Cantor Set Function $g(x)=1$ ($x\in{C}$), $g(x)=0$ ($x$ not in C) (Dis)Continuity

Question

Let $C$ be the Cantor (ternary) set. Define $g:[0,1]\rightarrow{\mathbb{R}}$ by $g(x)=1$ if $x\in{C}$ and $g(x)=0$ if x is not in C. How does one show that g is dicontinuous at any point $c\in{C}$ and continuous at every point c not in C. My idea is to use the fact that the Cantor set is a perfect set, so all its points are limit points, thus we can certainly construct a sequence $(x_n)\rightarrow{c}$ but I am not so sure now as to how this helps determine (dis)continuity. All help appreciated. Thank you.

Answer 1

Let $c\in C$. Since $c$ is a limit point of $C$, if $U$ is any neighborhood of $c$ then $U \cap (C\setminus\{c\}) \neq \emptyset$. On the other hand, $U$ contains also points of $[0,1]\setminus C$. Hence $g$ attains both values $0$ and $1$ in any neighborhood of $c$, so that it is discontinuous at $c$.

The continuity of $g$ on $[0,1]\setminus C$ follows from the fact that $[0,1]\setminus C$ is relatively open in $[0,1]$ (being $C$ closed).