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Bounded set

In mathematics, a bounded set is a fundamental concept in metric spaces, where a subset S of a metric space (X, d) is defined as bounded if there exists a point x \in X and a finite radius r > 0 such that every element of S lies within the open ball B(x, r) = \{ y \in X \mid d(x, y) < r \}.[^[1]] Equivalently, S is bounded if its diameter, defined as \sup \{ d(y, z) \mid y, z \in S \}, is finite.^[2] This property ensures that the set does not "extend infinitely" in any direction according to the metric, distinguishing it from unbounded sets like the natural numbers in the real line. In the specific context of the real numbers \mathbb{R} equipped with the standard metric d(x, y) = |x - y|, a set S \subseteq \mathbb{R} is bounded if it has both an upper bound (a number M such that s \leq M for all s \in S) and a lower bound (a number m such that s \geq m for all s \in S), meaning S is contained within some finite interval [m, M].^[3] The least upper bound is called the supremum \sup S, and the greatest lower bound is the infimum \inf S; a set is bounded if and only if both exist and are finite.^[4] Examples include closed intervals like [0, 1] or open balls in \mathbb{R}^n, while the integers \mathbb{Z} form an unbounded set. Bounded sets play a crucial role in real analysis and topology, underpinning theorems such as the Heine-Borel theorem, which states that in \mathbb{R}^n with the Euclidean metric, a set is compact if and only if it is closed and bounded.^[5] They also relate to convergence properties, as bounded sequences in complete metric spaces may have convergent subsequences under additional conditions like total boundedness. In normed vector spaces, boundedness aligns with sets of finite diameter, facilitating applications in functional analysis and optimization.^[2]

Core Definitions in Analysis

In the Real Numbers

In the real numbers, a subset S \subseteq \mathbb{R} is defined as bounded if there exists a positive real number M > 0 such that |x| \leq M for all x \in S. This condition ensures that all elements of S lie within a finite distance from the origin on the number line. Equivalently, S is bounded if it is both bounded above—meaning there exists some real number U such that x \leq U for all x \in S—and bounded below, with some real number L such that x \geq L for all x \in S. In this case, the supremum \sup S (the least upper bound) and infimum \inf S (the greatest lower bound) both exist and are finite real numbers, and the difference \sup S - \inf S < \infty measures the "length" of the interval containing S.^[4]^[6] The equivalence between the absolute value condition and the existence of finite supremum and infimum follows directly from the order properties of \mathbb{R}. Suppose |x| \leq M for all x \in S; then -M \leq x \leq M, so S is bounded below by -M and above by M. By the completeness axiom of the real numbers, every nonempty set bounded above has a finite supremum, and every nonempty set bounded below has a finite infimum; thus, \inf S \geq -M and \sup S \leq M, implying \sup S - \inf S \leq 2M < \infty. Conversely, if S is bounded above by U and below by L, then for all x \in S, x \in [L, U], so |x| \leq \max(|L|, |U|); letting M = \max(|L|, |U|), we have |x| \leq M. Moreover, since \sup S and \inf S are finite, their difference is necessarily finite, confirming boundedness. This equivalence underscores the tight connection between metric and order-based views in \mathbb{R}.^[4] Basic examples illustrate this definition. Closed intervals [a, b] with a \leq b are bounded, with \inf [a, b] = a and \sup [a, b] = b, fitting within [-|a|, b] if a \geq 0, or adjusted accordingly. Open intervals (a, b) are also bounded, sharing the same finite infimum and supremum despite not attaining them. Unions of finitely many such intervals, like [0, 1] \cup [2, 3], remain bounded by taking the overall lower bound as the smallest endpoint and upper bound as the largest. In contrast, the natural numbers \mathbb{N} are unbounded above (no finite supremum exists), and the positive reals \mathbb{R}^+ lack an upper bound. Geometrically, bounded sets in \mathbb{R} can always be enclosed within a finite-length segment of the number line, providing an intuitive visualization of their containment.^[3] The concept of bounded sets emerged in the early 19th century as part of efforts to rigorize calculus. Augustin-Louis Cauchy introduced foundational ideas in his 1821 textbook Cours d'analyse de l'École Royale Polytechnique, where boundedness played a key role in defining limits, continuity, and the completeness of the reals through Cauchy sequences and upper bounds for convergent series. This work laid the groundwork for modern real analysis by emphasizing precise conditions on sets to ensure the existence of limits and suprema.^[7]

