I am asked to find a closed convex set in $\Bbb R^2$ that has no extreme point. I know that the set $$S = \left\{ (x,y) \in \Bbb R^2 \mid 0 \leq x \leq 1, y \in \Bbb R \right\}$$ does not contain extreme points as it contains the lines $x=t$ where $0 \leq t \leq 1$, but I am wondering if I can prove that this set does not contain extreme points with the definition of extreme points:
$x$ is an extreme point if $\nexists x_1,x_2 \in S, x_1\neq x,x_2\neq x $ such that $x=\tfrac12(x_1+x_2)$.
On
Note that your definition of extreme point can be extended to any scalar $\lambda \in (0,1)$. That is, $z^*$ is an extreme point of $S$ if there does not exist $z_1,z_2 \neq z^*$ in $S$ and a scalar $\lambda \in (0,1)$ such that $$z^* = \lambda z_1 + (1-\lambda)z_2.$$ The idea is that a point $z^*$ is an extreme point of a set $S$ if it is not contained in the interior of a line between any two points within $S$. Perhaps you can prove this by contradiction? Suppose $(x^*,y^*) \in S$ is an extreme point. By your observation, $(x^*,y^*)$ must be of the form $(t,y^*)$ where $t \in [0,1]$ and $y^* \in \mathbb{R}$. If $t \notin \{0,1\}$, then let $(x_1,y_1) = (1,y^*)$ and $(x_2,y_2) = (0, y^*)$. Setting $\lambda = t$, we have that $$\lambda (x_1,y_1) + (1-\lambda)(x_2,y_2) = (t,ty^*) + (0,(1-t)y^*) = (t,y^*) = (x^*,y^*)$$ so $(x^*,y^*)$ is not an extreme point of $S$. Can you derive a similar contradiction if $t = 0,1$?
Edit: $\lambda$ lives in OPEN $(0,1)$.
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 $S$.