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It is well-known that bijection between $\textbf{R}^p$ and $\textbf{R}$ exist (e.g. here, though many other examples exist).

The problem with all these examples of bijections is that typically the mapping does not have any structure (obviously, it cannot be continuous); and most examples do not even make an attempt at providing an $f$ that has some structure.

My question is: What bijections $f:\textbf{R}^p \to \textbf{R}^q$, where you are allowed to specify natural numbers $p\neq q$, do you know that are "nice"?
"Nice" can mean that some basic geometric property is preserved, it can mean that $f$ is interpretable in some way, or it can mean that it is in some other way appealing.

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    $\begingroup$ A natural assumption is to require $f$ and $f^{-1}$ to be Borel and this is indeed possible. The classical one would be the "split digits" trick. There's a minor issue that it's not exactly bijective but the correction is only on a countable set. $\endgroup$
    – YCor
    Commented Sep 18, 2023 at 15:04
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    $\begingroup$ @YCor Care to make a short write-up of this and make a start on what will (hopefully) become a community wiki? :) $\endgroup$
    – user7088941
    Commented Sep 18, 2023 at 15:06
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    $\begingroup$ You can make $f$ $\mathbb Q$-linear. $\endgroup$
    – Emil Jeřábek
    Commented Sep 18, 2023 at 15:09
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    $\begingroup$ A bijection between $\mathbb R$ and $\mathbb R^2$ can be (amost) constructed by interlacing the binary expansions. (Mentioned by @YCor ) But to get an actual bijection you have to make picky adjustments to account for reals with two different binary expansions. Altermatively, you could follow logicians and use the term "reals" for ${\mathbb N}^{\mathbb N}$ and then get easy bijections. See also mathoverflow.net/q/65002/454 $\endgroup$
    – Gerald Edgar
    Commented Sep 18, 2023 at 15:13
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    $\begingroup$ However, I don't think $f$ can be made both $\mathbf{Q}$-linear and Borel. $\endgroup$
    – YCor
    Commented Sep 18, 2023 at 15:26

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