What is the Collatz Problem and how can it be solved?

[Hurkyl]

Matt's talking about the rows, not the columns. I think we all agree there is a bijection between the set of columns and N.

 

[matt grime]

Originally posted by Organic
Matt,

What is finite in {10... ,1100... ,11110000... ,...} set?

I'm not saying anything there is a finite set. it's the same proof as before that you don't understand.

take the row at position r. let us examine what set it corresponds to in the power set.

now the c'th column starts with 2^c 1s doesn't it? that is the pattern of the columns.the c'th entry in row r is the r'th entry in column c. Now, 2^c>c and if c>r we see the r'th entry in column c is a 1.

therefore all entries after the r'th in row r are 1.look at the finite portion of the diagram - notice how everything above the diagonal is a 1? that's just what we've proved above.

and what that tells us is that the element in the power set correspoding to that row r contains every element in N greater than r. thus it's complement only contains some subset of the numbers 1,2,..r.

Sets like this are called cofinite.

This is the same method of proof that shows the other list only contained elements in ther power set that had a finite number of elements in them because there the any row eventually becomes zero by the same argument. thus you've only enumerated the finite and cofinite sets and none of the (uncountable) remaining sets.The cut off point is different in each row but that doesn't matter, which is what i think you thought i meant earlier - that there was a vertical line you could draw in the diagram and all to the left of the vertical line is 0 (or 1) I am not saying that and i don't need to say that.

read the above proof carefully, and try and understand we're just formally stating that everythin above the diagonal is a 1 (in this case and 0 in the other diagram)

 

[matt grime]

also note i am not saying a row is finite, but that a row corresponds to a cofinite (or finite in the old case) element in the power set of N

 

[Organic]

Matt and Hurkyl,

What is finite in {10... ,1100... ,11110000... ,...} set?

This set is bijective to N.

Now, let us go back to look at this set like this:

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b] 
 ...,1,1,1,0  
 ...,1,1,0,1  
 ...,1,1,0,0  
 ...,1,0,1,1  
 ...,1,0,1,0  
 ...,1,0,0,1  
 ...,1,0,0,0  
 ...,0,1,1,1  
 ...,0,1,1,0  
 ...,0,1,0,1  
 ...,0,1,0,0  
 ...,0,0,1,1  
 ...,0,0,1,0  
 ...,0,0,0,1  
 ...,0,0,0,0  
 ...

By taking the n-th notation from each member we can build a list of unique sequences which its magnitude=2^aleph0.

 

[matt grime]

n'th notation from what?

did you even bother to try to understand that argument? No i didn't think so.

we aren't saying the set of columns is finite. why would we? have you even read the proof i just offered? attempted to understand what it said? or did you just presume to know it was wrong? It isn't.

 

[Organic]

Matt, if you write this: "n'th notation from what?" it is simply shows us that you did not read carefully what I write to you and to Hurkyl, so here it is again.



{10... ,1100... ,11110000... ,...} set is bijective to N.

By taking the n-th notation from each member we can build a list of unique sequences which its magnitude=2^aleph0:

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b] 
 ...,1,1,1,0  
 ...,1,1,0,1  
 ...,1,1,0,0  
 ...,1,0,1,1  
 ...,1,0,1,0  
 ...,1,0,0,1  
 ...,1,0,0,0  
 ...,0,1,1,1  
 ...,0,1,1,0  
 ...,0,1,0,1  
 ...,0,1,0,0  
 ...,0,0,1,1  
 ...,0,0,1,0  
 ...,0,0,0,1  
 ...,0,0,0,0  
 ...

 

[matt grime]

i've read and reread that post and you don't explain what the n'th notation is.

the indvidual words make sense but not when put together like that

what is a notation, what is the nth one? you didn't state what they are.

which members are you taking these notations from? members of rows, columns, power sets, N?

so illuminate it for the stupid round herenow have you actually read my proof that any row in the diagram corresponds to a confinite element of the power set?

we;ve got this diagram enumerating the cofinite sets, what's the point you're trying to make

 

[Organic]

Matt,

First, thank you for reading an rereading.


