Exercise involving light cones and light rays in special relativity I'm having some trouble in understanding (and solving) a particular exercise given in Gregory L. Naber's book "The Geometry of Minkowski Spacetime". First, let me supply the required definitions. This exercise involves the quadratic form $\mathcal{Q}(v) = (v^1)^2 + (v^2)^2 + (v^3)^2 - (v^4)^2$ which is induced by the Lorentz inner product on $\mathcal{M}$. 
We consider two distinct events $x_0$ and $x$ for which $\mathcal{Q}(x - x_0) = 0$, that is $(x^1 - x_0^1)^2 + (x^2 - x_0^2)^2 + (x^3 - x_0^3)^2 - (x^4 - x_0^4)^2 = 0$ and interpret this as the relationship of two events lying on the world line of some photon. With this in mind we define the null cone (or light cone) $C_N(x_0)$ at $x_0$ by
 $C_N(x_0) = \{\,x \in \mathcal{M}\,|\, \mathcal{Q}(x - x_0) = 0\, \}$.
Further, for any $x \in C_N(x_0)$ except $x_0$ we define the null world line (or light ray) containing both $x_0$ and $x$ by $R_{x_0,x} = \{\,x_0 + t(x - x_0) \,| \,t \in \mathbb{R}\, \}$. Now for the exercise in question:

Exercise 1.2.3:  Show that if $\mathcal{Q}(x - x_0) = 0$, then $R_{x_0,x} = R_{x,x_0}$.

By definition the $R$'s are just lines in $\mathcal{M}$, or equivalently one dimensional affine subspaces. Obviously the two subspaces $U_{x_0,x} = \{\,t(x - x_0)\,|\, t \in \mathbb{R}\,\}$ and $U_{x,x_0} = \{\,t(x_0 - x)\,|\, t \in \mathbb{R}\,\}$ are equal. Lets call this subspace $U$ for now. We know by a standard result about affine subspaces that either $x + U$ equals $x_0 + U$ or they are disjoint. That is we only need to find one element included in both sets and we are done. If we set $t = 1$ in $R_{x_0,x}$ and $t = 0$ in $R_{x,x_0}$ we get the element $\{x\}$ in both cases, thus $R_{x_0,x} = R_{x,x_0}$.
In my proof I didn't use the condition that their difference vector evaluates to zero with $\mathcal{Q}$. In euclidean space for any two distinct points there is exactly one line through them. Is this not true in $\mathcal{M}$? I understand, that sometimes two events in Minkowski space might not be connectable by a light ray, for example if they are two far apart in space. However, in this exercise the antecedent seems to be completely unnecessary. Where is the mistake?

Condition $\cal{Q}(x-x_0)=0$ is written there just because $R_{x_0,x}$ is defined only when $x\in C_N(x_0)$. Your proof is correct.
