Every TVS Has a Base of Closed and Balanced Nbds. of the Origin

# Every Topological Vector Space Has a Base of Closed and Balanced Neighbourhoods of the Origin

Proposition 1: Let $E$ be a topological vector space. Then $E$ has a base of closed and balanced neighbourhoods of the origin. |

**Proof:**Let $\mathcal U$ be a base of neighbourhoods of the origin. By the proposition on the Bases of Neighbourhoods for a Point in a Topological Vector Space, for each $U \in \mathcal U$ there exists a balanced neighbourhood $V_U$ of the origin such that $V_U \subseteq V_U + V_U \subseteq U$ (by the proposition on The Closure of a Balanced Set in a TVS page). Let:

\begin{align} \quad \mathcal V := \{ V_U : U \in \mathcal U \} \end{align}

- Then $\mathcal V$ is a collection of balanced sets, that is also a base of neighbourhoods of the origin, since given a neighbourhood $W$ of the origin, since $\mathcal U$ is a base of neighbourhoods of the origin there exists $U \in \mathcal U$ with $U \subseteq W$. Then:

\begin{align} \quad o \in V_U \subseteq U \subseteq W \end{align}

- Let:

\begin{align} \quad \overline{\mathcal V} := \{ \overline{V_U} : U \in \mathcal U \} \end{align}

- Since $E$ is a topological vector space and each $V_U$ is balanced, we see that each $\overline{V_U}$ is closed and balanced. Thus $\overline{\mathcal V}$ is a collection of closed and balanced neighbourhoods of the origin. All that remains to show is that $\overline{V}$ is a base of neighbourhoods of the origin.

- Let $W$ be any neighbourhood of the origin . Let $U$ be any neighbourhood of the origin such that $U \subseteq W$. Since $E$ is a topological space, there exists a balanced neighbourhood $V_U$ of the origin with $V_U \subseteq V_U + V_U \subseteq U$.

- Let $x \in \overline{V_U}$. Since $x + V_U$ is a neighbourhood of $x$, we have that $\overline{V_U} \cap (x + V_U) \neq \emptyset$. So there exists a point $y \in \overline{V_U} \cap (x + V_U)$. Then $y = v'$ and $y = x + v''$ for some $v', v'' \in V_U$. Combining these equations yields $v' = x + v''$, or equivalently:

\begin{align} \quad x = v' - v'' = v' - v'' \end{align}

- Thus $x \in \overline{V_U}$ implies that:

\begin{align} \quad x \in V_U \subseteq V_U - V_U = V_U + V_U \subseteq U \subseteq W \end{align}

- (Note that since $V_U$ is a balanced set, we have that $-V_U = V_U$).

- Since this holds for all $x \in \overline{V_U}$ we have that:

\begin{align} \quad \overline{V_U} \subseteq W \end{align}

- Thus $\mathcal V$ is a base of closed and balanced neighbourhoods of the origin. $\blacksquare$