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?

Martin Sleziak
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math student
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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 http://mathforum.org/library/drmath/view/62395.html – lhf Oct 25 '11 at 00:55
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    this 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
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    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 01:16
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    @PZZ for instance the map wrapping [0,1) around the unit circle. – JSchlather Oct 25 '11 at 01:37
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    @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 Answers3


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!)

Brian M. Scott
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    Just curious: Is the axiom of choice used anywhere in your proof? – YoTengoUnLCD Jan 17 '17 at 08:14
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    I think I'm going to start calling $\sim$ the "Vitali equivalence relation"... $x$ and $y$ are Vitali equivalent iff $x-y \in \mathbb{Q}$, etc. Honestly, this thing is useful enough to deserve a name. – goblin GONE Mar 02 '17 at 13:54
  • is $f$ injective? – David Feng Feb 16 '19 at 21:37
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    @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
  • Thank you I understood it. – Mukil Feb 24 '21 at 02:54
  • @Mukil: You’re welcome. – Brian M. Scott Feb 24 '21 at 02:56
  • 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
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    @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.

Aksel Bergfeldt
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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 non-empty open sets $A⊆X$, we have $f(A)=Y$.

Here's the basic facts:


  1. Every strongly Darboux function is an open function.
  2. If $X$ is non-empty, every Darboux function $X \rightarrow Y$ is surjective.
  3. If $X$ is non-empty and $f : X \rightarrow Y$ is a continuous Darboux mapping, then $Y$ carries the indiscrete topology.


  1. Trivial.

  2. Since $X$ is open and non-empty, hence $f(X)=Y.$ That is, $f$ is surjective.

  3. Let $B \subseteq Y$ denote a non-empty open set. Our goal is to show that $B=Y$. Since $f$ is surjective, $f^{-1}(B)$ is non-empty. 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 non-empty 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.

goblin GONE
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