Nowhere dense set

In topology, a nowhere dense set is a subset A of a topological space X such that the interior of the closure of A is empty, denoted \operatorname{int}(\overline{A}) = \emptyset.^[1] This condition implies that A fails to be dense in any nonempty open subset of X, as every open set contains a nonempty open subset disjoint from A.^[2] Nowhere dense sets are considered "topologically small" because they lack substantial presence in the space, with properties such as: any subset of a nowhere dense set is nowhere dense; the union of finitely many nowhere dense sets is nowhere dense; and the closure of a nowhere dense set remains nowhere dense.^[3] Classic examples include finite subsets of \mathbb{[R](/page/R)} with the standard topology, the set of integers \mathbb{Z} in \mathbb{R}, and the Cantor set, whose closure is itself compact and has empty interior.^[4] In contrast, dense sets like the rationals \mathbb{[Q](/page/Q)} in \mathbb{R} are not nowhere dense, as their closure is all of \mathbb{R} with nonempty interior.^[5] These sets play a central role in the Baire category theorem, which states that in a complete metric space, the countable union of nowhere dense sets—known as a set of first category or meagre set—cannot cover the entire space, as such spaces are of second category. This theorem has broad applications in analysis, such as proving the existence of nowhere differentiable continuous functions on \mathbb{R} and establishing that generic sets in Banach spaces exhibit certain regularity properties. Extensions of the concept appear in generalized topologies, including ideals generated by σ-nowhere dense sets (countable unions of nowhere dense sets) and their role in studying Baire-like properties in non-complete spaces.^[6]

Definition

Via Closure and Interior

In a topological space (X, \tau), a subset A \subseteq X is nowhere dense if the interior of its closure is empty, that is, \operatorname{int}(\operatorname{cl}(A)) = \emptyset^[1]. This formulation captures the intuitive notion of topological "smallness" by ensuring that A, together with all its limit points, fails to occupy any nonempty open region in X^[2]. To understand this definition, recall that the closure \operatorname{cl}(A) of a subset A is the smallest closed set containing A, equivalently the intersection of all closed sets containing A, or the set of all points x \in X such that every neighborhood of x intersects A nonemptily^[7]. The interior \operatorname{int}(B) of a subset B \subseteq X is the largest open set contained in B, or the union of all open subsets of B^[8]. The condition \operatorname{int}(\operatorname{cl}(A)) = \emptyset thus implies that \operatorname{cl}(A) contains no nonempty open set, meaning A is not dense in any nonempty open subset of X—if it were dense in some open U \neq \emptyset, then U \subseteq \operatorname{cl}(A), contradicting the empty interior^[2]. This property reflects the absence of clustering: the closure \operatorname{cl}(A) incorporates all accumulation points of A, yet the resulting set has no "interior" structure, so neither A nor its limit points fill out any open neighborhood completely^[1]. In other words, around every point in X, there exists a nonempty open neighborhood that avoids \operatorname{cl}(A) entirely, preventing A from being "thick" or prevalent in any local region^[2]. For verification, consider a finite subset A of the real line \mathbb{R} equipped with the standard topology. Finite sets in \mathbb{R} are closed, so \operatorname{cl}(A) = A, and \operatorname{int}(A) = \emptyset because no nonempty open interval (the basic open sets in \mathbb{R}) is contained in a finite collection of points^[3]. Thus, finite sets are nowhere dense in \mathbb{R}^[1].

