---
tag: YENN31BFZA
kind: definition
name: von Neumann bounded set
msc: [46A17, 46A03]
defines: 3BAJ8ZTVPX
uses: [0ZCY0Z80B6]
sources:
  - url: https://en.wikipedia.org/wiki/Bounded_set_(topological_vector_space)
---

## Definition

Let $\mathbb{K}$ be $\R$ or $\C$, and let $E$ be a topological vector space over $\mathbb{K}$.
A subset $B \subseteq E$ is *von Neumann bounded*, or *bounded*, if the following holds.
For every [[neighbourhood|0ZCY0Z80B6]] $U$ of $0$ in $E$ there is $s > 0$ such that $B \subseteq t U$ for every $t \in \mathbb{K}$ with $\abs{t} > s$.

## 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.