# Cutting an $n$ dimensional vector space into two parts with an $n-1$ dimensional subspace.

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.

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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

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.
Thanks! ${}{}{}$ –  Noomi Holloway Nov 27 '12 at 5:40