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November 20th, 2017, 03:42 PM  #11  
Senior Member Joined: Aug 2012 Posts: 1,971 Thanks: 550  Quote:
Now the union of all $p(x)$ for all real numbers $x$ is still a countable set of finite decimals. You are not picking up the limits, only the finite strings. But we suspect that you can't leave any real numbers out without missing some strings. That's the tricky part here. So in your terminology, $T$ can be the set of all terminating decimals. Then for my function $p$, the union of all the sets $p(t)$ for $t \in T$ is the same set of all terminating strings. Right? In other words I don't need to take $p(\pi 3)$, all I need is $p(.1), p(.14), \dots$ which gives me the same set of initial segments as $pi  3$. Last edited by Maschke; November 20th, 2017 at 03:48 PM.  
November 20th, 2017, 03:55 PM  #12 
Senior Member Joined: Aug 2012 Posts: 1,971 Thanks: 550 
ps  My tree proof must be wrong. In this case $T$ is the set of nodes that continually branch to $0$ after some point. This set contains every node in every path. So $T$ covers the entire tree. Is that right? So I'm no longer sure what the great mystery is, except for finding the flaw in my proof. pps  Yes my proof is wrong. If you have an infinite path, every one of its nodes is already in the 0chain below that node. Your idea about $T$ is exactly right. Now there's no mystery any more. Is that correct? In the end all we did was convince ourselves that the set of finite strings is the set of finite strings. But we took the scenic route to get there! Last edited by Maschke; November 20th, 2017 at 04:18 PM. 
November 20th, 2017, 06:09 PM  #13 
Senior Member Joined: Jun 2014 From: USA Posts: 364 Thanks: 26  
November 20th, 2017, 08:08 PM  #14 
Senior Member Joined: Aug 2012 Posts: 1,971 Thanks: 550  

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ordering, question, set, uncountable 
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