In Metric Spaces

In a metric space (X, d), a subset S \subseteq X is bounded if there exists a nonnegative real number M such that d(x, y) \leq M for all x, y \in S.^[8] Equivalently, S is bounded if its diameter \operatorname{diam}(S) = \sup \{ d(x, y) \mid x, y \in S \} < \infty.^[8] This definition abstracts the notion of boundedness from ordered structures like the real numbers to arbitrary distance functions, emphasizing pairwise distances rather than order bounds.^[9] A key characterization is that a nonempty set S is bounded if and only if it is contained in some open ball of finite radius.^[8] To see that finite diameter implies containment in a ball, assume \operatorname{diam}(S) = D < \infty and select any x_0 \in S; then for all y \in S, d(x_0, y) \leq D, so S \subseteq B(x_0, D), the open ball of radius D centered at x_0.^[8] The converse holds because if S \subseteq B(x, r) for some x \in X and r < \infty, then d(x', y') \leq d(x', x) + d(x, y') < 2r for all x', y' \in S, yielding \operatorname{diam}(S) \leq 2r < \infty.^[8] Examples illustrate boundedness in specific metrics. In \mathbb{R}^n with the Euclidean metric, open or closed balls and ellipsoids are bounded, as their diameters are finite (e.g., the unit ball has diameter 2).^[8] In contrast, the entire space \mathbb{R}^n is unbounded under this metric.^[8] For the discrete metric on any set X, where d(x, y) = 1 if x \neq y and d(x, x) = 0, every subset S \subseteq X (finite or infinite) has \operatorname{diam}(S) \leq 1 < \infty and is thus bounded.^[8] Boundedness exhibits several metric-specific properties. It is hereditary under subsets, as \operatorname{diam}(T) \leq \operatorname{diam}(S) for T \subseteq S, and preserved under finite unions, since \operatorname{diam}(\bigcup_{i=1}^k S_i) \leq \max_i \operatorname{diam}(S_i).^[8] However, bounded sets need not be closed or compact; for instance, the open unit ball \{ x \in \mathbb{R} \mid |x| < 1 \} is bounded (diameter 2) but open, hence neither closed nor compact in \mathbb{R}.^[8] Regarding sequences, every Cauchy sequence in a metric space is bounded: if \{x_n\} is Cauchy, there exists N such that d(x_m, x_n) < 1 for all m, n \geq N, so the tail lies in B(x_N, 1) and the finite initial segment is contained in some ball, making the entire sequence bounded.^[10] Conversely, boundedness does not imply completeness, as a bounded set may contain a Cauchy sequence without a limit point in the space; for example, the rational numbers in (0, 1) form a bounded incomplete metric subspace containing non-convergent Cauchy sequences.^[9]

Generalizations in Vector Spaces

In Normed Linear Spaces

In a normed linear space (X, \|\cdot\|), a subset S \subseteq X is bounded if there exists M < \infty such that \|x\| \leq M for all x \in S, or equivalently, if \sup_{x \in S} \|x\| < \infty. This condition ensures that S is contained in some ball of finite radius centered at the origin. Since the metric d(x, y) = \|x - y\| is induced by the norm, boundedness in this sense is equivalent to S having finite diameter, \operatorname{diam} S = \sup_{x, y \in S} \|x - y\| < \infty, as \|x - y\| \leq \|x\| + \|y\| \leq 2M.^[11] A key property of bounded sets in normed spaces arises from the linear structure: they can be absorbed by scalar multiples of the unit ball. Specifically, for any \varepsilon > 0, there exists t > 0 such that tS \subseteq \varepsilon \overline{B}(0, 1), where \overline{B}(0, 1) = \{x \in X : \|x\| \leq 1\} is the closed unit ball; if \sup_{x \in S} \|x\| = M < \infty, then t = \varepsilon / M suffices, since \|t x\| = t \|x\| \leq t M = \varepsilon. This absorption reflects the homogeneity of the norm, \|\lambda x\| = |\lambda| \|x\| for \lambda \in \mathbb{R}, which implies that if S is bounded, then \lambda S is bounded for any scalar \lambda, with \sup_{z \in \lambda S} \|z\| = |\lambda| M < \infty. Unlike purely metric spaces, this algebraic scaling distinguishes boundedness in normed spaces. The norm-induced metric aligns with the general metric definition of boundedness, reducing to containment in a finite-radius ball.^[11]^[12] Examples illustrate these concepts clearly. In finite-dimensional spaces like \mathbb{R}^n with the Euclidean norm \|\cdot\|_2, bounded sets are precisely those with finite diameter, coinciding with the metric notion in the induced topology. In contrast, the infinite-dimensional space \ell^2 of square-summable sequences, equipped with \|x\|_2 = \sqrt{\sum_{n=1}^\infty |x_n|^2}, admits the closed unit ball \{x \in \ell^2 : \|x\|_2 \leq 1\} as a bounded set, since \sup \|x\|_2 = 1 < \infty, yet this ball is not compact due to the lack of total boundedness in infinite dimensions.^[11] A fundamental theorem linking bounded sets to operator theory is the uniform boundedness principle: in a normed space X, if \{T_\alpha\} is a family of continuous linear operators from X to another normed space that is pointwise bounded—meaning \sup_\alpha \|T_\alpha x\| < \infty for each fixed x \in X—then the family is uniformly bounded, \sup_\alpha \|T_\alpha\| < \infty, so each T_\alpha maps bounded sets in X to bounded sets in the codomain. This result, also known as the Banach-Steinhaus theorem, underscores how boundedness controls the behavior of pointwise limits of operators.^[11]

