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I need to prove that the complement in $\mathbb{R}^{n}$, $n>1$ of a limited subset $X \subset \mathbb{R}^{n}$ has only one ilimited connected component. This lies in prove that $\mathbb{R}^{n} - B\left[ 0;r\right]$ (where $B\left[0;r\right]$ is the closed ball with center in the origin and radius $r$) is connected, since it will be the only one ilimited connected component (any another component different from this will be in the closed ball).

How to prove that? There is another way to prove the statement whitout the usage of this? Thanks in advance!

big_GolfUniformIndia
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2 Answers2

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$ \mathbb R \setminus B[0,r]$ is not connected, because it is union of two (cl)open sets, $(-\infty, - r) \cup (r, +\infty) $. For $n\ge 2 $, $\mathbb R^n \setminus B[0,r]$ is connected because it is path-connected.

Adriano Banchieri
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    Yes, I forgot to write that $n>1$; there is some reference with the prove of the "path-connectedness" of $\mathbb{R}^{n} - B[0;r]$? Thanks in advance. – big_GolfUniformIndia Nov 27 '20 at 18:28
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    I'm quite sure you can find one, but I don't know right now. The idea is quite like this: $\mathbb R ^n \setminus B[0, r] $ is an open set. Thus, for every point in it, there exists an open ball entirely contained in it, and you can connect the point in the centre of the ball to any other point in the ball by a continuous path. In such a way, you will be able to join every couple of points, because the result of "gluing" the paths is still a path. – Adriano Banchieri Nov 27 '20 at 18:42
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    @GuilhermeNetto The boundary of $B[0; r]$ i.e. the sphere $S^{n - 1}(r)$ is path-connected as long as $n > 1$. So if the straight line $\ell(t)$ between any two points $p_1, p_2 \in \mathbb R^n - B[0; r]$ intersects the sphere at two points other points $p_3, p_4$, just modify $\ell(t)$ between $p_3, p_4$ to "go on top of the sphere". If it doesn't intersect the sphere or does so at one point, just use $\ell(t)$. – balddraz Nov 27 '20 at 18:46
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    Give a look also to this one. It's about polygonally-connectedness and can be useful. Polygonally-connected sets are path-connected, and path-connected open-sets are polygonally-connected sets. https://math.stackexchange.com/questions/2161210/polygonally-connected-open-sets – Adriano Banchieri Nov 27 '20 at 19:01
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It is not hard to check that $\mathbb{R}^2 \backslash B[0,r]$ is connected.

Let $p$, $q$ be points in $\mathbb{R}^n\backslash B[0,r]$. Let $V$ a two dimensional subspace of $\mathbb{R}^n$ containing $p$, $q$. By the above, $V\backslash B[0,r]$ is connected. We conclude that $p$, $q$ are in the same component of $\mathbb{R}^n \backslash B[0,r]$.

Let $X$ be a bounded subset of $\mathbb{R}^n$. Consider $r>0$ such that $X \subset B[0,r]$. Now $\mathbb{R}^n \backslash B[0,r]$ is contained in one component of $\mathbb{R}^n\backslash X$. The other components (if they exist) are contained in $B[0,r]$, so bounded.

orangeskid
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