For a Cantor set $\mathcal{C} \subset S^3$ such that $\pi_1(S^3 \setminus \mathcal{C})=0$, prove $S^3 \setminus \mathcal{C}$ can be split by a sphere.

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I'm working from the paper Cantor Sets in $S^3$ with Simply Connected Complements by Richard Skora.

On page 184 the second sentence states that any Cantor set $\mathcal{C} \subset S^3$ such that $\pi_1(S^3 \setminus \mathcal{C}) = 0$, prove $S^3 \setminus \mathcal{C}$ can be split by a sphere (i.e. for any distinct $p,q \in S^3 \setminus \mathcal{C}$, there is a pl $2$-sphere $S$ embedded into $S^3 \setminus \mathcal{C}$, such that only one of $p$ or $q$ lies in the interior of $S$, and the other in its exterior).

The paper provides two hints; that we should apply the Hurewicz Isomorphism Theorem and the Sphere Theorem.

By the hypothesis, and the Hurewicz Isomorphism Theorem we can see that $\tilde{H}_i(S^3 \setminus \mathcal{C}) = 0$ for $i \in \{ 0,1 \}$. We also have that $\tilde{H}_2(S^3 \setminus \mathcal{C}) \cong \pi_2 (S^3 \setminus \mathcal{C})$.

If I can prove that $\pi_2 (S^3 \setminus \mathcal{C}) \neq 0$, then use of the Sphere Theorem of Papakyriakopoulos enables us to split the two distinct points $p$ and $q$ by an embedding of a $2$-sphere.

My question then is how we prove that $\tilde{H}_2(S^3 \setminus \mathcal{C}) \neq 0$; I had two separate ideas for this - none of which have come to any fruition:

  1. If we consider $S^3 \setminus \{p_0, p_1, p_2, ..., p_n \}$ then this is homotopy equivalent to a wedge of $n$ $2$-spheres. I'm not therefore sure whether we are able to stretch this to a Cantor set to claim it is homotopy equivalent to an infinite wedge of circles, so has a non trivial $\tilde{H}_2(S^3 \setminus \mathcal{C})$; seeing as, for example, the Cantor set could have positive Lebesgue measure?

  2. To use a Mayer-Vietoris sequence - I'm not sure quite how the sequence would arise; I've tried a few different approaches, but I thought maybe we could take an $n$-cell which contains $\mathcal{C}$, however I'm not sure we can ensure this exists, because I think $\mathcal{C}$ isn't necessarily bound in $S^3$?

Thanks for your time reading this post!

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Actually, you do not need the Chech cohomology, just a little bit of analysis. Namely, start with the characteristic function $\chi: C\to \{0,1\}$ of a proper clopen subset of $C$. Then, by the Tietze-Urysohn extension theorem, the function $\chi$ will extend continuously to a map $S^3\to [0,1]$. Then you can modify the extension to make it smooth outside of $C$: $f: S^3\to {\mathbb R}$. Now, use Sard's theorem to find a regular value $t\in (0,1)$ of $f$. The hypersurface $f^{-1}(t)$ will separate $C$ and, hence, define a nontrivial element of $H_2(S^3\setminus C)$.