von Neumann bounded set

Remarks

Since the balanced neighbourhoods of $0$ form a neighbourhood base, an equivalent condition is: for every neighbourhood $U$ of $0$ there is $t > 0$ with $B \subseteq t U$. If the topology of $E$ is generated by a family $\mathcal{P}$ of seminorms, then $B$ is bounded if and only if $\sup_{x \in B} p(x) < \infty$ for every $p \in \mathcal{P}$. Example: in a normed space, the closed unit ball $B$ is bounded; a neighbourhood $U$ of $0$ contains a ball of some radius $\varepsilon > 0$, and then $B \subseteq tU$ for all $\abs{t} > 1 / \varepsilon$. Non-example: in $\R^2$ with the Euclidean norm, the line $L = \R \times \set{0}$ is not bounded; for $U$ the open unit disc, $tU$ is the disc of radius $\abs{t}$, which never contains $L$. Subsets of bounded sets, finite unions of bounded sets, closures of bounded sets and compact sets are bounded.