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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 nontivial in $H_1(\mathbb{S}^3\setminus B)$ --- a contradiction.

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.

By Alexander duality, $\mathbb{R}^3\setminus S$ has two connected components.

Let $u\colon \mathbb{R}^3\to \mathbb{S}^3\setminus (A\cup B)$ be the universal covering. Denote by $\tilde S$ a lift of $S$.

Again, by Alexander duality $\mathbb{R}^3\setminus \tilde S$ has two connected components. One of them, say $\Omega$ is bounded.

Observe that $u(\Omega)$ is one of the components of $\mathbb{R^3}\setminus S$ and it does not contain $A$ nor $B$. It follows that both $A$ and $B$ lie in the other connected component — a contradiction.

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

By Alexander duality, $\mathbb{R^3}\setminus S$ has two connected components.

Let $u\colon \mathbb{R^3}\to \mathbb{S^3}\setminus (A\cup B)$ be the universal covering. Denote it by $\tilde S$ a lift of $S$.

Again, by Alexander duality $\mathbb{R^3}\setminus \tilde S$ has two connected components, one of them, say $\Omega$ is bounded.

Observe that $u(\Omega)$ is one of the components of $\mathbb{R^3}\setminus S$ and it does not contain $A$ nor $B$. It follows that both $A$ and $B$ lie in the other connected component --- a contradiction.

By Alexander duality, $\mathbb{R}^3\setminus S$ has two connected components.

Let $u\colon \mathbb{R}^3\to \mathbb{S}^3\setminus (A\cup B)$ be the universal covering. Denote by $\tilde S$ a lift of $S$.

Again, by Alexander duality $\mathbb{R}^3\setminus \tilde S$ has two connected components. One of them, say $\Omega$ is bounded.

Observe that $u(\Omega)$ is one of the components of $\mathbb{R^3}\setminus S$ and it does not contain $A$ nor $B$. It follows that both $A$ and $B$ lie in the other connected component — a contradiction.