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

 A: It can be proved by following three steps.
(a) Let $\left\{{\Omega_{\alpha}}\right\} (\alpha \in I)$ be a collection of convex subsets of $\mathbb{R}^n$. Then $\bigcap_{\alpha \in I}\Omega_{\alpha}$ is also a convex.
proof: Taking any $x_1,x_2\in \bigcap_{\alpha\in I}\Omega_{\alpha}$. We get that $x_1,x_2\in \Omega_{\alpha}$ for $\alpha\in I$. And then we have $\theta x_1+(1-\theta)x_2\in \Omega_{\alpha}$ for any $\theta\in [0,1]$ since $\Omega_{\alpha}$ are convex sets. Thus $\theta x_1+(1-\theta)x_2\in \bigcap_{\alpha\in I}{\Omega_{\alpha}}$.
(b) Hyperplanes are convex and halfsapces are also convex.
\begin{equation}
\text{Hyperplanes}: \left\{{x|a^Tx=b}\right\} \quad \text{Halfspaces}: \left\{{x|a^Tx\leq b}\right\}
\end{equation}
proof: Assume that $x_1,x_2\in \Omega$, and we have $a^Tx_1=b, a^Tx_2=b$. Hence we can get
\begin{equation}
a^T(\theta x_1+(1-\theta)x_2)=\theta a^T x_1+(1-\theta)a^Tx_2=b
\end{equation}
 i.e., $(\theta x_1+(1-\theta)x_2)\in \Omega$. similarly, we also can prove that halfspaces are convex.
(c) As we observed from the definition of polyhedra. A polyhedron is defined as the solution set of a finite number of linear equalities and inequalities. It mean that a ployhedron is the intersection of a finite number of halfspaces and hyperplanes. Based on (b), we know that halfspaces and hyperplanes are convex. Furthermore, we know polyhedron is convex based on (a).
A: 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. 
A: Look at Boyd's book Section 2.2.4
http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf
A: A polyhedron can be defined as a finite intersection of halfspaces and hyperplanes. Halfspaces and hyperplanes are convex sets, the intersection of convex sets is a convex set, and thus all polyhedrons are convex.
