In functional analysis and related areas of mathematics, a set in a topological vector space is called bounded or von Neumann bounded, if every neighborhood of the zero vector can be inflated to include the set. A set that is not bounded is called unbounded. Bounded sets are a natural way to define locally convex polar topologies on the vector spaces in a dual pair, as the polar set of a bounded set is an absolutely convex and absorbing set. The concept was first introduced by John von Neumann and Andrey Kolmogorov in 1935.

Definition

Suppose X is a topological vector space (TVS) over a field \mathbb{K}. A subset B of X is called **' or just **' in X if any of the following equivalent conditions are satisfied:

<ol> <li>: For every [neighborhood](https://bliptext.com/articles/neighborhood-mathematics) V of the origin there exists a real r > 0 such that for all scalars s satisfying |s| \geq r. <li>B is [absorbed](https://bliptext.com/articles/absorbing-set) by every [neighborhood](https://bliptext.com/articles/neighborhood-mathematics) of the origin.</li> <li>For every neighborhood V of the origin there exists a scalar s such that </li> <li>For every neighborhood V of the origin there exists a real r > 0 such that for all scalars s satisfying |s| \leq r.</li> <li>For every neighborhood V of the origin there exists a real r > 0 such that for all real </li> <li>Any one of statements (1) through (5) above but with the word "neighborhood" replaced by any of the following: "[balanced](https://bliptext.com/articles/balanced-set) neighborhood," "open [balanced](https://bliptext.com/articles/balanced-set) neighborhood," "closed [balanced](https://bliptext.com/articles/balanced-set) neighborhood," "open neighborhood," "closed neighborhood". <li>For every sequence of scalars that converges to 0 and every sequence in B, the sequence converges to 0 in X. <li>For every sequence in B, the sequence converges to 0 in X.</li> <li>Every [countable](https://bliptext.com/articles/countable-set) subset of B is bounded (according to any defining condition other than this one).</li> </ol> If \mathcal{B} is a [neighborhood basis](https://bliptext.com/articles/neighborhood-basis) for X at the origin then this list may be extended to include: <li>Any one of statements (1) through (5) above but with the neighborhoods limited to those belonging to </ol> If X is a [locally convex](https://bliptext.com/articles/locally-convex-topological-vector-space) space whose topology is defined by a family \mathcal{P} of continuous [seminorms](https://bliptext.com/articles/seminorm), then this list may be extended to include: <li>p(B) is bounded for all </li> <li>There exists a sequence of non-zero scalars such that for every sequence in B, the sequence is bounded in X (according to any defining condition other than this one).</li> <li>For all B is bounded (according to any defining condition other than this one) in the [semi normed space](https://bliptext.com/articles/semi-normed-space) (X, p).</li> <li>B is weakly bounded, i.e. every continuous linear functional is bounded on B </li> </ol> If X is a [normed space](https://bliptext.com/articles/[norm](https://bliptext.com/articles/norm-mathematics)ed-space) with [norm](https://bliptext.com/articles/norm-mathematics) \|\cdot\| (or more generally, if it is a semi[normed space](https://bliptext.com/articles/[norm](https://bliptext.com/articles/norm-mathematics)ed-space) and \|\cdot\| is merely a semi[norm](https://bliptext.com/articles/norm-mathematics)), then this list may be extended to include: <li>B is a norm bounded subset of By definition, this means that there exists a real number r > 0 such that for all b \in B.</li> <li> <li>B is a subset of some (open or closed) ball. </ol> If B is a vector subspace of the TVS X then this list may be extended to include: <li>B is contained in the closure of \{0\}. </ol> A subset that is not bounded is called.

Bornology and fundamental systems of bounded sets

The collection of all bounded sets on a topological vector space X is called the or the A or of X is a set \mathcal{B} of bounded subsets of X such that every bounded subset of X is a subset of some The set of all bounded subsets of X trivially forms a fundamental system of bounded sets of X.

Examples

In any locally convex TVS, the set of closed and bounded disks are a base of bounded set.

Examples and sufficient conditions

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

<ul> <li>Finite sets are bounded.</li> <li>Every [totally bounded](https://bliptext.com/articles/totally-bounded) subset of a TVS is bounded.</li> <li>Every [relatively compact set](https://bliptext.com/articles/relatively-compact-set) in a topological vector space is bounded. If the space is equipped with the [weak topology](https://bliptext.com/articles/weak-topology-polar-topology) the converse is also true.</li> <li>The set of points of a [Cauchy) sequence](https://bliptext.com/articles/cauchy-sequence) is bounded, the set of points of a [Cauchy](https://bliptext.com/articles/cauchy-[net](https://bliptext.com/articles/net-mathematics)) [net](https://bliptext.com/articles/net-mathematics) need not be bounded.</li> <li>The closure of the origin (referring to the closure of the set \{0\}) is always a bounded closed vector subspace. This set is the unique largest (with respect to set inclusion ) bounded vector subspace of X. In particular, if is a bounded subset of X then so is </li> </ul> **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 \{0\} is unbounded There exists a [Fréchet space](https://bliptext.com/articles/fr-chet-space) X having a bounded subset B and also a dense vector subspace M such that B is contained in the closure (in X) of any bounded subset of M.