{10... ,1100... ,11110000... ,...} set is bijective to N.

The n-th notation of each member is:

n-th 1 2 3 . .  1 2 3 . . . .  1 2 3 . . . . . . . .  1 2 3
    {1 0 . . . ,1 1 0 0 . . . ,1 1 1 1 0 0 0 0 . . . ,. . .}

By taking the n-th notation from each member we can build a list of unique sequences which its magnitude=2^aleph0:

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]<--> n-th 1
 ...,1,1,1,0 <--> n-th 2
 ...,1,1,0,1 <--> n-th 3 
 ...,1,1,0,0 <--> n-th 4 
 ...,1,0,1,1 <--> n-th 5 
 ...,1,0,1,0 <--> n-th 6 
 ...,1,0,0,1 <--> n-th 7 
 ...,1,0,0,0 <--> n-th 8 
 ...,0,1,1,1 <--> n-th 9 
 ...,0,1,1,0 <--> n-th 10
 ...,0,1,0,1 <--> n-th 11
 ...,0,1,0,0 <--> n-th 12
 ...,0,0,1,1 <--> n-th 13
 ...,0,0,1,0 <--> n-th 14
 ...,0,0,0,1 <--> n-th 15
 ...,0,0,0,0 <--> n-th 16
 ...

 

[matt grime]

ok got it. you take the n'th element in each column as the elements in the nth row.

That does not give you "2^aleph-0" possibilities, there are exactly as many rows as there are natural numbers. How can you claim that there are 2^aleph-0 rows. Why? What set of cardinality 2^aleph-0 is it in bijection with and how?

 

[Organic]

Matt,

A Binary Tree with infinitely many nodes has a width of magnitude aleph0 and a length of 2^aleph0.

Simple as that.

 

[matt grime]

That would depend on how you defined binary tree - the rooted binary tree with countably infinite branch set has only a countable number of nodes - it is the countable union of finite sets.

Besides that has nothing to do with this issue.

Of course you don't think that binary tree means what it really means.

So what is your binary tree? You seem to think it is a Cantor set. There fore you seem to think that the rows of your matrix are in bijection with a cantor set. How? Please demonstrate? As I can prove that a cantor set is not in bijection with N, that the rows are in bijection with N, don't you think that that is a little problematic. So please exhibit this notional bijection with the Cantor set. Or explicitly state what your binary tree is. This will all involve taking (inverse/direct) limits in the correct sense, but of course you know all about thatedit: and seeing as you don't think there are an uncountable number of points in the real line, don't you think taking an uncountable subset is wrong? It is conceptually impossible in your world, and if you're in mine you need to prove all your assertions.

 

[Hurkyl]

Fun fact.

A (balanced) binary tree with infinitely many nodes has zero leaves.

 

[Organic]

Matt and Hurkyl,

Besides that has nothing to do with this issue (the Binary Tree).

Let us take again our set:

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]<--> n-th 1
 ...,1,1,1,0 <--> n-th 2
 ...,1,1,0,1 <--> n-th 3 
 ...,1,1,0,0 <--> n-th 4 
 ...,1,0,1,1 <--> n-th 5 
 ...,1,0,1,0 <--> n-th 6 
 ...,1,0,0,1 <--> n-th 7 
 ...,1,0,0,0 <--> n-th 8 
 ...,0,1,1,1 <--> n-th 9 
 ...,0,1,1,0 <--> n-th 10
 ...,0,1,0,1 <--> n-th 11
 ...,0,1,0,0 <--> n-th 12
 ...,0,0,1,1 <--> n-th 13
 ...,0,0,1,0 <--> n-th 14
 ...,0,0,0,1 <--> n-th 15
 ...,0,0,0,0 <--> n-th 16
 ...

