Proof. Let $A \subset E$ be a convex and circled set that absorbs every bounded set of $E$, and $\seq{U_n}\subset \cn_{E}(0)$ be a decreasing fundamental system of neighbourhoods at $0$. If $A \not\in \cn_{E}(0)$, then $U_{n} \setminus nA \ne \emptyset$ for all $n \in \natp$. For each $n \in \natp$, let $x_{n} \in U_{n} \setminus nA$, then $x_{n} \to 0$ as $n \to \infty$, and $\seq{x_n}$ is bounded. However, since $A$ absorbs every bounded set of $E$, there exists $n \in \natp$ such that $nA \supset \seq{x_n}$, which contradicts the assumption that $A \not\in \cn_{E}(0)$.$\square$