Via Open Sets

A subset A of a topological space X is nowhere dense if and only if for every nonempty open set U \subseteq X, there exists a nonempty open set V \subseteq U such that V \cap \overline{A} = \emptyset, where \overline{A} denotes the closure of A.^[9] This characterization emphasizes that A fails to be dense in any nonempty open subset of X, meaning the closure of A does not intersect every nonempty open subset within U. Equivalently, A is nowhere dense if and only if \overline{A} is not dense in any nonempty open set U \subseteq X, or, stated differently, the complement X \setminus \overline{A} is dense in X.^[9] In this formulation, the density of the complement ensures that no nonempty open set is entirely contained within \overline{A}, aligning with the absence of interior points in the closure. These open-set characterizations are equivalent to the primary definition that the interior of \overline{A} is empty. To see this, suppose \operatorname{int}(\overline{A}) = \emptyset; then for any nonempty open U \subseteq X, the set U \setminus \overline{A} is open and nonempty (since otherwise U \subseteq \overline{A}, contradicting the empty interior), providing the required V = U \setminus \overline{A}. Conversely, if \operatorname{int}(\overline{A}) \neq \emptyset, let U = \operatorname{int}(\overline{A}); then \overline{A} is dense in U (as U \subseteq \overline{A}), so no such V exists within U.^[9] In topological spaces equipped with a basis \mathcal{B}, the nowhere dense property can be verified by checking the condition solely on basis elements: A is nowhere dense if and only if for every B \in \mathcal{B}, there exists V \in \mathcal{B} with V \subseteq B and V \cap \overline{A} = \emptyset. This local check simplifies computations in spaces where a convenient basis, such as open balls in metric spaces, is available.^[9]

Properties

Invariance and Heredity

A nowhere dense set in a topological space exhibits invariance under closure operations. If A is nowhere dense, then its closure \overline{A} is also nowhere dense, as the interior of \overline{A} coincides with the interior of the closure of \overline{A}, which remains empty.^[2] This property underscores the stability of the notion under taking limits in the topological sense. The collection of nowhere dense sets is hereditary with respect to subsets. That is, if B \subseteq A and A is nowhere dense, then B is nowhere dense, since the closure of B is contained in the closure of A, implying that the interior of the closure of B is contained in the empty interior of the closure of A.^[10] Nowhere density is preserved under restrictions to open subsets, reflecting its relative nature. A subset A of a topological space X is nowhere dense in X if and only if, for every nonempty open subset U \subseteq X, the set A \cap U is nowhere dense in the subspace U equipped with the relative topology. In one direction, if A is nowhere dense in U, then its closure in X has empty interior, ensuring the same in X; conversely, the relative closure aligns with the intersection of the ambient closure, preserving the empty interior condition.^[11]

Finite Operations

The union of finitely many nowhere dense sets is nowhere dense. To see this, consider two nowhere dense sets A, B \subset X in a topological space X; the general case follows by induction. The closure satisfies \overline{A \cup B} = \overline{A} \cup \overline{B}. Suppose U \subset \overline{A \cup B} is a nonempty open set. Then U \setminus (U \cap \overline{B}) is a nonempty open subset of \overline{A}, contradicting that A is nowhere dense (as \operatorname{int}(\overline{A}) = \emptyset). Similarly, interchanging A and B yields a contradiction. Thus, \operatorname{int}(\overline{A \cup B}) = \emptyset.^[12] Arbitrary intersections of nowhere dense sets are nowhere dense, as the intersection of any collection is a subset of each set in the collection, and subsets of nowhere dense sets are nowhere dense.^[3] The collection of nowhere dense subsets of a topological space X forms a proper ideal in the power set Boolean algebra \mathcal{P}(X), meaning it is closed under taking subsets and finite unions but contains neither X nor \emptyset^c. It is not a \sigma-ideal, however, as countable unions of nowhere dense sets need not be nowhere dense; for example, the rational numbers \mathbb{Q} in \mathbb{R} form a dense set that is a countable union of singletons, each of which is nowhere dense.^[3]

