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 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.

<snip>
--
Ben.