If [open sets](/page/Open%20Set) are the primitive data of a [topology](/page/Topology), closed sets are their complementary counterpart — the [sets](/page/Set) whose complements are open. The theory of closed sets is entirely dual to that of open sets, but the duality is not symmetric in its consequences: closed sets are stable under arbitrary *intersections* (but only finite unions), while open sets are stable under arbitrary *unions* (but only finite intersections). This reversal of roles, mediated by [De Morgan's laws](/theorems/622), gives closed sets a character distinct from that of open sets — they are the natural home for limit points, convergent [sequences](/page/Sequence), and the zero sets of continuous [functions](/page/Function). This page develops closed sets in depth: the duality with open sets, the closure operator and limit points, the sequential characterisation in [metric spaces](/page/Metric%20Space), the behaviour of closed sets under intersections (including the Cantor Intersection Theorem), the role of closed sets in detecting [continuity](/page/Continuity), and closed maps. The abstract framework is developed on the [Topology](/page/Topology) page; the companion treatment of open sets is on the [Open Set](/page/Open%20Set) page. [motivation] ## Motivation ### The Metric Picture In a metric space $(X, d)$, a set $F$ is closed if it "contains all its limit points" — whenever a sequence in $F$ converges, the limit belongs to $F$. Equivalently, $F$ is closed if its complement $X \setminus F$ is [open](/page/Open%20Set): every point outside $F$ has a ball that misses $F$ entirely. The closed interval $[0, 1] \subset \mathbb{R}$ is the prototype: any convergent sequence $x_n \in [0,1]$ has its limit in $[0,1]$, because $0 \le x_n \le 1$ for all $n$ implies $0 \le \lim x_n \le 1$. The open interval $(0, 1)$ is not closed: the sequence $1/n \in (0,1)$ converges to $0 \notin (0,1)$. ### Why the Complement Definition? In general topological spaces, sequences may not suffice to detect all topological properties (in non-first-countable spaces, nets or filters are needed). The correct general definition of "closed" is therefore not "contains all sequential [limits](/page/Limit)" but "has an open complement." This complement-based definition works uniformly across all topological spaces and automatically produces the right closure properties via De Morgan's laws: since arbitrary unions of open sets are open, arbitrary intersections of closed sets (whose complements are arbitrary unions of open sets) are closed. ### The Asymmetry Open sets are closed under arbitrary unions and finite intersections. Closed sets, by duality, are closed under arbitrary *intersections* and finite *unions*. This asymmetry matters: the infinite union $\bigcup_{n=1}^\infty [1/n, 1] = (0, 1]$ is not closed in $\mathbb{R}$ (the limit point $0$ is missing), just as the infinite intersection $\bigcap_{n=1}^\infty (-1/n, 1/n) = \{0\}$ is not open. Finite operations preserve both openness and closedness; infinite operations preserve only one. [/motivation] ## Definition In a topological space, closed sets are defined by complementation. In a metric space, this is equivalent to the sequential characterisation. [definition:Closed Set] Let $(X, \tau)$ be a [topological space](/page/Topology). A subset $F \subseteq X$ is **closed** if its complement is open: \begin{align*} F \text{ is closed} \iff X \setminus F \in \tau. \end{align*} [/definition] By the axioms of a topology, $\varnothing = X \setminus X$ and $X = X \setminus \varnothing$ are closed (since $X$ and $\varnothing$ are open). The further closure properties follow from dualising the open-set axioms via [De Morgan's laws](/theorems/622): [quotetheorem:306] The proof is a direct application of De Morgan: the complement of $\bigcap_{i \in I} F_i$ is $\bigcup_{i \in I}(X \setminus F_i)$, which is an arbitrary union of open sets, hence open; the complement of $F_1 \cup \cdots \cup F_n$ is $(X \setminus F_1) \cap \cdots \cap (X \setminus F_n)$, a finite intersection of open sets, hence open. The asymmetry — arbitrary intersections but only finite unions — is the exact dual of the open-set axioms. The infinite union $\bigcup_{n=1}^\infty [1/n, 1] = (0, 1]$ shows that the finiteness restriction on unions is necessary: $(0, 1]$ is not closed in $\mathbb{R}$ because its complement $(-\infty, 0] \cup (1, \infty)$ is not open — the point $0$ has no $\varepsilon$-ball contained entirely in $(-\infty, 0] \cup (1, \infty)$, since every ball around $0$ contains positive numbers not in the complement. A set can be both open and closed (**clopen**): $\varnothing$ and $X$ always are, and in the discrete topology every set is clopen. A set