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January 13th, 2016, 02:47 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 834 Thanks: 67  Cantor's Diagonal Argument https://en.wikipedia.org/wiki/Cantor's_diagonal_argument Let T be the set of all infinite sequences of binary digits. Each such sequence represents a positive integer or 0.* T is countable. *1010 = 1x2^0 + 0x2^1 + 1x2^2 + 0x2^3 = 5 Every real number can be represented by an infinite binary sequence. The real numbers are countable. Last edited by skipjack; January 13th, 2016 at 03:17 PM. 
January 13th, 2016, 03:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 16,226 Thanks: 1150  
January 13th, 2016, 04:20 PM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 834 Thanks: 67 
Every sequence of zeros and ones represents a positive integer or 0. Then the set T of all infinite sequences of zeros and ones is countable in the sense that the set of all the integers is countable. Cantor maintains T is not countable. It's possible that 3 appears way, way, down the line. But it's there. What are we talking about? 1) T={s1,s2,..}={{1001...},{01110...}.....}, T is the set of all sets of infinite sequences. (T is not the set of infinite sequences.) Countable or 2) T={1001...,0110,....}, sequentially without the commas. (T is the set of all infinite sequences.) How do you create this set? How do you know you have created the set if you can't distinguish between infinite member sets? I presume the former. The latter is nevernever land, and you can pretty much say anything you like about it. Alright I'll think about it: List infinite sets of zeros and ones in sequence. Then create one set which contains all these sequences. So you are talking about an infinite arrangement of infinite sequences. But what does that have to do with countability of binary or decimal numbers? By what rule are you selecting infinite series from this infinite series to count? You are just playing word games with symbols. Ref https://en.wikipedia.org/wiki/Cantor's_diagonal_argument Last edited by skipjack; January 13th, 2016 at 08:39 PM. 
January 13th, 2016, 08:44 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,226 Thanks: 1150 
Are you discussing these things with yourself? You seem to be answering your own questions.

January 13th, 2016, 10:08 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 834 Thanks: 67 
"In his 1891 article*, Cantor considered the set T of all infinite sequences of binary digits (i.e. consisting only of zeros and ones). He begins with a constructive proof of the following theorem: If s1, s2, â€¦ , sn, â€¦ is any enumeration of elements from T, then there is always an element s of T which corresponds to no sn in the enumeration. To prove this, given an enumeration of arbitrary members from T, like e.g. s1 = (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0, 1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0, 1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...) s7 = (1, 0, 0, 0, 1, 0, 0, ...)"  But that's ridiculous. You don't count a set with an infinite number of elements by selecting subsets of infinite elements and counting the number of such subsets. Of course the set will be uncountable because that's not how you count. No matter what the sequence of '0's and '1's is, you count the set sequentially. S=0,1,0,1,1, .... N=1,2,3,4,5,...... That's countable. * https://en.wikipedia.org/wiki/Cantor's_diagonal_argument EDIT: Furthermore, how are the digits arranged? You can't even define the set of all binary sequences because you can't even start to write it down. Is it {0,1,0,0...} or {1,1,0.1....},.. No matter how many digits you add these are different sets. Or is the assumption that no matter how you start, any arbitrary sequence, such as the second one, will be found somewhere in the sequence? Last edited by zylo; January 13th, 2016 at 10:34 PM. 
January 13th, 2016, 10:26 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 16,226 Thanks: 1150 
Is it mathematically correct? You can call it ridiculous if you wish, and it might not do what you would do, but it's correct.

January 14th, 2016, 05:31 AM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,357 Thanks: 2086 Math Focus: Mainly analysis and algebra  Quote:
Quote:
The proof uses the assumption that we can complete the list to produce a contradiction.  
January 14th, 2016, 09:36 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 834 Thanks: 67 
To count the elements in the set T of all binary numbers arrange them in a column in any order and start counting diagonally: 1,0,0,1,0,0,....... 0,1,0,0,1,0...... 1,0,1,0,1,0...... 1,1,0,1,1,0...... ..................... 1, 0,0, 1,1,0, 1,0,0,1, ...... 1, 2,3, 4,5,6, ................. This is the same argument used to count the rationals. If it doesn't work here, it doesn't work there. Cantor's argument that T is uncountable is wrong. https://en.wikipedia.org/wiki/Cantor's_diagonal_argument Last edited by skipjack; January 14th, 2016 at 01:20 PM. 
January 14th, 2016, 09:43 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,357 Thanks: 2086 Math Focus: Mainly analysis and algebra 
You are just clutching at straws. You have no understanding of what you are doing. In this case, you have no idea whether you will have every sequence. And indeed, Cantor's argument shows that you won't because the list you require can't be constructed. 
January 14th, 2016, 10:28 AM  #10  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 834 Thanks: 67  Quote:
If the list can't be constructed then T can't be constructed. If you can't construct a list of all binary numbers, how can you construct the set of all binary numbers?  

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