# open maps which are not continuous

What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?

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Since $(0,1)$ and $\mathbb R$ are homeomorphic via a linear map composed with $\arctan$, it suffices to find a map $\mathbb R \to \mathbb R$ that is open but not continuous. Googling that gives you mathforum.org/library/drmath/view/62395.html –  lhf Oct 25 '11 at 0:55
this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. –  user12014 Oct 25 '11 at 1:13
One can build such a function from a Cantor set $C$ (the usual "middle thirds" set will do). Send each point in $C$ to $0$, and map each connected component of the complement of $C$ homeomorphically to the interval $(-1,1)$. Then the image of any open set intersecting $C$ will be $(-1,1)$ (thus open), and the image of any open set not meeting $C$ will also be open, since it's a union of homeomorphic images of open sets. Of course, each point of $C$ will be a discontinuity. –  user83827 Oct 25 '11 at 1:16
@PZZ for instance the map wrapping [0,1) around the unit circle. –  JSchlather Oct 25 '11 at 1:37
@PZZ: In fact there are no counterexamples of the type you're suggesting: if $I$ and $J$ are intervals in $\mathbb{R}$ and $f: I \rightarrow J$ is a continuous bijection, then $f^{-1}$ is necessarily continuous. By coincidence this is exactly where I am in my Spivak calculus course, so see e.g. Theorem 37 in $\S 6.4$ of math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) –  Pete L. Clark Oct 25 '11 at 3:18

## 1 Answer

Explicit examples are moderately difficult to construct, but it’s not too hard to come up with non-constructive examples; here’s one such.

For $x,y\in\mathbb{R}$ define $x\sim y$ iff $x-y\in \mathbb{Q}$; it’s easy to check that $\sim$ is an equivalence relation on $\mathbb{R}$. For any $x\in\mathbb{R}$, $[x] = \{x+q:q\in\mathbb{Q}\}$, where $[x]$ is the $\sim$-equivalence class of $x$. In particular, each equivalence class is countable. For any infinite cardinal $\kappa$, the union of $\kappa$ pairwise disjoint countably infinite sets has cardinality $\kappa$, so there must be exactly as many equivalence classes as there are real numbers. Let $h$ be a bijection from $\mathbb{R}/\sim$, the set of equivalence classes, to $\mathbb{R}$. Finally, define $$f:(0,1)\to\mathbb{R}:x\mapsto h([x])\;.$$

I claim that if $V$ is any non-empty open subset of $(0,1)$, $f[V]=\mathbb{R}$, which of course ensures that $f$ is open. To see this, just observe that every open interval in $(0,1)$ intersects every equivalence class. (It should be no trouble at all to see that $f$ is wildly discontinuous!)

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