can be neither open nor closed: the half-open interval $[0, 1) \subset \mathbb{R}$ is neither (its complement $(-\infty, 0) \cup [1, \infty)$ is not open because of $1$, so $[0,1)$ is not closed; and $[0,1)$ is not open because of $0$). In a connected space, $\varnothing$ and $X$ are the *only* clopen sets. ## Closure, Limit Points, and the Sequential Characterisation The passage from a set to its closure — the smallest closed set containing it — is one of the most fundamental operations in topology. It can be described in three equivalent ways: as an intersection of closed sets, via limit points, or (in metric spaces) via convergent sequences. These descriptions are the working tools for proving that specific sets are closed. ### The Closure Operator The closure of $A$ adds to $A$ all points that are "infinitely close" to $A$ in the topological sense: every neighbourhood of such a point meets $A$. [definition:Closure] Let $(X, \tau)$ be a topological space and $A \subseteq X$. The **closure** of $A$ is \begin{align*} \overline{A} := \bigcap \{F \subseteq X : F \text{ is closed and } A \subseteq F\}, \end{align*} the smallest closed set containing $A$. [/definition] A set is closed if and only if $F = \overline{F}$. The closure operator satisfies the **Kuratowski axioms**: $\overline{\varnothing} = \varnothing$, $A \subseteq \overline{A}$, $\overline{\overline{A}} = \overline{A}$ (idempotence), and $\overline{A \cup B} = \overline{A} \cup \overline{B}$. These four properties characterise the closure operator: any function $\mathcal{P}(X) \to \mathcal{P}(X)$ satisfying them determines a unique topology on $X$ whose closed sets are the fixed points of the function. This gives an alternative axiomatisation of topology in terms of closure rather than open sets. ### Limit Points and the Derived Set A point $x$ belongs to $\overline{A} \setminus A$ if every neighbourhood of $x$ meets $A$ but $x$ itself is not in $A$. The concept that captures this is the limit point (or accumulation point). [definition:Limit Point] Let $(X, \tau)$ be a topological space and $A \subseteq X$. A point $x \in X$ is a **limit point** (or **accumulation point**) of $A$ if every open set $U$ containing $x$ satisfies \begin{align*} (U \setminus \{x\}) \cap A \neq \varnothing. \end{align*} The set of all limit points of $A$ is the **derived set** $A'$. [/definition] The key distinction is between $x \in \overline{A}$ (every neighbourhood of $x$ meets $A$) and $x \in A'$ (every neighbourhood of $x$ meets $A$ in a point *other than $x$*). An isolated point of $A$ — a point $a \in A$ with a neighbourhood $U$ satisfying $U \cap A = \{a\}$ — belongs to $\overline{A}$ but not to $A'$. The closure decomposes as $\overline{A} = A \cup A'$, so a set is closed if and only if it contains all its limit points. [example:Derived Set Of The Rationals] In $\mathbb{R}$ with the standard topology, $\mathbb{Q}' = \mathbb{R}$: every real number is a limit point of $\mathbb{Q}$, since every open interval contains infinitely many rationals. Consequently $\overline{\mathbb{Q}} = \mathbb{Q} \cup \mathbb{Q}' = \mathbb{R}$, which recovers the density of $\mathbb{Q}$ in $\mathbb{R}$. By contrast, the derived set of $\mathbb{Z}$ is $\mathbb{Z}' = \varnothing$ — every integer $n$ has the neighbourhood $(n - 1/2, n + 1/2)$ containing no other integer — so $\overline{\mathbb{Z}} = \mathbb{Z}$, confirming that $\mathbb{Z}$ is closed. [/example] ### Sequential Characterisation in Metric Spaces In a metric space, convergent sequences provide a complete description of the closure and hence of closed sets. This is because metric spaces are first countable: every point has a countable neighbourhood basis $\{B(x, 1/n)\}_{n \in \mathbb{N}}$. [definition:Sequentially Closed Set] A subset $F$ of a topological space $X$ is **sequentially closed** if whenever $\{x_n\}$ is a sequence in $F$ with $x_n \to x$, the limit $x$ belongs to $F$. [/definition] In a metric space (or more generally in any first-countable space), a set is closed if and only if it is sequentially closed. The proof in the forward direction is immediate: if $F$ is closed and $x_n \in F$ with $x_n \to x$, then $x \in \overline{F} = F$. For the reverse, if $F$ is not closed then there exists $x \in \overline{F} \setminus F$; for each $n$, the ball $B(x, 1/n)$ meets $F$ in some point $x_n$, and $x_n \to x$ with $x \notin F$, contradicting sequential closedness. In general topological spaces, sequentially closed does not imply closed. Nets or filters are needed to recover the equivalence — a set is closed if and only if it is closed under convergence of *nets*. [example:Closed Set Via Sequential Criterion] Define $F = \{(x, y) \in \mathbb{R}^2 : y \ge x^2\}$ — the