> Am Mittwoch, 5. Februar 2014 20:20:51 UTC+1 schrieb Ben Bacarisse: >> WM <email@example.com> writes: >> >> > Am Mittwoch, 5. Februar 2014 17:41:13 UTC+1 schrieb Ben Bacarisse: >> >> >> >> 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? >> > >> > They would know that also the other way is already realized, for every >> > n, in a rationals-complete list. And they would know that this >> > rationals-complete liste is realized by the Binary Tree. You cannot >> > cope with them. >> >> You don't teach them how to tell if two infinite sequences are the same >> or not? > > You cannot tell it either unless you have a finite definition of both > of them. But you did not know that or even don't know yet - they know > it.
Oh you are now being very silly. The properties of a path defined by a supposed bijection can be argued about perfectly well. If f exists, it has a provable consequence -- that there is a path not in the image of f. I do not believe your students don't know this. You claim they don't just because you are stuck for any other answer.
>> If >> they argued as I suggested you'd tell them they are wrong. > > No, I would tell them that they are right, but that the contrary also > is right.
Let's be clear -- if they argue that for any set of paths that are in bijection with N, that bijection defines a path not in the image of the bijection they are right. What exactly is the "contrary"? That there exist a path set, in bijection with N such that all paths are in the image of the bijection?
> That's what we call an antinomy. It is a well-known paradox >that matheologians cannot see the other side.
Well it would be a problem except that the contrary is not true.