# f is continuous at uncountably many points of $[0,1]$ and discontinuous at uncountably many points of $[0,1].$

Justify: Let $K\subset [0,1]$ be the cantor set. Then the function $$f:[0,1]\to\mathbb R:x\mapsto \begin{cases}1,~\text{if}~~x\in K\\0,~\text{if}~~x\notin K\end{cases}$$ is continuous at uncountably many points of $[0,1]$ and discontinuous at uncountably many points of $[0,1].$

HINT: Let $U=[0,1]\setminus K$; then $U$ is a dense open subset of $[0,1]$, so every point of $[0,1]$ is a limit of points of $U$. Give this some thought, but if it isn’t enough, I’ve added a bigger hint in the spoiler-protected field below; mouse-over to see it.
Show that $f$ is continuous at points of $U$ and discontinuous at points of $K$.
Since $[0, 1]\setminus K$ contains intervals it is clearly continuous on uncountably many points. Show that it is discountinuous on $K$, which is uncountable, and you are done. For this you should use the fact that $K$ is nowhere dense.