Convex cones in $\mathbb{R}^n$ - Orthants

Question

I am trying to understand the notion of convex cones. So, here are my questions.

I can understand that the non-negative orthant, $\mathbb{R^n_+}$, defined as $\left\{ (x_1, \ldots, x_n) \in \mathbb{R}^n | x_i \geq 0, i = 1, \ldots,n \right\}$ is a convex cone. But I do not understand what is special about this.

This raises the following questions:

1. In $\mathbb{R}^2$, is every quadrant a convex cone?

2. What can we say about the octants in $\mathbb{R}^n$ in general? For example, the non-positive octant in $\mathbb{R}^3$?

I have split the question into two cases to avoid any special cases that might only be valid in $\mathbb{R}^2$.

Answer 1

Every orthant in $\mathbb{R}^n$ can be expressed as the intersection of a finite number of closed halfspaces containing the origin. i.e. an orthant in $\mathbb{R}^n$ can be described as the set $\mathcal{S} = \lbrace x \in \mathbb{R}^n \vert a_i^Tx \leq 0, i = 1, \cdots, p\rbrace$.

Clearly, from the above definition, we can see that $\forall \alpha \in \mathbb{R}_+, \alpha x \in \mathcal{S}$. Thus, every orthant is a closed convex cone.