
map
Suppose $\displaystyle f(x) = \begin{cases} 0 \ \ \ \text{if} \ \ x \notin \mathbb{Q} \\ 1/q \ \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = p/q \ \text{in lowest terms} \end{cases} $
1. Fix $\displaystyle t \in \mathbb{R} $ and $\displaystyle n \in \mathbb{N} $. Show that $\displaystyle f $ maps only finitely many elements of $\displaystyle (t1/2, t+1/2) $ to $\displaystyle 1/n $.
So the length of this interval is $\displaystyle 1 $. I think its $\displaystyle n1 $. But that is intuition. How do you make it rigorous?
2. Prove that $\displaystyle f $ is continuous at every irrational number. So $\displaystyle \forall \epsilon >0, \exists \delta >0 $ such that $\displaystyle xa < \delta \implies f(x) < \epsilon $. Do we want to elements that dont map to $\displaystyle 1/n $?

Basically we are trying to show that $\displaystyle f $ maps a finite number of rational numbers in lowest terms in the interval $\displaystyle I = (t \frac{1}{2}, t+\frac{1}{2}) $ to $\displaystyle \frac{1}{n} $. Let $\displaystyle x = p/n \in I $ and in lowest terms. Then $\displaystyle p $ and $\displaystyle n $ are coprime. Can we somehow use this to deduce that the maximum number is $\displaystyle n1 $? number of fractions $\displaystyle p/n $ in interval is $\displaystyle n $. number in lowest terms is $\displaystyle \leq n1 $.