April 4th, 2016, 02:52 PM  #11  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,615 Thanks: 2604 Math Focus: Mainly analysis and algebra  Quote:
Instead, your ego demands that you claim to be the only mathematician to understand this stuff and to have solved it in a couple of weeks. You have no understanding and seek only to confuse students who genuinely wish to master the subject by continually posting falsehoods, inaccurate logic and other assorted nonsense because you either can't or won't admit that you are wrong. You clearly don't even understand the concept of Proof by Contradiction. You certainly don't understand even the basic structure of Cantor's proof which you keep posting a link to. If you had ever read it properly you would know that the Diagonal argument doesn't once refer to the list of all sequences, which you in your blind arrogance claim to have constructed. You utterly refuse to discuss any of the proofs as written, instead preferring to shoot down you own imaginary nonsense versions, which bear no more than a superficial resemblance to the original.  
April 4th, 2016, 04:47 PM  #12  
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  Quote:
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He begins with a constructive proof of the following theorem:He uses infinite sequences from the start and throughout.If $s_1, s_2, â€¦ , s_n, â€¦$ is any enumeration of elements from $T$, then there is always an element $s$ of $T$ which corresponds to no $s_n$ in the enumeration. The article then explains Cantor's diagonal procedure by which $s$ is constructed and explains why $s$ is not already in the enumeration, but is in $T$. All that is set out before a second theorem is explained that uses proof by contradiction. You've posted several times that he's already incorrect, because his constructive (diagonalisation) method is valid only for finite sequences, but you've never stated where precisely Cantor does something wrong in his preliminary definitions and proof, and what exactly is wrong. You then stated "whatever Sn is, S has "1" digits past the last "1" digit of Sn", ignoring that for an infinite Sn (Cantor doesn't use finite sequences), it needn't have a last "1" digit.  
April 6th, 2016, 08:04 AM  #13 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
Let T be set of all infinite sequences of binary digits. If infinite means countably infinite, Cantor's proof fails because countably infinite sequences of binary digits are natural numbers which are obviously countable. If infinite means not countably infinite, then you obviously can't enumerate them (Cantor's assumption) because each place in the sequence gives a member of the enumeration. Last edited by zylo; April 6th, 2016 at 08:36 AM. Reason: clarify 
April 6th, 2016, 09:19 AM  #14 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,615 Thanks: 2604 Math Focus: Mainly analysis and algebra 
Do you not understand that whether the sequences themselves are finite or not, an enumeration (list) is automatically countable? Your set T does not exist. There is no bijection by which "all infinite binary sequences" is mapped to the natural numbers. This too is something you are incapable of understanding. And yet, despite such trivial errors you persist in claiming to have sufficient mathematical skill to dispute the work of one of the greatest mathematicians to have lived. You aren't even intelligent enough to follow his proof properly and you are incapable of answering Skipjack's simple questions. 
April 6th, 2016, 10:15 AM  #15 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  It does in this context, as all sequences are countable (so your alternative of uncountably infinite needn't be considered). How do you justify generalizing from finite sequences or sequences with only a finite number of "1" digits to all countably infinite sequences? You haven't provided any way of putting all the countably infinite binary sequences in 1 to 1 correspondence with the natural numbers. Your example of 0x2^{0} + 0x2^{1} + 1x2^{2} + ... (corresponding to the sequence 0, 0, 1, 1, ...) doesn't do that, because 0x2^{0} + 0x2^{1} + 1x2^{2} + ... is in general a divergent summation, not a natural number. On one occasion, you stated the contrary (that they're obviously uncountable), so why are you now contradicting your own earlier assertion? 
April 7th, 2016, 06:43 AM  #16  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
T = {0011..., 1101..., 0100..., 1010,......} List each marked sequence. If infinite means not countably infinite, you obviously can't assume a countably infinite list and Cantor's argument doesn't apply. If you nevertheless go through the motions of Cantor's argument, you will conclude: If infinite means not countably infinite, infinite is not countably infinite. __________________________ Response to post #15 Every natural number is an infinite binary sequence and every infinite binary sequence is a natural number. 2=01000000000000000000000........=2 64=0000010000000000000000........=64 What does convergence have to do with this? It's the binary number system. Every real number in [0,1) has a natural number correspondence: .00110000000..... > 001100000000000.....  If you don't like the natural number argument, if you assume T consists of all countably infinite binary sequences, then they will form a countably infinite enumeration (11 correspondence with the natural numbers), and Cantors argument that the sequence is uncountable makes no sense: https://en.wikipedia.org/wiki/Cantor...gonal_argument "Cantor's diagonal argument... was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into onetoone correspondence with the infinite set of natural numbers." ..... "In his 1891 article, Cantor considered the set T of all infinite sequences of binary digits (i.e. consisting only of zeroes and ones)." Last edited by skipjack; April 7th, 2016 at 11:47 AM.  
April 7th, 2016, 12:12 PM  #17  
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  Quote:
Not necessarily... your condition "if you assume" is redundant, as T is defined as the set of all countably infinite binary sequences, but you've given no reason for your assertion that they will form a countably infinite enumeration. That assertion contradicts the unconditional assertion you made elsewhere that T is obviously uncountable.  
April 7th, 2016, 12:22 PM  #18  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,615 Thanks: 2604 Math Focus: Mainly analysis and algebra  Quote:
Of course, as usual, you are confused by the phrase "infinite binary sequences" which means that the sequences are infinite, not that there are uncountably infinitely many of them. The set $T$ being the set of "all countably infinite binary sequences" means that each of the sequences is countably infinite in length, not that there are a countably infinite number of them. Of course, the word "countable" is tautological here because as sequence cannot have an uncountably infinite length since it is not more than a list of elements.  
April 7th, 2016, 12:34 PM  #19 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007 
The condition "If infinite means not countably infinite" doesn't apply, so there's nothing to be considered anyway.

April 11th, 2016, 07:49 AM  #20 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Mark a place in each sequence, never duplicating a place. Then put every marked sequence in a list. There are as many sequences in the list as there are places in a sequence.
Last edited by zylo; April 11th, 2016 at 07:53 AM. 

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