```Date: Feb 5, 2014 6:19 PM
Author: Ben Bacarisse
Subject: Re: Wm mis-explains what he means by a Binary Tree

WM <wolfgang.mueckenheim@hs-augsburg.de> writes:> Am Mittwoch, 5. Februar 2014 20:20:51 UTC+1 schrieb Ben Bacarisse:>> WM <wolfgang.mueckenheim@hs-augsburg.de> 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 asupposed bijection can be argued about perfectly well.  If f exists, ithas a provable consequence -- that there is a path not in the image off.  I do not believe your students don't know this.  You claim theydon'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 inbijection with N, that bijection defines a path not in the image of thebijection they are right.  What exactly is the "contrary"?  That thereexist a path set, in bijection with N such that all paths are in theimage 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.<snip>-- Ben.
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