Let $C\colon=C(e)$ be the future time-cone of a Lorentz space $E$. Then the sets $V(x,x^\prime)\colon=(x^\prime-C)\cap(x+C)$, $x,x^\prime\in E$, form a basis of a topology ${\cal T}$ on $E$. Prove that $\pa Z=L\cup\{0\}$.
We define a euclidean norm $|.|$ and show that the corresponding norm-topology coincides with ${\cal T}$: Suppose $x=-\la x,e\ra e+u$ for some $u\in e^\perp$, then put (cf. exam):
$$
|x|^2\colon=\la x,e\ra^2+\Vert u\Vert^2~.
$$
Thus for $u\in e^\perp$ we have: $|u|=\Vert u\Vert$. For $x=-\la x,e\ra e+u$ and $y=-\la y,e\ra e+v$ we have
$$
|\la x,y\ra|
=|-\la x,e\ra\la y,e\ra+\la u,v\ra|
\leq|\la x,e\ra\la y,e\ra|+|\la u,v\ra|
\leq|x||y|
$$
which shows that $\la.,.\ra$ is continuous with respect to the norm-topology, the sets $V(x,x^\prime)$ are open in the norm-topology and thus the norm-topology is finer than ${\cal T}$. On the other hand the ball $B\colon=\{x\in E: |x| < r\}$ contains the set $(re-C)\cap(-re+C)$: Indeed for every $x=ae+u\in(re-C)\cap(-re+C)$, $u\in e^\perp$ we get:
\begin{eqnarray*}
&&-x+re,x+re\in C\\
\Lrar&&-u+(r-a)e,u+(r+a)e\in C\\
\Lrar&&\Vert u\Vert^2 < (r-|a|)^2
\quad\mbox{and}\quad
r-|a| > 0~.
\end{eqnarray*}
It follows that
$$
|x|^2
=a^2+\Vert u\Vert^2
< a^2+(r-|a|)^2
=a^2+r^2+a^2-2r|a|
< r^2~.
$$
i.e. $x\in B$ and therefore both topologies coincide. Since the translates of open balls form a basis, the sets $V(x,x^\prime)$ form a basis as well.
2. Obviously $\bar Z=Z\cup L\cup\{0\}$ and thus $\pa Z=L\cup\{0\}$.