> Am Mittwoch, 5. Februar 2014 16:01:41 UTC+1 schrieb Ben Bacarisse: >> WM <firstname.lastname@example.org> writes: >> >> > Am Mittwoch, 5. Februar 2014 02:40:36 UTC+1 schrieb Ben Bacarisse: >> >> WM <email@example.com> writes: >> >> >> >> > Am Dienstag, 4. Februar 2014 15:04:56 UTC+1 schrieb Ben Bacarisse: >> >> > >> >> >> > "It" differs? What makes it differ? Only your belief that after the >> >> >> > Binary Tree has been constructed other paths may sneak in. >> >> >> >> >> >> It's provable. You should get your students to prove it because it's >> >> >> not hard. Given a countable set of paths that "constructs" the tree -- >> >> >> i.e. p(1), p(2), ... such that every edge is included in some p(k), ask >> >> >> them to construct a path that has sneaked in: one that is in the graph >> >> >> but that is not in the set p(i). Do you think they would not be able t >> >> >> do this? >> >> >> >> > Of course. >> >> <snip> >> >> So the construction need not define the set of paths. >> > >> > The construction defines all sets of nodes. If every path is a set of >> > nodes, then the construction defines every path. >> >> It defines the tree so, implicitly it defines the paths, but there are >> paths not in the set used in the construction. You've agree that this >> is true regardless of what countable set it used to construct the tree.
> No. It is true because I have used for simplicity and in order to > disprove the possibility of infinite definitions only one finite > definition of the tails.
No, it is true regardless of the countable set used to construct the tree. Are you now retracting your agreement that your students would be able to construct such a path from a set of paths that are in bijection with N?
> If I use all possible finite definitions then there remains nothing to > diagonalize. That would be the same as the list containing all > terminating rationals.
So your students would not be able to make the construction that you previously agreed they could? Despite it being a simple construction based on the fact the set of paths used in the construction is in bijection with N?
>> Thus no countable set of paths includes all the paths in the tree that >> it constructs (provided it covers every node). >> > Look, every finite definition of the tails can be used. In its > simplest form use the tree or list of all terminating rational > numbers. Then Cantor's diagonal argument fails at every level with > finite index. And further levels are not subject to diagonalization.
Whatever this means, I take it that you are now retracting your agreement about what your students could prove? If they gave the "obvious" construction based on the bijection f: N -> P that the path p(n) "goes the other way" to the path f(n)(n) does would you mark them down? In all conscience?
>> We are in agreement: your students can prove what you claim to be false. > > You are wrong. My claim is not that my set of paths covering or > constructing the Binary Tree contains all possible paths.
Oh. Excellent. So any countable set of paths you use to construct the set leaves some out? That's exactly what "uncountable" means.