Now let us make a little redundancy diet

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
...  1-1-1-1 <--> n-th 1
     \  \ \0 <--> n-th 2
      \  0-1 <--> n-th 3 
       \  \0 <--> n-th 4 
       0-1-1 <--> n-th 5 
        \ \0 <--> n-th 6 
         0-1 <--> n-th 7 
          \0 <--> n-th 8 
 ... 0-1-1-1 <--> n-th 9 
     \  \ \0 <--> n-th 10
      \  0-1 <--> n-th 11
       \  \0 <--> n-th 12
       0-1-1 <--> n-th 13
        \ \0 <--> n-th 14
         0-1 <--> n-th 15
          \0 <--> n-th 16
 ...

and we got

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
          /1 <--> n-th 1
         1 
        / \0 <--> n-th 2
       1   
       /\ /1 <--> n-th 3 
      /  0
     /    \0 <--> n-th 4 
 ... 1    
     \    /1 <--> n-th 5 
      \  1 
       \/ \0 <--> n-th 6
       0  
        \ /1 <--> n-th 7
         0
          \0 <--> n-th 8
          
          /1 <--> n-th 9 
         1
        / \0 <--> n-th 10
       1  
       /\ /1 <--> n-th 11
      /  0 
     /    \0 <--> n-th 12
 ... 0    
     \    /1 <--> n-th 13
      \  1
       \/ \0 <--> n-th 14
       0  
        \ /1 <--> n-th 15
         0
          \0 <--> n-th 16
 ...

 

[matt grime]

Full credit to him, it takes some doing to talk about binary trees (uniquely path connected), even drawing them so they appear to be what you I might think, and then after much pressing claim it is actually a Cantor Set (totally disconnected), and further that as this is TD he can do whatever he damn well pleases. (Despite the fact it was your decision to move it here, Hurkyl, and that he wanted to post it in the maths forum.) That's some cheek

 

[matt grime]

And is that supposed to demonstrate a bijection with the Cantor set? How?

If it helps, a Cantor set is in bijection with the set of base 3 expansions of all reals between 0 and 1 with no 2 in their expansions, with the obivious dyadic representations identified.

 

[Organic]

Matt and Hurkyl,

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
          /1 <--> n-th 1
         1 
        / \0 <--> n-th 2
       1   
       /\ /1 <--> n-th 3 
      /  0
     /    \0 <--> n-th 4 
 ... 1    
     \    /1 <--> n-th 5 
      \  1 
       \/ \0 <--> n-th 6
       0  
        \ /1 <--> n-th 7
         0
          \0 <--> n-th 8
          
          /1 <--> n-th 9 
         1
        / \0 <--> n-th 10
       1  
       /\ /1 <--> n-th 11
      /  0 
     /    \0 <--> n-th 12
 ... 0    
     \    /1 <--> n-th 13
      \  1
       \/ \0 <--> n-th 14
       0  
        \ /1 <--> n-th 15
         0
          \0 <--> n-th 16
 ...

 

[Organic]

Matt,

Please look at http://www.mathacademy.com/pr/prime/articles/cantset/

Cantor set is a hybrid of base 2 and base 3 ratios.

Therefore it is easy to show that 2^aleph0 is Cantor set where Cantor set is a Binary Tree:

                ?
__________________________________

      1                    0
_____________        _____________

  1       0            1       0
_____   _____        _____   _____

1  0    1  0         1  0    1  0  
__ __   __ __        __ __   __ __

 

[matt grime]

And? That is not a bijection from the (countable) set of rows corresponding to cofinite elements the power set of N to the Cantor Set, or any thing that has cardinality 2^aleph-0It tells you how to superimpose a binary tree onto another picture. that does not define a bijection. Nor does it imply that the binary tree is 'uncountable' (what aspect of it is uncountable in your opinion)

 

[matt grime]

Organic, I know what a Cantor set is.
You claimied the binary tree was a Cantor set in this thread earlier. It isn't. One wonders what you think a Cantor set is, or a binary tree.

Or why that defines a bijection.

 

[Organic]

Matt,

You forget my proof |R|>|N| but both N and R are countable.

 

[matt grime]

Evidently you don't understand the maths.

The Cantor set is not a binary tree. You have your own odd definition for binary tree, don't you. Care to explain what it is?

