I am currently reading through Lectures on convex geometry and have come across an argument that I cannot quite wrap my head around.
In the Proof of Theorem 1.22, the following construction is used:
Let $x \in ext(P)$ be an extreme point of $P := \bigcap_{i=1}^k H_i$, where $H_i$ are closed half-spaces. Now, we construct $D := \bigcap_{i=1}^k A_i$ with $E_i$ being the hyperplanes that are used to construct the closed half-spaces $H_i$: $$ A_i = \left\{\begin{matrix}E_i,&x\in E_i\\int(H_i),&x\notin E_i\end{matrix}\right. $$ The intersection of only $E_i$ yields an affine subspace. The intersection of only $int(H_i)$ ($int$ yields the interior of the set) yields an open set. Hence, $D$ is relatively open. Also, $x \in D \subseteq P$.
Now to the part I have yet to understand:
The text says, that because $x$ is an extreme point of $P$ and $D$ is relatively open, it must follow that $dim(D)=0$, whereby $D = \{x\}$.
Why must the dimension ($dim(D)$ refers to the dimension of the affine hull of $D$) be zero? Does the set being open necessitate that?
The reasoning goes as follows:
Since $D$ is relatively open, $x$ must be an inner point. If $D$ was not 0-dimensional, $x$ would be the inner point of a line, which contradicts $x$ being an extreme point. Therefore, $D$ is 0-dimensional.