Im learning and we have a theorem that says:
Let $C$ be a non-empty, convex subset of $\mathbb R^d$ and let $p \in \mathbb R^d$ be a point which is not in the closure of $C$. Then there exists a strict separating hyperplane, which means there exists an $\eta \in \mathbb R^d\backslash \{0\}$ such that $p\cdot\eta<c\cdot \eta$ for all $c\in C$.
My Question is the following statement from our professor and what its proof is.
He said said that for a random non-empty convex subset $C \subset \mathbb R^d$ and a point $p \in \mathbb R^d$ with $p\notin C$ there generally doesnt exist a $\eta \in \mathbb R^d\backslash \{0\}$ such that $p\cdot\eta<c\cdot \eta$ for all $c\in C$.
The difference between the theorem and statement is obviously that the point has to be not in the closure of $C$ if there should be a strict separating hyperplane.
A proof or an example would be nice.
Try $C=\{(x,y)\in\mathbb R^2\mid y\gt0\}\cup\{(0,0)\}$ (can you check that $C$ is indeed convex?).
Then, for $\eta\ne(0,0)$, the set $C_\eta=\{c\cdot\eta\mid c\in C\}$ has no lower bound except if $\eta$ is a positive multiple of $(0,1)$. And if $\eta=(0,1)$, then $C_\eta=[0,+\infty)$ hence the property that $p\cdot\eta\lt c\cdot\eta$ for every $c$ in $C$ is equivalent to $p\cdot\eta\lt0$.
Thus $\eta$ does not exist for the points $p=(x,0)$, $x\ne0$, although these points are not in $C$. Obviously, all these points $p$ are in the closure of $C$, as was to be expected.