The cantor Set is the ***limit*** of that process.

So how does being able to draw a rooted binary tree on top of some other diagram provide you with a bijection with the Cantor set?

 

[Organic]

Matt,

Cantor set is a hybrid of base 2 and base 3 ratios.

Therefore Cantor set can be a base 2 fractal and base 3 fractal.

Your Cantor set is disjoint because you don't look at it from the complex level.

 

[Hurkyl]

Full credit to him, it takes some doing to talk about binary trees (uniquely path connected), even drawing them so they appear to be what you I might think, and then after much pressing claim it is actually a Cantor Set (totally disconnected), and further that as this is TD he can do whatever he damn well pleases. (Despite the fact it was your decision to move it here, Hurkyl, and that he wanted to post it in the maths forum.) That's some cheek

You're right. He's dropped the attitude he had for a while that caused me to start moving his posts over here in the first place.



Anyways, Organic, the trick to the "binary tree" construction of the Cantor set is that it's not enough; it only generates the points in the Cantor set with terminating decimal expansions. In order to get the whole Cantor set, you have to take the (topological) closure of this set of points.


Given a set S, the closure of S is the set of all points that are limits of sequences of elements of S. For example, the closure of (0, 1) is [0, 1]. (the sequence 0.1, 0.01, 0.001, 0.0001, ... is entirely in (0, 1), and the limit is 0, so 0 is in the closures of (0, 1))

 

[Organic]

Hurkyl,

You can't contradict it.

{10... ,1100... ,11110000... ,...} set is bijective to N.

The n-th notation of each member is:

n-th 1 2 3 . .  1 2 3 . . . .  1 2 3 . . . . . . . .  1 2 3
    {1 0 . . . ,1 1 0 0 . . . ,1 1 1 1 0 0 0 0 . . . ,. . .}

By taking the n-th notation from each member we can build a list of unique sequences which its magnitude=2^aleph0:

      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]<--> n-th 1
 ...,1,1,1,0 <--> n-th 2
 ...,1,1,0,1 <--> n-th 3 
 ...,1,1,0,0 <--> n-th 4 
 ...,1,0,1,1 <--> n-th 5 
 ...,1,0,1,0 <--> n-th 6 
 ...,1,0,0,1 <--> n-th 7 
 ...,1,0,0,0 <--> n-th 8 
 ...,0,1,1,1 <--> n-th 9 
 ...,0,1,1,0 <--> n-th 10
 ...,0,1,0,1 <--> n-th 11
 ...,0,1,0,0 <--> n-th 12
 ...,0,0,1,1 <--> n-th 13
 ...,0,0,1,0 <--> n-th 14
 ...,0,0,0,1 <--> n-th 15
 ...,0,0,0,0 <--> n-th 16
 ...

If by closure you mean:

The collection of all points such that every neighborhood of these points intersects the original set in a nonempty set.

The result of this definition cannot be but at least Binary Tree where each point of it is a connaction (an intersection) with at least three neighbors (one above it and two below it).

 

[Hurkyl]

The problem is that each of the
10101010...
11001100...
11110000...
...

are only aleph0 long.

 

[Organic]

Hurkyl,

It is 2^aleph0 long because it "grows" in geometric series and not arithmetic series.

<---arithmetic series
      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]<--> n-th 1      geometric series 
 ...,1,1,1,0 <--> n-th 2      |
 ...,1,1,0,1 <--> n-th 3      |
 ...,1,1,0,0 <--> n-th 4      |
 ...,1,0,1,1 <--> n-th 5      |
 ...,1,0,1,0 <--> n-th 6      |
 ...,1,0,0,1 <--> n-th 7      |
 ...,1,0,0,0 <--> n-th 8      |
 ...,0,1,1,1 <--> n-th 9      |
 ...,0,1,1,0 <--> n-th 10     |
 ...,0,1,0,1 <--> n-th 11     |
 ...,0,1,0,0 <--> n-th 12     |
 ...,0,0,1,1 <--> n-th 13     |
 ...,0,0,1,0 <--> n-th 14     |
 ...,0,0,0,1 <--> n-th 15     |
 ...,0,0,0,0 <--> n-th 16     |
 ...                          V

 

[Hurkyl]

The problem is that the geometric growth only exists in the finite realm. You have to do some other operation (such as take a limit, or a nested union) to transfer to the infinite case, and in this case, that operation doesnt' catapault you to 2^aleph0.

