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Let $A \subset \mathbb{R}^3$ denote Antoine's necklace. It is well-known that $A$ is a Cantor space and that $\mathbb{R}^3 \backslash A$ is not simply connected. Futhermore, $\pi_1(\mathbb{R}^3 \backslash A)$ is not even finitely-generated.

But what is really known this group? Is countable? Is it isomorphic to some classical group such that $\mathbb{R}$ or $\mathbb{Q}$? Is it torsion-free or abelian?

Seirios
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    It is not one of the familiar groups: It contains abelian subgroups of rank 2, but also free subgroups; it is countable, as fundamental group of any manifold. – Moishe Kohan Oct 18 '14 at 20:18

1 Answers1

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In his book Knots and Links, Rolfsen shows that, writting Antoine's necklace $A$ as an intersection of chains of tori $\bigcap\limits_{n \geq 0} C_n$ as usual, the inclusions $\mathbb{R}^3 \backslash C_n \hookrightarrow \mathbb{R}^3 \backslash C_{n+1}$ are $\pi_1$-injective, so that the fundamental group

$$ \pi_1( \mathbb{R}^3 \backslash A) = \bigcup\limits_{n \geq 0} \pi_1( \mathbb{R}^3 \backslash C_n)$$

is a limit of knot groups. In particular, we deduce that the group is countable, non-abelian and torsion-free (thanks to studiosus).

Moreover, it is an exercice in Rolfsen's book to show that $H_1(\mathbb{R}^3 \backslash A)=0$, so $\pi_1(\mathbb{R}^3 \backslash A)$ is a perfect group (ie. equal to its own commutator subgroup).

Seirios
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