continuous open mappings are monotonic This is a problem in baby Rudin. Chapter 4, problem 15. I'm just looking at examples at the moment. It says that any continuous open map is monotone. Recall, an open map $f: X \to Y$ is such that any open set $U$ in $X$ will haves its image, $f(U)$ open in $Y$.
So,
Let $$ f: \mathbb{R} \to [-1, 1] $$ $$f: x \mapsto \sin(x) $$ and consider the open interval $$ (\frac{\pi}{3}, \frac{7\pi}{12}) $$ The image of this interval in $[-1, 1]$ is $$ (\frac{\sqrt{3}}{2}, 1] $$ which is open in $[-1, 1]$ (its complement is $[-1, \frac{\sqrt3}{2}]$, certainly closed). The problem is, the function is not monotonic on this interval. It is continuous, so it must not be an open map. I'm pretty sure I have the image right, so, where am I making a mistake?
I really hope I'm not making a dumb error!
 A: The mistake is misinterpreting for which codomain the statement holds.  Continuous open maps from $\mathbb{R}$ to $\mathbb{R}$ are monotone.  You have changed the codomain to the image with subspace topology, and this changes the meaning of open map.  Since the interval $\left(\frac{\sqrt{3}}{2},1\right]$ is not open in $\mathbb{R}$, your example demonstrates the fact that this non-monotone continuous map cannot be an open map.
Athough in this case you have a map that is not open, note that if you are trying to show that a map is open, it is not enough to give an example of an open set that is mapped to an open set.  You would have to prove that every open set is mapped to an open set.  On the other hand, to show that a map is not open, you only need to demonstrate that there exists an open set that is not mapped to an open set, and this fact will come in handy if you approach this problem using contraposition.
A side remark: You will at some point have to invoke completeness of $\mathbb{R}$ (or a theorem that uses completeness).  The function $f(x)=x^3-6x$ as a map from $\mathbb{Q}$ to $\mathbb{Q}$ is continuous and open but not monotone.
