TITLE: Prove that the set $S=\{(x,y)\in R^2\,|\,0\leq x\leq 1,y\in R \}$ has no extreme points
QUESTION [5 upvotes]: I am asked to find a closed convex set in $R^2$ that has no extreme point.
I know that the set $S=\{(x,y)\in R^2|0\leq x\leq 1,y\in R \}$ does not contain extreme points as it contains the the lines $x=t$ where $0 \leq t \leq 1$, but I am wondering if I can prove that this set does not contain extrem points with the definition of extreme points.
Where x is an extreme point if $\nexists x_1,x_2 \in S, x_1\neq x,x_2\neq x $ such that $x=1/2(x_1+x_2)$

REPLY [3 votes]: Consider the subset of the plane, $S = \{ (x,y) \mid 0\le x \le 1 \}$, a closed vertical strip of width $1$.
The negation of $(x,y)$ being an extreme point of $S$ is that there do exist two points in $S$ such that $(x,y)$ is their midpoint.
An easy way to show this (since the strip extends infinitely up and down) is to adjust the vertical coordinate plus and minus by an equal amount.  That is, taking any $(x,y) \in S$, we see:
$$ (x,y) = \frac{1}{2} ((x,y+1) + (x,y-1)) $$
So no point $(x,y) \in S$ is an extreme point of $S$.