Corollary 1

— The interior $\overset{\circ}{A}$ of a convex set $\overline{A}$, is itself a convex set; if $\overset{\circ}{A}$ is not empty, then it coincides with the interior of $\overline{A}$, and $\overline{A}$ is a convex set that coincides with the closure of $\overset{\circ}{A}$.

It follows from prop. 16, that if $\overset{\circ}{A}$ is not empty, then it is a convex set and every point of $\overline{A}$ is a cluster point of $\overset{\circ}{A}$. Next we show that every interior point of $\overline{A}$ belongs to $\overset{\circ}{A}$. Let x be an interior point of $\overline{A}$ and suppose, for definiteness that $x = 0$. Let V be a symmetric neighbourhood of 0 that is contained in $\overline{A}$ and let $y \in \overset{\circ}{A} \cap V$; now $-y \in \overline{A}$, and therefore, by prop. 16, we see that $0 \in \overset{\circ}{A}$, if $y \neq 0$; this is obviously true if $y = 0$.