On page 56 of John Ratcliffe's Foundations of Hyperbolic Manifolds, it says that
Corollary 1. The set of all positive (resp. negative) time-like vectors is a convex subset of $\mathbb{R}^n$.
I'm confused about this statement. In $\mathbb{R}^{1, 2}$, the light cone is a double cone (i.e. the solution set of $|x_1| = \sqrt{x_2^2 + \vdots + x_n^2}$); I thought a cone is not convex because if we pick two points on the same level on the cone, a line connecting those two points wouldn't be on the cone since the cone is "empty".
Specifically, $(\sqrt{2}, 1, 1)^T$ and $(\sqrt{2}, 1, -1)^T$ are two points on the light cone of $\mathbb{R}^{1, 2}$. Their midpoint is $(\sqrt{2}, 1, 0)^T$ which is not light-like.
Can someone tell me what I'm missing? Thanks in advance!
Spelling out exactly what all these objects are should clear things up.
You've correctly described the light cone as the solution set of the equation $|x_1| = \sqrt{x_2^2 + \cdots + x_n^2}$.
The positive time like vectors are the solution set of the inequality $$x_1 > \sqrt{x_2^2 + \cdots + x_n^2} $$ which forms a convex set.
The negative time like vectors are the solutions of $$x_1 < - \sqrt{x_2^2 + \cdots + x_n^2} $$ which also forms a convex set.
And the space like vectors are the solutions of $$-\sqrt{x_2^2 + \cdots + x_n^2} < x_1 < \sqrt{x_2^2 + \cdots + x_n^2} $$ which does not form a convex set.