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

by Organic
Tags: arguments, cantor, diagonalization, point, view
 PF Patron Sci Advisor Emeritus P: 16,094 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 ...
 PF Patron Sci Advisor Emeritus P: 16,094 How exactly are you using it for output?
 PF Patron Sci Advisor Emeritus P: 16,094 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: 1,572 here it is organic, plain and simple. these guys just aren't going to buy into your theories until you can show them how cantor's arguments fail using their language. you have to start with the axioms of set theory. you have to define functions and such. you have to define onto functions. you have to look at powersets. you have to use their language or else they won't believe you. and they're not going to necessarily try to learn your language, which ain't math (no offense intended), so you have to come to their level and do the following: write out cantor's argument as he wrote it and tell them exactly which line or axiom or whatever you think is wrong. and you may not convince them until you give a countexample they can believe. it's just not credible to draw a tree with dots on it and call that a proof. mathematicians eschew such "proofs." they worry a heck of a lot about what hidden assumptions you might be making when you write three little dots. my problem with your three little dots is that each "dot" represents an infinite enumerable set. i on the other hand have written them something very similar to what you are intending. in my article, i show the following: 1. there is a set x such that there is a function f that maps x onto the powerset of x. 2. if there is a function that maps x onto its powerset then its powerset contains at least one "fuzzy set". 3. if a set's powerset contains no fuzzy sets then there is no function that maps x onto its powerset. (this and 2 are logically equivalent) i spell out all my assumptions and all that good stuff. i also claim that what i do fits with set theory rather than being a replacement that no one should bother looking at. (i could, for example, have a set theory where there is only one axiom, the universal set axiom, but that wouldn't be too interesting.) i contend that my tuzfc is an interesting set theory and it has some cool implications. my problem is that no one reads it, for whatever reason, and gives me feedback. so for all i know it's complete trash. i've stared at it so many times i don't know heads from tails. it looks fine to me but what do i know? so these three results and my article i think are what you want to accomplish: an ammendment to the cantor argument. a revision. i fully agree that cantor needs revision but you're not going to convince anyone the way you're trying to do it. but that shouldn't be the point. you do it because you enjoy discovering mathematics as do i. i don't honestly really care if my theory is right or publishable because it was so fun to create. if it was a waste of time, then c'est la vie. not the first time i've had a set (lol) back. i urge all of you, organic and hurkyl especially, to really give my paper a chance and read it. all feedback is welcome. hurkyl, i hope i've given organic enough feedback so that i can throw that plug in for my thread; hope this won't be considered trolling his thread.
 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.
 PF Patron Sci Advisor Emeritus P: 16,094 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.