The Closure of an Absolutely Convex Set in a TVS

# The Closure of an Absolutely Convex Set in a TVS

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

**Proof:**Let $A \subseteq E$ be absolutely convex. Then, for all $x, y \in A$ and for all $\lambda, \mu \in \mathbf{F}$ with $|\lambda| + |\mu| \leq 1$ we have that $\lambda x + \mu y \in A$.

- Let $a, b \in \overline{A}$ and let $\lambda, \mu \in \mathbf{F}$ be such that $|\lambda| + |\mu| \leq 1$.

- Since $E$ is a topological vector space, for each neighbourhood $U$ of the origin, there exists a balanced neighbourhood of the origin, $V$, such that:

\begin{align} \quad V + V \subseteq U \end{align}

- (See the proposition on the Bases of Neighbourhoods for a Point in a Topological Vector Space page).

- Since $E$ is a topological vector space and $V$ is a neighbourhood of the origin, we have that $a + V$ is neighbourhood of $a$ and $b + V$ is a neighbourhood of $b$. Since $a \in \overline{A}$, we have that $A \cap (a + V) \neq \emptyset$. Similarly, since $b \in \overline{A}$, we have that $A \cap (b + V) \neq \emptyset$. So take $x \in A \cap (a + V)$ and take $y \in A \cap (b + V)$. Then:

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

- Hence:

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

- But since $V$ is balanced, and since $|\lambda| \leq |\lambda| + |\mu| \leq 1$ we see that $\lambda V \subseteq V$. A similar argument goes for observing that $\mu V \subseteq V$. Thus:

\begin{align} \quad \lambda x + \mu y & \in (\lambda A + \mu A) \cap (\lambda a + \mu b + V + V) \\ & \in (\lambda A + \mu A) \cap (\lambda a + \mu b + W) \end{align}

- But observe that $\lambda A + \mu A \subseteq A$. Indeed, if $z \in \lambda A + \mu A$ then $z = \lambda a' + \mu b'$ for some $a', b' \in A$. But since $A$ is absolutely convex and since $|\lambda| + |\mu| \leq 1$ we have that $\lambda a' + \mu b' \in A$. Thus:

\begin{align} \quad \lambda x + \mu y \subseteq A \cap (\lambda a + \mu b + W) \end{align}

- Since every neighbourhood of the point $\lambda a + \mu b$ is of the form $\lambda a + \mu b + U$ where $U$ is a neighbourhood of the origin, the above inclusion shows that:

\begin{align} \quad A \cap (\lambda a + \mu b + U) \neq \emptyset \end{align}

- Thus $\lambda a + \mu b \in \overline{A}$ and consequently, $\overline{A}$ is absolutely convex. $\blacksquare$