TITLE: Complement of a simply connected set is simply connected
QUESTION [8 upvotes]: 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 $D^c$?

REPLY [6 votes]: I consider path-connectedness to be part of "simply connected". As a counterexample to your question when the set is not closed, take the Warsaw circle (a closed up topologist's sine curve), delete a point on the "reasonable" part of the curve, and take the complement of the result. The punctured Warsaw circle is not path-connected, but its complement is simply connected. 
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.