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?
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
and even more explicit
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
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 $\omega(a),\omega(b)<0$.