$A$ 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".



Is it possible to separate two linked (geometric) circles in $\Bbb R^3$ by a set homeomorphic to the 2-sphere (with arbitrarily “bad” homeomorphism)?
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.