Stability properties

<ul> <li>In any TVS, finite [unions](https://bliptext.com/articles/union-set-theory), finite [Minkowski sums](https://bliptext.com/articles/minkowski-sum), scalar multiples, translations, subsets, [closures](https://bliptext.com/articles/closure-topology), [interiors](https://bliptext.com/articles/interior-topology), and [balanced hulls](https://bliptext.com/articles/balanced-set) of bounded sets are again bounded.</li> <li>In any [locally convex TVS](https://bliptext.com/articles/locally-convex-topological-vector-space), the [convex hull](https://bliptext.com/articles/convex-set) (also called the [convex envelope](https://bliptext.com/articles/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](https://bliptext.com/articles/lp-space) L^p spaces for 0 < p < 1 have no nontrivial open convex subsets.</li> <li>The image of a bounded set under a [continuous linear map](https://bliptext.com/articles/continuous-linear-map) is a bounded subset of the codomain.</li> <li>A subset of an arbitrary [(Cartesian) product](https://bliptext.com/articles/product-space) of TVSs is bounded if and only if its image under every coordinate projections is bounded.</li> <li>If and X is a topological vector subspace of Y, then S is bounded in X if and only if S is bounded in Y. </ul>

Properties

A locally convex topological vector space has a bounded neighborhood of zero if and only if its topology can be defined by a seminorm. The polar of a bounded set is an absolutely convex and absorbing set. Using the definition of uniformly bounded sets given below, Mackey's countability condition can be restated as: If are bounded subsets of a metrizable locally convex space then there exists a sequence of positive real numbers such that 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 \mathcal{B} of subsets of a topological vector space Y is said to be in Y, if there exists some bounded subset D of Y such that which happens if and only if its union is a bounded subset of Y. In the case of a normed (or seminormed) space, a family \mathcal{B} is uniformly bounded if and only if its union is norm bounded, meaning that there exists some real M \geq 0 such that for every or equivalently, if and only if A set H of maps from X to Y is said to be if the family is uniformly bounded in Y, which by definition means that there exists some bounded subset D of Y such that or equivalently, if and only if is a bounded subset of Y. A set H of linear maps between two normed (or seminormed) spaces X and Y is uniformly bounded on some (or equivalently, every) open ball (and/or non-degenerate closed ball) in X if and only if their operator norms are uniformly bounded; that is, if and only if Assume H is equicontinuous and let W be a neighborhood of the origin in Y. Since H is equicontinuous, there exists a neighborhood U of the origin in X such that for every h \in H. Because C is bounded in X, there exists some real r > 0 such that if t \geq r then So for every h \in H and every t \geq r, which implies that Thus is bounded in Y. Q.E.D. Let W be a balanced neighborhood of the origin in Y and let V be a closed balanced neighborhood of the origin in Y such that Define which is a closed subset of X (since V is closed while every h : X \to Y is continuous) that satisfies for every h \in H. Note that for every non-zero scalar n \neq 0, the set n E is closed in X (since scalar multiplication by n \neq 0 is a homeomorphism) and so every C \cap n E is closed in C. It will now be shown that from which follows. If c \in C then H(c) being bounded guarantees the existence of some positive integer such that where the linearity of every h \in H now implies thus and hence as desired. Thus expresses C as a countable union of closed (in C) sets. Since C is a nonmeager subset of itself (as it is a Baire space by the Baire category theorem), this is only possible if there is some integer n \in \N such that C \cap n E has non-empty interior in C. Let be any point belonging to this open subset of C. Let U be any balanced open neighborhood of the origin in X such that The sets form an increasing (meaning p \leq q implies ) cover of the compact space C, so there exists some p > 1 such that (and thus ). It will be shown that for every h \in H, thus demonstrating that is uniformly bounded in Y and completing the proof. So fix h \in H and c \in C. Let The convexity of C guarantees z \in C and moreover, z \in k + U since Thus which is a subset of Since n V is balanced and we have which combined with gives Finally, and imply as desired. Q.E.D. Since every singleton subset of X is also a bounded subset, it follows that if is an equicontinuous set of continuous linear operators between two topological vector spaces X and Y (not necessarily Hausdorff or locally convex), then the orbit of every x \in X is a bounded subset of Y.

Bounded subsets of topological modules

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