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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.

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

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$ Jan 30, 2019 at 23:21
  • $\begingroup$ @MatheusAndrade, yes, they are line segments, unless $\bar x \to \bar y$ passes through the origin, but that's easily fixed. $\endgroup$
    – lhf
    Jan 31, 2019 at 0:01
  • $\begingroup$ 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 $\mathbb{R}^n \setminus E$? $\endgroup$ Jan 31, 2019 at 0:05
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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: $|P(v)-P(v')|=|\langle (v-v')|e_n\rangle|\,\le \|v-v'\|\cdot \|e_n\|=\|v-v'\|.$

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  • $\begingroup$ 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. $\endgroup$ Jan 31, 2019 at 0:01

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