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Precompact set

In topology, a precompact set, also known as a relatively compact set, is a subset A of a topological space X such that the closure \overline{A} of A in X is a compact subset.^[1] This notion captures sets that are "almost compact" in the sense that adjoining their limit points yields compactness, distinguishing it from the set itself being compact.^[2] In metric spaces, precompactness admits a concrete characterization: a subset is precompact if and only if it is totally bounded, meaning that for every \varepsilon > 0, it can be covered by finitely many balls of radius \varepsilon.^[3] In complete metric spaces, this is equivalent to relative compactness, and every sequence in such a set has a Cauchy subsequence converging in the closure.^[4] More generally, in uniform spaces, precompact sets coincide with totally bounded sets, where the uniformity allows for a notion of "finite nets" analogous to the metric case.^[5] Precompact sets play a central role in various mathematical contexts, such as the study of topological groups, where a group is precompact if it is isomorphic to a dense subgroup of a compact group, implying total boundedness in its uniformity.^[6] They also appear in functional analysis, for instance, in the Arzelà–Ascoli theorem, which characterizes precompact subsets of function spaces by equicontinuity and pointwise boundedness.^[3] In locally compact Hausdorff spaces, precompact open neighborhoods exist around each point, facilitating constructions like countable bases in second countable settings.^[7]

Definition and properties

Definition

In topology, a subset of a topological space is called compact if every open cover of the subset has a finite subcover.^[8] A subset Y of a topological space X is precompact (or relatively compact) if its closure \operatorname{cl}_X(Y) is a compact subset of X.^[9]^[10] Formally, Y is precompact in X if and only if \operatorname{cl}_X(Y) is compact in X.^[11] The terms "precompact set" and "relatively compact set" are synonymous in this context, referring to relative compactness with respect to the ambient space X.^[11]^[3] Note that in some contexts, particularly uniform spaces, "precompact" may instead denote total boundedness, but the usage here aligns specifically with relative compactness.^[3]

Basic properties

A precompact set, also known as a relatively compact set, in a topological space is characterized by the compactness of its closure. One fundamental property is that every compact subset of a topological space is precompact, as the closure of a compact set coincides with the set itself, which is compact by definition.^[12] Subsets of precompact sets inherit this property: if A \subseteq B and B is precompact, then the closure of A is contained within the closure of B, which is compact, making A precompact as well.^[12] Furthermore, the collection of precompact sets is closed under finite unions and finite intersections. For finite unions, the closure of the union of finitely many precompact sets is contained in the finite union of their compact closures, which is compact. For finite intersections, the closure of the intersection is contained in the intersection of the compact closures, and the intersection of finitely many compact sets is compact.^[12] In Hausdorff spaces, the closure of a precompact set is not only compact but also closed, since every compact subset of a Hausdorff space is closed.^[13] This follows from the separation properties of Hausdorff spaces, where points can be separated by disjoint open neighborhoods, ensuring that compact sets contain all their limit points.^[13] The closure of a precompact set serves as the smallest compact set containing it, as any compact set containing the precompact set must contain its closure, and the closure itself is compact.^[12] This minimal property underscores the role of precompact sets in approximations to compact structures within broader topological spaces.^[12]

Characterizations in specific spaces

Note: While this article defines "precompact" as relatively compact (closure is compact), the term is sometimes used in the literature on metric and uniform spaces to mean "totally bounded." The characterizations below reflect the article's definition, with qualifications for completeness where the notions coincide.

