I'm new to affine spaces, and had a question I'm not quite certain about. Let $E$ be an affine space of dimension $d < \infty$, and $L, H$ affine subspaces of $E$ with $H$ a hyperplane. When is $L \cap H$ a hyperplane of $L$?
I believe this is the case if and only if $L \cap H$ is nonempty and properly contained in $L$. The implication $\Leftarrow$ is clear, and for the converse, my reasoning is as follows:
Under the assumption $L \cap H \neq \emptyset$, if $\vec{L}, \vec{H}$ are the translation spaces corresponding to $L, H$, the general theory of affine spaces tells us that
$$\textrm{Dim}(L \cap H) = \textrm{Dim}(L) + \textrm{Dim}(H) - \textrm{Dim}(\vec{L} + \vec{H})$$
We want to show that $L \cap H$ has dimension one less than that of $L$, which is the same as saying that $\vec{L}$ is not contained in $\vec{H}$. Suppose $\vec{L}$ were contained in $\vec{H}$. Choose an element $e \in L \cap H$. and let $l$ be any element of $L$. The translation $l - e$ lies in $\vec{L}$, hence in $\vec{H}$. But then $$l = e + (l-e) \in H$$ because $e \in H$ and $H$ is an orbit under the action of $\vec{H}$. This implies that $L \subseteq H$, so $L \cap H = L$, contradicting our assumption that $L \cap H \subsetneq L$. Is this reasoning correct?