Is an unit-cube polyhedron? What about other platonic solids?

Question

Definitions

According to my linear programming course and screenshot here (Finnish), a polyhedron is such that it can be constrained by a finite amount of inequalities such that $$P=\{\bar x\in \mathbb R^n | A \bar x\geq \bar b\}, A\in \mathbb R^{m\times n},\bar b\in \mathbb R^m$$ and a convex function $f(x)$ must satisfy $$f(\lambda \bar x+(1-\lambda)\bar y)\leq \lambda f(\bar x)+(1-\lambda) f(\bar y) \text{, } \forall \bar x, \bar y, \lambda \in [0,1]$$ and a convex set $C$ is such that $$\bar x, \bar y \in C\rightarrow \lambda \bar x+(1-\lambda)\bar y\in C.$$

By this calculation, I think an unit-cube is not a polyhedron but an open half-space such as $\{x>0,y>0,z>0| x,y,z\in \mathbb R\}$ is a polyhedron. This strikes my intuition because I have earlier assumed polyhedron to be a closed geometrical object such as platonic shapes, terminology used in my geometry class but not in my linear optimization class! I want to recheck this:

Is an unit-cube a polyhedron? What about other platonic shapes, are they polyhedrons if they are closed? Do the definitions change between areas such as geometry and linear programming?

Answer 1

A unit cube is delimited by six inequalities. You have listed three: $x \ge 0, y \ge 0, z \ge 0$. There are also $x \le 1, y \le 1, z \le 1$ You can get inequalities of this form by making $A = \begin {pmatrix} -1&0&0 \end {pmatrix}$ and $b=1$

Answer 2

I cannot yet fully understand Ross Millikan's method but below how I would show that an unit-cube is a polyhedron. I showed that it is possible to have a finite amount of $Ax\geq b$-form inequalities for an unit-cube.

As for the general question about other platonic solids, a similar method should prove them to be polyhedrons. I think the definitions are the same between linear programming and geometry -- the earlier definition is just succint formalism, a bit tricky one.