# 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

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 $\lambda \not\equiv 0$.