Connectedness of complements of vector subspaces of $\mathbb{R}^n$

Question

I want to prove the following:

Let $E \subset \mathbb{R}^n$ be a vector subspace of $\mathbb{R}^n$ which is not equal to $\mathbb{R}^n$. Then $\mathbb{R}^n \setminus E$ is connected if, and only if, the dimension of $E$ is at most $n - 2$ (or equivalently, codimension at least $2$).

For the $\implies$ direction, here's what I've got: Let $E$ be a vector subspace with dimension $n - 1$. Then: $$\mathbb{R}^n \setminus E = A \cup B$$

where $A = \{v \in \mathbb{R}^n \ \vert \ \langle v, n \rangle < 0 \}$ and $B = \{v \in \mathbb{R}^n \ \vert \ \langle v, n \rangle > 0 \}$ and $n$ is a normal vector to $E$ (which certainly exists). Now, intuitively, I know that $A$ and $B$ are disjoint open sets, but I haven't been able to prove it.

I couldn't think of anything yet regarding the $\impliedby$ direction. My strategy was proving that any two points in $\mathbb{R}^n \setminus E$ can be joined by a line or a union of broken lines, but I have been unsuccessful in getting anywhere with that approach.

Answer 1

Hint: Let $F$ be a linear complement for $E$, that is, $\mathbb{R}^n = E \oplus F$. Take $x,y \in \mathbb{R}^n \setminus E$. One path joining $x$ and $y$ is $ x \to \bar x \to \bar y \to y$, where the bar means projection onto $F$. In the path $\bar x \to \bar y$, you need to avoid the origin.

Answer 2

If $\dim (E)=n-1$ let $\{e_j:1\le j\le n-1\}$ be an orthonormal basis for $E$ and let $E\cup \{e_n\}$ be an orthonormal basis for $\Bbb R^n.$ In other words $E^{\perp}=\{re_n:r\in \Bbb R\}.$ The projection $P(v)=\langle v|e_n\rangle$ is continuous from $\Bbb R^n$ to $\Bbb R$ so $A=P^{-1}(-\infty,0)$ and $B=P^{-1}(0,\infty)$ are open.

$P$ is Lipschitz-continuous: $|P(v)-P(v')|=|\langle (v-v')|e_n\rangle|\,\le \|v-v'\|\cdot \|e_n\|=\|v-v'\|.$