The Closure of a Balanced Set in a TVS

Proposition 1: Let $E$ be a topological vector space. If $A \subseteq E$ is balanced then $\overline{A}$ is balanced.

• Proof: Let $A \subseteq E$ be balanced. Then for all $\lambda \in \mathbf{F}$ with $|\lambda| \leq 1$ we have that $\lambda A \subseteq A$.

• Let $\lambda \in \mathbf{F}$ be such that $|\lambda| \leq 1$ and let $a \in \overline{A}$. Let $U$ be a neighbourhood of the origin. Since $E$ is a topological vector space, there exists a balanced neighbourhood of the origin, $V$, such that $V \subseteq U$ (see the proposition on the Bases of Neighbourhoods for a Point in a Topological Vector Space page). Then $a + V$ is a neighbourhood of $a$. Since $a \in \overline{A}$ we have that $A \cap (a + V) \neq \emptyset$. So let $x \in A \cap (a + V)$. Then:

(1)

\begin{align} \quad \lambda x \in \lambda (A \cap (a + V)) = (\lambda A) \cap (\lambda a + \lambda V) \end{align}

• Since $A$ and $V$ are balanced and since $|\lambda| \leq 1$ we have that $\lambda A \subseteq A$ and $\lambda V \subseteq V$ Thus:

(2)

\begin{align} \quad \lambda x &\in A \cap (\lambda a + V) \\ & \in A \cap (\lambda a + U) \end{align}

• Since every neighbourhood of $\lambda a$ is of the form $\lambda a + U$ (where $U$ is a neighbourhood of the origin) and since the above inclusion holds, we have that $A \cap (\lambda a + U) \neq \emptyset$. Thus $\lambda a \in \overline{A}$. Since this holds for all $a \in \overline{A}$, we see that $\lambda \overline{A} \subseteq \overline{A}$ whenever $\lambda \in \mathbf{F}$ is such that $|\lambda| \leq 1$. Thus $\overline{A}$ is balanced. $\blacksquare$