# Complement of a simply connected set is simply connected

I saw the following surprising statement in Wikipedia:

When $D\subseteq\Bbb C$ is a simply connected compact set, then its complement $E=D^c$ is a simply connected domain in the Riemann sphere that contains $\infty$, ...

This property sounds like something specific to $S^2$. Is it true that the complement of a compact simply connected subset of $S^n$ is simply connected? Is this true if $D$ is not necessarily compact/closed? How do you prove this?

A related question (which may need to be moved elsewhere): is there a subset $D\subseteq S^2$ such that there are no nontrivial paths in either $D$ or $D^c$?

• what do you mean by $D$ not necessarily closed? in a Hausdorff space (like $S^n$) compact sets are closed Jun 19 '16 at 23:10
• @Riccardo I mean not necessarily compact or closed, just simply connected. Jun 19 '16 at 23:10
• re the related question, I think the answer is yes. There are continuum many paths in $S^2$, say $\{P_\alpha:\alpha<\frak c\}$ so by transfinite recursion one may pick two points $a_\alpha,b_\alpha$ in $P_\alpha$ with $\{a_\alpha,b_\alpha\}\cap\{a_\beta,b_\beta\}=\emptyset$ for all $\beta<\alpha$, and let $D=\{a_\alpha:\alpha<\frak c\}$. Dec 21 '17 at 2:29

The complement of the obvious $S^2$ in $S^n$ is homotopy equivalent to $S^{n - 3}$, so no (take $n=3,4$). You can do arbitrarily badly; many many groups appear as the fundamental group of knotted 2-spheres in $S^4$, and the same for n-spheres in $S^{n+2}$ etc.
Indeed, fix $n \geq 3$; Kervaire proved that a finitely presented group $G$ is the fundamental group of the complement of some smoothly embedded $S^n \hookrightarrow S^{n+2}$ iff the abelianization of $G$ is $\Bbb Z$, $H_2(G;\Bbb Z) = 0$, and $G$ is normally generated by some element $m$. These are not too violent to ask.