Bounded set (topological vector space)

Examples and sufficient conditions

Unless indicated otherwise, a topological vector space (TVS) need not be Hausdorff nor locally convex.

• Finite sets are bounded.
• Every totally bounded subset of a TVS is bounded.
• Every relatively compact set in a topological vector space is bounded. If the space is equipped with the weak topology the converse is also true.
• The set of points of a Cauchy sequence is bounded, the set of points of a Cauchy net need not be bounded.
• The closure of the origin (referring to the closure of the set ${\displaystyle \{0\}}$) is always a bounded closed vector subspace. This set ${\displaystyle {\mathrm{cl}}_X\{0\}}$ is the unique largest (with respect to set inclusion ${\displaystyle \subseteq }$) bounded vector subspace of ${\displaystyle X.}$ In particular, if ${\displaystyle B\subseteq X}$ is a bounded subset of ${\displaystyle X}$ then so is ${\displaystyle B+{\mathrm{cl}}_X\{0\}.}$

Unbounded sets

A set that is not bounded is said to be unbounded.

Any vector subspace of a TVS that is not a contained in the closure of ${\displaystyle \{0\}}$ is unbounded

There exists a Fréchet space ${\displaystyle X}$ having a bounded subset ${\displaystyle B}$ and also a dense vector subspace ${\displaystyle M}$ such that ${\displaystyle B}$ is not contained in the closure (in ${\displaystyle X}$) of any bounded subset of ${\displaystyle M.}$

Stability properties

• In any TVS, finite unions, finite Minkowski sums, scalar multiples, translations, subsets, closures, interiors, and balanced hulls of bounded sets are again bounded.
• In any locally convex TVS, the convex hull (also called the convex envelope) of a bounded set is again bounded. However, this may be false if the space is not locally convex, as the (non-locally convex) Lp space ${\displaystyle L^p}$ spaces for ${\displaystyle 0<p<1}$ have no nontrivial open convex subsets.
• The image of a bounded set under a continuous linear map is a bounded subset of the codomain.
• A subset of an arbitrary (Cartesian) product of TVSs is bounded if and only if its image under every coordinate projections is bounded.
• If ${\displaystyle S\subseteq X\subseteq Y}$ and ${\displaystyle X}$ is a topological vector subspace of ${\displaystyle Y,}$ then ${\displaystyle S}$ is bounded in ${\displaystyle X}$ if and only if ${\displaystyle S}$ is bounded in ${\displaystyle Y.}$
  • In other words, a subset ${\displaystyle S\subseteq X}$ is bounded in ${\displaystyle X}$ if and only if it is bounded in every (or equivalently, in some) topological vector superspace of ${\displaystyle X.}$

Properties

A locally convex topological vector space has a bounded neighborhood of zero if and only if its topology can be defined by a single seminorm.

The polar of a bounded set is an absolutely convex and absorbing set.

Mackey's countability condition—If ${\displaystyle B_1,B_2,B_3,\dots }$ is a countable sequence of bounded subsets of a metrizable locally convex topological vector space ${\displaystyle X,}$ then there exists a bounded subset ${\displaystyle B}$ of ${\displaystyle X}$ and a sequence ${\displaystyle r_1,r_2,r_3,\dots }$ of positive real numbers such that ${\displaystyle B_i\subseteq r_{i}B}$ for all ${\displaystyle i\in \mathbb{N}}$ (or equivalently, such that ${\displaystyle \frac{1}{r_1}{B}_1\cup \frac{1}{r_2}{B}_2\cup \frac{1}{r_3}{B}_3\cup \cdots \subseteq B}$).

Using the definition of uniformly bounded sets given below, Mackey's countability condition can be restated as: If ${\displaystyle B_1,B_2,B_3,\dots }$ are bounded subsets of a metrizable locally convex space then there exists a sequence ${\displaystyle t_1,t_2,t_3,\dots }$ of positive real numbers such that ${\displaystyle t_1{B}_1,t_2{B}_2,t_3{B}_3,\dots }$ are uniformly bounded. In words, given any countable family of bounded sets in a metrizable locally convex space, it is possible to scale each set by its own positive real so that they become uniformly bounded.

