Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
14  
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 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 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 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 at 18:47

1 Answer 1

up vote 3 down vote accepted

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 $\lambda \not\equiv 0$.

– Daniel Fischer

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.