# Explain `All polyhedrons are convex sets´

My teacher in course in Mat-2.3140 of Aalto University claims that 'All polyhedrons are convex sets' here. This premise was in a false-or-not-problem 'The feasible set of linear integer problem is polyhedron'. You can see below the screenshot of the solution.

Wikipedia shows nonconvex polyhedrons such as orthogonal polyhedron here.

What should I now believe? Is polyhedron convex or not?

Definitions on the lecture slides (p.8, L4)

Polyhedron is 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 $$\bar x, \bar y \in C\rightarrow \lambda \bar x+(1-\lambda)\bar y\in C.$$

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Just so you know, your first link is only available to Aalto students. As to your question - does the course contain a definition of polyhedron? It may simply be a question of two different definitions, one allowing non-convex things and the other not. –  Matthew Pressland Jan 11 '13 at 10:46
@MattPressland shared relevant information in the q. –  hhh Mar 4 '13 at 14:34

I suspect you are confused with the definition. Usually a a polyhedron is defined by specifying a finite subset of $n-1$ dimensional affine subspaces in $\mathbb{R}^{n}$. In this way what you get is always convex. This is the definition people use when work on combinatorical topology or algebraic combinatorics. You should confirm this with your teacher though.
Is this "[u]sually a polyhedron is defined by specifying a finite subset of $n-1$ dimensional affine subspaces in $\mathbb R^n$" the method Ross uses here? –  hhh Feb 26 '13 at 22:45