Equivalence relation on the complement of a subspace.

Question

I'm trying to solve this problem: Let $V$ be a real vector space with dimension $n$ and $S$ a subspace of $V$ with dimension $n-1$. Define an equivalence relation $\equiv$ on the set $V\setminus S$ by $u\equiv v$ if the line segment

$$L(u,v)=\{(1-t)u+tv: t\in [0,1]\}$$ satisfies that $L(u,v)\cap S=\emptyset$. Prove that $\equiv$ is an equivalence relation and it has exactly two equivalence classes.

I proved that $\equiv$ is reflexive and symmetric since for $u,v\in V\setminus S$ we have that $L(u,u)=\{u\}$ and $L(u,v)=L(v,u)$. But I couldn't prove that $\equiv$ is transitive. I saw with some examples in $\mathbb{R}^3$ using lines that the hypothesis of $\dim S=n-1$ is necessary for this but in the general case I don't know how to use this. And for the last part I think that for every $v\in V\setminus S$ the unique two equivalence classes are $[v]$ and $[-v]$. Could you please give me some suggestions for this? Thanks.

Answer 1

Let $\{e_1,\ldots,e_n\}$ be a basis of $V$ such that $\{e_1,\ldots,e_{n-1}\}\subset S$. For $x\in V$, let $x_k$ be the k-th component of $x$ in that basis. Then we have $V\setminus S = \{x\in V | x_n \neq 0\}$ and for $x,y\in V\setminus S$ we have $$x\equiv y \iff tx_n+(1-t)y_n\neq0 \text{ for all } t\in [0,1]$$ $$\iff (x_n\gt0\land y_n\gt0) \lor (x_n\lt0\land y_n\lt0) $$ In other words, $x\equiv y \iff $ the $e_n$ componnents of $x$ and $y$ have the same sign. That is clearly a reflexive, symmetric and transitive relation, and the equivalence classes are just the two sets determined by the sign of that component.

Answer 2

Suppose $u \equiv v, v \equiv w$, and there are two points $x, y$ on the line segment $u, w$ such that $x, y \in S$. Then since $u$ is a linear combination of $x, y$, $u$ is in $S$, which is a contradiction to the domain of $U$.

So there can be at most one point on $L(u, w)$ which is in the subspace $S$. Note that the subspace $S$ divides points in $V \backslash S$ into two halves (since the dimension is $n-1$, the orthogonal complement of $S$ is a line, and these two halves are decided by the direction of the vector from a point $p$ to its projection on $S$). $v$ is in one of these subspaces. Then note that at least one of $u, w$ (wlog $u$) are in the other half as $v$, because there is a point on $L(u, w)$ in $S$. Hence, there is a point in $L(u, v)$ which is in $S$, and this is a contradiction to $u \equiv v$.

This also shows that the two equivalence classes are in fact the ones corresponding to the two halves that $S$ divides $V$ into.

Answer 3

Let $N$ be a normal vector to $S$.

Since the dimension of $S$ is $n-1$, it follows that $$S=\{x\in V\mid N\cdot x = 0\}$$ hence $V{\setminus}S=A\cup B$, where \begin{align*} A&=\{x\in V\mid N\cdot x > 0\}\\[4pt] B&=\{x\in V\mid N\cdot x < 0\}\\[4pt] \end{align*} Let $u,v\in V{\setminus}S$.

Consider $3$ cases . . .

Case $1$:$\;u,v\in A$.

Then for all $t\in [0,1]$, $$ N\cdot \bigl((1-t)u+tv)\bigr) = (1-t)(N\cdot u)+t(N\cdot v) $$ which is positive since it's a convex combination of the two positive real numbers $N\cdot u$ and $N\cdot v$.

Hence for case $1$, we have $L(u,v)\subset A$, so $u\equiv v$.

Case $2$:$\;u,v\in B$.

Then for all $t\in [0,1]$, $$ N\cdot \bigl((1-t)u+tv)\bigr) = (1-t)(N\cdot u)+t(N\cdot v) $$ which is negative since it's a convex combination of the two negative real numbers $N\cdot u$ and $N\cdot v$.

Hence for case $2$, we have $L(u,v)\subset B$, so $u\equiv v$.

Case $3$:$\;$One of $u,v$ is in $A$, and the other is in $B$.

Since $N\cdot u$ and $N\cdot v$ have opposite signs, it follows that $$N\cdot \bigl((1-t)u+tv)\bigr)=0$$ for some $t\in (0,1)$.

Hence for case $3$, $L(u,v)$ does not lie entirely in $V{\setminus}S$, so $u\not\equiv v$.

Based on the above case analysis, it follows that $u\equiv v\;$iff $u,v$ are either both in $A$ or both in $B$.

It's now immediate that $\equiv$ is an equivalence relation.