In Topological Vector Spaces

In topological vector spaces, the notion of boundedness generalizes the metric or norm-based definitions by relying on the absorption property with respect to neighborhoods of the origin, accommodating topologies that may not arise from norms or metrics. A subset S of a topological vector space X is bounded if for every neighborhood U of the zero vector, there exists t > 0 such that tS \subset U.^[11] This condition captures the intuitive idea that S is "small" in the topological sense, as multiples of neighborhoods can eventually absorb the entire set. When X is a normed space, this topological definition of boundedness is equivalent to the classical one: S is bounded if and only if \sup_{x \in S} \|x\| < \infty. To see this, note that balanced neighborhoods in normed spaces include scalar multiples of the unit ball, so the absorption property implies the set lies within a finite multiple of the unit ball, bounding the norms; conversely, if the supremum norm is finite, then for any neighborhood U containing the unit ball scaled by some factor, a suitable t absorbs S.^[11] Examples illustrate this concept across different topologies. In the space \mathbb{C} equipped with its usual topology (induced by the modulus norm), compact sets are bounded, as their closed and bounded nature ensures absorption by neighborhoods of 0.^[11] In the space \mathcal{M}(\mathbb{R}) of Radon measures on \mathbb{R} under the weak* topology (as the dual of C_0(\mathbb{R})), sets of measures with bounded total variation are bounded, since the total variation seminorm controls absorption in this dual topology.^[13] Bounded sets in topological vector spaces exhibit several key properties, particularly in locally convex settings. In a locally convex topological vector space, every bounded set is contained in the closed convex hull of a compact set, reflecting the role of convexity in controlling topological size.^[11] However, boundedness does not imply precompactness in general; for instance, in the weak topology on a Banach space, the closed unit ball is bounded but not precompact, as its closure is not compact.^[11] In Fréchet spaces, which are complete metrizable locally convex spaces defined by a countable family of seminorms \{p_n\}, a set S is bounded if and only if \sup_{x \in S} p_n(x) < \infty for each n, corresponding to equicontinuity with respect to the defining seminorms.^[11] This characterization connects directly to the uniform boundedness principle, which states that a pointwise bounded family of continuous linear operators on a barrelled space (such as a Fréchet space) is equicontinuous, implying the family maps bounded sets to bounded sets uniformly.^[11]

