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?

$\begingroup$ of course not ! $\endgroup$– Denis SerreNov 7 '11 at 7:07

1$\begingroup$ No. (Do you want a specific counterexample? This feels a bit like a homework question...) $\endgroup$– Tom SmithNov 7 '11 at 7:11

$\begingroup$ I am sorry to have omitted the condition $U\supset K$, this makes the question looks rather stupid. But I must say that this is not a homework question, this is a claim (without proof) in a proof of a paper I am reading. I don't think it is completely trivial, as this is false if we replace Rn with some other connected spaces, such as the torus. Therefore somehow we must use the property of $\mathbb{R}^n$ (e.g. Jordan's theorem), but I have no idea how to. Perhaps it's also interesting to see if we can replace $\mathbb{R}^n$ with other spaces, e.g. spheres. $\endgroup$– KwongNov 7 '11 at 11:27
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.

$\begingroup$ I am not very familiar with relative homology, but I think we need more than that. I can only see that if $H_0(U, UK)=0$, the inclusion induces a surjective map from $H_0(UK)$ to $H_0(U)$, by the long exact sequence of relative homology. (Correct me if I am wrong. ) Also, if your argument is correct, doesn't it imply that if $K$ is the equator of the torus $T^2$, and $U$ is a tubular neighborhood of $K$, then $UK$ is connected? $\endgroup$– KwongNov 7 '11 at 10:24

$\begingroup$ you also have $H_1(\mathbb{R}, \mathbb{R} K)=0$ using that $H_1(\mathbb{R})=0$ (crucially!), so by excision $H_1(U, UK)=0$, too. The last part of my answer was incorrect, let me edit (sorry i answered too quickly). $\endgroup$– PierreNov 7 '11 at 11:43

$\begingroup$ This answer has the advantage of answering his extension question (that is, on what manifolds this is still true), so bravo $\endgroup$ Nov 7 '11 at 14:37

$\begingroup$ Thanks for the neat answer. So we can replace $\mathbb{R}^n$ by a connected space with vanishing $H_1$, and $K$ to be closed subset. Nice. $\endgroup$– KwongNov 8 '11 at 5:19
No. $\;\;$ Let $n=2$, $\; K = [0,1]^2 \;$, $\;$ and $\; U = (0,1)\times (\infty,\scriptsize+\normalsize\infty) \;$.

$\begingroup$ $K$ is not a subset of $U$ (as e.g. $(0,0) \in K \setminus U$). $\endgroup$ Nov 7 '11 at 18:12