I am looking for some examples of open functions from $\mathbb{R}$ (or $\mathbb{R}^n$) to $\mathbb{R}$ that are not continuous. I know that the classic example for an open, discontinuous function is mapping the points on the unit circle to the interval $[0, 2\pi)$, but the domain is not reals. Maybe there is a way to compose this function with another one to get a function that is from $\mathbb{R}$ to $\mathbb{R}$.
For clarity, a function $f:\mathbb{R}^k \rightarrow \mathbb{R}^n$ is open if for each $A \subset \mathbb{R}^k$ with $A$ open, $f(A)$ is open. In other words, the image of every open subset is open.
I know that if the preimage of every open set is open, then the function is continuous. So I thought a discontinuous function whose inverse is continuous might work. However, I couldn't find such a function.
Best Answer
Define an equivalence relation $\sim$ on $\Bbb R$ by $x\sim y$ iff $x-y\in\Bbb Q$. For each $x\in\Bbb R$ the $\sim$-class of $x$ is $[x]=x+\Bbb Q=\{x+q:q\in\Bbb Q\}$, which is clearly dense in $\Bbb R$. Let $\mathscr{C}=\{[x]:x\in\Bbb R\}$, the partition of $\Bbb R$ into $\sim$-classes; each member of $\mathscr{C}$ is countable, so $|\mathscr{C}|=|\Bbb R|$, and there is a bijection $\varphi:\mathscr{C}\to\Bbb R$. Define
$$f:\Bbb R\to\Bbb R:x\mapsto\varphi([x])\,.$$
Let $U$ be any non-empty open set in $\Bbb R$, and let $y\in\Bbb R$; there is a $C\in\mathscr{C}$ such that $\varphi(C)=y$. $C$ is dense in $\Bbb R$, so there is an $x\in U\cap C$, and $$f(x)=\varphi([x])=\varphi(C)=y\,,$$ so $f[U]=\Bbb R$. That is, $f[U]=\Bbb R$ for each non-empty open $U\subseteq\Bbb R$, so $f$ is open.
Clearly, however, $f$ is not continuous at any point: for each $x\in\Bbb R$ and $C\in\mathscr{C}$ there is a sequence in $C$ converging to $x$, so that continuity of $f$ at $x$ would require that $f(x)=\varphi(C)$ for each $C\in\mathscr{C}$, i.e., that $f(x)=y$ for each $y\in\Bbb R$.