In metric spaces

In metric spaces, precompactness admits a sequential characterization: a subset Y of a metric space X is precompact if and only if every sequence in Y has a subsequence that converges to a point in the closure \overline{Y} of Y in X.^[14] This property leverages the fact that metric spaces are first-countable, allowing compactness to be tested via sequences, and extends to precompact sets by considering limits in the closure.^[3] Precompact subsets of metric spaces are always totally bounded. Moreover, in complete metric spaces, the converse holds: Y \subseteq X is precompact if and only if it is totally bounded, meaning that for every \varepsilon > 0, Y can be covered by finitely many open balls of radius \varepsilon centered in X.^[3] Total boundedness ensures that Y is "small" in a uniform sense, preventing the existence of sequences without Cauchy subsequences.^[15] Specifically, a set is totally bounded if it admits a finite \varepsilon-net for any \varepsilon > 0, meaning a finite collection of points such that every element of Y lies within \varepsilon of one of these points; this finite covering property directly implies the sequential condition, as any sequence in Y must accumulate near the net points, yielding a Cauchy subsequence.^[3] Conversely, the absence of total boundedness allows construction of a sequence with no Cauchy subsequence, hence no convergent one in the closure.^[16] In complete metric spaces, this equivalence yields a Heine-Borel-like property: a subset Y is precompact if and only if its closure \overline{Y} is compact.^[17] Since completeness ensures that every Cauchy sequence converges, total boundedness of Y implies \overline{Y} is complete and totally bounded, hence compact; the converse holds as subsets of compact sets inherit total boundedness.^[3] A classic example is the open unit ball B(0,1) = \{ x \in \mathbb{R}^n : \|x\| < 1 \} in the Euclidean metric space \mathbb{R}^n with n \geq 1. This set is totally bounded, as it can be covered by finitely many balls of radius \varepsilon for any \varepsilon > 0, but it is not compact since its closure is the closed unit ball, which includes the boundary points essential for compactness in finite-dimensional Euclidean spaces.^[3]

In uniform spaces

In uniform spaces, the notion of precompactness for a subset Y of a uniform space (X, \mathcal{U}) can be characterized using the induced uniformity: Y is precompact if its closure \overline{Y} (in the induced topology) is compact. Equivalently, in terms of total boundedness with respect to the induced uniformity, Y is totally bounded if, for every entourage V \in \mathcal{U}, there exists a finite family of subsets \{Y_1, \dots, Y_n\} of Y such that Y_i \times Y_i \subseteq V for each i and Y = \bigcup_{i=1}^n Y_i. This condition ensures that Y can be finitely covered by "V-small" sets, generalizing the epsilon-net covering from metric spaces to the entourage-based structure of uniformities. In complete uniform spaces, total boundedness is equivalent to precompactness.^[18] In the context of the uniform topology induced by \mathcal{U}, a totally bounded subset Y has the property that its completion \hat{Y} (with the extended uniformity) is compact. Since a uniform space is compact if and only if it is complete and totally bounded, the total boundedness of Y implies that \hat{Y} satisfies both conditions, with completeness inherited from the completion process and total boundedness from the definition. This aligns total boundedness with the existence of a compact uniform completion, distinguishing it from purely topological compactness. In complete uniform spaces, this implies relative compactness (topological compactness of \overline{Y}); in general, total boundedness implies the existence of a compact uniform completion, but \overline{Y} may not be topologically compact if the space is incomplete.^[5] Relative compactness in uniform spaces implies total boundedness only if the ambient uniform space (X, \mathcal{U}) is complete, as incomplete spaces may have compact closures that are not totally bounded. Unlike metric spaces, uniform spaces admit non-metrizable examples; for instance, in the indiscrete uniformity on an infinite set X (generated by the single entourage X \times X), every subset has compact closure (since the indiscrete topology is compact), and is totally bounded, as it admits a trivial finite cover by itself, which is V-small, and the completion remains the compact indiscrete space.

Examples and counterexamples

Examples

In Euclidean space \mathbb{R}^n equipped with the standard topology, any closed and bounded subset is compact by the Heine-Borel theorem, and thus precompact since its closure coincides with itself.^[12] For instance, the closed unit ball \{ x \in \mathbb{R}^n : \|x\| \leq 1 \} is a precompact set.^[12] Precompact sets need not be closed or compact themselves; their closures suffice to be compact. The open unit disk \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 < 1 \} in the plane has a closure that is the closed unit disk, which is compact by the Heine-Borel theorem, making the open disk precompact.^[12] Similarly, the open interval (0,1) in \mathbb{R} is precompact because its closure [0,1] is compact.^[12] In spaces of continuous functions, such as C(K) where K is a compact subset of a metric space with the supremum norm, a family of functions that is pointwise bounded and equicontinuous forms a precompact set, as established by the Arzelà–Ascoli theorem.^[19] For example, the set of all continuous functions on [0,1] with Lipschitz constant at most 1 and values bounded by 1 is precompact in C([0,1]).^[19] In the context of topological groups, a dense subgroup of a compact group is precompact.^[6] The rational numbers \mathbb{Q} with the subspace topology induced from the circle group \mathbb{R}/\mathbb{Z} (which is compact) form a dense subgroup and hence a precompact set.^[6]

