Prove that a non-empty closed convex $S$ has an extreme point iff it does not contain any line

Question

I am wondering how to prove the following theorem:

Let $S \subset \mathbf{R}^{n}$ be a non-empty closed convex set. Then $S$ has at least one extreme point iff $S$ does not contain any line.

Extreme point $x \in S$ means that there is no couple of points $y, z \in S$ and $\lambda \in (0;1)$ such that $x = \lambda y + (1-\lambda)z$.

I know that every non-empty closed convex set can be represented as an intersection of all closed half-spaces containing the set but I do not know if it is helpful in proving the theorem.

Answer 1

Assume that $S$ contains an extreme point $x$ and a line $L$. By convexity, the closure of the convex hull of $L\cup\{x\}$ is a subset of $S$. But the closure of the convex hull of $L\cup\{x\}$ contains the line parallel to $L$ passing through $x$. This is a contradiction to $x$ being an extreme point. Therefore, $S$ cannot contain an extreme point and a line at the same time.

Now assume that $S$ contains no line and no extreme points. Then $S$ is actually the empty set; a contradiction. (See Corollary 18.5.3 for the result and the beginning of Section 17 for the notions in R. T. Rockafellar: Convex Analysis. Princeton University Press, Princeton (1970).)