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December 19th, 2017, 07:41 PM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 332 Thanks: 26  Oh Boy, A Crazy Challenge to Cantor's Argument
Let g(1) = 0 and g(0) = 1. For any infinite binary string $x = x_1x_2x_3\dots$, let $f( \,x) \, = g(x_1)g(x_2)g(x_3)\dots$. If for each element $x$ of a countable set $X$ of infinite binary strings, $f( \,x) \,$ is also in $X$, then let $X$ be considered â€˜balanced.â€™ Let V be a collection of countable sets of infinite binary strings that partition the set of infinite binary strings where each element of V is balanced. Given a countable list of infinite binary strings, Cantorâ€™s diagonal argument can be used to create an infinite binary string that is not in the list (an antidiagonal). Let each element of V be well ordered in a fashion such that the antidiagonal of the ordering is equal to $a = 111\dots$. By definition, no element of V can contain $a$, so the elements of V cannot partition the set of infinite binary strings. Now, either no such collection V can exist where each element of V is balanced or Cantorâ€™s diagonal argument is faulty. On a side note, December 20th is my birthday. Yay! 
December 19th, 2017, 08:07 PM  #2 
Senior Member Joined: Jun 2014 From: USA Posts: 332 Thanks: 26 
PS  I used V for the collection because I'm thinking along the lines of Vitali sets... It would also be easy to exclude 000... and 111... from the set of infinite binary strings with this argument in order to poke at the continuum hypothesis. 
December 20th, 2017, 02:35 AM  #3  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 126 Thanks: 39 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
PS happy birthday! Last edited by cjem; December 20th, 2017 at 03:18 AM.  
December 21st, 2017, 02:48 PM  #4  
Senior Member Joined: Jun 2014 From: USA Posts: 332 Thanks: 26  Quote:
I have no interesting result when trying to invoke such an ordering across the elements of V. Knowing that each element of V is balanced and of an ordering that produces a similar antidiagonal across all the elements isn't enough to provide us any additional insight, seemingly.  
December 21st, 2017, 02:56 PM  #5  
Senior Member Joined: Aug 2012 Posts: 1,765 Thanks: 480  Quote:
In any event, in your original argument if you have any partition whatsoever into countable sets (of binary strings), since it's a partition some element of the partition contains $11111...$, therefore any antidiagonal must contain a $0$. Balanced sets don't seem to make any difference at all, if I'm understanding your argument correctly. Last edited by Maschke; December 21st, 2017 at 03:15 PM.  
December 21st, 2017, 03:30 PM  #6  
Senior Member Joined: Jun 2014 From: USA Posts: 332 Thanks: 26  Quote:
Quote:
For example, what if the partition, being balanced and with each element of a certain ordering, was enough for us to do something like try and enumerate the elements of the partition with the exception of $a = 111\dots$? I don't think that's the case or anything, but that certainly would be interesting to say the least. But, what I'm saying is that my logic was flawed and not meaningful. I have no argument for you to misinterpret.  
December 21st, 2017, 04:20 PM  #7  
Senior Member Joined: Aug 2012 Posts: 1,765 Thanks: 480  I think my mind glazed over that part. Quote:
But if this is all irrelevant now, that's ok too. Haha. You should never drink and derive! Last edited by Maschke; December 21st, 2017 at 04:28 PM.  

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