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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3Since $(0,1)$ 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 http://mathforum.org/library/drmath/view/62395.html – lhf Oct 25 '11 at 00:55

3this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{1}$ will do. – Oct 25 '11 at 01:13

7One 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 01:16

3@PZZ for instance the map wrapping [0,1) around the unit circle. – JSchlather Oct 25 '11 at 01:37

2@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 http://math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) – Pete L. Clark Oct 25 '11 at 03:18

@Pete, oh that is embarrassing. Since your reply is actually helpful, I will leave the comment where it is. – Oct 25 '11 at 03:25
3 Answers
Explicit examples are moderately difficult to construct, but it’s not too hard to come up with nonconstructive examples; here’s one such.
For $x,y\in\mathbb{R}$ define $x\sim y$ iff $xy\in \mathbb{Q}$; 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 nonempty 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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1Just curious: Is the axiom of choice used anywhere in your proof? – YoTengoUnLCD Jan 17 '17 at 08:14

6I think I'm going to start calling $\sim$ the "Vitali equivalence relation"... $x$ and $y$ are Vitali equivalent iff $xy \in \mathbb{Q}$, etc. Honestly, this thing is useful enough to deserve a name. – goblin GONE Mar 02 '17 at 13:54


1@DavidFeng: No. All $x$ from the same equivalence class give the same value. For example, $f(\frac12)=f(\frac13)$ since $\frac12\frac13\in\mathbb Q$ – celtschk Mar 02 '19 at 21:55

Can someone explain how's h defined. It is a bijection from where to where? – Mukil Feb 23 '21 at 11:59

@Mukil: From $\Bbb R/\sim$ to $\Bbb R$, exactly as it says. These two sets have the same cardinality, so there is some bijection between them. We don't have to know exactly what it looks like (which is a good thing, since we can’t: we’re definitely using the axiom of choice here). – Brian M. Scott Feb 24 '21 at 00:23



Not related to maths, but to math.SE. It's wonderful that 10 years after posting his answer, Brian is still available here to discuss it. – Giuseppe Negro May 28 '21 at 10:03

@BrianM.Scott Can you please give some hints why is this function f discontinuous? – Sanajit Patra Apr 30 '22 at 01:31

1@SanajitPatra: For any $x\in(0,1)$, $[x]$ is a dense subset of $\Bbb R$ (since it’s just a translate of $\Bbb Q$), so $[x]\cap(0,1)$ is a dense subset of $(0,1)$. Fix $x\in(0,1)\setminus\Bbb Q$, and let $y=h(x)$; $x$ is irrational, so $x\not\sim\frac12$. Now show that there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $[x]$ converging to $\frac12$, and observe that $h(x_n)=y$ for each $n\in\Bbb N$, but $h\left(\frac12\right)\ne y$, so $h$ is discontinuous at $\frac12$. The argument can be generalized to show that $h$ is nowhere continuous. – Brian M. Scott Apr 30 '22 at 20:33
There is in fact a rather easy example of a function $\mathbb R \to \mathbb R$ such that the image of every open set is $\mathbb R$: Let $(x_i)_{i\in\mathbb Z_+}$ be the binary expansion of $x$, so that each $x_i \in \{0,1\}$. Let then $$f(x) = \sum_{k=1}^\infty\frac{(1)^{x_k}}k\quad \textrm{if the series converges}$$ $$f(x) = 0\quad \textrm{otherwise.}$$ Since the harmonic series (or a tail of it) can be made to converge to any real number by changing signs in the appropriate way, this function has $f((a,b)) = \mathbb R$ for any real $a < b$. Hence this function is open, though clearly not continuous at any point.
The harmonic series can be substituted with any other absolutely unbounded series where the summand goes to zero.
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Let me conceptualize around Brian's answer a bit.
Definition 0. If $X$ and $Y$ are topological spaces, a function $f:X→Y$ is said to be strongly Darboux iff for all nonempty open sets $A⊆X$, we have $f(A)=Y$.
Here's the basic facts:
Proposition.
 Every strongly Darboux function is an open function.
 If $X$ is nonempty, every Darboux function $X \rightarrow Y$ is surjective.
 If $X$ is nonempty and $f : X \rightarrow Y$ is a continuous Darboux mapping, then $Y$ carries the indiscrete topology.
Proofs.
Trivial.
Since $X$ is open and nonempty, hence $f(X)=Y.$ That is, $f$ is surjective.
Let $B \subseteq Y$ denote a nonempty open set. Our goal is to show that $B=Y$. Since $f$ is surjective, $f^{1}(B)$ is nonempty. Since $f$ is continuous, $f^{1}(B)$ is open. Hence $f(f^{1}(B))=Y$. But since $f$ is surjecive, hence $f(f^{1}(B))=B.$ So $B=Y$.
Putting these together, we see that every strongly Darboux function $f:\mathbb{R} \rightarrow \mathbb{R}$ is a discontinuous open mapping.
$f$ is an open mapping by (1).
$f$ is discontinuous by (3), because it's domain is nonempty and it's codomain doesn't carry the indiscrete topology.
And, of course, Brian's answer guarantees the existence of a strongly Darboux function $\mathbb{R} \rightarrow \mathbb{R}$. This completes the proof.
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