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 9:45This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. 


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 $\; U = (0,1)\times (\infty,\scriptsize+\normalsize\infty) \;$. 

