Let $n \ge 3$ and let $L$ be a compact line segment i.e. of the form $[a,b]$ in $\mathbb R^n$. Then is $\mathbb R^n \setminus L$ simply connected ?
I can only see it is path connected.
Please help.
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1https://math.stackexchange.com/questions/287062/the-complement-of-jordan-arc – Madara Uchiha Apr 05 '18 at 17:37
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Yes, it is simply connected. Given a loop $\gamma$, consider a translation of $\gamma$ parallel to the line segment $L$ which takes $\gamma$ far enough from $L$. That is, if $a$ and $b$ are distinct elements of $L$, let $\gamma_t=\gamma+t(b-a)$. If $t\gg0$, then there is a hyperplane separting $\mathbb{R}^n$ in two parts, such that the image of $\gamma_t$ is in one of them and $L$ in the other one. Then contract $\gamma$ to a point.
José Carlos Santos
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Could you write it down explicitly ? the "Parallel translation" thing ... – user Apr 05 '18 at 17:30
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I've edited my answer. I hope that everything is clear now. – José Carlos Santos Apr 05 '18 at 17:34
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You have shown that all loops are freely homotopic to a point, which is slightly different: you have to take care of the basepoint somehow. – Mariano Suárez-Álvarez Apr 05 '18 at 17:36
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@users Take $t_0$ such that $t_0(b-a)$ is no longer in the segmen $L$. Then take the hyperplane orthogonal to $L$ passing through $t_0(b-a)$. – José Carlos Santos Apr 05 '18 at 17:39
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A priori, to show simple connectedness, you need to show every loop is contractible, fixing the base point. – Cheerful Parsnip Apr 05 '18 at 17:39
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1@MarianoSuárez-Álvarez But if $\mathbb{R}^n - L$ is path-connected, then basepoints can be ignored, no? – mathphys Apr 05 '18 at 17:39
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Sure you can, but you have to say it at some point. Or rather, no, you cannot ignore the base point: you have to take care of it by noting that in a path connected space a loop is contractible iff it is freely contractible: that is not ignoring the base point but taking care of it. – Mariano Suárez-Álvarez Apr 05 '18 at 17:41