Theoretical Connections

To Baire Category Theorem

A set is meager, also known as of the first category, if it can be expressed as a countable union of nowhere dense sets.^[3] Nowhere dense sets thus serve as the fundamental building blocks, or "atoms," in the category-theoretic sense of topological smallness.^[3] The Baire category theorem, introduced by René Baire in his 1899 doctoral thesis Sur les fonctions de variables réelles, establishes that in a complete metric space, the intersection of countably many dense open sets is dense.^[13] Equivalently, no nonempty open set in such a space is meager.^[3] Baire originally developed the concepts of nowhere dense sets and category to classify the discontinuities of real functions and prove the theorem specifically for \mathbb{R}^n.^[13] The theorem extends to locally compact Hausdorff spaces, where every nonempty open set is of the second category.^[14] The complements of meager sets are termed comeager, or residual, and are of the second category; in Baire spaces—those satisfying the conclusion of the theorem—such comeager sets are dense.^[3] This framework has key implications for existence results, highlighting points that avoid pathological behaviors; for instance, it aids in analyzing the continuity points of functions, such as confirming the possibility of functions continuous precisely at irrational points while discontinuous at rationals, by leveraging the density of comeager sets in spaces like the irrationals.^[3] A standard proof outline for the complete metric space case proceeds by showing that finite intersections of dense open sets remain dense, then uses completeness to construct a nested sequence of nonempty closed balls with diameters tending to zero, ensuring their intersection is nonempty and thus the countable intersection of the dense opens is dense.^[3] For the locally compact Hausdorff version, the argument adapts by covering open sets with compact subsets and applying a similar finite intersection density property within those compacts.^[14]

To Measure Theory

Nowhere dense sets in \mathbb{R} equipped with the Lebesgue measure exhibit a fundamental independence from measure-theoretic size. Such sets can have Lebesgue measure zero, as exemplified by the middle-thirds Cantor set, which is uncountable, closed, and nowhere dense yet possesses measure zero.^[15] Conversely, nowhere dense sets can also have positive Lebesgue measure, such as fat Cantor sets, which are perfect, nowhere dense subsets of [0,1] constructed by removing intervals of lengths summing to less than 1, resulting in positive measure while maintaining empty interior in their closure.^[16] This duality highlights that topological "smallness" in the sense of category does not dictate measure-theoretic smallness. A key consequence is that no nowhere dense set can attain full Lebesgue measure within any open interval. Since the closure of a nowhere dense set has empty interior, every open interval contains a subinterval disjoint from this closure, implying that the set's measure in that interval is strictly less than the interval's length. Moreover, while countable unions of measure-zero nowhere dense sets remain meager (first category) and have measure zero, the broader class of meager sets—countable unions of nowhere dense sets—need not be measure zero; for instance, a suitable countable union of disjoint fat Cantor sets of decreasing positive measures can yield a meager set with positive total measure. This underscores the failure of category notions to align perfectly as a \sigma-ideal with respect to Lebesgue measure. In the density topology on \mathbb{R}, where the open sets are those subsets U such that every point x \in U is a density point (i.e., the upper density \limsup_{h \to 0^+} \frac{m(U \cap (x-h, x+h))}{2h} = 1, with m denoting Lebesgue measure), the nowhere dense sets coincide precisely with the sets of Lebesgue measure zero.^[17] This equivalence, first established in foundational works on the density topology, bridges topological category and measure theory by reinterpreting "nowhere dense" through the lens of asymptotic density, revealing a deeper structural analogy between the two frameworks.^[17]

