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I have some hardship to conceive the intuitive meaning of a polyhedron which does not contain a line.

Suppose polyhedron $P = \{ x\in \mathbb{R^n : Ax \leq b, A\in \mathbb{R^{ m\times n }}} \}$ is not empty.

It is said that if P does not contain a line, then it has at least one extreme point, n independent active constraints exist and and therefore a basic feasible solution exists. Why is here the argument of a polyhedron not containing a line leading those following assertions? Thanks!

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    Consider the polyhedron $P$ in $\mathbb{R^2}$ given by $x + y \le 1$. $P$ is not empty, but has no extreme points. So to guarantee extreme points, some additional conditions are needed. Nonempty is not sufficient. – 2017-02-09
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    ... and in fact, a closed convex polyhedron $P$ that contains a line in direction ${\bf v}$ can't have extreme points, because if it contains ${\bf x}$ then it also contains ${\bf x} \pm \bf v$. – 2017-02-09
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    @RobertIsrael it can not contain a line in direction v because this implies unboundedness of the feasibility region? Can we say this? – 2017-02-09
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    @RobertIsrael ah no! because if this feasibility region contains x,x+v and x-v then we can write x as a convex combination of x+v and x-v such that x = 1/2 (x+v) + 1/2 (x-v) which is a contradiction because an extreme point can not be a convex combination of two points which are contained in the polyhedron. many thanks for the idea! – 2017-02-09

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