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

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.

• This cannot be strictly true because $\mathbb{R}^n \setminus E$ is connected when $E=\mathbb{R}^n$.
– lhf
Jan 30, 2019 at 23:03
• Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $\varnothing$ is not connected? Jan 30, 2019 at 23:04
• @GEdgar Notice I stated codimension $\geq 2$. Jan 30, 2019 at 23:07
• So: the same question I asked becomes: Why state codimension ${} \ge 2$ instead of codimension ${}\ne 1$? Jan 30, 2019 at 23:12
• No particular reason. They mean the same thing. Jan 30, 2019 at 23:14

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.

• Thanks! Is my $\implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x \to \bar{x} \to \bar{y} \to y$. Does each $\to$ mean the line joining the two points? Jan 30, 2019 at 23:21
• @MatheusAndrade, yes, they are line segments, unless $\bar x \to \bar y$ passes through the origin, but that's easily fixed.
– lhf
Jan 31, 2019 at 0:01
• Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $\alpha$ and take another line passing through $y$ with inclination $\beta \neq \alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $\to$ contained in $\mathbb{R}^n \setminus E$? Jan 31, 2019 at 0:05

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

• An intuitive explanation of why this does not apply when $\dim E\le n-2$ is that $E^{\perp}$ minus the origin is still connected so the projection of $\Bbb R^n$ onto $E^{\perp}$ maps $\Bbb R^n\setminus E$ onto a connected set. Jan 31, 2019 at 0:01