# Showing that $\mathbb{R}^n - D^n$ is path-connected [closed]

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.

• It's not hard to write down an explicit path in cases. Commented Jan 19, 2021 at 21:39
• "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) Commented Jan 19, 2021 at 21:39
• 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.
– user325968
Commented Jan 20, 2021 at 0:24
• @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. :) Commented Jan 20, 2021 at 3:07
• "I solved this problem." Commented Jan 20, 2021 at 14:08

Solution 1:

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

Now,

• 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 $$q$$.