Boundedness in Ordered Structures

In Partially Ordered Sets

In a partially ordered set (poset) (P, \leq), a subset S \subseteq P is bounded above if there exists an element u \in P, called an upper bound, such that s \leq u for every s \in S; similarly, S is bounded below if there exists a lower bound \ell \in P with \ell \leq s for every s \in S. A subset is bounded if it possesses both an upper and a lower bound, though the terms "bounded above" and "bounded below" are often used specifically when only one type of bound is relevant. This definition relies solely on the order relation and does not invoke any metric or topological structure.^[14]^[15] Examples illustrate the concept clearly in familiar posets. In the totally ordered set (\mathbb{R}, \leq), the closed interval [a, b] with a \leq b is bounded, as a serves as a lower bound and b as an upper bound. In the power set \mathcal{P}(X) of a nonempty set X, ordered by inclusion \subseteq, any finite subset F \subseteq \mathcal{P}(X) is bounded above by X itself, since A \subseteq X for all A \in F, and bounded below by the empty set \emptyset, as \emptyset \subseteq A for all A \in F. These cases highlight how boundedness captures containment within order-theoretic "intervals" without requiring numerical measurement.^[16]^[17] Key properties emerge in structured posets. In a complete lattice—a poset where every subset has a supremum (least upper bound) and infimum (greatest lower bound)—any bounded subset necessarily possesses these extrema, as the completeness ensures their existence for all subsets. The Dedekind-MacNeille completion of a poset embeds it order-isomorphically into the smallest complete lattice containing it, preserving the boundedness of all subsets through the embedding. In chains (totally ordered sets), the notion reduces to the classical boundedness in the real numbers via an order embedding, where upper and lower bounds correspond directly to those in \mathbb{R}.^[18]^[15] The roots of boundedness in posets lie in Richard Dedekind's foundational work on cuts from the late 19th century, particularly his 1872 essay "Continuity and Irrational Numbers," which introduced order-theoretic partitions to construct the reals and influenced the development of lattice theory by emphasizing bounds and completeness in ordered structures.^[19] In a lattice L, a subset S \subseteq L is bounded if it has both a lower bound l \in L (such that l \leq s for all s \in S) and an upper bound u \in L (such that s \leq u for all s \in S). In a bounded lattice, which possesses a global bottom element \bot and top element \top, any subset S satisfying \bot \leq s \leq \top for all s \in S is bounded by these extremal elements. This notion extends the concept of boundedness from partially ordered sets by leveraging the lattice's join (\vee) and meet (\wedge) operations, though the definition itself relies on the underlying order.^[20]^[21] In distributive lattices, bounded subsets exhibit additional algebraic structure when closed under specific operations: a downward-closed bounded subset closed under finite joins forms an ideal, while an upward-closed bounded subset closed under finite meets forms a filter. These ideals and filters preserve the distributive property of the lattice, enabling the ideal lattice I(L) to itself be distributive. Furthermore, Stone duality for Boolean algebras—a special class of distributive lattices—establishes a contravariant equivalence between the category of Boolean algebras and Stone spaces (compact, totally disconnected Hausdorff spaces), where the clopen sets of the Stone space form the dual Boolean algebra.^[21]^[22] Representative examples illustrate these concepts. In the lattice of subspaces of a vector space, ordered by inclusion, every subspace is bounded above by the full space and below by the zero subspace. Boundedness plays a key role in applications to logic and computation. In domain theory for denotational semantics, domains are often modeled as complete partial orders or lattices where bounded sets (pairs of elements with a common upper bound) ensure consistency of approximations, facilitating the interpretation of recursive programs via least fixed points. In rough set theory, bounded approximations arise through lower and upper approximations of concepts, where the lower approximation provides a definite bound and the upper a possible bound, enabling uncertainty modeling in data analysis without probabilistic assumptions.^[23]^[24] A fundamental theorem in complete lattices states that every subset, including bounded ones, possesses a supremum (arbitrary join \bigvee S) and infimum (arbitrary meet \bigwedge S), providing a closure under these operations; for bounded S, these closures refine the bounds while preserving lattice structure. This property underpins fixed-point theorems, such as Tarski's theorem, which asserts that any monotone function f on a complete lattice has a complete lattice of fixed points, with the least fixed point given by the join of the iterated images \bigvee_{n \in \mathbb{N}} f^n(\bot) and the greatest by the meet of the pre-images \bigwedge_{n \in \mathbb{N}} f^n(\top).^[21]^[25]

