Re: Another set with cardinality |Z|

From: Robert Israel (israel_at_math.ubc.ca)
Date: 09/23/04 

Date: 23 Sep 2004 02:42:47 GMT

In article <fa69ae35.0409221823.682ae186@posting.google.com>,
Eray Ozkural exa <erayo@bilkent.edu.tr> wrote:
>Let's have an algorithm that starts with
>0.1 in binary, and constructs a tree in breadth-first fashion

> 0.1
> 0.01 0.11
>0.001 0.011....

>You get the idea... It's obvious that this tree has the same
>cardinality as Z, since this is a nonhalting algorithm (or since I can
>give an integer to every node, etc.) Now, I want to prove that such a
>subdivision procedure cannot generate all x in (0,1) in an intuitive
>way. Is the easiest method proof by contradiction?

Hint: can you generate 1/3?

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada