This problem has me stumped. I'm not sure how to proceed.

Let $A = (0,\infty)$ and let $k: A \to \mathbb{R}$ be defined as follows: $$ k(x) = \begin{cases} 0 & \text{for } x \in \mathbb{R}_+\setminus\mathbb{Q} \\\\ n & \text{for } x = \frac mn \in \mathbb{Q}_+ \text{ with } (m,n) = 1 \end{cases}$$ Prove that $k$ is unbounded on every open interval in $A$. Conclude that $k$ is not continuous at any point of $A$.

  • $\begingroup$ You have to show that for every open interval $(a,b)$, $k$ is unbounded on $(a,b)$. That is, for every $M$ there exists $x \in (a,b)$ with $|k(x)| \ge M$. Can you see how to show this? $\endgroup$ – Nate Eldredge Oct 27 '11 at 2:34
  • $\begingroup$ So am I proving the absences of an upper and lower bound? $\endgroup$ – Jon Oct 27 '11 at 2:38
  • $\begingroup$ Since $k(x)\ge 0$ for all $x$, you need only show that given an interval $(a,b)$ and a number $M$, there is some $x\in(a,b)$ such that $k(x)\ge M$. $\endgroup$ – Brian M. Scott Oct 27 '11 at 2:42
  • $\begingroup$ @Jon: Yes, that's equivalent. $\endgroup$ – Nate Eldredge Oct 27 '11 at 2:43

For an interval $(a,b)$ lets show that for every natural $M$ there are only a finite number of rationals $\frac{p}{q}$ such that $a < \frac{p}{q} < b$ and $0 \leq q \leq M$.
The second inequality gives us a finite number of values for $q$, so for each of these values we have from the first inequality that $aq < p < qb$ so $p$ can only take a finite amount of values for each value of $q$, this gives us a finite number of possible combinations for $p$ and $q$.

To apply this, just notice that for every interval $(a,b)$ if $k(x)$ was bounded then for every rational $\frac{p}{q} \in (a,b)$ with $(p,q) = 1$, then $k(\frac{p}{q}) = q \leq M$. Using the above we will end with only a finite number of rationals in $(a,b)$ which is false. Then $k(x)$ is unbounded on $(a,b)$.

  • $\begingroup$ Great. I think the issue I was having was the application of the second inequality to the interval. Thank you for the help :) $\endgroup$ – Jon Oct 27 '11 at 3:15

HINT: If $p$ is prime, $n$ is an integer, and $n/p$ is not an integer, what is $k(n/p)$?

  • $\begingroup$ I'm not sure I follow. $\endgroup$ – Jon Oct 27 '11 at 2:49
  • $\begingroup$ Is n/p necessarily rational or irrational? Per the problem, given x irrational, then k(x) = 0, so if t = n/p, then k(t) = 0. $\endgroup$ – Jon Oct 27 '11 at 2:50
  • 1
    $\begingroup$ @Jon: Here $n$ is intended to represent an integer, so $n/p$ is rational. I’ve now clarified the question. $\endgroup$ – Brian M. Scott Oct 27 '11 at 3:10
  • $\begingroup$ Thanks for the clarification. $\endgroup$ – Jon Oct 27 '11 at 3:16

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