References

[1]
Bounded Set -- from Wolfram MathWorld
A set S in a metric space (S,d) is bounded if it has a finite generalized diameter, ie, there is an R<infty such that d(x,y)<=R for all x,y in S.
[2]
Real Numbers:Bounded Subsets - UTSA
Nov 14, 2021 · A set S is bounded if it has both upper and lower bounds. Therefore, a set of real numbers is bounded if it is contained in a finite interval.
[3]
[PDF] The supremum and infimum - UC Davis Math
A set is bounded if it is bounded both from above and below. The supremum of a set is its least upper bound and the infimum is its greatest upper bound.
[4]
[PDF] 5. ABSOLUTE EXTREMA Definition, Existence & Calculation
The remaining terms in the hypothesis of the theorem are “closed bounded set”. Meaning of BOUNDED. A set of S real numbers is said to be bounded if there exists ...
[5]
Mathematical Treasure: Cauchy's Cours d'Analyse
In this book, Cauchy used series and sequences extensively to prove his results, introduced the δ−ε (delta-epsilon) notation, and provided the definitions used ...
[6]
https://www.math.toronto.edu/igelfeld/mat133suppsects/Sec5.pdf
[7]
[PDF] Overview of Real Analysis (Folland) - Brett Saiki
a metric space. Some examples: (i) The Euclidean distance ρ(x, y) = |x − y ... Every totally bounded set is bounded. (The converse is false in general ...
[8]
[PDF] Chapter 1. Metric spaces - Proofs covered in class
Theorem 1.12 – Cauchy implies bounded. In a metric space, every Cauchy sequence is bounded. Proof. Suppose {xn} is a Cauchy sequence in a metric space X ...
[9]
[PDF] FUNCTIONAL ANALYSIS | Second Edition Walter Rudin
... bounded set E c X. This definition conflicts with the usual notion of a bounded function as being one whose range is a bounded set. In that sense, no linear ...
[10]
[PDF] TOPOLOGICAL VECTOR SPACES1 1. Definitions and basic facts.
(When τ is defined by a countable collection of seminorms.) Note on the different notions of 'bounded set'. On a normed vector space, both notions coincide ...
[11]
[PDF] The Measure Problem - Purdue Math
The set of all bounded measures is a Banach space M(S) with total variation as norm. If π is a measurable mapping of S onto Z, then for every bounded measure.
[12]
[PDF] Section 8.6 - Transparencies for Rosen, Discrete Mathematics & Its ...
Definition: Let S be a subset of A in the poset (A, R). If there exists an element a in A such that sRa for all s in S, then a is called an upper bound.
[13]
[PDF] Notes on Ordered Sets
Sep 22, 2009 · Definition 1.3 We say that a subset E ⊆ S is bounded (from) above, if. U(E) , ∅, i.e., when there exists at least one element y ∈ S satisfying ( ...
[14]
[PDF] 4. Partial Orderings
Definition of a Partial Order. Definition 4.1.1. (1) A relation R on a set A is a partial order iff R is. • reflexive,. • antisymmetric, and. • transitive.
[15]
[PDF] Lattice Theory Lecture 1 Basics - nmsu math
It is a complete lattice if every subset A ⊆ L has a least upper bound ⋁A and a greatest lower bound ⋀A. Example The rational unit interval [0,1] ∩ Q is a ...<|control11|><|separator|>
[16]
On complete ordered fields - Mathematics and Computation
Sep 9, 2019 · A poset P is (Dedekind-MacNeille) complete when every inhabited bounded subset has a supremum (for the classically trained, S ⊆ P is ...<|control11|><|separator|>
[17]
Dedekind's Contributions to the Foundations of Mathematics
Apr 22, 2008 · In his next step—and proceeding further along set-theoretic and structuralist lines—Dedekind introduces the set of arbitrary cuts on his initial ...
[18]
lattice in nLab
Feb 23, 2024 · Then one may call a lattice that does have a top and a bottom a bounded lattice; in general, a bounded poset is a poset that has top and bottom ...Definition · Bounded lattices and... · Lattice homomorphisms
[19]
[PDF] Lattice Theory Lecture 2 Distributive lattices - nmsu math
Ideals and Filters. Definition An ideal of a lattice L is a subset I ⊆ L where. 1. if y ∈ I and x ≤ y, then x ∈ I. 2. if x,y ∈ I then x ∨ y ∈ I.
[20]
[PDF] Stone Duality for Boolean Algebras - The University of Manchester
Definition. A Boolean space is a topological space X that is compact. Hausdorff and such that every open set is a union of clopen sets (clopen means.
[21]
[PDF] On the Lattice of Subspaces of a Vector Space - Mizar
In this paper F denotes a field and V1 denotes a strict vector space over F. Let us consider F, V1. The functor (V1 ) yields a strict bounded lattice and is ...
[22]
[PDF] AN INTRODUCTION TO BOOLEAN ALGEBRAS
This is Stone's representation theorem for finite Boolean algebras. Proof. Let B be a finite Boolean algebra and let S be the set of all atoms in B. We want.
[23]
[PDF] Tutorial on Semantics Part II - Domain Theory
Jun 20, 2011 · A pair of elements x,y in a dcpo are said to be bounded or consistent if there is some z such that x,y ≤ z. Definition. A Scott domain is an ω- ...<|separator|>
[24]
[PDF] ROUGH SET APPROXIMATIONS: A CONCEPT ANALYSIS POINT ...
Rough set theory was proposed by Pawlak for analyzing data and reasoning about data. From a concept analysis point of view, we review and reformu- late main ...
[25]
[PDF] Tarski's Fixed Point Theorem*
Tarski's theorem states that a monotone function on a complete lattice has a complete lattice of fixed points, in particular a least and greatest. A useful ...