Let $f: \Bbb R\to [0,\infty)$ be bijective, show that $f$ has infinite many discontinuities.
Clearly, there exists one and only one $x_0$ such that $f(x_0)=0$. And $\forall\ x\neq x_0, f(x)>0$. I have no idea. Thinking more than two days.
Argue by contradiction, only finitely many discontinuities, then ...