 

[Organic]

Hurkyl,

Forget about the word "growth".

I am talking about infintley long arithmetic series = aleph0
and infinitely long geometric series = 2^aleph0.

More than that both serieses are depended.

 

[Hurkyl]

There is no ω-th term in a geometric sequence, nor in an arithmetic sequence. An infinitely long geometric (or arithmetic) sequence has aleph0 terms, one for each finite ordinal.

 

[Organic]

Hurkyl,

Thank you, it is just something that I used to explain Matt my point of view in some particular case.

So here it is with no -th things:

<---arithmetic series
      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]   geometric series 
 ...,1,1,1,0                  |
 ...,1,1,0,1                  |
 ...,1,1,0,0                  |
 ...,1,0,1,1                  |
 ...,1,0,1,0                  |
 ...,1,0,0,1                  |
 ...,1,0,0,0                  |
 ...,0,1,1,1                  |
 ...,0,1,1,0                  |
 ...,0,1,0,1                  |
 ...,0,1,0,0                  |
 ...,0,0,1,1                  |
 ...,0,0,1,0                  |
 ...,0,0,0,1                  |
 ...,0,0,0,0                  |
 ...                          V

 

[Hurkyl]

There is no infinite term in a geometric series. Each of the aleph0 terms are finite.

 

[Organic]

Hurkyl,

Now you have it in front of your eyes, two depended serieses.


The first one (arithmetic) cannot be but with aleph0 cardinality,
therefore the other one (geometric) cannot be but with 2^aleph0 cardinality.

<---arithmetic series
      3 2 1 0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]   geometric series 
 ...,1,1,1,0                  |
 ...,1,1,0,1                  |
 ...,1,1,0,0                  |
 ...,1,0,1,1                  |
 ...,1,0,1,0                  |
 ...,1,0,0,1                  |
 ...,1,0,0,0                  |
 ...,0,1,1,1                  |
 ...,0,1,1,0                  |
 ...,0,1,0,1                  |
 ...,0,1,0,0                  |
 ...,0,0,1,1                  |
 ...,0,0,1,0                  |
 ...,0,0,0,1                  |
 ...,0,0,0,0                  |
 ...                          V

 

[Hurkyl]

First off, the rows and columns are not in an arithmetic nor a geometric sequence; they're not even numbers!

It is the width and the height that are in those sequences... however, geometric and arithmetic sequences only have finite terms. The width and height of the infinite matrix cannot be in said sequence!


Let me put it this way. Except for the first, every term in an arithmetic or geometric sequence has a term immediately before it. So if the width and height of your infinite matrix are part of these sequences, then what are the terms that come immediately before them?

 

[Organic]

Hurkyl,

then what are the terms that come immediately before them?

Please give some simple example in plain English.

 

[Hurkyl]

You're "sequences" arise from looking at pieces of the whole array. You look at a 2x1 piece, then a 4x2, then an 8x3, 16x4, ...

The thing is, each term in an arithmetic (geometric) sequence has a finite index. These sequences can only describe finite pieces of your whole array.


One of the many forms of the fundamental problem with progressing to the infinite case is this:

Each term (but the first) of an arithmetic sequence is defined to be a fixed constant plus the term before it.

You assert aleph0 is in the arithmetic sequence of widths. Well, what is the term before it?

The same complaint applies to the geometric sequence.



This fundamental problem rears its ugly head for any iterative process where each step is built from the step immediately before it. Such a process only works for terms that can be generated in a finite number of steps. There are, of course, aleph0 of these terms, but none of these terms correspond to a ω-th step of the process (if one even exists)

 

[Organic]

Hurkyl,

First off, the rows and columns are not in an arithmetic nor a geometric sequence; they're not even numbers!

