# Elementary proof that $\mathbb{R}^n$ minus hyperplane is not connected

I was wondering if an elementary proof is possible of the following fact, i.e. without using Invariance of Domain, Jordan Curve Theorem, etc.

Prove that if $H$ be a hyperplane in $\Bbb R^n,$ then $\Bbb R^n\setminus H$ is not connected.

Thank you for any help.

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A hyperplane in $\mathbb{R}^n$ is given by an equation $\lambda(x) = c$ for a nonzero linear form $\lambda$. Then $\mathbb{R}^n\setminus H = \lambda^{-1}((-\infty,c)) \cup \lambda^{-1}((c,+\infty))$ is a decomposition into two disjoint open sets, both of which are nonempty since $\lambda \not\equiv 0$. – Daniel Fischer Aug 25 '13 at 14:17
@DanielFischer The converse is also true: If $E\subsetneq\mathbb{R}^n$ is a disconnected subspace, then $E$ is a hyperplane. In other words, a subspace $E\subsetneq\mathbb{R}^n$ is connected if, and only if, $\dim E\leq n-2$. Could you help me to prove it? – Pedro Mar 22 '15 at 13:47
@Pedro You're only considering linear (or affine) subspaces, right? And you mean that $\mathbb{R}^n\setminus E$ is connected if and only if $\dim E \leqslant n-2$, presumably. Can you show it if $E$ is of the form $E = \{ x\in \mathbb{R}^n : x_{d+1} = \dotsc = x_n = 0\}$, where $d = \dim E$? Then adapt that proof to the general case. – Daniel Fischer Mar 22 '15 at 13:53
@DanielFischer Yes, $E$ is a linear subspace and I should have typed "$\mathbb{R}^n\setminus E$ is connected". I started with that particular form of $E$, but then I got this solution that seems to be valid for any $E$: Let $a,x\in \mathbb{R}^n\setminus E:=C$. Since $\dim E\leq n-2$, there exists $z_x\in\mathbb{R}^n\setminus \mathrm{span}\big(\{x\}\cup E\big)$. Define $C_x=[a,x]$ if $[a,x]\cap E=\varnothing$ and $C_x=[a,z_x]\cup [z_x,x]$ if $[a,x]\cap E\neq \varnothing$. The set $C_x$ is a connected subset of $C$ that contains $a$. Thus $C=\bigcup_{x\in C} C_x$ is connected. Is it right? – Pedro Mar 22 '15 at 18:22
@Pedro Yes, that works. You don't need to distinguish the cases if you don't want to, you can always take the path passing through $z_x$. (And what we have so far forgotten to mention: we need the restriction $\dim E < n$, for if $\dim E = n$, we have $\mathbb{R}^n\setminus E = \varnothing$, which is connected [unless you're using a different definition of connected spaces].) – Daniel Fischer Mar 22 '15 at 18:47

A hyperplane in $\mathbb R^n$ is given by an equaiton $\lambda(x)=c$ for a nonzero linear form $\lambda$. Then $\mathbb{R}^n\setminus H = \lambda^{-1}((-\infty,c)) \cup \lambda^{-1}((c,+\infty))$ is a decomposition into two disjoint open sets, both of which are nonempty since $\lambda \not\equiv 0$.