Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\Bbb F$.

Let $E$ be a von Neumann-bounded subset of $X$.

Then the closure $E^-$ of $E$ is von Neumann-bounded.

Proof
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.

From Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood: Corollary, there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that:


 * $W^- \subseteq V$

Since $E$ is von Neumann-bounded, there exists $s > 0$ such that:


 * $E \subseteq t W$ for $t \in \Bbb F$ with $t > s$.

From Topological Closure of Subset is Subset of Topological Closure, we have:


 * $E^- \subseteq \paren {t W}^-$ for $t \in \Bbb F$ with $t > s$.

From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have:


 * $\paren {t W}^- = t W^-$

Then we have:


 * $E^- \subseteq t W^- \subseteq t V$ for $t \in \Bbb F$ with $t > s$.

Since $V$ was arbitrary, we have that $E^-$ is von Neumann-bounded.