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I was looking at the question "Are there any valid continuous Sudoku grids?" and that lead me to this question.

Let $U$ be the unit interval $[0,1]$ or $[0,1)$. (pick which is easier for you)

Does there exist a bijective function $f: U \to U$ such that for every subinterval $I$ of $U$, the set $\{f(x)\ |\ x \in I\}$ is dense in $U$?

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  • $\begingroup$ This is shown in one of the answers of the question you are citing... $\endgroup$ – Mushu Nrek Oct 28 '20 at 19:31
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Choose open subintervals $I_1,I_2,\dots$ of $U$ such that if $J$ is an open subinterval of $U,$ then $J$ contains $I_n$ for some $n.$ (For example we could let $I_1,I_2,\dots$ be the open subintervals of $U$ with rational endpoints.)

We can then choose pairwise disjoint countable sets $D_n\subset I_n$ such that $D_n$ is dense in $I_n$ for each $n.$

And we can choose pairwise disjoint countable sets $E_n\subset U$ such that $E_n$ is dense in $U$ for each $n.$

Now define $f:U\to U$ as follows: For each $n=1,2,\dots$ let $f$ be a bijection of $D_n$ onto $E_n.$ Then $f$ is a bijection of $\cup D_n$ onto $\cup E_n.$ Now $U\setminus \cup D_n$ has the same cardinality as $U\setminus \cup E_n.$ So to finish the definition of $f,$ let $f$ be any bijection of $U\setminus \cup D_n$ onto $U\setminus \cup E_n.$

It follows that $f:U\to U$ is a bijection. Let $I\subset U$ be a subinterval. Then $I_n\subset I$ for some $n.$ Since $f(I_n) = E_n,$ $f$ is the desired function.

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  • $\begingroup$ Perfect! I'm thinking we could use $D_n = (\sqrt{p_n} \cdot \mathbb{Q}) \cap I_n$ and $E_n = (\sqrt{p_n} \cdot \mathbb{Q}) \cap U$, where $p_n$ is the $n$-th prime number. $\endgroup$ – Paul Oct 28 '20 at 22:33

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