Exercise 4.3.7

Recall Dirichlet’s function is

And Thomae’s function is

(a)

Let $a\in Q$ and set $𝜖=1$. For any $\delta >0$ there will exist points $x\in (a-\delta ,a+\delta )\cap I$ by the density of $I$ in $R$ with $|f(x)-f(a)|=|0-1|=1$ not less then $𝜖$, therefore there does not exist a $\delta$ to match $𝜖=1$ and so $f$ is discontinuous at $a$. Since $a$ was arbitrary (the $a\in I$ case is identical) $g$ must be discontinuous at all of $R$.

(b)

By the same argument as in (a) for any $m∕n\in Q$ no matter how small $\delta$ is, we can find an irrational number within $\delta$ of $m∕n$ meaning $𝜖$ cannot be made smaller then $|f(m∕n)-f(x)|=1∕n$.

(c)

Let $a\in I$, we want to show $t(x)<𝜖$ for $|x-a|<\delta$. I claim the set $\{x\in V_1(a):t(x)\ge 𝜖\}$ is finite, this can be seen since the requirement that $t(x)\ge 𝜖$ is the same as $x=m∕n$ and $1∕n\ge 𝜖$. It is easy to see there are finitely many points like this (consider how there are finitely many $n$ and finitely many $m$ given $n$) thus we can say $\{x\in V_1(a):t(x)\ge 𝜖\}=\{x_1,\dots ,x_n\}$ and set $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\{|x_i-a|:i\in \{1,\dots <!--\; FUNCTION\; APPLICATION\; -->,n\}\}$ to ensure every $x\in V_{\delta }(a)$ has $t(x)<𝜖$.