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

 A: 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 $b=1$
A: 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.
A: Definition of polyhedron,  "A polyhedron is a close solid having certain no. of (regular or irregular) polygonal (flat) faces such that the sum of the solid angles subtended by all its faces at any point exactly inside it must be $4\pi$ Ste-radian". Mathematically, it is given as $$\bbox [4pt, border: 1px solid blue;] {\omega _{total}=4\pi \space sr} $$
Where, $\omega _{total} \space \to$ total solid angle subtended by all the faces of any existing polyhedron at any arbitrary point inside it.
This definition is valid for all the regular and irregular polyhedrons having polygonal faces either regular or irregular or both.  
To prove that a cube is a polyhedron, let's consider the center of the cube as an internal point (we can select any other point as well) only for simplification of calculations, 
Now, let a be the edge length of a cube then we know that each face of a square is at a normal distance $ \frac{a}{2} $ . Now the solid angle ($\omega$) subtended by any square of side $a$ ant any point ying at a normal distance $d$ is given as
$$\omega =4sin^{-1}\left(\frac{a^2}{a^2+4d^2}\right)$$
Above formula, for a square plane, is derived from basic formula of solid angle $\iint_Ssin\theta d \theta d \phi $
now, substituting the corresponding values in the formula, we get the solid angle subtended by each square face of the cube at its center, as follows
$$\omega =4sin^{-1}\left(\frac{a^2}{a^2+4\left(\frac{a}{2}\right)^2}\right)=4sin^{-1}\left(\frac{1}{2} \right)=4\left(\frac{\pi}{6}\right)=\frac{2\pi}{3} sr$$
Now, the total solid angle ($\omega _{total}$) subtended by all 6 square faces at the center of cube 
$$\omega _{total}=(no. \space of \space faces) \cdot (solid \space angle \space subtended \space by \space each \space face)=6\left(\frac{2\pi}{3}\right)=4\pi \space sr$$
Hence, the cube is a polyhedron (also called as regular hexahedron). It is also a (closed) solid. 
