the set of all positive (resp. negative) time-like vectors is convex

Question

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!

Answer 1

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.