication, then we get
2*2*2*... =/= 2*2*2*...
Regards, WM
WM
Jan 3, 2023, 11:32:12 AM1/3/23
to
Ben Bacarisse schrieb am Dienstag, 3. Januar 2023 um 02:40:24 UTC+1:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> If you assume your conclusion, you can't be wrong. You might as well
> cut out all the rest of the argument as simply state that every real in
> [0,1] must have from finite "instructions" (a formula, or whether else
> you might choose to permit).
As "the list" you may assume all real numbers which fit into the enumerated lines.
> cut out all the rest of the argument as simply state that every real in
> [0,1] must have from finite "instructions" (a formula, or whether else
> you might choose to permit).
Of course the complete enumeration is already nonsense because every line has infinite distance from the completion.
But diagonals need finite instructions.
Regards, WM
WM
Jan 3, 2023, 11:57:45 AM1/3/23
to
Ben Bacarisse schrieb am Dienstag, 3. Januar 2023 um 02:24:23 UTC+1:
> Maybe they all just
> keep silent and write this incantation down ready for the exam. But if
> your were ever asked to define it, what would you say?
The set of paths cannot be written as a sequence. After all elements of a sequence have been applied, there is something remaining. All nodes can be removed from the Binary Tree in the sequence. But paths will remain. There are more than nodes.
> It is also possible to buy a path containing no new
> nodes at every step.
Paths are defined by nodes. Paths without nodes cannot exist is the Binary Tree. Note that paths there are defined extensionally, by nodes, not by a formula.
> Do your students ever ask you to define (i.e. no waffle about) what
> "most" means in the context of an infinite collection? If someone were
> brave enough to ask you to say what you mean, what would you tell them?
There are ℵo nodes. There are more than ℵo*ℵo*ℵo... paths.
Regards, WM
> WM <askas...@gmail.com> writes:
> > 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.
>
> The player can't become bankrupt. Their wealth is a non-decreasing
> function of the (numbered) steps in the game, though if they always buy
> a sub-path of the previously chosen path then they will not hold on to
> any particular cent coin indefinitely (since they will never earn any
> more coins). This is true in WMaths as well as in proper mathematics.
But not in set theory.
> > 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.
>
> The player can't become bankrupt. Their wealth is a non-decreasing
> function of the (numbered) steps in the game, though if they always buy
> a sub-path of the previously chosen path then they will not hold on to
> any particular cent coin indefinitely (since they will never earn any
> more coins). This is true in WMaths as well as in proper mathematics.
>
> > If no player gets bankrupt, the number of paths cannot surpass the
> > number of nodes.
>
> Has a student never asked you to define (i.e. not waffle about) what you
> mean by one infinite number "surpassing" another?
I mean what is usually understood: There are many more elements in an uncountable set than in a countable set.
> > If no player gets bankrupt, the number of paths cannot surpass the
> > number of nodes.
>
> Has a student never asked you to define (i.e. not waffle about) what you
> mean by one infinite number "surpassing" another?
> Maybe they all just
> keep silent and write this incantation down ready for the exam. But if
> your were ever asked to define it, what would you say?
>
> > Note: If set theory is right, then most paths that you can buy do not
> > contain new nodes.
>
> At every step it is possible to buy a new path containing infinitely
> many new nodes.
But the sequence can be completed. Otherwise the Cantorlist would not have a diagonal.
> > Note: If set theory is right, then most paths that you can buy do not
> > contain new nodes.
>
> At every step it is possible to buy a new path containing infinitely
> many new nodes.
> It is also possible to buy a path containing no new
> nodes at every step.
> Do your students ever ask you to define (i.e. no waffle about) what
> "most" means in the context of an infinite collection? If someone were
> brave enough to ask you to say what you mean, what would you tell them?
Regards, WM
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