A question about connectedness in Euclidean space - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T02:01:33Z http://mathoverflow.net/feeds/question/80272 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80272/a-question-about-connectedness-in-euclidean-space A question about connectedness in Euclidean space Kwong 2011-11-07T06:53:56Z 2011-11-07T11:49:20Z <p>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?</p> http://mathoverflow.net/questions/80272/a-question-about-connectedness-in-euclidean-space/80274#80274 Answer by Ricky Demer for A question about connectedness in Euclidean space Ricky Demer 2011-11-07T07:07:33Z 2011-11-07T07:07:33Z <p>No. $\;\;$ Let $n=2$, $\; K = [0,1]^2 \;$, $\;$ and $\; U = (0,1)\times (-\infty,\scriptsize+\normalsize\infty) \;$.</p> http://mathoverflow.net/questions/80272/a-question-about-connectedness-in-euclidean-space/80277#80277 Answer by Pierre for A question about connectedness in Euclidean space Pierre 2011-11-07T08:17:12Z 2011-11-07T11:49:20Z <p>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 $H_1(\mathbb{R}, \mathbb{R}-K)= H_1(U, U-K)= 0$ when $\mathbb{R}-K$ is connected. </p> <p>So $H_0(U-K)$ injects into $H_0(U)$ and $U-K$ must be connected.</p>