March 13th, 2016, 04:19 AM  #21 
Global Moderator Joined: Dec 2006 Posts: 16,376 Thanks: 1174  When you do that for a list of the natural numbers, each sequence has a finite number of nonzero digits, but the "trick" defines a sequence with an infinite number of nonzero digits  which is a sequence not already listed, but not a sequence that corresponds to a natural number.

March 13th, 2016, 04:33 AM  #22  
Senior Member Joined: Mar 2016 From: UK Posts: 101 Thanks: 2  Quote:
Remove the words "natural numbers" from the sentence. Any sequence indexed by the natural numbers mustbe finite, and therefore must have a finite number of nonzero elements. Using the sequence to represent the natural or real numbers don't even come into it at this point. There are an infinite number of natural numbers, so a sequence can have an infinite number of elements but it can't be infinitely long if it is indexed on the Natural numbers since all Natural numbers are finite  there is no Natural number available to represent the sequence's length. This statement of course directly contradicts itself, and I think that is where my problem is. Or possibly yours  
March 13th, 2016, 04:42 AM  #23 
Senior Member Joined: Mar 2016 From: UK Posts: 101 Thanks: 2 
In fact I can go further  the indices used for the elements of the set of sequences is a tuple  two Natural numbers together. Tuples are countable. Therefore the complete set of elements of all of the set of sequences is countable. That's a real problem if the set itself is not countable. Last edited by Jimbo; March 13th, 2016 at 04:51 AM. 
March 13th, 2016, 06:19 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra  
March 13th, 2016, 06:21 AM  #25 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra  
March 13th, 2016, 07:30 AM  #26  
Senior Member Joined: Mar 2016 From: UK Posts: 101 Thanks: 2  Quote:
Yes it is. Then the complete set of sequences must also be countable, since it must be smaller than the set of all the elements of all the sequences. We can always find a new value but the resultant set is still countable.  
March 13th, 2016, 07:41 AM  #27 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra 
It's not a contradiction. Refer to the post you read yesterday which demonstrated that every natural number is finite but that the set of natural numbers is of infinite size. The correct conclusion is not that the complete set of sequences is countable, but that any list of sequences is countable 'this is the definition of "countable"). But the diagonal argument shows that no list of sequences is complete. Since no countable set of sequences is complete, the complete set of sequences cannot be countable. Last edited by v8archie; March 13th, 2016 at 07:49 AM. 
March 13th, 2016, 07:49 AM  #28 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra  
March 13th, 2016, 09:56 AM  #29  
Senior Member Joined: Mar 2016 From: UK Posts: 101 Thanks: 2  Quote:
Which directly implies that the complete set of sets is countable. Quote:
Last edited by Jimbo; March 13th, 2016 at 10:10 AM.  
March 13th, 2016, 10:25 AM  #30 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra  

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