Lemma 9.1.2. Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^{o}$, and $y \in \ol{A}$, then
\[\bracs{tx + (1 - t)y|t \in (0, 1)}\subset A^{o}\]
Proof. Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^{o} \in \cn^{o}(y)$. Since $y \in \ol{A}$, $\alpha A^{o} \cap A \ne \emptyset$, so there exists $z \in A^{o}$ such that $\alpha z \in A$.
Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
\[\mu z + (1 - \mu)\alpha z = \frac{\alpha z}{\alpha - 1}+ \frac{(\alpha - 1 - \alpha)\alpha z}{\alpha - 1}= 0\]
By (TVS1) and (TVS2),
\[U = \underbrace{\bracs{\mu w + (1 - \mu)\alpha z|w \in A^o}}_{\subset A}\in \cn^{o}(0)\]
so $0 \in A^{o}$.$\square$