Generalizations

Uniformly bounded sets

A family of sets ${\displaystyle \mathcal{B}}$ of subsets of a topological vector space ${\displaystyle Y}$ is said to be uniformly bounded in ${\displaystyle Y,}$ if there exists some bounded subset ${\displaystyle D}$ of ${\displaystyle Y}$ such that $${\displaystyle B\subseteq D\text{ for every }B\in \mathcal{B},}$$ which happens if and only if its union $${\displaystyle \cup \mathcal{B}:=\bigcup _{B\in \mathcal{B}}B}$$ is a bounded subset of ${\displaystyle Y.}$ In the case of a normed (or seminormed) space, a family ${\displaystyle \mathcal{B}}$ is uniformly bounded if and only if its union ${\displaystyle \cup \mathcal{B}}$ is norm bounded, meaning that there exists some real ${\displaystyle M\ge 0}$ such that $\Vert b\Vert \le M$ for every ${\displaystyle b\in \cup \mathcal{B},}$ or equivalently, if and only if $\underset{\stackrel{b\in B}{B\in \mathcal{B}}}{sup}\Vert b\Vert <\mathrm{\infty }.$

A set ${\displaystyle H}$ of maps from ${\displaystyle X}$ to ${\displaystyle Y}$ is said to be uniformly bounded on a given set ${\displaystyle C\subseteq X}$ if the family ${\displaystyle H(C):=\{h(C):h\in H\}}$ is uniformly bounded in ${\displaystyle Y,}$ which by definition means that there exists some bounded subset ${\displaystyle D}$ of ${\displaystyle Y}$ such that ${\displaystyle h(C)\subseteq D\text{ for all }h\in H,}$ or equivalently, if and only if $\cup H(C):=\bigcup _{h\in H}h(C)$ is a bounded subset of ${\displaystyle Y.}$ A set ${\displaystyle H}$ of linear maps between two normed (or seminormed) spaces ${\displaystyle X}$ and ${\displaystyle Y}$ is uniformly bounded on some (or equivalently, every) open ball (and/or non-degenerate closed ball) in ${\displaystyle X}$ if and only if their operator norms are uniformly bounded; that is, if and only if $\underset{h\in H}{sup}\Vert h\Vert <\mathrm{\infty }.$

Proposition—Let ${\displaystyle H\subseteq L(X,Y)}$ be a set of continuous linear operators between two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ and let ${\displaystyle C\subseteq X}$ be any bounded subset of ${\displaystyle X.}$ Then ${\displaystyle H}$ is uniformly bounded on ${\displaystyle C}$ (that is, the family ${\displaystyle \{h(C):h\in H\}}$ is uniformly bounded in ${\displaystyle Y}$) if any of the following conditions are satisfied:

1. ${\displaystyle H}$ is equicontinuous.
2. ${\displaystyle C}$ is a convex compact Hausdorff subspace of ${\displaystyle X}$ and for every ${\displaystyle c\in C,}$ the orbit ${\displaystyle H(c):=\{h(c):h\in H\}}$ is a bounded subset of ${\displaystyle Y.}$

Since every singleton subset of ${\displaystyle X}$ is also a bounded subset, it follows that if ${\displaystyle H\subseteq L(X,Y)}$ is an equicontinuous set of continuous linear operators between two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ (not necessarily Hausdorff or locally convex), then the orbit $H(x):=\{h(x):h\in H\}$ of every ${\displaystyle x\in X}$ is a bounded subset of ${\displaystyle Y.}$

Bounded subsets of topological modules

The definition of bounded sets can be generalized to topological modules. A subset ${\displaystyle A}$ of a topological module ${\displaystyle M}$ over a topological ring ${\displaystyle R}$ is bounded if for any neighborhood ${\displaystyle N}$ of ${\displaystyle 0_M}$ there exists a neighborhood ${\displaystyle w}$ of ${\displaystyle 0_R}$ such that ${\displaystyle wA\subseteq B.}$