Counterexamples

In the real line \mathbb{R} equipped with the standard topology, the entire space \mathbb{R} serves as a basic counterexample to precompactness, as its closure is itself and \mathbb{R} is not compact due to the existence of open covers without finite subcovers, such as the collection of open intervals (n, n+2) for all integers n.^[20] Similarly, the set of natural numbers \mathbb{N} embedded in \mathbb{R} is not precompact, since its closure in \mathbb{R} is \mathbb{N} itself, which fails to be compact as it contains no convergent subsequences and admits the open cover by singletons without finite subcover.^[20] In non-Hausdorff spaces, the distinction between compactness and closedness further illustrates failures of precompactness. For instance, consider the particular point topology on an infinite set X with designated point p \in X, where the open sets are the empty set and all subsets containing p. Here, the singleton \{p\} is compact, as any open cover has a finite subcover consisting of a single set containing p, but its closure is the entire space X, which is not compact for infinite X since open covers excluding sets containing p may require infinitely many to cover X \setminus \{p\}.^[20] This example highlights how compact sets need not be closed in non-Hausdorff spaces, thereby preventing such sets from being relatively compact if their closures exceed compact subsets.^[20] It is important to distinguish the topological notion of a precompact set from the related but distinct concept in topological group theory, where a precompact group is one whose completion is compact, often applied to subgroups embeddable into compact groups.^[21]

Applications

In functional analysis

In Banach spaces, which are complete normed linear spaces, a subset is precompact if and only if it is totally bounded, meaning that for every \epsilon > 0, it can be covered by finitely many balls of radius \epsilon; the closure of such a set is then compact in the norm topology.^[22] This equivalence holds because Banach spaces are complete metric spaces, and precompactness aligns with total boundedness in complete metric settings.^[23] Compact linear operators between Banach spaces exemplify this property, as they map bounded sets—such as the unit ball—into precompact sets, reflecting their ability to "approximate" infinite-dimensional spaces by finite-dimensional ones.^[24] For instance, the closed unit ball in a finite-dimensional Banach space is itself compact (hence precompact), by the Heine-Borel theorem adapted to Euclidean-like structures in finite dimensions.^[23] Precompactness in the norm topology contrasts with behavior in weaker topologies on Banach spaces. While the unit ball is precompact in the norm topology only for finite-dimensional spaces, Alaoglu's theorem establishes that the closed unit ball of the dual space is compact in the weak* topology, providing a form of compactness without total boundedness in the stronger norm.^[25] This distinction underscores how precompact sets in functional analysis often rely on completeness and operator properties rather than uniform boundedness alone.^[24]

Arzelà–Ascoli theorem

The Arzelà–Ascoli theorem provides a fundamental characterization of precompact subsets in the space of continuous functions on a compact metric space. Let K be a compact metric space equipped with metric d, and let C(K) denote the space of all real-valued continuous functions on K endowed with the supremum norm \|f\|_\infty = \sup_{x \in K} |f(x)|. A subset F \subseteq C(K) is precompact if and only if it is bounded (i.e., there exists M > 0 such that \|f\|_\infty \leq M for all f \in F) and equicontinuous (i.e., for every \varepsilon > 0, there exists \delta > 0 such that d(x, y) < \delta implies |f(x) - f(y)| < \varepsilon for all f \in F and all x, y \in K).^[26] Equicontinuity strengthens the individual uniform continuity of each f \in F by ensuring that the modulus of continuity is uniform across the family, which is crucial for controlling the behavior of functions simultaneously. Boundedness prevents the functions from diverging in the supremum norm, while equicontinuity guarantees that the family does not oscillate wildly between points. A proof sketch proceeds by first establishing total boundedness in the supremum norm: the separability of the compact metric space K allows construction of a countable dense subset S \subseteq K; a diagonal argument then yields a subsequence converging pointwise on S, and equicontinuity extends this to uniform convergence on all of K via the density of S. The closure of such an F is thus compact in C(K), confirming precompactness.^[26] A key corollary is that a family of functions in C(K) that is uniformly bounded and equicontinuous has compact closure in C(K). This follows directly from the theorem, as the closure inherits boundedness and equicontinuity and is closed by definition.^[26] The theorem is named after the Italian mathematicians Cesare Arzelà and Giulio Ascoli, who around 1883 independently established necessary and sufficient conditions for uniform convergence of subsequences in families of continuous functions on closed bounded intervals.^[27] It has significant applications in partial differential equations, such as proving existence of solutions via compactness of approximating sequences in the Peano theorem for ordinary differential equations, and in approximation theory, where it facilitates uniform limits of polynomial or spline approximations on compact domains.^[28]

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