# If there always a supporting hyperplane $H$ through an extreme point $x$ of a convex set $X$ such that $H\cap X=\{x\}$?

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 $x = \lambda y + (1-\lambda)z$.

No - $x$ could be a $C^1$ boundary point separating a straight section of the boundary from a curved section, as in this example constructed from a square and a circle quadrant:
Here the extreme point $x$ of $X$ has as its only supporting hyperplane the tangent line $H$, which intersects $X$ along a nontrivial interval.