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Does there exist a Borel (or even continuous) function $f:\mathcal{C}\to\mathcal{C}$, where $\mathcal{C}$ is the Cantor set (or Cantor space $2^\omega$) such that for every nonempty closed perfect set $P\subseteq\mathcal{C}$, $f|P$ maps surjectively onto $\mathcal{C}$?

Such functions (on $\mathbb{R}$) are called perfectly everywhere surjective here: https://core.ac.uk/download/pdf/83599431.pdf, but the maps constructed there rely on, in essence, a well-ordering of $\mathbb{R}$ and are likely far from any kind of measurability.

Hoping that the self-similarity of the Cantor set could be exploited here.

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    $\begingroup$ There is no continuous example. If $p$ is any element of the Cantor set, then if $f^{-1}(p)$ is uncountable, it contains a closed perfect set which maps only onto $p$, and if $f^{-1}(p)$ is countable, its complement contains a closed perfect set whose image does not contain $p$. $\endgroup$
    – Anonymous
    Commented Jun 28, 2021 at 23:41
  • $\begingroup$ Ah, yes. I suspect then if $f$ is Borel (or measurable), then since it must be continuous on a large set, it cannot have this property either. $\endgroup$
    – Iian Smythe
    Commented Jun 28, 2021 at 23:42

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As suggested by the comment above, the answer is no. The relevant fact is that every nonmeager subset of $\mathcal{C}$ with the Baire property (in particular, any Borel set) contains a nonempty closed perfect set.

Suppose that there was a Borel function $f$ with this property. Observe that for every $p\in\mathcal{C}$, $f^{-1}(\{p\})$ is a comeager Borel set: $f^{-1}(\{p\})$ is a Borel set which meets every nonmeager set, since nonmeager subsets of $\mathcal{C}$ must contain perfect sets, so its complement is meager.

But then, if we take $p\in\mathcal{C}$, $f^{-1}(\{p\})$ contains a perfect set which maps only onto $p$, a contradiction.

This argument works for any function which is Baire measurable, a similar argument can be given for Lebesgue measurable.

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  • $\begingroup$ In other words, to say that $f:\mathcal C\to\mathcal C$ is "perfectly everywhere surjective" simply means that, for every point $p\in\mathcal C$, the inverse image $f^{-1}(p)$ has the Bernstein property: both it and (of course) its complement meet every nonempty perfect subset of $\mathcal C$. So of course $f^{-1}(p)$ can't be a Borel set or even analytic, since it's an uncountable set which contains no nonempty perfect set. How do you define Lebesgue measurability for subsets of $\mathcal C$? $\endgroup$
    – bof
    Commented Jun 30, 2021 at 2:08
  • $\begingroup$ It's the product "coin-flip" measure on $\mathcal{C}$. Alternatively, you can pull back the measure from the unit interval via the decimal expansion map which is a bijection, except on a countable (and thus measure zero) set. $\endgroup$
    – Iian Smythe
    Commented Jun 30, 2021 at 2:44
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I think that recursion theory gives a clearer way to answer the question. If $f$ is a Borel function, then it is a hyperarithmetic reduction relative to a real, say $x$. Then fix any "regular" forcing (random forcing, Cohen forcing etc) which always produce ``powerless" generic reals, we may pick up a perfect set $P$ of such generic reals. Now restricted to $P$, $f$ cannot range over the whole Cantor space. For example, $\mathscr{O}^x$, the hyperjump of $x$, does not belong the range.

The method can be push up to more set theoretical. For example, by almost the same argument, it can be shown that in Solovay model, there is no such function.

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  • $\begingroup$ Do you mean there is no such function in the Solovay model? $\endgroup$
    – Iian Smythe
    Commented Jun 29, 2021 at 15:39
  • $\begingroup$ Ah, sorry for the typos. Yes, you are right. $\endgroup$
    – 喻 良
    Commented Jun 29, 2021 at 23:49

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