Can this be generalized to higher dimension? Here is a famous theorem in complex analysis:

If $G$ is an open connected subset of $\mathbb{C}$, then $G$ is simply connected iff $\hat{\mathbb{C}}\setminus G$ is connected.

I'm curious if this result can be generalized to higher dimension.
That is,

Let $G$ be an open connected subset of $\mathbb{R}^n$. Then is $H_{n-1}(G)=0$ iff $\hat{\mathbb{R}^n}\setminus G$ is connected? (Hat denotes the one-point compactification)


This follows from Alexander duality, at least if $G$ is not pathological.  Using the notation from the Wikipedia page, let $Y = G$ and $X = \hat{\mathbb{R}^n} \setminus G$.  Then there is an isomorphism $$\tilde{H}_q(X) \cong \tilde{H}^{n-q-1}(Y).$$  If $\hat{\mathbb{R}^n} \setminus G$ is path-connected, then $\tilde{H}_0(\hat{\mathbb{R}^n} \setminus G) = 0$, so $\tilde{H}^{n-1}(G) = 0$.  Then by the universal coefficients theorem, we have $\tilde{H}_{n-1}(G)^\vee = 0$.  Because this holds for any coefficients, we deduce that $\tilde{H}_{n-1}(G) = 0$.  This argument is also reversible, so you also have the other implication.  
