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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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    $\begingroup$ 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 $\endgroup$ – lhf Oct 25 '11 at 0:55
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    $\begingroup$ this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. $\endgroup$ – user12014 Oct 25 '11 at 1:13
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    $\begingroup$ 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. $\endgroup$ – user83827 Oct 25 '11 at 1:16
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    $\begingroup$ @PZZ for instance the map wrapping [0,1) around the unit circle. $\endgroup$ – JSchlather Oct 25 '11 at 1:37
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    $\begingroup$ @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!) $\endgroup$ – Pete L. Clark Oct 25 '11 at 3:18
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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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  • $\begingroup$ Just curious: Is the axiom of choice used anywhere in your proof? $\endgroup$ – YoTengoUnLCD Jan 17 '17 at 8:14
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    $\begingroup$ 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. $\endgroup$ – goblin Mar 2 '17 at 13:54
  • $\begingroup$ is $f$ injective? $\endgroup$ – David Feng Feb 16 at 21:37
  • $\begingroup$ @DavidFeng: No. All $x$ from the same equivalence class give the same value. For example, $f(\frac12)=f(\frac13)$ since $\frac12-\frac13\in\mathbb Q$ $\endgroup$ – celtschk Mar 2 at 21:55
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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:

Proposition.

  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.

Proofs.

  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.

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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 decimal 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 unbounded series where the summand goes to zero.

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