March 24th, 2016, 09:32 AM  #41 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
For any enumeration of a countably infinite set of nonterminating binary sequences, Cantor used a "diagonal" construction to produce a nonterminating binary sequence s that is not in that enumeration. If you think that part of his argument contains a flaw, explain in this thread what the flaw is. 
March 24th, 2016, 10:12 AM  #42  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
Let T be the set of all infinite binary sequences. T is obviously uncountable, so you can't assume, as Cantor does, that it is countable. Given: T={1,2,3,4} Assume T={1,2,3} > 4 does not belong to T Conclusion T={1,2,3.4} If you assume infinite means countably infinite, Cantor's argument leads to the conclusion that the natural numbers are uncountable. Last edited by skipjack; March 24th, 2016 at 01:10 PM.  
March 24th, 2016, 10:25 AM  #43 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
That doesn't identify a flaw in the part of Cantor's argument that I specified, in which the enumeration considered is countably infinite by definition, and each binary sequence is countably infinite by the definition of a nonterminating sequence. Can you identify a flaw in that particular part of Cantor's argument? 
March 24th, 2016, 01:38 PM  #44 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Cantor's s is infinite. Infinity is not a natural number.
Last edited by zylo; March 24th, 2016 at 01:51 PM. Reason: add: first sentence. 
March 24th, 2016, 02:34 PM  #45 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
Given that the sequence s is intended to be infinitely long, as is each of the sequences used, why is that a flaw?
Last edited by skipjack; March 25th, 2016 at 05:46 AM. 
March 25th, 2016, 03:35 AM  #46 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,618 Thanks: 2608 Math Focus: Mainly analysis and algebra  
March 25th, 2016, 10:53 AM  #47  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
The argument of post #42 holds no matter how you interpret infinite. Given an infinite binary sequence, let the last nonzero binary digit be in the nth place. Then this sequence is a natural number. It follows that all the sequences with finite n are a countable subset of the set of all infinite binary sequences. The argument of post #42 now follows. You can't assume T is countable to arrive at a contradiction. Given T={1,2,3,4}, you can't assume T={1,2,3}. EDIT: The above seems to prove that the set of all infinite binary sequences is uncountable, since the natural numbers are a subset. Without Cantor's diagonal argument. The resolution depends on the answer to the question: Is a countably infinite binary sequence always a natural number? If the answer is yes, and I suspect it is, then "Cantor's s is infinite. Infinity is not a natural number" may still be valid. Last edited by skipjack; April 4th, 2016 at 11:40 AM.  
March 25th, 2016, 12:43 PM  #48  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
Assume T is countably infinite and list all members: There are countably infinite many digit places. There are always more rows than digit places: 0 0 1 0 0 1 1 0 for example. s has a digit for each row. Therefore s has more than countably infinite many digits and is therefore not a member of T and there is no contradiction. Cantor's argument fails. EDIT This replaces previous post. Last edited by skipjack; April 4th, 2016 at 11:41 AM. Reason: add countably and binary  
March 25th, 2016, 01:00 PM  #49  
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010  Quote:
Quote:
Quote:
You're now clearly going beyond the scope of my request. I referred to the first part of Cantor's proof. In that part, Cantor isn't making the supposition you mention and isn't obtaining a contradiction; he is just constructing a sequence s that is derived from the sequences he's using and is different from every one of them.  
March 25th, 2016, 02:57 PM  #50 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
Post #47 is obsolete replaced by my previous post, #48. A countably infinite binary sequence exists for every value of n and is therefore always a natural number. Therefore the set of all countably infinite binary sequences is a set of natural numbers and therefore countable. The flaw in Cantor's proof to the contrary is given in my previous post, #48. _______________________________________ 0 1 1 0..... =0x2^0+1x2^1+1x2^2+........ 

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