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$A$ and $B$ are two linked (geometric) circles in $\Bbb{R}^3$. (Let, for definiteness, both have radius = 1, the first lies in the $z=0$ plane and its center is the origin of coordinates $(0,0,0)$, the second lies in the $y=0$ plane and its center is the point $(1,0,0)$.)

Is it true that the circles $A$ and $B$ cannot be separated by a set that is homeomorphic to the $2$-sphere? (The homeomorphism can be arbitrarily “bad”, as in the case of the horned Alexander sphere.)

(A set $S$ separates $A$ and $B$ iff $A$ and $B$ are subsets of different сonnected components of $\mathbb{R}^3\setminus S$.)

I know how to solve this problem in a "smooth" case using knot theory. But this solution doesn't work when the embedding of the $2$-sphere is arbitrarily "bad".

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  • $\begingroup$ I think a useful lemma should be: any simple closed curve in $\mathbb R^3$ is isotopic to a plane closed simple curve. $\endgroup$
    – Pietro Majer
    Commented Jul 26, 2022 at 8:24

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By Alexander duality, $\mathbb{S}^3\setminus S$ has two connected components with trivial homologies. On the other hand, $A$ is nontrivial in $H_1(\mathbb{S}^3\setminus B)$ — a contradiction.

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  • $\begingroup$ @MikhailPatrakeev $\tilde S$ is compact, so it lies in a ball, say $B$. The complement $\mathbb{R}^3\setminus B$ is connected --- so the other connected component have to be in $B$. $\endgroup$
    – Anton Petrunin
    Commented Jul 26, 2022 at 18:52
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    $\begingroup$ @MikhailPatrakeev I simplified the argument considerably. $\endgroup$
    – Anton Petrunin
    Commented Jul 26, 2022 at 19:11

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