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Suppose that $f: \mathbb R^n \to \mathbb R^n$ is a bijection and $n\geq2$. Can $f$ send every open set onto non-open set?

I do not know what exactly to write about this problem. Did I try anything? No, I do not have a good idea. Why am I asking this? Because here is this very highly upvoted question of Willie Wong that I think of, and, it seems to me as a good start to investigate what exactly can bijections "do" and what they can´t, so, as a starting point, I decided to ask this question.

Edit: Thomas wrote a useful comment that $\emptyset$ and $\mathbb R^n$ are mapped onto open sets. So to exclude trivialities, suppose that from consideration we exclude the empty set and the whole space $\mathbb R^n$, to make this more interesting.

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  • $\begingroup$ @ThomasAndrews I made an edit, to make this a little more interesting. But even now it could be trivial. $\endgroup$ – Grešnik Jun 18 at 16:20
  • $\begingroup$ Consider the map which maps tuples of irrational elements $(i_1, \ldots, i_n)$ to $(i_1 + \sqrt{2}, \ldots, i_n + \sqrt{2})$, tuples of rational numbers $(q_1, \ldots, q_n)$ to $(q_1 + 1, \ldots, q_n + 1)$ and the other 'mixed' elements to themselves. It already does a good job at not mapping opens to opens, although there exist opens which do get mapped to opens. $\endgroup$ – Alexander Geldhof Jun 18 at 16:21
  • $\begingroup$ @AlexanderGeldhof You can find any open set $U$ such that if $x\in U$ then $x+(\alpha,\alpha,\dots,\alpha)\in U$ for any real $\alpha.$ Then your $f$ would send $U$ to $U,$ so it doesn't work. $\endgroup$ – Thomas Andrews Jun 18 at 16:24
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    $\begingroup$ The set $\mathbb R^n\setminus\{\mathbf v\}$ is always open, and always goes to the open set, $\mathbb R^n\setminus\{f(\mathbf v)\}.$ @AnteP. $\endgroup$ – Thomas Andrews Jun 18 at 16:27
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    $\begingroup$ There exist bijections which map no open ball of finite radius to an open set. $\endgroup$ – Alexander Geldhof Jun 18 at 16:31
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The set $\mathbb R^n\setminus\{\mathbf v\}$ is always open, and always goes to the open set, $\mathbb R^n\setminus\{f(\mathbf v)\}.$

More generally, if $F\subseteq \mathbb R^n$ is finite then $\mathbb R^n\setminus F$ is open, and its image $\mathbb R^n\setminus f(F)$ is open.

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  • $\begingroup$ Can we generalize that to: If $S$ is closed and bounded and such that $\mathbb R^n \setminus S$ is open then the image of $\mathbb R^n \setminus S$ is open? $\endgroup$ – Grešnik Jun 18 at 16:32
  • $\begingroup$ @AnteP. No, you can't, unless $f$ is open. $\endgroup$ – user10354138 Jun 18 at 16:36
  • $\begingroup$ Actually, for any closed set $S$, bounded or otherwise, $\mathbb R^n\setminus S$ is open. I n general topology, that's basically the definition of closed. But it is not true that $f(S)$ will be closed. You can rephrase your question to be for a bijection $f$ such that for each closed $S$ (other than $\emptyset,\mathbb R^n,$) $f(S)$ is not closed. $\endgroup$ – Thomas Andrews Jun 18 at 16:36
  • $\begingroup$ Is there any book where there is material only about bijections? $\endgroup$ – Grešnik Jun 18 at 16:42
  • $\begingroup$ I know of no books solely or even predominantly about bijections. Sorry. @AnteP. $\endgroup$ – Thomas Andrews Jun 18 at 18:15
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Reading over the OP's question and their comments, it becomes evident that they are interested in 'scramble-it-up' bijections and @AlexanderGeldhof's comment. Here we want to offer some food for thought to the OP's curiosity; we'll only be examining bijections $f: \Bbb R \to \Bbb R$.

The first thing to note, is that if $\tau: \Bbb R \to G$ is a bijection onto a set $G$, then every bijection $g: G \to G$ can be mapped to a bijection

$${\tau}^{-1} \circ g \circ \tau: \Bbb R \to \Bbb R$$

and every bijection mapping $\Bbb R$ to $\Bbb R$ has this form.

A second point of interest is that by using the axiom of choice, the existence of bijections can be postulated yet you can't specify an algorithim to 'pin things down'; see

Hamel Basis

You can 'accept' the existence of bijective linear transformations of $\Bbb R$ over $\Bbb Q$ by matching up any two different Hamel Bases, but don't 'strain your brain' trying to 'see them' (c.f. this).

Finally, it seems only fair to define a transformation on $\Bbb R$ that 'rips apart' bounded open intervals.

Define $f\colon\mathbb{R}\to\mathbb{R}$ as follows:

$f(x) = \left\{\begin{array}{lr} \;\;\;x+1\, \;\;\;\text{ |} & \text{when } x \text{ is a rational number}\\ \;\;\;x-1\, \;\;\;\text{ |} & \text{when } x \text{ is an irrational number} \end{array}\right\}$

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