Re: The complete infinite binary tree has only countably many infinite paths.

Happens every couple of months here. The informational or educational value
of such threads is zero for anybody with the slightest knowledge of set
theory.

You could have learned the following: There is no path 0.111... in the
tree, namely a path that is distinct by a certain number (aleph_0) of
ones from all paths used to construct the tree. There is only, for
every path with n bits of value 1, another path with one more bits of
value 1. That is the whole infinite story.

But the same holds for Cantor's list. There is no line that shows you
that the diagonal number is not in the list. There is only a line that
shows you that the diagonal number in not in the first n lines of the
list but that does not exclude that it could be in the next line.

Regards, WM

I see. You don't agree with Cantor's proof, either, and for the same reason. You could have skipped all this stuff about infinite binary trees.

Cantor's diagonal proof shows, if you like, that the number is not in the first n lines for ALL n. More specifically, it shows that it is not the nth entry for any n. It cannot appear in the 1st position, or in the 2nd, or in the 3rd position, ... it is missing from the list entirely.

If you try and list all the Reals in [0,1], you will see how hard it is even without using Cantor to find a missing Real. Very few constructions even manage to put 1/3 on the list.





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