A new point of view on Cantor's diagonalization arguments

[Hurkyl]

You can find any finite-length binary sequence in the tree. You miss most infinite-length sequences.

 

[Organic]

Prove it.

But first you have to prove that |Z*| < aleph0

 

[phoenixthoth]

in standard math, is there anything wrong with cantor's diagonal argument?

 

[Organic]

In standard Math The opposite of any given diagonal has to be added to the list, therefore no list of magnitude aleph0 can be in a bijection with R members which means that |R| is uncountable.

But look at this:

...0101 and ...1010 are in the list, for example:

Let us take again our set:

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

Now let us make a little redundancy diet:

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

and we get:

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

 

[phoenixthoth]

can you turn that into a rigorous argument? most people eschew "proofs by picture."

 

[Organic]

Hurkyl wrote,

You can find any finite-length binary sequence in the tree. You miss most infinite-length sequences.

My answer is:

First you have to prove that |Z*| < aleph0

 

[phoenixthoth]

this is a dumb question, but isn't |Z*| defined to be alpeh0 so aleph0=|Z*|? i can prove that |Z*| has the smallest infinite cardinal number if you like...

 

[Hurkyl]

Well, seeing how you only tell us about a small portion of the tree, and keep changing it anytime one of us asks you about a specific sequence, I do have to admit I'm only presuming it does not contain every sequence.

So while I don't have a mathematical proof of my claim, it doesn't matter since we're not talking about a mathematically described object. In most situations, when one person keeps changing their idea every time an objection is raised, it is taken as pretty solid proof that the idea does not cover all objections.


Were you asking me to prove an infinite length sequence exists? I can do that mathematically.


You want me to prove |Z*| < aleph0? What do you mean by Z*? All the meanings I could imagine you mean satisfy |Z*| = aleph0, nor do I see how this statement relates to my assertions.

 

[Organic]

phoenixthoth,

You missed the point, because |Z*| = aleph0 and I use all Z* members to construct my list, than Hurkyl argument does not hold, see for youself:

...0101 and ...1010 are in the list, for example:

Let us take again our set:

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

Now let us make a little redundancy diet:

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

and we get:

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

 

[phoenixthoth]

yeah, but won't you be using standard math?

i think organic thinks that nothing is wrong with cantor in standard math. i think organic thinks that it is standard math that is wrong. if so, good luck proving that one. one way to do it is to develop another consistent system...

 

[phoenixthoth]

this is known as an "ad nausum" fallacy. repeating the same argument doesn't make it correct.

 

[Organic]

yeah, but won't you be using standard math?

i think organic thinks that nothing is wrong with cantor in standard math. i think organic thinks that it is standard math that is wrong. if so, good luck proving that one. one way to do it is to develop another consistent system...

Did you read my paper here: http://www.geocities.com/complementarytheory/NewDiagonalView.pdf

 

[Hurkyl]

I find it hard to see how a matrix with 16 rows and 4 columns can contain every infinite binary sequence.

 

[Organic]

Hurkyl,

But you don't find it hard to see how Z* = {0,1,2,3,...} isn't it?

 

[Hurkyl]

Nope. But, you see, Z* comes with a definition. (or axioms, if you prefer)

 

[phoenixthoth]

three dots are not a complete list of R={0,1,...}. as i said earlier, each dot has too much information in it for that to be more than an *infinitesimal* partial list.

three dots don't work in proofs. (they only help you see. so while we may see what you see, or not, that is *not* a proof.)

proof:
let x equal 1+(-1)+1+(-1)+...
1+(-1)+1+(-1)+...=
(1+(-1))+(1+(-1))+...=
0+0+...=0.
therefore, x=0.

also, x=
1+(-1)+1+(-1)+...=
1+((-1)+1)+((-1)+1)+...=
1+0+0+...=1.
therefore, x=1.

therefore, 0=1.

therfore, if we allow three dots to be a proof then we will have to sacrifice the law of identity. do you see this? do you believe 0=1?

