Let X be a convex set. Prove that $x$ is an extreme point of X if and only if $X \setminus \{ x\}$ is convex.
I have already shown the $\Rightarrow$ direction. That is, take $x_1, x_2 \in X$ s.t. $x_1 \neq x_2$, then by convexity of $X$, $\alpha x_1 + (1-\alpha)x_2 \in X$. Since $x$ is an extreme point of $X$, there is no combination of $0< \alpha <1$ and $x_1, x_2 \in X$ (with $x_1 \neq x_2$) such that $x = \alpha x_1 + (1-\alpha)x_2$, then $\alpha x_1 + (1-\alpha)x_2 \in X \setminus \{ x\}$. Thus, $X \setminus \{ x\}$ is convex.
How can I prove that the converse is also true?
Best Answer
Suppose $X \setminus \{x\}$ is convex and $x=\alpha x_1+(1-\alpha) x_2$ with $0<\alpha<1$ and $x_1,x_2 \in X$. If $x_1=x$ we get $(1-\alpha)x=(1-\alpha)x_2$ so $x_2=x$. Thus $x=x_1=x_2$. Similarly, $x=x_2$ implies $x=x_1=x_2$. If $x \neq x_1$ and $x \neq x_2$ then $x_1,x_2 \in X \setminus \{x\}$ contradicting convexity of $X \setminus \{x\}$. We have proved that $x$ is an extreme point of $X$.