April 28th, 2017, 07:32 AM  #21 
Global Moderator Joined: Dec 2006 Posts: 17,911 Thanks: 1382  
April 28th, 2017, 08:39 AM  #22 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra 
The reason that 101010.......... (continuing without end) isn't a number is that $$\sum_{n=0}^\infty a_n 2^n \quad a_n \in \{0,1\}$$ doesn't converge unless $a_n = 0$ for all $n \gt N \in \mathbb N$. Last edited by v8archie; April 28th, 2017 at 08:48 AM. 
April 28th, 2017, 09:10 AM  #23  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Quote:
Let me count the infinite binary sequences for you. Map them as follows: 1000000...... > 1 0100000...... > 2 1100000...... > 3 0010000...... > 4 . . .  
April 28th, 2017, 09:34 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra 
Of course convergence has something to do with it! So what is the value of your natural number represented by the sequence $$a_0a_1a_2\ldots \quad a_n \in \{0,1\}$$ when the series $$\sum_{n=0}^\infty a_n2^n$$ does not converge? This is exactly the definition of your supposed map, so without an answer your argument collapses. Since natural numbers have no binary point, your attempted obfuscation is clearly irrelevant. Last edited by v8archie; April 28th, 2017 at 09:57 AM. 
April 28th, 2017, 10:23 AM  #25  
Senior Member Joined: Jun 2014 From: USA Posts: 308 Thanks: 21  Quote:
You seem to be an uncountably many binary sequences short of a full listing... The binary sequences can be put into bijection with $P( \,\mathbb{N}) \,$ quite easily. If that mapping is 1to1 from $\mathbb{N}$ onto the set of infinite binary sequences, then we'd also have a mapping that is 1to1 onto $P( \,\mathbb{N}) \,$. Since you don't like Cantor's diagonal argument, consider his Theorem: Let $f : \mathbb{N} \longrightarrow P( \,\mathbb{N}) \,$ Let $A = \{ n : n \notin f( \,n) \, \}$ If $f$ is surjective, then there exists a $k \in \mathbb{N}$ such that $f( \,k) \, = A$: $k \in A \rightarrow k \notin f( \,k) \, = A$ $k \notin A \rightarrow k \in f( \,k) \, = A$ No $k$ can exist such that $f( \,k) \, = A$, so the assumption that $f$ was surjective was faulty.  
April 28th, 2017, 10:42 AM  #26 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
I note $\displaystyle p=a{_n}2{^n}+......+a{_1}2{^1}+a{_0}2{^0}$ $\displaystyle =a{_0}2{^0}+a{_1}2{^1}......+a{_n}2{^n}$ for all n. So the count of infinite binary sequences 1000000...... > 1 0100000...... > 2 1100000...... > 3 0010000...... > 4 . . can be interpreted in either direction (map or actual number). The "last" number in the sequence is 1111111.............., which is the inverse of 000000000. Note any natural number can be expressed as a unique infinite binary sequence and vice versa. If the binary sequence wasn't infinite, the count of natural numbers would stop at some number. Cantor's set of infinite binary sequences is simply the set of natural numbers in binary notation. 
April 28th, 2017, 10:42 AM  #27 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra  
April 28th, 2017, 10:52 AM  #28  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra  Quote:
As usual, you don't understand the difference between the concepts "arbitrarily large" and "infinite". Last edited by v8archie; April 28th, 2017 at 10:56 AM.  
April 28th, 2017, 11:01 AM  #29 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra  Since every natural number has a successor, what is the successor to 1111111..............? Or do you claim that there is now a greatest natural number? And that the natural numbers are not closed under addition or multiplication?

April 28th, 2017, 12:05 PM  #30 
Senior Member Joined: Aug 2012 Posts: 1,521 Thanks: 364  

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