Date: Feb 5, 2014 6:19 PM
Author: Ben Bacarisse
Subject: Re: Wm mis-explains what he means by a Binary Tree
WM <firstname.lastname@example.org> writes:
> 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
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.
>> 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.