Timeline for Show that $\mathbb{R}^n\setminus K$ is connected
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 10 at 17:51 | vote | accept | Philipp | ||
| Mar 6 at 19:44 | answer | added | copper.hat | timeline score: 1 | |
| Mar 6 at 18:13 | comment | added | user8675309 | The real issue is that your official problem statement should be obviously false to you, yet you "proved" it to be true in your first couple postings. I would spend time figuring out your error. | |
| Mar 6 at 18:08 | comment | added | Severin Schraven | Wlog $0\in K$ (otherwise $K=\emptyset$ or we can just translate). Pick $x\in K^c$, then $\lambda x\in K^c$ for all $\lambda\geq 1$. Let $R=2 diam(K)$. To connect $x,y\in K^c$ first walk along a straight line until $(R/\vert x\vert)x$, rotate to $(R/\vert y\vert)y$ and walk down to $y$. | |
| Mar 6 at 17:59 | history | edited | Philipp | CC BY-SA 4.0 |
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| Mar 6 at 17:58 | comment | added | Philipp | @user8675309, no there is nothing else missing. I guess the problem is simply not well stated. As Plutoro suggested, let's assume that $K$ is convex! | |
| Mar 6 at 17:52 | comment | added | user8675309 | Let $n=2$ and $K$ be a circle... contradiction. In fact if OP's statement were true, every winding number for every closed curve would have to be 0. | |
| Mar 6 at 17:47 | comment | added | Philipp | @Plutoro, yes you are right, it should be $n>1$. | |
| Mar 6 at 17:46 | history | edited | Philipp | CC BY-SA 4.0 |
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| Mar 6 at 17:41 | comment | added | Plutoro | I wonder if something is missing from the statement. What if $n=1$, and $K=\{0,1\}$. Then $K$ is non-empty and compact, but $\mathbb R\setminus K=(-\infty,0)\cup(0,1)\cup(1,\infty)$ is not connected. Perhaps $K$ must also be convex? | |
| Mar 6 at 17:05 | history | asked | Philipp | CC BY-SA 4.0 |