My Math Forum Well Ordering of an Uncountable Set Question

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November 20th, 2017, 03:42 PM   #11
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 Originally Posted by AplanisTophet I found ordinals potentially useful for trying to pinpoint where all subtrings of a path become covered despite the path itself being omitted. We need not use them. Because all finite strings appear in the initial segments of elements of $T$ (all substrings to mirror your node example), we can have a subset of $I$ that covers all of $F$ (all the substrings) despite omitting the vast majority of the elements of $I$ (omitting the vast majority of the paths).
Ah. I don't think it covers $F$ in that sense. For example, associate with each real number the set of initial segments of its decimal representation. This is like my $q$ idea except that for example, ok let's pick a new function letter, $p(\pi - 3) = \{.1, .14, .141, \dots\}$

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: 1,887 Thanks: 525 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 0-chain 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
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 Originally Posted by Maschke 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!
Agreed. Sometimes it's good to take the scenic route though...

Thank you Maschke

November 20th, 2017, 08:08 PM   #14
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 Originally Posted by AplanisTophet Thank you Maschke
You're very welcome. That was fun.

 Tags ordering, question, set, uncountable

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