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.

This was a scenario I was trying to set up today. Suppose $V$ is an $n$-dimensional $\mathbb{R}$-vector space, and let $S$ be an $n-1$ dimensional subspace.

Then we can define a relation $\equiv$ on the set $V\setminus S$ by saying $u\equiv v$ if the 'segment' connecting them $$ L(u,v)=\{\lambda u+(1-\lambda)v: \lambda\in[0,1]\} $$ is such that $L(u,v)\cap S=\emptyset$.

It's not hard to see that $\equiv$ is reflexive and symmetric, but I can't show it is transitive. I believe that $V\setminus S=\{s+dv: s\in S, d\in \mathbb{R}^\times\}$, where $v$ is some fixed vector in $V\setminus S$, and $d$ is a nonzero scalar. I assumed that $u\equiv w$ and $w\equiv v$, so that $\lambda u+(1-\lambda)w\notin S$ and $\mu w+(1-\mu)z\notin S$ for any $\lambda,\mu\in[0,1]$, but I couldn't derive that $\rho u+(1-\rho)v\notin S$ for all $\rho\in[0,1]$.

I had the same difficulty showing that $\equiv$ partitions $V\setminus S$, into exactly two classes, corresponding to the two opposite 'sides' of $S$ in $V$.

Is there a way to show that $\equiv$ is transitive, and thus an equivalence relation that partitions $V\setminus S$ into two congruence classes? Thanks.

share|improve this question
1  
Given $v$ not in $S$, let $A=\{{s+dv:s\in S,d\gt0\}}$, $B=\{{s+dv:s\in S,d\lt0\}}$. Can you prove $A\cup B=S^c$, $A\cap B$ is empty, $L(x,y)$ misses $S$ if $x,y$ both in $A$ or both in $B$, $L(x,y)$ meets $S$ if $x$ in $A$, $y$ in $B$? –  Gerry Myerson Nov 27 '12 at 5:06
    
@GerryMyerson That's not how I'm used to verifying equivalence relations, but it worked great. Thanks. –  Noomi Holloway Nov 27 '12 at 5:37

1 Answer 1

up vote 2 down vote accepted

We have $S=\ker f$ for some linear functional $f$.

Note that $u\equiv v$ if $f(\lambda u+(1-\lambda)v)\ne0$ for all $\lambda\in[0,1]$. As $f(\lambda u+(1-\lambda)v)=\lambda f(u)+(1-\lambda)f(v)$, this is equivalent to $f(u)f(v)>0$ (both need to be on the same side of $0$).

Now, f $u\equiv v$ and $v\equiv w$, we know that $f(u)f(v)>0$ and $f(v)f(w)>0$. Then $f(u)f(w)>0$, as both have the same sign as $f(v)$. So the relation is transitive.

The equivalence classes are precisely the sets $\{f>0\}$ and $\{f<0\}$, which are the two sides of the hyperplane you are looking for.

share|improve this answer
    
Thanks! ${}{}{}$ –  Noomi Holloway Nov 27 '12 at 5:40

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.