September 18th, 2018, 12:57 PM  #11 
Senior Member Joined: Dec 2015 From: France Posts: 103 Thanks: 1  
September 18th, 2018, 02:19 PM  #12 
Senior Member Joined: Aug 2012 Posts: 2,427 Thanks: 760  You haven't demonstrated to anyone's satisfaction that you understand it. If you're going to claim a refutation of a universally accepted 140 year old theorem, at least you could state it and give the standard proof, and show where you think it's wrong.

September 19th, 2018, 07:50 AM  #13 
Senior Member Joined: Dec 2015 From: France Posts: 103 Thanks: 1  In fact, this theorem was not accepted at the beginning. I quote Wiki:"Henri Poincaré referred to his ideas as a "grave disease" infecting the discipline of mathematics, and Leopold Kronecker's public opposition and personal attacks included describing Cantor as a "scientific charlatan", a "renegade" and a "corrupter of youth." Kronecker objected to Cantor's proofs that the algebraic numbers are countable, and that the transcendental numbers are uncountable... Writing decades after Cantor's death, Wittgenstein lamented that mathematics is "ridden through and through with the pernicious idioms of set theory", which he dismissed as "utter nonsense" that is "laughable" and "wrong"".
Last edited by skipjack; September 20th, 2018 at 07:54 PM. 
September 19th, 2018, 08:02 AM  #14  
Senior Member Joined: Oct 2009 Posts: 912 Thanks: 354  Quote:
See, right now Cantor's proof is based entirely on a few axioms, called the ZF axioms (together with some logical axioms). The entire proof can be given using only those axioms and the modus ponens inference rule. If you claim there is a logical error in the proof (which you seem to be doing, forgive me if this is not right), then you must find out which step in the rigorous formal proof is not acceptable. This is NOT what Kronecker did. Kronecker just didn't accept the ZF axioms as true. They didn't exist when he was alive of course, but it is clear from his writings that he would find the axiom of infinity totally meaningless and he wouldn't accept it. Poincare, I'm not sure what his issue was, I didn't look into it. So sure, if you don't accept either the ZF axioms, or the logical axioms, or if you are against axiomatizing mathematics in this way, then you have a point. But then you need to state THIS clearly. You have not stated anything so far, so I guess this is not your point. If you do accept the ZF axioms, and the entire philosophy of axiomatizations, then you have to find out the exact error in the fully rigorous proof as can be found for example on metamath: canth  Metamath Proof Explorer I'm sure if you succeed in going to this site and point out the exact line that an error happens, then mathematicians will immediately believe you (I sure would) since you brought it back to the fundamentals. If you do find such an error which does invalidate Cantor, then I am pretty sure a Fields medal is waiting for you (depending on your age). So if you believe you are right, you have nothing to lose. Of course, if you don't find the exact flaw in the above site, and just come with vague arguments, you will not find ANY mathematician listening to you. Maybe this is because we are arrogant, but I'm just stating what will happen. So yeah, if you think you are right. I have given you exactly the path to get recognition. Last edited by skipjack; September 20th, 2018 at 07:56 PM.  
September 19th, 2018, 02:14 PM  #15  
Senior Member Joined: Dec 2015 From: France Posts: 103 Thanks: 1  Quote:
Below is a short summary of the first error I have found in Cantor’s theorem for discussion here. Cantor uses a proof by contradiction, which is the following. The proposition to be proved P: A list cannot contain all subsets of ℕ. 1. Assume that P is false. That is: a list L contains all subsets of ℕ. 2. A subset K is created. It is shown that K is not in L, which shows P is true. 3. P cannot be true and false at the same time, So, P is true. See section “Recall of the proof” of « Analysis of the proof of Cantor's theorem » The scheme of logic is: 1. P is assumed false 2. P is shown true. 3. Then P must be true. This logic holds if P is correctly stated. What if the statement of P is incorrect? In this case, will the proof fall apart? The error I have found is that the used list in P is limited in length by the creation of K. In section “Case of infinite set” of « Analysis of the proof of Cantor's theorem » I have shown that for a subset to be selfish or non selfish, the binary string of the subset must have the diagonal bit which equals 1 or 0. As the list L contains only subsets that are selfish or non selfish, the binary list L’ contains the diagonal. The length of the diagonal equals the length of the list L’ and the length of ℕ which is the width of the list L’. So, the length of L equals the length of ℕ. In this case, the proposition “A list cannot contain all subsets of ℕ” is not correct, because in the derivation the use of the property “selfish or non selfish” imposes the length of the list L to equal the length of ℕ. So, the length of the list in the proposition P is not free of constraint, but is imposed by another element in the same context. One can argue that because all subsets of ℕ cannot be put in the list L with the length of ℕ, subsets of ℕ are uncountable. But is ℚ countable? ℚ is created by one ℕ in xaxis and another in yaxis. In this case, ℚ is a list whose length is a free variable not limited by ℕ of its axis. Must the length of ℚ equal the length of its axis? PK  
September 19th, 2018, 05:34 PM  #16 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
Every finite binary sequence represents a natural number. Can you take it from there? Last edited by skipjack; September 20th, 2018 at 07:58 PM. 
September 19th, 2018, 08:10 PM  #17 
Senior Member Joined: Oct 2009 Posts: 912 Thanks: 354  Sure, but the ONLY way to convince the mathematical world of your findings is to learn the mathematical language and to find an error in the mathematical language. If you're going to use natural language, people will just ignore you.

September 20th, 2018, 04:16 AM  #18 
Senior Member Joined: Dec 2015 From: France Posts: 103 Thanks: 1  Halas, you are right. I will discuss this subject in this language when I will master it.

September 20th, 2018, 04:20 AM  #19 
Senior Member Joined: Dec 2015 From: France Posts: 103 Thanks: 1  Sorry, I do not understand the sense of your phrase "take it from there". Below is the table 2 of my article which uses binary sequences to represent subsets. The diagonal is written in red.
Last edited by skipjack; September 20th, 2018 at 07:58 PM. 
September 20th, 2018, 05:31 AM  #20 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
Is an infinite binary sequence a natural number?
Last edited by skipjack; September 20th, 2018 at 07:59 PM. 

Tags 
binaryexpressed, cardinality, continuum hypothesis, diagonal argument, numbers, real, set 
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