SOLUTION TO PROBLEM H-12

There are no such functions. Indeed, assume by contradiction that such a function exists. Observe that it must be one-to-one since if $x\neq y$, $\vert f(x) - f(y)\vert \ge 1 > 0$, i.e. $f(x) \neq f(y)$. On the other hand the range $f({\mathbb R})$ of f is a set of isolated points since for each f(x) the only point in $f({\mathbb R})$ within $1\over 2$ from f(x) is f(x) itself. Therefore $f({\mathbb R})$ is a countable set. This is a contradiction because it means that f is a bijective transform of ${\mathbb R}$ onto the countable set $f({\mathbb R})$, and ${\mathbb R}$ is known to be uncountable.

Questions and/or comments should be directed to Judy Downey or Griff Elder

    Back to the Problem of the Week Page
 
 

Last modified:   Thu Apr 11 20:14:27 CDT 2002