If I get a list of 2^aleph0 unique elements, then their name is not important, first: they exists, second: give them a name.

This fundamental problem rears its ugly head for any iterative process where each step is built from the step immediately before it. Such a process only works for terms that can be generated in a finite number of steps. There are, of course, aleph0 of these terms, but none of these terms correspond to a ù-th step of the process (if one even exists)

Hurkyl, there is no process here but an immediate existence based on ZF axiom of infinity iterations of the power_value of the matrix:

<---arithmetic series
      3 2 1 0 [b]<---The power_value of the matrix[/b] = aleph0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]   geometric series 
 ...,1,1,1,0                  |
 ...,1,1,0,1                  |
 ...,1,1,0,0                  |
 ...,1,0,1,1                  |
 ...,1,0,1,0                  |
 ...,1,0,0,1                  |
 ...,1,0,0,0                  |
 ...,0,1,1,1                  |
 ...,0,1,1,0                  |
 ...,0,1,0,1                  |
 ...,0,1,0,0                  |
 ...,0,0,1,1                  |
 ...,0,0,1,0                  |
 ...,0,0,0,1                  |
 ...,0,0,0,0                  |
 ...                          V

 

[moshek]

Gaifman lecture

Organic !

please do read the abstract of Haim Gaifman lecture at:

http://www.cs.tau.ac.il/~nachumd/semester.html

thank you
Moshek

 

[matt grime]

How many times do we have to explain induction to you? There is a set of statements labelled by n in N, if the r'th implies the r+1st and if the 1'st is true then all the statements are ture by induction.

That is the statement is true for all n in N. It does not state the statement (ugly turn of phrase, sorry) labelled by N is true; that statement might not even make sense. You cannot write down something labelled by N and say it is true because it is true for each n in N, that is not induction, it is plain wrong.

COUNTER EXAMPLES, again.

the set of rows in the finite diagram is finite, therefore the set of rows for N should be finite as the first set is finite, and if the n'th set is finite the n+1st set only has twice as many and is hence finite. You can't just pick and choose the properties that you think pass through
Do you ever actually think about the counter arguments to your claims or do you just blithely turn the handle and crank out another piece of 'complementary theory'?The infinite case must be taken as a limit of something in the proper sense. You do not do this, and I am certain you do not even know how to do so. Listen to the people who do.

The 'limit' of a nested union of sets of cardinality 2^n is not 2^aleph-0. There is no substantiation for this claim other than your misuse of the axiom of infinity, which just states that the elements of N form an inductive set. |N| is not in N so you can't claim anything using it and induction.

And to make it absolutely clear, no one is saying that the power set of N does not have card 2^aleph-0, but that the list of elements you construct does not in anyway correspond to any set of that cardinality. The only reason you have for saying it is is that it follows from the finite cases - but it doesn't, otherwise it follows from the finite cases that the set of rows is finite - induction does not allow you to claim the things you are claiming *must* be true.

 

[Organic]

Matt,

aleph0 magnitude exists in ({},{__}) open interval where {} content is the weak limit and {__} is the strong limit.

To understand this please look at:

http://www.geocities.com/complementarytheory/Theory.pdf

Again aleph0 cannot be something which is beyond "infinitely many" because no potential infinity can be an actual infinity and the reason is very simple.

Actual infinity is the limit of any information system including Math language.

Therefore only potential infinity which defined as "infinitely many objects" can be used as an input to Math language.

Therefore the "transfinite" universes that trying to "eat the cake" (to construct a solid line by using the model of points or segments) and also to keep it untouched (to insist that this "pointed/segmented" line is beyond its points/segments) is nothing but a conceptual mistake, when the logic is Boolean logic.

In Boolean Logic x and the negation of x is false, x=solid_line AND x=infinitely many points or segments, does not hold.

