Prove that $\mathbb{R}^n\setminus \{0\} $ is connected for $n > 1$

Question

Prove that $\mathbb{R}^n\setminus \{0\} $ is connected for $n > 1$.

I don't understand where to start proving this since

$$\mathbb{R}^n \setminus \{0\} = (-\infty,0)^n \cup (0, \infty)^n $$

Which is the union of two disjoint nonempty open sets, so it can't be connected. Obviously I won't be told to prove something is true that isn't so I know I'm missing something.

We have proven that $\mathbb{R}^n$ is connected using this theorem:

Let S be a topological space, and let $T_0$ and $\{T_w\}_{w\in W} $ be connected subsets of S. Assume that $T_0 \cap T_w \neq \emptyset $ for each w. Then $T_0 \cup \left( \cup_{w \in W} T_w \right)$ is connected.

Using the first connected set {0} and the indexed ones as lines that go through the point {0} indexed by the unit sphere.

I was hoping to do something similar with this problem, but I can't see a way to do that. Help would be appreciated. And apologies for any Latex mistakes. I'll try to fix them but I'm on vacation and only have my phone currently.

Answer 1

Hint: It's usually way easier to show path connected than connected. Path connected implies connected, it's stronger. So take two arbitrary points and show you can connect them with a path. There are two cases: The points are not on the same line as the origin on opposite sides (The easy case), and they are: (The almost as easy case)

Can you manage?

Answer 2

Your "formula" for $\mathbb{R}^n\backslash \{0\}$ is wrong. Draw $\mathbb{R}^2$ and you will understand why.

If you want to avoid usage of path-connectedness, do the following. For $n>1$, take $A,B \subset \mathbb{R}^n \backslash \{0\}$ given by $A:=\{x \in \mathbb{R}^n \mid x_n > 0\}$ and $B:=\{x \in \mathbb{R}^n \mid x_n<0\}$. $A$ and $B$ are clearly connected (both in fact homeomorphic to $\mathbb{R}^n$). Since the closure of connected sets is connected, $\overline{A}$ and $\overline{B}$ are connected. But $\overline{A}$ and $\overline{B}$ have points in common. Therefore, their union is connected. But their union is the entire $\mathbb{R}^n\backslash \{0\}$ .

Fun exercise: Where does this fail for $\mathbb{R}$?

"But $\overline{A}$ and $\overline{B}$ have points in common" - They don't, for $n=1$.

Answer 3

There is a path from any point to the unit disc, $r(t)= ((1-t)+{t \over \|x\|}) x$

Now all that remains is to show that the unit disk is connected. Pick $u_1,u_2$ on the unit disk. Let $v_1=u_1$ and $v_2$ on the unit disc in the span of $u_1,u_2$ such that $v_1 \bot v_2$.

Now consider the path $\rho(\theta) = (\cos \theta) v_1 + (\sin \theta) v_2$, and show that $\rho(\theta)$ is on the unit disk for all $\theta$ and that there is some $\theta'$ such that $\rho(\theta') = u_2$.

Here is an explicit construction: Let $v_2 = { u_2 -\langle u_1, u_2 \rangle u_1 \over \| u_2 -\langle u_1, u_2 \rangle u_1 \| }$. Note that $v_2$ is on the unit disk and we have $u_2 = \| u_2 -\langle u_1, u_2 \rangle u_1 \| v_2 + \langle u_1, u_2 \rangle u_1$. Now check that $\| u_2 -\langle u_1, u_2 \rangle u_1 \|^2 + (\langle u_1, u_2 \rangle u_1)^2 = 1$, hence there is some $\theta'$ such that $\sin \theta' = \langle u_1, u_2 \rangle u_1$ and $\cos \theta' = \| u_2 -\langle u_1, u_2 \rangle u_1 \|$, from which we have $\rho(\theta') = u_2$. Hence the map $\rho:[0,\theta'] \to S^{n-1}$ is a path connecting $u_1,u_2$ in $S^{n-1}$.

Answer 4

$n$-dimensional polar coordinates give rise to a continuous surjection from the cartesian product of connected intervals to $\mathbb R^n \setminus \{0\}$.