Exercise 4.3.14

If $x\in F^o$ (which is open) then we can find $V_{𝜖}(x)\subseteq F^o$ where there will be both irrational and rational numbers, indicating that $f$ is discontinuous over $F^o$.

If $x\in \overline{F^c}\setminus F^c$, $x$ must be a limit point of $F^c$, and therefore all $V_{𝜖}(x)$ will intersect $F^c$ at some point, and thus $f(y)=3$ for some $y\in V_{𝜖}(x)$, preventing $f$ from being continuous in $\overline{F^c}\setminus F^c$.

If $x\in F^c$ (which is open) then we can find $V_{𝜖}(x)\subseteq F^c$ which is a constant 3, and therefore $f$ is continuous over $F^c$. Thus, $f$ is discontinous only over $F$.

$f(x)=0$ for $x\in F^c$ and by choosing $\delta =𝜖>0$ we will have $\inf <!--\; FUNCTION\; APPLICATION\; -->\{|y-a|:a\in F^c\}<𝜖$ for $y\in V_{\delta }(x)$ (simply consider $a=x$) implying $f$ is continuous over $F^c$.

Since $F$ is open, for any given $x\in F$ we can find $\alpha >0$ so that $\inf <!--\; FUNCTION\; APPLICATION\; -->\{|y-a|:a\in F^c\}>\gamma >0$ for all $y\in V_{\alpha }(x)$. (One way to do this is by choosing $\beta$ so that $V_{\beta }(x)\subseteq F$, taking $\alpha =\beta ∕2$, noting that $\{a:\exists <!--\; FUNCTION\; APPLICATION\; -->y\in V_{\alpha }(x)\text{ such that }|y-a|<\alpha \}=V_{\beta }(x)$, and concluding that $a\in F^c⟹a\notin V_{\beta }(x)⟹\forall <!--\; FUNCTION\; APPLICATION\; -->y\in V_{\alpha }(x),|y-a|\ge \alpha$.) Then since for any $V_{\delta }(x)$, there must be points $y_1,y_2$ where $d(y_1)=1,d(y_2)=0$, it must be impossible to satisfy the definition of continuity for $𝜖<\gamma$ (since in the $\delta$-neighbourhood of $x$, $f(x)$ will jump by at least that amount between rational and irrational numbers), and therefore $f$ is discontinuous for any $x\in F$.