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This is another special case of this question.

Recall that we call a subset $Z$ of $\mathbb{S}^n$ ambiently-reversible, if there is an orientation-reversing homeomorphism $h: \mathbb{S}^n \to \mathbb{S}^n$ fixing $Z$ pointwise.

Question 1: Is there an ambiently-reversible wild Cantor set in $\mathbb{S}^n, n\geq 3$? Is there a non-ambiently-reversible one?

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  • $\begingroup$ To produce a reversible one is quite trivial. Just take $K$ wild Cantor with $K\cap -K$ empty, and consider $K\cup -K$. $\endgroup$
    – YCor
    Commented Aug 22, 2022 at 21:19
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    $\begingroup$ @YCor: we want to fix $Z$ pointwise, so I don't see how your example is reversible. $\endgroup$
    – Agelos
    Commented Aug 23, 2022 at 8:47
  • $\begingroup$ Ah, indeed, I didn't notice "pointwise" $\endgroup$
    – YCor
    Commented Aug 23, 2022 at 8:53
  • $\begingroup$ BTW, one can ask which subsets of $\mathbb{S}^n$ are setwise reversible, but I don't expect the answer to be nice. $\endgroup$
    – Agelos
    Commented Aug 23, 2022 at 8:57
  • $\begingroup$ I think there is an ambiently-reversible wild Canter set. Recall that R. H. Bing was the first to produce exotic involution of S^3 having wild fixed points. In particular, he showed that AH $\cup_{id}$ AH is homeomorphic to S^3, where AH denotes the crumpled 3-cube bounded by the Alexander horned sphere. One may have more examples by requiring that the crumpled cubes satisfy the Disjoint Disk Property. $\endgroup$
    – Shijie Gu
    Commented Aug 28, 2022 at 7:15

1 Answer 1

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Similar to the proof of the exercise problem in Part 1 of the OP Question, one may generalize it to higher dimensions. However, since the direct extension of Jordan-Schoenflies theorem is impossible, the generalization is not obvious. I think the OP is related to the problem of exotic involution of $S^n$. When $Z$ is a Cantor set, $Z$ can be considered as a closed subset of a wildly embedded sphere in $S^3$. the Alexander horned sphere $\Sigma$. $\Sigma$ separates $S^3$ and the closures of the corresponding components in the complement are crumpled 3-cubes. Denote the crumpled 3-cube by $AH$. As I made in the comment, R. H. Bing stunned the community by showing that $AH \cup_{id} AH$ $\approx S^3$, the argument relies on his shrinking criterion. The point is that his result shows that $S^3$ admits an involution fixing a wild 2-sphere, the Alexander horned sphere $\Sigma$. Note that $Z \subset \Sigma$. Then $Z$ is ambiently-reversible.

This might be a rudimentary example to the OP. Because it seems that the OP requires that the Cantor set $Z$ is wild in $S^3$. The one in the Alexander horned sphere is not wild in $S^3$. Indeed, $Z$ is twice flat in the sense that it is flat both as a subset of the Alexander horned sphere and as a subset of $S^3$. However, there are results may handle Cantor sets which are wildly embedded in the ambient space. For instance, for any crumpled $n$-cube $C$, $n \geq 5$, $C ∪_{Id} C \approx S^n$ iff $C$ satisfies the Disjoint Disks Property. See Thm 7.10.6, P. 413, Daverman-Venema.

Using the involution of spheres alone might not answer whether there exists a non-ambiently-reversible Cantor set, a potential example is produced by Daverman. Utilizing a sticky Cantor set in $S^n$ ($n\geq 4$) constructed by Krushkal, he showed that the sewings of some crumpled cubes $C \cup_{Id} C$ is not homeomorphic to sphere. Furthermore, there is no homeo $H: Bd C \to Bd C$ close to $Id$ such that $C \cup_{H} C \approx S^n$.

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