 

[matt grime]

Hmm, looks like garbage. Seeing as you are attempting to prove things in mathematics are not sufficient your personal issues are not important - we can prove mathematics is perfectly consistent in this issue, and it is your ignorance of it that is causing you to misundertand it. Did I say aleph-0 is 'beyond n in N?' No I didn't. It isn't a natural number. Lots of things aren't natural numbers. Now, get back to the point in hand which is that you are now trying to prove that the rows of your diagram are in bijection with a Cantor set. Where is the proof of this? Their construction bijects them with N, but at no point is there a bijection witha Cantor Set. Your only indication as to why this is true is the patently false assertion that it must follow from the finite case because the rows there biject with a set of cardinality 2^n for n columns. This is not true.

 

[moshek]

How can you use the word Garbage to somone who creat interest of more than 24 pages on this web-site ?

Don't you feel that Organic is trying to sow you somting that you still can't recognize as new interpetation to almost everyting in mathematics?

Moshek

 

[Organic]

Moshek,

It is great to see that There is onther life in Math world which is not limited to Matt's understanding abilities.


Matt,

aleph0 magnitude exists in ({},{__}) open interval where {} content is the weak limit and {__} is the strong limit.

To understand this please look at:

http://www.geocities.com/complementarytheory/Theory.pdf

Again aleph0 cannot be something which is beyond "infinitely many" because no potential infinity can be an actual infinity and the reason is very simple.

Actual infinity is the limit of any information system including Math language.

Therefore only potential infinity which defined as "infinitely many objects" can be used as an input to Math language.

Therefore the "transfinite" universes that trying to "eat the cake" (to construct a solid line by using the model of points or segments) and also to keep it untouched ( to insist that this "pointed/segmented" line is beyond its points/segments (it means a solid_line) ) is nothing but a conceptual mistake, when the logic is Boolean logic.

In Boolean Logic x and the negation of x is false, x=solid_line AND x=infinitely many points or segments, does not hold.

 

[moshek]

Organic:

I did not convins yet that collatz problem 3n+1 is undisideabl by the self similarity fractal type to the infinit axiom in set theory but your arguments are really beutifulls:

Please look on:

www.as.huji.ac.il/midrasha04.htm[/URL]

Yours

Moshek

 

[Hurkyl]

If I get a list of 2^aleph0 unique elements, then their name is not important, first: they exists, second: give them a name.

There's no good reason to believe the list has 2^aleph0 unique elements.

 

[Organic]

Hi Moshek,

If the result of "to be eqivanet to an axiom of some system" is true then 3n+1 is true in ZF.

This is what I write at the end of:

http://www.geocities.com/complementarytheory/3n1proof.pdf

"An axiom of some Mathematical system cannot be proved by definition.

Therefore Collatz sequences are true but cannot be proved within ZF axiomatic system."

 

[Organic]

Hi Hurkyl,

This fundamental problem rears its ugly head for any iterative process where each step is built from the step immediately before it. Such a process only works for terms that can be generated in a finite number of steps. There are, of course, aleph0 of these terms, but none of these terms correspond to a ù-th step of the process (if one even exists)

Hurkyl, there is no process here but an immediate existence based on ZF axiom of infinity iterations of the power_value of the matrix:

<---arithmetic series
      3 2 1 0 [b]<---The power_value of the matrix[/b] = aleph0
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]   geometric series 
 ...,1,1,1,0                  |
 ...,1,1,0,1                  |
 ...,1,1,0,0                  |
 ...,1,0,1,1                  |
 ...,1,0,1,0                  |
 ...,1,0,0,1                  |
 ...,1,0,0,0                  |
 ...,0,1,1,1                  |
 ...,0,1,1,0                  |
 ...,0,1,0,1                  |
 ...,0,1,0,0                  |
 ...,0,0,1,1                  |
 ...,0,0,1,0                  |
 ...,0,0,0,1                  |
 ...,0,0,0,0                  |
 ...                          V

 

[Organic]

Dear Moshek Hurkyl and Matt,

I did not sleep fore more then 36 hours, so please if you want we will continue after I take some good sleep.

Bye.

 

[moshek]

Organic:

I really don't know yet if you are rigth, maybe.

