Question:
Let $f: \mathbb R \to \mathbb R$ be continuous and open, that is if $A \subset \mathbb R$ is open then $f(A) \subset \mathbb R$ is open. Prove that $f$ is injective.
Attempt:
Suppose $f$ is not injective then there exit $x, y \in \mathbb R$ such that $$x < y \implies f(x) = f(y) = c$$
Take the closed inteval $[x,y] \subset \mathbb R$, as $f$ is continuous then $\displaystyle {f|_{[x,y]}}$ is continuous, by Weierstrass Theorem we have that $f$ has a maximum or a minimum point.
Let's assume $m = \max \{f(a) ; a \in [x,y]\}$. Now there is $x' \in [x,y]$ such that $f(x') = m$. If we take the open $(x'-\delta, x'+\delta)$ centered at $x'$ and we have
(1) If $m = c$ then $f$ is constant on the interval $[x,y]$ and $f((x'-\delta, x'+\delta)) = \{c\}$ which is closed, thus a contradiction;
(2) If $m \neq c$ then we would have $f((x'-\delta, x'+\delta)) = (b, m]$, where $b$ can also be $b = \infty$. Again a contradiction.
Well, this is my least embarassing attempt. I'm not sure how to show $(2)$ $100 \%$.
I have also tried to show $f^{-1}f(A) = A$ for any $A \subset \mathbb R$, tried to work on the connected space $\mathbb R$ by finding a contradiction using the intervals $E_{[f > c]}$ and $E_{[f < c]}$ open when $A$ is open.
Any thoughts?
Note: I've already seen this to try something out, but the fact that $f$ is monotone on this exercise comes as a consequence.
Best Answer
You could use that $f((x,y))$ is open, so that neither the maximum nor the minimum are attained in the interior $(x,y)$ (why exactly?).
Use this to derive a contradiction to $f(x)=f(y)$.
This uses your main idea, but makes the case distinction superfluous.