Let $X \subset \mathbb{R}^n$ be a nonempty convex set, and suppose $x$ is an extreme point of $X$. Does there always exist a supporting hyperplane $H$ to $X$ at $x$ (i.e., $x$ lies on $H$) such that $H \cap X = \{x\}$?
Thoughts: My intuition says that the above statement holds. The above statement is equivalent to the following: "does there always exist a vector $c \in \mathbb{R}^n$ such that $x$ is the unique solution to the Problem $ \underset{z \in X}{\min} c^{\text{T}}z $?" From Chapter 2 of Bertsimas and Tsitsiklis' Introduction to to Linear Optimization, I know that the above result is true when $X$ is a polyhedron, but the generalization to general convex sets doesn't seem to follow directly.
For reference, the definition of extreme point I use is as follows: $x$ is an extreme point of a convex set $X$ if there do not exist two vectors $y, z \in X$, with $y \neq x$ and $z \neq x$, and a scalar $\lambda \in [0,1]$ such that $x = \lambda y + (1-\lambda)z$.