But if you rally can show by this that the 3n+1 problem is undisidable problem in number theory, as a simple and natural Godel theorem. then it will consider as the greatest discovery in the history of the euclidian mathematics !

Good luck !
Moshek

 

[Hurkyl]

Hrm, ω is supposed to be the omega (\omega), the symbol typically used for the ordinal that is the well-order type of the natural numbers (which is equal to the set of natural numbers in the Von Neumann model). Is it not showing up on your computer?



Ok, let's try a different tact. Let's suppose I don't understand how to make the entire array based on the sample entries you've given. Let's also suppose that I still won't get it if you give me more sample entries.

So the question is, can you explain what is your array without simply writing sample entries and presuming the reader can fill in the details?

For example, instead of saying:

...010101

You could say

"The sequence whose n-th term is 1 if n is even and 0 if n is odd"

Instead of

...10000001000010011

you could say

"The sequence whose n-th term is f(n) where:
f(n) := 1 if n = p^2 for some integer p
f(n) := 0 otherwise"

Instead of

...1000100100101

you could say

"The sequence whose n-th term is f(n) where:
f(n) := 1 if n = g(p) for some integer p
f(n) := 0 otherwise
where g is recursively defined as
g(1) := 1
g(i) = g(i-1) + i for all integers i > 1"

Instead of

...000011110000111100001111

you could say

"The sequence whose n-th term is f(n) where:
f(n) := 1 if floor(n / 4)¹ is even
f(n) := 0 otherwise"

Hurkyl

1: The floor function, floor(x), \lfloor x \rfloor, is the function that "rounds down" a real number. So, floor(1) = 1, floor(2.5) = 2, floor(-3.7) = -4.

PS: It's not an entirely unreasonable supposition; you disagree with the statements I make about the array as I think it's supposed to be!

 

[matt grime]

I can write that Organic's musings are garbage because he does not attempt to define anything he uses and just invents some notation at random. He has claimed to have defined a multiplication on N which does not even send natural numbers to natural numbers, he makes plainly wrong statements about the axiom of infinity, he misuses the phrase binary tree, does not understand the concept of countable, thinks that mathematics implies there is a bijection from a set of cardinality aleph-0 to one of card 2^aleph-0. Very often the sentences he writes do not make sense as sentences, never mind mathematically. He also claims frequently to have a cast iron proof that is so cast iron that he has to keep rewriting it to remove errors. He has also demonstrated that he does not know what a bijection is despite claiming their use in a 'proof', he did not understand the n maps to n+1 is a bijection from N to a proper subset of itself, claiming seom bizarre things about this fact. He frequently contradicts himself and does not extend to others the courtesy he demands from them by not reading and understanding their counter arguments.

Moreover he does not seem to grasp that it is not mathematics that cannot handle infinity but his philosphy. None of the constructions and results he claims must be true are true in mathematics, and are only true in his philosophy. With these presumptions he then tells us our mathematics is faulty, when the only inconsistencies arise if we believe his unfounded assertions.

He often cites the axiom of infinity induction, yet no such thing exists without his philosophy, and it is something he refuses to explain.One needs only to see that he believes that axioms cannot be proven true to see that he doesn't understand that which he claims is wrong - axioms are proven true trivally in an axiomatic theory.

The collatz conjecture he insisted was undecidable in ZF, despite not evidently knowing what that meant as he asserted it was equivalent to the axiom of infinity (something he was unable to prove). Of course if it is equivalent to the axiom of the infinite set then it is trivally provable in ZF. He appears now to have changed his position on that despite repeatedly berating me for being too unimaginative to see how he was correct. Another example of him contradicting himself.

Proving the Collatz conjecture is undecidable in ZF is not that big a deal really. Conway showed there are Collatz type conjectures that are undecidable. It is known the conjecture is true for all numbers less than 10^53 I believe.If you wish to see why Organic provokes much ire from mathematicians just read all his amny posts where he passionately argues against the blindingly obvious, and cannot even understand the simple objections raised against his argument. One needs only read some of his respsonses to the challenge to prove his construction has the properties he claims to see that he doesn't understand what's going on.