Let $$f:\mathbb R\rightarrow \mathbb R$$ be a continuous function . $f$ maps open sets to open sets then prove that $f$ must be monotonic.
Now if I take $(0,1)\cup (3,4)$ such that and $f$ linearly maps $(0,1)$ increasingly to $(5,6)$ and $(3,4)$ decreasingly onto $(6,5)$(I did not know how else to write this , hope you people understand) i.e. $$f((0,1)\cup (3,4))=(5,6)$$ then $f$ satisfies the given criteria without being monotone . So ,I guess it should have been "intervals" instead of "sets". In that case , I need some hints as to how to begin the proof.
Best Answer
The statement you need to prove is correct: it applies to open sets in general, not just intervals. Here's a hint on how to proceed:
Consider the case where $f$ is not monotonic. That means it must have a local maximum/minimum at a point, say, $x_0$. Consider a small open interval that contains $x_0$. What can you say about its image under $f$?