Show that $\mathbb{R}^n - D^n$ is path-connected, for $n>1$. Here $D^n$ is the closed ball centred at the origin $O$ with radius 1.

To be honest, I solved this problem. I spent around 30 minutes to solve this. I thought about many ways, but failed in all but one. I am curious to know other ways to solve this problem.

I won't post my solution intentionally. I am curious about other ways to solve this. This problem seems trivial for $\mathbb{R}^2$ and $\mathbb{R}^3$ but gets difficult for $n>4$.

Thank you.

  • $\begingroup$ It's not hard to write down an explicit path in cases. $\endgroup$
    – Randall
    Commented Jan 19, 2021 at 21:39
  • 6
    $\begingroup$ "I won't post my solution intentionally" - I really don't want to waste time writing a solution if it's not the one that is useful to you.... (p.s. I did not downvote, though) $\endgroup$
    – qualcuno
    Commented Jan 19, 2021 at 21:39
  • $\begingroup$ The problem is indeed trivial for n = 2. For n > 2, pick any two points in your space. Consider the two-dimensional subspace through the origin and the two points and apply the case n = 2 in this subspace. Handle the special case when the origin and your two points are collinear. $\endgroup$
    – user325968
    Commented Jan 20, 2021 at 0:24
  • $\begingroup$ @guidoar, Why won't it be useful? I said that there are different ways I tried but couldn't complete. I will learn new ways for solving this problem. :) $\endgroup$ Commented Jan 20, 2021 at 3:07
  • $\begingroup$ "I solved this problem." $\endgroup$
    – Randall
    Commented Jan 20, 2021 at 14:08

2 Answers 2


Solution 1:

Just for convencience sake, via a dilatation we can wlog assume that the space is rather $X = \mathbb{R}^n \setminus \frac{1}{2} D^n$.


  • The sphere $S^n$ is path connected: if $x,y \in S^n$ are not antipodal and $c(t)$ is the segment joining $x$ with $y$, then $c/\|c\|$ is a path from $x$ to $y$ contained in the sphere. If $x = -y$, pick a third point, use transitivity of path connectedness.

  • Once again; path connectedness is transitive so it is enough to note that any point can be connected to a point in $S^n$. If $x \in X$, so is $x/\|x\|$, and you can check that the segment joining $x$ and $x/\|x\|$ is contained in $X$.

Solution 2:

A techonological argument: path-connectedness is a homotopy invariant, hence it's sufficient to prove it for $Y = \mathbb{R}^n \setminus \{0\} \simeq X$. Now pick $x,y \in Y$. If $x \not \in \langle y\rangle$, the segment $\vec{xy}$ is contained in $Y$. Otherwise pick $z \not \in \langle y\rangle$ and by the exact same argument connect $x$ and $y$ to $z$ via their respective segments. Now use transitivity.


The problem boils down to parameterizing a path on the boundary of $D^n$ that connects two points $a, b$ on that boundary. This is not difficult; let's assume that $\gamma_{a, b}(t)$ is such a parameterization, where $t$ increases from $0$ to $1$ and $\gamma_{a, b}(t)$ equals $a$ for $t=0$ and equals $b$ for $t=1$.

Now, if $p, q$, are two arbitrary points in $X = \mathbb{R}^n \setminus D^n$, two cases are possible:

  1. The rectilinear segment from $p$ to $q$ is contained entirely in $X$, hence serves as a suitable path from $p$ to $q$.
  2. The rectilinear segment from $p$ to $q$ intersects the boundary $D^n$ at two points, $a$ and $b$. Here we would be tempted to concatenate: the rectilinear segment $p$ to $a$, then the path $\gamma_{a,b}$, then the segment $b$ to $q$. The problem with this is that $\gamma_{a, b}$ does not lie in the open set $X$. However, this is an easy fix: in the first paragraph of this answer, instead of $D^n$, use the closed ball centered at the origin and having radius $(1+\epsilon)$, where the positive $\epsilon$ is small enough to make sure that the latter ball does not contain $p$ or $q$.

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