Open and closed balanced convex zero-neighborhood refinements

Choose a symmetric open zero-neighborhood $O$ with $O+O\subseteq U$, then an open convex zero-neighborhood $C\subseteq O$. Joint scalar continuity at $(0,0)$ supplies $\delta >0$ and an open zero-neighborhood $W$ with $\{a:\mathrm{\mid }a\mathrm{\mid }<\delta \}W\subseteq C$. Put $W_0=(\delta \mathrm{/}2)W$, which is open and contains zero.

Define $B={\bigcup }_{\mathrm{\mid }a\mathrm{\mid }\le 1}a{W}_0$. If $aw\in B$ and $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, then $\lambda (aw)=(\lambda a)w\in B$, so $B$ is balanced. It contains $W_0$. For $w=(\delta \mathrm{/}2)v\in W_0$, $aw=(a\delta \mathrm{/}2)v\in C$ since $\mathrm{\mid }a\delta \mathrm{/}2\mathrm{\mid }<\delta$, so $B\subseteq C$. Moreover $B$ is open: all nonzero dilates are open, and the zero dilate contributes only zero, which already belongs to $W_0$.

Put $V=\mathrm{co}(B)$. Then $W_0\subseteq B\subseteq V\subseteq C\subseteq O$, and $V$ is convex. For $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, distributing $\lambda$ through a finite convex sum leaves its summands in $B$, proving balance of $V$. To prove openness, represent $v={\sum }_j{t}_j{b}_j\in V$. Some $t_k>0$, and $t_{k}B+{\sum }_{j\mathrm{\ne }k}{t}_j{b}_j$ is an open subset of $V$ containing $v$. Thus $V$ is an open balanced convex zero-neighborhood. Zero coefficients cause no problem because the sum of the coefficients is one.

If $x\in \bar{V}$, then $x-O$ is an open neighborhood of $x$ and meets $V$. Write $v=x-o\in V$ with $o\in O$; hence $x=v+o\in V+O\subseteq O+O\subseteq U$. Thus $\bar{V}\subseteq U$. By closure preservation, $\bar{V}$ is convex and balanced; it is closed and contains the open zero-neighborhood $V$. Finally balance implies $-V\subseteq V$ and applying negation again gives $V=-V$; the same applies to every balanced set, including the empty set.