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.
Best Answer
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 $\mathbb{R}$?