> 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 to >> >> 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. Thus no countable set of paths includes all the paths in the tree that it constructs (provided it covers every node).
We are in agreement: your students can prove what you claim to be false.