 

[Organic]

Hurkyl,

Because I use Z* members to construct my tree, it stands on Z* definitions.

see for yourself:
...0101 and ...1010 are in the list, for example:

Let us take again our set:

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

Now let us make a little redundancy diet:

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

and we get:

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

 

[Hurkyl]

Because I use Z* members to construct my tree, it stands on Z* definitions.

Yes. You used exactly 16 Z* members to label rows, and exactly 4 Z* members to label columns, and some mysterious ellipses which don't come with definitions or axioms to describe them.

 

[Organic]

Hurkyl,

Please don't do that, you know exactly how my list is constructed so why are you playing this game?

 

[phoenixthoth]

three dots don't work for definitons either. can you define real numbers using three dots?

 

[Hurkyl]

I know how I think the list is constructed, and that construction doesn't have any of the magical properties you ascribe to your list.

 

[Organic]

Hurkyl,

Ok, please read in your rigorous way how my list is constructed.

 

[phoenixthoth]

as an aside, this reminds me of the battle between darth maul, obi-wan and qui-gon in star wars episode i. i love this!

organic, I'm not saying you're darth maul or bad or anything and i admire your spirit. will you reply to my posts about three dots not being a proof nor a definition?

 

[Organic]

phoenixthoth,

three dots don't work for definitons either. can you define real numbers using three dots?

Please write the full represetation of pi in base 2.

 

[Hurkyl]

Ok, please read in your rigorous way how my list is constructed.

It is an array whose rows and columns are labelled by the natural numbers.

The $s_{i,j}$ entry (where the first index is the row and the second index is the column) is a 1 if and only if $\lfloor i / 2^j \rfloor$ is an even number.

 

[phoenixthoth]

how does that address what real numbers are? is your definition of "real number" "pi written in base 2?" examples prove nothing. you can give 10,000 examples, each with three dots, and not have a definition. so, tell me, what is the definition of real numbers, or anything, using three dots?

for example, you say that Z*:={0,1,2,...}.

well, how do i know what's hidden in those three dots? is 3.14 in the list or not? that's something a definition can decide for you. with just three dots, you have to just shrug your shoulders and say, "well, don't you know what i mean?" well, to play devil's advocate, i don't know what you mean. only a rigorous definition will elucidate what you mean in a way acceptable to my mathematical standards; anything less will not be tolerated. it is a closed minded approach but view it this way. suppose you're in karate class. you punch in a way not in accordance with teaching and your sensei scolds you (or hits you with a stick!) and says, "NO! YOU SHOULD PUNCH THIS WAY!" you're thinking, what the heck? my punch would hurt someone, so why can't i punch this way? if you were courageous enough to ask, the sensei would reply with this: "we know the best way already. this way of punching is the most effective."

if you want to prove the sensei wrong you need to go to another dojo because most senseis are too caught up in their own ego to listen to you and your what they call ignorance. only once in a blue moon is a new martial art created and not everyone is cut out to make their own style. it's even harder to get people to follow your lead, if anyone even listens to you.

do you understand the metaphor?

so it is closed minded in a way to do what we are asking but it's also the most effective way. how do i know this is more effective? well, because your proof would mean that all infinite sets have the same size, which detracts from the richness of the infinities of sets; that is aestecally unappealling and can be disproven by couterexample (in standard math, of course). it says that there are as many real numbers as there are natural numbers, right? well, can't you see by my picture that the two are *differnent* sizes:
...
___

?

 

[Organic]

Hurkyl,

It is an array whose rows and columns are labelled by the natural numbers.

Now please write what is the magnitude of the length and what is the magnitude of the width of this matrix? , but first pay attention that the length of the matrix depends on all Z* members that are used as power_values of this matrix and determine its length, as we can see here:

...0101 and ...1010 are in the list, for example:

Let us take again our set:

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

Now let us make a little redundancy diet:

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

and we get:

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

 

[Hurkyl]

The rows and the columns are indexed by the natural numbers. Thus, they are both of cardinality aleph0.

 

[Organic]

Hurkyl,

Ok, use Cantor's diagonal method on my list and prove that its opposite not in the list.

 

[phoenixthoth]

but we've already told you what's wrong with your proof; therefore, there is no need to do that.