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.
(1)
\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:
(2)
\begin{align} \quad o \in V_U \subseteq U \subseteq W \end{align}
• Let:
(3)
\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:
(4)
\begin{align} \quad x = v' - v'' = v' - v'' \end{align}
• Thus $x \in \overline{V_U}$ implies that:
(5)
\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:
(6)
\begin{align} \quad \overline{V_U} \subseteq W \end{align}
• Thus $\mathcal V$ is a base of closed and balanced neighbourhoods of the origin. $\blacksquare$
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