Examples

Nowhere Dense Sets in Metric Spaces

In metric spaces, particularly the real line \mathbb{R} equipped with the standard Euclidean metric, nowhere dense sets exemplify subsets that are "scattered" in the sense that they and their closures avoid containing any nonempty open intervals. These sets play a key role in understanding category in topological spaces, where their closures have empty interiors. Classic examples include discrete or closed sets without interior points, illustrating how even uncountable such sets can lack substantial topological structure.^[14] Singletons provide the simplest examples of nowhere dense sets in \mathbb{R}. For any rational or irrational number q \in \mathbb{R}, the set \{q\} is closed, so its closure is itself, and it contains no open interval, hence \operatorname{int}(\{q\}) = \emptyset. Finite unions of singletons, such as any finite set F \subset \mathbb{R}, are also nowhere dense, as their closure remains finite and discrete, with empty interior.^[18] The set of integers \mathbb{Z} in \mathbb{R} extends this idea to an infinite discrete set. Here, \mathbb{Z} is closed because its complement is open, and \operatorname{int}(\mathbb{Z}) = \emptyset since no open interval is contained within it. Thus, \mathbb{Z} is nowhere dense in \mathbb{R}. The rational numbers \mathbb{Q} in \mathbb{R}, while dense (with \operatorname{cl}(\mathbb{Q}) = \mathbb{R}), are not themselves nowhere dense but form a meager set as a countable union of singletons, each of which is nowhere dense.^[14] A more sophisticated uncountable example is the middle-thirds Cantor set C \subset [0,1], constructed by iteratively removing open middle-third intervals from [0,1]. The set C is closed and perfect (every point is a limit point), yet \operatorname{int}(C) = \emptyset because it contains no nonempty open intervals. Consequently, C is nowhere dense in \mathbb{R}.^[19] Another illustrative closed nowhere dense set is A = \{1/n : n \in \mathbb{N}\} \cup \{0\} \subset \mathbb{R}, the limit points of the sequence $1/n. This set is compact and closed, with its only limit point at 0, but it includes no open intervals, so \operatorname{int}(A) = \emptyset.^[20] In contrast, non-examples highlight sets that fail to be nowhere dense. An open interval like (0,1) has nonempty interior, so it is not nowhere dense. Similarly, the irrationals \mathbb{R} \setminus \mathbb{Q} have closure \mathbb{R} with nonempty interior, making them dense rather than nowhere dense.^[14]

Nowhere Dense Sets with Positive Lebesgue Measure

In the unit interval [0,1], nowhere dense sets can possess positive Lebesgue measure, though this measure is strictly less than 1. Constructions exist achieving measures arbitrarily close to 1 by optimizing the removal of intervals (e.g., around rationals) to account for overlaps, allowing the remaining set to have measure exceeding 1/2 and approaching 1.^[21]^[22] A prominent example is the Smith–Volterra–Cantor set, also known as the fat Cantor set. This closed set is constructed iteratively from [0,1] by removing the open middle interval of length $1/4 at the first stage, leaving two closed intervals of length $3/8 each. At the second stage, the open middle interval of length $1/16 is removed from each of these, and this process continues, with $2^{n-1} open intervals of length $1/4^n removed at stage n. The total measure removed is \sum_{n=1}^\infty 2^{n-1}/4^n = 1/2, so the resulting set has Lebesgue measure $1/2. Despite this positive measure, the set is nowhere dense because it contains no open intervals, as the construction ensures the remaining pieces shrink sufficiently without leaving contiguous segments.^[23] More generally, nowhere dense sets of positive measure in [0,1] can be formed by enumerating the rationals \{q_k\}_{k=1}^\infty and removing, around each q_k, an open interval of length \epsilon / 2^k for small \epsilon > 0, ensuring the intervals are contained within [0,1]. The total length removed is at most \epsilon < 1, leaving a complement of measure greater than $1 - \epsilon. The closure of this complement has empty interior because the removed set is dense (as the rationals are dense), making every subinterval intersect the removed portions. Due to possible overlaps in these intervals, the actual measure removed may be less than \epsilon, allowing the remaining set to have measure exceeding $1/2 in optimized cases and arbitrarily close to 1.^[24] Such sets are closed (upon taking closure if needed), nowhere dense by construction, and have positive Lebesgue measure. As a single nowhere dense set, each is meager in the Baire category sense, yet its complement has full measure in [0,1]. These examples highlight the independence of topological category and Lebesgue measure. Historically, Henry J. S. Smith first constructed a nowhere dense set of positive measure in 1875, followed by Vito Volterra's similar example in 1881, with variants by Georg Cantor in 1883 aimed at demonstrating that notions of "smallness" in category and measure differ. Cantor's 1883 variant further illustrated such sets in the context of monotonic functions.^[23]^[25]^[26]