Proposition 9.1.4. Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^{o} \ne \emptyset$, then $\ol{A}= \ol{A^o}$.
Proof. Since $A^{o} \subset A$, $\ol{A^o}\subset \ol{A}$. Let $x \in A^{o}$, then for any $y \in \ol{A}$,
\[y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}}\subset \ol{A^o}\]
by Lemma 9.1.2.$\square$