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.
 A: 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. 
A: Suppose $\mathcal{B}=\{v_1,v_2,\cdots,v_{n-1}\}$ is a basis for $S$. Extend this to $\mathcal{B}\cup\{v_n\}$ for $V$. Then $$V\setminus S=\{s+dv_n: s\in S, d\in \mathbb{R}^\times\}$$
$$u,v \in V\setminus S\Rightarrow \exists s_1,s_2 \in S: \begin{cases}u=s_1+d_uv_n\\v=s_2+d_vv_n\end{cases}$$
I am going to show that $L(u,v) \cap S =\emptyset$ is equivalent to $d_ud_v >0$:
$$L(u,v) \cap S =\emptyset\\ \ \Leftrightarrow\ \ \forall \lambda \in [0,1]:\ \ \lambda u +(1-\lambda)v \not \in S\\ \ \Leftrightarrow\ \  \forall \lambda \in [0,1]:\ \ \lambda s_1+(1-\lambda)s_2 +(\lambda d_u+(1-\lambda)d_v)v_n \not \in S\\ \ \Leftrightarrow\ \ \forall \lambda \in [0,1]: \lambda d_u+(1-\lambda)d_v \ne 0\\ \ \Leftrightarrow\ \ d_u d_v>0$$
Now, $u\equiv v$ and $v\equiv w$ imply $d_ud_v>0$ and $d_vd_w>0$, which, in turn, implies $d_ud_w>0$, and so $u \equiv w$.
