Prove that $R$ given by $L(a,b)\cap W =\varnothing$ is an equivalence relation. Let $V$ be a real vector space and suppose $W$ is a subspace of $V$ with $\dim(W)=n-1$. Now define a relation $R$ on $V-W$ s.t. $$aRb\iff L(a,b)=\{ra+(1-r)b | 0\leq r \leq 1\}\text{ has the property that }L(a,b) \cap W = \varnothing$$ Prove that $R$ is an equivalence relation.
I am having problem to check transitivity. If $aRb$ and $bRc$ suppose $a$ is not related to $c$, then $\exists t \in (0,1)$ s.t $ta+(1-t)c \in W$. How do I get a contradiction?
 A: The intuition why this is an equivalence relation is, as explained by almagest, that the hyperplane $W$ seperates $V-W$ into two connected components. Two points $a,b\in V-W$ are in the same component if and only if the line segment $L(a,b)$ does not intersect the hyperplane $W$.
For a formal approach, describe $W$ as the kernel of some non-trivial linear-form $\omega\colon V\to\mathbb R$. Now let's translate $aRb$ using this description. Instead of $L(a,b)\cap W=\varnothing$ we can say

For all $x\in L(a,b)$ we have $\omega(x)\neq 0$.

and even more explicit

For all $0\le r\le 1$ we have $r \omega(a) + (1-r)\omega(b)\neq 0$.

Now this is a statement solely about the two real numbers $\omega(a)$ and $\omega(b)$. In fact, using the intermediate value theorem, you can check that this is equivalent to

The two non-zero real numbers $\omega(a)$ and $\omega(b)$ have the same sign.

From this the fact that this relation is an equivalent relation becomes obvious.

Let me elaborate on the equivalence claimed above.
Suppose $\omega(a)$ and $\omega(b)$ are non-zero real numbers with different signs. Consider the real function $f\colon [0,1]\to\mathbb R$ with $r\mapsto r\omega(a)+(1-r)\omega(b)$. This function is continuous with $f(0)=\omega(b)$ and $f(1)=\omega(a)$. Since $f(1)$ and $f(0)$ are non-zero of different signs, the intermediate value theorem says there exists $r\in(0,1)$ such that $f(r)=0$, hence $r\omega(a)+(1-r)\omega(b)=0$.
Now suppose that $\omega(a)$ and $\omega(b)$ are non-zero real numbers with the same sign. When $\omega(a),\omega(b)>0$, you can conclude that $r\omega(a)+(1-r)\omega(b)>0$ for all $r\in[0,1]$ since products and sums of positive numbers are positive. The same approach works for $\omega(a),\omega(b)<0$.
