Here is a question which seems true to me but I can't rigorously show. Suppose $K$ is a compact subset of $\mathbb{R}^n$ such that $\mathbb{R}^n\setminus K$ is connected, does it follow that for any connected open set $U\subset \mathbb{R}^n$ such that $U\supset K$, $U\setminus K$ is also connected?

closed as too localized by Denis Serre, Ryan Budney, Will Jagy, Matthew Daws, Joel David Hamkins Nov 7 '11 at 9:45
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Yes. Let $C$ be the closed complement of $U$, then by excision of $C$ we have $H_1(\mathbb{R}, \mathbb{R}  K) = H_1(U, U  K)$; since $H_1(\mathbb{R})=0$, you also have in fact $H_1(\mathbb{R}, \mathbb{R}K)= H_1(U, UK)= 0$ when $\mathbb{R}K$ is connected. So $H_0(UK)$ injects into $H_0(U)$ and $UK$ must be connected. 


No. $\;\;$ Let $n=2$, $\; K = [0,1]^2 \;$, $\;$ and $\; U = (0,1)\times (\infty,\scriptsize+\normalsize\infty) \;$. 

