$C$ is Cantor Set. Define $f(x)=−\pi$ if $x\not\in C$ and $f(x)=\pi$ if $x\in C$. Prove that $f$ is discontinuous at every element of $C$ and continuous otherwise.
The element in Cantor Set can be any point in $[0,1]$, then how does $f$ discontinuous at these points since $f$ is always $\pi$?
On
The Cantor set $C$ has the property that for any $x \in C$ there is a sequence $x_n$ of points outside of $C$ that converges to $x$ (as $C$ is nowhere dense) and this contradicts (sequential) continuity at $x$. Any $x \notin C$ has an open neighbourhood missing $C$ ($C$ is compact hence closed), and so $f$ is locally constant aroudn $x$, hence continuous.
The Cantor set has empty interior. Therefore, if $x\in C$ you have $f(x)=\pi$, but there are points as close to $x$ as you want which are out of $C$. For each such point $y$, $f(y)=-\pi$. Therefore, $f$ is discontinuous at $x$.