I already have an Idea, but im not completely sure if that works. Probably you could tell me, if I am right?
Suppose $f$ is not constant, because $\mathbb{R}$ and $\emptyset$ are both open.
Let $M := \lbrace x : f(x) \neq 0 \rbrace$ and let x $\in M$. WLOG $f(x) > 0$. Define $\varepsilon := f(x) – 0 = f(x)$. Because $f$ is continuous the existence of an $\delta > 0$ such that $\vert x – y \vert < \delta \Rightarrow \vert f(x) – f(y) \vert < \varepsilon = f(x)$. Now $B_\delta(x) \subseteq M$ because $\vert f(x) \vert > \vert f(y) \vert$, especially $f(y) \neq 0$.
Is that right? Tank you already for your help.
Best Answer
Other than the fact that $f(x)-0=f(x)$ rather than $0$, it is correct.
Alternatively, you can say that that set is open since it is equal to $f^{-1}(\mathbb{R}\setminus\{0\})$, $\mathbb{R}\setminus\{0\}$ is open and $f$ is continuous.