# A new point of view on Cantor's diagonalization arguments

by Organic
Tags: arguments, cantor, diagonalization, point, view
 Emeritus Sci Advisor PF Gold P: 16,091 You said it was a list. (which, by definition, has only aleph0 rows) And yes, if you output this list, I don't see why I cannot use it as an input. I moved it here because you're not doing mathematics. You may be intent on studying the topics that mathematics likes to study, but you're not doing it in a mathematical fashion. I don't remember the circumstances, but you seemed to prefer theory development to philosophy, so I move your posts here once I think it's clear that you don't want to do things in a mathematical fashion.
 P: 1,210 Hurkyl, Please look again on this tree and tell me exactly how to you want to use it as an input.  {..4,3,2,1,0}=Z* 2 2 2 2 2 ^ ^ ^ ^ ^ | | | | | v v v v v /1 1 / \0 1 /\ /1 / 0 / \0 1 |\ /1 | \ 1 | \/ \0 / 0 | \ /1 | 0 | \0 ..1 | /1 | 1 | / \0 \ 1 | /\ /1 | / 0 |/ \0 0 \ /1 \ 1 \/ \0 0 \ /1 0 \0 /1 1 / \0 1 /\ /1 / 0 / \0 1 |\ /1 | \ 1 | \/ \0 / 0 | \ /1 | 0 | \0 ..0 | /1 | 1 | / \0 \ 1 | /\ /1 | / 0 |/ \0 0 \ /1 \ 1 \/ \0 0 \ /1 0 \0 ...
 Emeritus Sci Advisor PF Gold P: 16,091 How exactly are you using it for output?
 Emeritus Sci Advisor PF Gold P: 16,091 A list of properties that uniquely specify it would be nice. Just to be entirely clear, let me ask this question: Can I label each leaf with a unique natural number?
 P: 1,210 Hurkyl, The property of my Binary tree is based on this invariant structure:  1 = child / / Father = ? \ \ 0 = child The number of the Childs depends on any existing Z* member = {0,1,2,3,...} used as the power_value of each level in the tree. Because |{0,1,2,3,...}| = aleph0, and these members are used as power_values for each level in the tree, the result can't be but a tree width with aleph0 magnitude and a tree length with 2^aleph0 magnitude. Each child is the beginning of infinitely long sequence of 01 unique combinations. We can label each child with a unique natural number but this is only an illusion of a bijection that can clearly shown here: http://www.geocities.com/complementa.../Countable.pdf
 P: 1,210 Dear phoenixthoth, One of the big problems of highly advanced systems is that some fundamental property was forgotten behind. For example, let us say that you finished building a house and then you discover that some fundamental calculations about the strength of the first floor are wrong, and it means that you can't let people to live in this house. You have no choice but to rebuild the house. The way I constructed the binary tree give it length of 2^aleph0 magnitude on width of aleph0 magnitude. If Standard Math using the word "all" when defines set Z* then aleph0=2^aleph0 because there is a bijection between N and P(N).
 P: 1,572 If Standard Math using the word "all" when it define set Z* then aleph0=2^aleph0 because there is a bijection between N and P(N). i simply don't understand why that should be or is the case. can you give me more details in your reasoning without overloading me? can you sketch your proof of that?
 P: 1,210 Dont you see that we can always find any given 01 sequence and its opposite in the tree?  {..4,3,2,1,0}=Z* 2 2 2 2 2 ^ ^ ^ ^ ^ | | | | | v v v v v /1 1 / \0 1 /\ /1 / 0 / \0 1 |\ /1 | \ 1 | \/ \0 / 0 | \ /1 | 0 | \0 ..1 | /1 | 1 | / \0 \ 1 | /\ /1 | / 0 |/ \0 0 \ /1 \ 1 \/ \0 0 \ /1 0 \0 /1 1 / \0 1 /\ /1 / 0 / \0 1 |\ /1 | \ 1 | \/ \0 / 0 | \ /1 | 0 | \0 ..0 | /1 | 1 | / \0 \ 1 | /\ /1 | / 0 |/ \0 0 \ /1 \ 1 \/ \0 0 \ /1 0 \0 ...
 P: 1,572 isn't "always" another word for "all?"
 P: 1,210 Yes they are the same, but now i am talking about Standard Math, so when we are using "always" or "all" then Cantor's diagonal method does not hold because we can always find any given 01 unique sequence and its opposite in the tree.
 Emeritus Sci Advisor PF Gold P: 16,091 You can find any finite-length binary sequence in the tree. You miss most infinite-length sequences.
 P: 1,210 Prove it. But first you have to prove that |Z*| < aleph0
 P: 1,572 in standard math, is there anything wrong with cantor's diagonal argument?
 P: 1,210 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 {...,1,1,1,1}<--> 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 ... 1-1-1-1 <--> 1 \ \ \0 <--> 2 \ 0-1 <--> 3 \ \0 <--> 4 0-1-1 <--> 5 \ \0 <--> 6 0-1 <--> 7 \0 <--> 8 ... 0-1-1-1 <--> 9 \ \ \0 <--> 10 \ 0-1 <--> 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 ... 1 \ /1 <--> 5 \ 1 \/ \0 <--> 6 0 \ /1 <--> 7 0 \0 <--> 8 /1 <--> 9 1 / \0 <--> 10 1 /\ /1 <--> 11 / 0 / \0 <--> 12 ... 0 \ /1 <--> 13 \ 1 \/ \0 <--> 14 0 \ /1 <--> 15 0 \0 <--> 16 ...
 P: 1,572 can you turn that into a rigorous argument? most people eschew "proofs by picture."
P: 1,210
Hurkyl wrote,
 You can find any finite-length binary sequence in the tree. You miss most infinite-length sequences.