My textbook says the following:
The nonnegative orthant is the set of points with nonnegative components, i.e.,
$$\mathbb{R}_+^n = \{ x \in \mathbb{R}^n \mid x_i \ge 0, i = 1, \dots, n\} = \{ x \in \mathbb{R}^n \mid x \succeq 0 \}.$$
(Here $\mathbb{R}_+$ denotes the set of nonnegative numbers: $\mathbb{R}_+ = \{ x \in \mathbb{R} \mid x \ge 0 \}.$) The nonnegative orthant is a polyhedron and a cone (and therefore called a polyhedral cone).
A cone is defined earlier in the textbook as follows:
A set $C$ is called a cone, or nonnegative homogeneous, if for every $x \in C$ and $\theta \ge 0$ we have $\theta x \in C$.
A polyhedron is defined earlier in the textbook as follows:
$$\mathcal{P} = \{ x \mid a_j^T x \le b_j, j = 1, \dots, m, c_j^T x = d_j, j = 1, \dots, p \}$$
I can see how a nonnegative orthant is a polyhedron, but I don't see why it is also a cone? For instance, according to the above definition of cone, there is no requirement that $x \ge 0$, unlike for a nonnegative orthant.
I would greatly appreciate it if people could please take the time to clarify this.
Best Answer
Let's pick an element from the nonnegative orthant, $x$, then we have $x \ge 0$.
We take an arbitrary nonnegative number $\theta$ and construct the vector $y=\theta x$.
Each component of $y$ is $y_i=\theta x_i$ where $\theta \ge 0$ and $x_i \ge 0$, hence we have $y_i \ge 0$.
Hence $y$ still resides in the nonnegative octant.
Hence the nonnegative orthant is a cone.
The nonnegative orthant is a cone, it doesn't mean that all cone is a nonegative orthant. It is a special case of a cone, it can has special property such as having componentwise nonnegativity for each element.