Nonempty interior feature of a proper cone One of the features of a proper cone is solid which means a proper cone has nonempty interior. What dose nonempty interior mean?
I was reading Boyd convex optimization, and I saw this term "Nonempty interior" 
I don't know what it means.
 A: Consider two sets:
$$C_1=\left\{(x_1,x_2)\,|\,x_1\geq 0,~x_2=0\right\}$$
$$C_2=\left\{(x_1,x_2)\,|\,x_1,x_2\geq 0\right\}$$
$C_1$ and $C_2$ are both closed, convex cones, but only $C_2$ has a non-empty interior. It's simply the standard set definition: a set has a non-empty interior if it includes points that are not on its boundary. The point $(1,1)$ is on the interior of $C_2$, but every point in $C_1$ is of the form $(x_1,0)$, which is on its boundary, so $C_1$ has no interior.
There are some good reasons why it is worthwhile to focus on cones with non-empty interiors. Boyd et. al. define a "proper" cone as a cone that is closed and convex, has a non-empty interior, and contains no straight lines. The dual of a proper cone is also proper. For example, the dual of $C_2$, which is proper, happens to be itself. The dual of $C_1$, on the other hand, is
$$C_1^*=\left\{(x_1,x_2)\,|\,x_1\geq 0\right\}$$
Note that $C_1$ has a non-empty interior; $C_1^*$ has a non-empty interior, but contains straight lines. Proper cones establish partial orderings that mimic a lot of the useful properties of standard inequalities. See the discussion of "generalized inequalities" in that book, or even elsewhere on Math.SE.
From a practical standpoint, algorithms to solve problems involving cone constraints (e.g., barrier methods) generally depend on the cones being proper. Cones that have an empty interior or contain straight lines can generally be represented by combinations of equations and constraints involving proper cones.
A: The interior of a set is the largest open set contained in the set. 
Another characterization is that $x$ is in the interior of a set $C$ iff there exists some $\epsilon>0$ such that $B(x,\epsilon) \subset C$.
For example, the cone $C_1 = [0,\infty) \subset \mathbb{R}$ is solid since $C_1^\circ = (0, \infty)$ is non-empty.
However, the cone $C_2 = C_1 \times \{0\} \subset \mathbb{R}^2$ is not solid as 
$C_2^\circ = \emptyset$.
One useful feature of convex sets is that if $x$ is in the interior and $y$ is in the set, then $\lambda x + (1-\lambda)y $ is in the interior for all $\lambda \in (0,1]$.
Perhaps more relevant to convex analysis is the relative interior, which is the interior relative to the affine hull of the set.
