Characterization of von Neumann-Boundedness in terms of Local Basis

Theorem

Let $\struct {X, \tau}$ be a topological vector space.

Let $\mathcal B$ be a local basis for $\mathbf 0_X$ in $\struct {X, \tau}$.

Let $E \subseteq X$.

Then $E$ is von Neumann-bounded if and only if:

for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$.

Proof

Necessary Condition

Suppose that $E$ is von Neumann-bounded.

Then:

for each open neighbourhood $V$ of ${\mathbf 0}_X$, there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$

Since $\mathcal B$ consists of open neighbourhoods of ${\mathbf 0}_X$, we in particular have:

for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$.

$\Box$

Sufficient Condition

Suppose that:

for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$.

Let $U$ be an open neighbourhood of ${\mathbf 0}_X$.

Then there exists some $V \in \mathcal V$ such that $V \subseteq U$ and:

there exists $s > 0$ such that $E \subseteq t V \subseteq t U$ for each $t > s$.

So $E$ is von Neumann-bounded.

$\blacksquare$