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EDIT 4: the impossibility of such an example for $\mathbb{R}$ has been confirmed positive in this paper.

EDIT 3: not a duplicate - the linked question has an answer only for dimensions $n>=3$. Considering the examples given in the comments, additional criteria:

  1. The map must be defined on all of $\mathbb{R^2}$
  2. It must be open and closed as a map to $\mathbb{R^2}$, not $f(\mathbb{R^2})$

EDIT 2: duplicate of this question, voting to close.

EDIT: It appears I've read wrong and the exercise asks for $f$ not necessarily defined on the whole plane. I'm leaving the question open for a possible example on the whole plane.

Finding an example of such a map is a part of an exercise from John Lee's "Introduction to Topological Manifolds", suggesting that it's possible, but I'm failing to do so. We are looking for a map $f: \mathbb{R^2} \rightarrow \mathbb{R^2}$ that is open, closed, but not continuous, with the usual topology on $\mathbb{R^2}$ spanned by open disks $B_r(x) = \{y \in \mathbb{R^2}: ||y-x|| < r\}$. So far it seems to me (argumentation below) that it's impossible in the case of $f: \mathbb{R} \rightarrow \mathbb{R}$, and that this argumentation can be extended to prove that it's also impossible in the plane case.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be open and closed, and let $x_0$ be a point of discontinuity. We distinguish the cases:

  1. (Isolated) $f_{x \rightarrow x_0-} = f_{x \rightarrow x_0+} = y_1, f(x_0) = y_2$. Then $f((a,b) \ni x_0) = (c_1, c_2) \cup\{y_2\} \cup (c_3, c_4)$ - not open.
  2. (Jump) $f_{x \rightarrow x_0-} = y_1, f(x_0) = f_{x \rightarrow x_0+} = y_2$.Then $f((a,b) \ni x_0) = (c_1, y_1) \cup\ [y_2, c_2)$ - not open.
  3. (One-sided infinity) $f_{x \rightarrow x_0-} = +\infty, f(x_0) = f_{x \rightarrow x_0+} = y$. Then for $(a,b) \ni x_0$ small enough so that $f(x) > y, \forall x \in (a,x_0)$ we have $f((a,b)) = (c_1, +\infty) \cup [y, c_2)$ - not open.
  4. (Double-sided infinity) $f_{x \rightarrow x_0-} = +\infty, f(x_0) = y, f_{x \rightarrow x_0+} = -\infty$. Then for $(a,b) \ni x_0$ small enough so that $f(x) > y, \forall x \in (a,x_0)$ and $f(x) < y, \forall x \in (x_0,b)$ we have $f((a,b)) = (c_1, +\infty) \cup \{y\} \cup (-\infty, c_2)$ - not open.

Other cases are essentially the same.(jump from the other side, infinity with the same sign, etc.). Am I missing some cases? Or is this correct, but there are significantly different cases for $\mathbb{R^2}$?

Nikita Dezhic
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    For the jump discontinuities, you have neglected the case $f(x_0^-) < f(x_0) < f(x_0^+)$ (or symmetric). But the important question is whether the map should be open/closed as a map to $\mathbb{R}^2$ or as a map to $f(\mathbb{R}^2)$. – Dermot Craddock Jun 22 '25 at 13:51
  • @Dermot Craddock It's worded "a map between subsets of the plane", so it seems like the latter. I personally would be satisfied with either one. – Nikita Dezhic Jun 22 '25 at 13:58
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    For the latter, an easy one-dimensional example is mapping $\mathbb{R}$ to $(-\infty,-1) \cup {0} \cup (1,+\infty)$ in the "obvious" way. From that, higher-dimensional examples can be obtained in different ways. – Dermot Craddock Jun 22 '25 at 14:02
  • @BrunoAndrades It's not closed in $\mathbb{R}$, but it's closed in $f(\mathbb{R})$. – Dermot Craddock Jun 22 '25 at 14:16
  • @BrunoAndrades That's why in the first comment I asked whether the map should be open/closed as a map to $\mathbb{R}^2$ or as a map to $f(\mathbb{R}^2)$. – Dermot Craddock Jun 22 '25 at 14:40
  • @Dermot Craddock Thanks, my question is answered by your example and remark about where the map should be considered opened/closed, I missed that. If you wish, you can post this as an answer and I will accept it. – Nikita Dezhic Jun 22 '25 at 15:11
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    Why should there exist left and right limit on any point? – Christophe Boilley Jun 22 '25 at 15:43
  • @Christophe Boilley not any, at least one. – Nikita Dezhic Jun 22 '25 at 16:05
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    There exist functions without left or right limit on any point, e.g. characteristic function of $\mathbb Q$, but it can be weirder. – Christophe Boilley Jun 22 '25 at 16:07
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    As far as I understood, the linked question does not provide an answer to the question here, but just one for higher dimensions. Therefore, I vote for reopen this one. – Ulli Jun 23 '25 at 21:17
  • @Ulli You are correct, I have missed that. – Nikita Dezhic Jun 24 '25 at 09:15
  • @Christophe Boilley You're right. Without a proof or even a conviction that it's true, I assumed that a nowhere continuos function would definitely not be either open or closed, but it would be even cooler if there is such an example. – Nikita Dezhic Jun 24 '25 at 15:27
  • Some (everywhere discontinuous) functions are surjective onto $\mathbb R$ on any non-degenerate interval so they are open. But the clopen property is harder to get. – Christophe Boilley Jun 24 '25 at 16:17
  • @Ulli No, only about the case of the plane, I'll clarify this in the question. – Nikita Dezhic Jun 24 '25 at 17:42
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    About EDIT 4: this paper proves that the answer is negative, i.e., every open, closed map $\mathbb R \rightarrow \mathbb R$ is continuous! – Ulli Jun 25 '25 at 07:41
  • @Ulli Poor wording on my part: I intended to say that my hypothesis about the impossibility of such example has been confirmed positive. Changed the wording of the edit without adding a new one to avoid (even more) clutter. – Nikita Dezhic Jun 27 '25 at 09:25

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