region above the parabola $y = x^2$. We show $F$ is closed using the sequential characterisation. Let $(x_n, y_n) \in F$ with $(x_n, y_n) \to (a, b)$. Then $x_n \to a$ and $y_n \to b$, and $y_n \ge x_n^2$ for all $n$. Since the function $(x, y) \mapsto y - x^2$ is continuous and $y_n - x_n^2 \ge 0$, we have $b - a^2 = \lim(y_n - x_n^2) \ge 0$, so $b \ge a^2$ and $(a, b) \in F$. [/example] ## Closed Sets and Intersections The stability of closed sets under arbitrary intersections — with no restriction on the number of sets — is the most powerful algebraic property that closed sets possess. It is dual to the stability of open sets under arbitrary unions and underpins the definitions of closure, the construction of generated $\sigma$-algebras, and the Cantor Intersection Theorem. ### Why Arbitrary Intersections Work The proof is immediate from De Morgan: the complement of $\bigcap_{i \in I} F_i$ is $\bigcup_{i \in I}(X \setminus F_i)$, which is an arbitrary union of open sets, hence open. This argument requires no restriction on the index set $I$ — it works for finite, countable, and uncountable intersections alike. This is the reason that many objects in analysis are defined as intersections: the closure $\overline{A}$ is the intersection of all closed sets containing $A$; the closed linear span $\overline{\operatorname{span}}(S)$ is the intersection of all closed subspaces containing $S$; the $\sigma$-algebra $\sigma(\mathcal{E})$ generated by a collection $\mathcal{E}$ is the intersection of all $\sigma$-algebras containing $\mathcal{E}$. In each case, arbitrary stability under intersections guarantees that the "smallest closed set with property $P$" exists and is itself closed. ### The Cantor Intersection Theorem The most important result about intersections of closed sets is that *nested* sequences of nonempty compact (or, in complete metric spaces, closed and shrinking) sets have nonempty intersection. This is the foundation for many existence arguments in analysis. [quotetheorem:624] The compact version applies in any Hausdorff space and uses the finite intersection property: if the intersection were empty, the complements would form an open cover of $K_1$ with no finite subcover, contradicting compactness. The complete metric version requires the diameter condition — without it, the intersection can be empty: the sets $K_n = [n, \infty)$ are closed, nonempty, and nested in $\mathbb{R}$, but $\bigcap_{n=1}^\infty [n, \infty) = \varnothing$ because the diameters do not shrink to zero. The Cantor Intersection Theorem is the engine behind many classical results: the proof of the [Baire Category Theorem](/theorems/630) (intersections of dense open sets in a complete metric space are dense), the construction of the Cantor set (an intersection of nested closed sets), the proof of the Bolzano-Weierstrass theorem (via nested bisection), and the existence of solutions to ordinary differential equations via the Picard iteration (the solution is the unique point in the intersection of nested closed balls). ### $F_\sigma$ Sets and the Borel Hierarchy A countable *union* of closed sets — dual to the $G_\delta$ (countable intersection of open) concept — is called an $F_\sigma$ **set** (from the French *fermé*, "closed," and *somme*, "sum"). Not every $F_\sigma$ set is closed, and not every $F_\sigma$ set is open — the class sits strictly between. [example:Q As An F Sigma Set] The set of rationals $\mathbb{Q} \subset \mathbb{R}$ is an $F_\sigma$ set: $\mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{q\}$, a countable union of closed singletons. But $\mathbb{Q}$ is neither open (no interval is contained in $\mathbb{Q}$) nor closed ($\overline{\mathbb{Q}} = \mathbb{R} \neq \mathbb{Q}$). Its complement $\mathbb{R} \setminus \mathbb{Q}$ (the irrationals) is a $G_\delta$ set. [/example] The $F_\sigma$ and $G_\delta$ classes are the first two levels of the **Borel hierarchy** — the iterated process of taking countable unions and intersections starting from open and closed sets. The full hierarchy produces the Borel $\sigma$-algebra $\mathcal{B}(X)$, the smallest $\sigma$-algebra containing all open sets (equivalently, all closed sets). This $\sigma$-algebra is the natural domain for Borel measures and is the foundation of measure theory on topological spaces. ## Closed Sets and Continuity One of the most useful characterisations of continuity is in terms of closed sets: a function is continuous if and only if preimages of closed sets are closed. This "closed-set criterion" is often easier to verify than the open-set definition and connects closed sets directly to the structure of continuous functions. ### The Closed-Set Criterion for Continuity A function $f: X \to Y$ between topological spaces is continuous if and only if $f^{-1}(V)$ is open for every open $V \subseteq Y$. Taking complements: $f^{-1}(Y \setminus F) = X \setminus f^{-1}(F)$, so $f^{-1}(V)$ is open for every open $V$ if and only if $f^{-1}(F)$ is closed for every closed $F$. This gives an equivalent characterisation: $f$ is continuous if and only if preimages of closed sets are closed. This characterisation is particularly natural for real-valued functions, because many important subsets of $\mathbb{R}$ are closed: $\{0\}$, $[a, b]$, $(-\infty, a]$, $[a, \infty)$. Continuity immediately implies that the zero set $f^{-1}(\{0\})$, the sublevel set $f^{-1}((-\infty, a])$, and the superlevel set $f^{-1}([a, \infty))$ are all closed. ### Zero Sets and Level Sets The preimage of a closed set under a continuous function is closed. The most important special case is the zero set: if $f: X \to \mathbb{R}$ is continuous, then $f^{-1}(\{0\}) = \{x \in X : f(x) = 0\}$ is closed, because $\{0\}$ is closed in $\mathbb{R}$. More generally, the level set $f^{-1}(\{c\})$ is closed for any $c \in \mathbb{R}$. [example:Level Sets Of Continuous Functions] The unit sphere $S^{n-1} = \{x \in \mathbb{R}^n : |x| = 1\}$ is the level set $f^{-1}(\{1\})$ of the continuous function $f: \mathbb{R}^n \to \mathbb{R}$ defined by $f(x) = |x| = (x_1^2 + \cdots + x_n^2)^{1/2}$. Since $f$ is continuous and $\{1\}$ is closed in $\mathbb{R}$, the sphere $S^{n-1}$ is closed in $\mathbb{R}^n$. Similarly, the closed unit ball $\overline{B}(0, 1) = \{x : |x| \le 1\} = f^{-1}((-\infty, 1])$ is a sublevel set of $f$, hence closed. And the set $\{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \le 4,\; y \ge x\}$ is the intersection $g^{-1}((-\infty, 4]) \cap h^{-1}([0, \infty))$ where $g(x,y) = x^2 + y^2$ and $h(x,y) = y - x$, a finite intersection of closed sets, hence closed. [/example] This technique — expressing a set as a preimage of a closed set under a continuous function, or as a finite intersection of such preimages — is the standard method for proving that a given set is closed. It is far more efficient than verifying the sequential characterisation or checking that the complement is open. ### Closed Sets Separate: Urysohn's Lemma In sufficiently nice spaces (normal spaces — those in which disjoint closed sets can be separated by open sets), closed sets can always be separated by continuous functions. This is the content of **Urysohn's Lemma**: if $F_0$ and $F_1$ are disjoint closed subsets of a normal space $X$, there exists a continuous function $f: X \to [0, 1]$ with $f|_{F_0} = 0$ and $f|_{F_1} = 1$. Every metric space is normal (the function $f(x) = d(x, F_0) / (d(x, F_0) + d(x, F_1))$ works), so Urysohn's lemma applies in all metric settings. This result is fundamental for partition-of-unity constructions, the Tietze extension theorem, and the metrization theorems. ## Closed Maps Continuity pulls closed sets backward (preimages of closed sets are closed). The complementary condition — pushing closed sets forward — defines a closed map, which plays a different role from that of an [open map](/page/Open%20Set). ### Definition and the Closed Map Lemma A function that sends closed sets to closed sets is called a closed map. Like openness, this is independent of continuity: a function can be continuous without being closed, and closed without being continuous. [definition:Closed Map] A function $f: X \to Y$ between topological spaces is a **closed map** if for every closed set $F \subseteq X$, the image $f(F)$ is closed in $Y$. [/definition] The most important source of closed maps is the interaction between compactness and the Hausdorff property: [quotetheorem:317] The proof uses the fact that compact subsets of Hausdorff spaces are closed ([Theorem 307](/theorems/307)): if $C \subseteq X$ is closed and $X$ is compact, then $C$ is compact (a closed subset of a compact space); the continuous image $f(C)$ is compact; and a compact subset of a Hausdorff space is closed. This chain — closed $\to$ compact $\to$ compact image $\to$ closed — gives the result. The Closed Map Lemma is the key ingredient in the proof of the [Topological Inverse Function Theorem](/theorems/318): a continuous closed bijection has a continuous inverse (because the inverse maps open sets to open sets, which is equivalent to the forward map being closed). [example:Projection Is Not Closed] The projection $\pi_1: \mathbb{R}^2 \to \mathbb{R}$ defined by $\pi_1(x, y) = x$ is continuous and [open](/page/Open%20Set) but not closed. The hyperbola $H = \{(x, y) : xy = 1\}$ is closed in $\mathbb{R}^2$ (it is the level set of the continuous function $(x,y) \mapsto xy$ at value $1$), but its image $\pi_1(H) = \mathbb{R} \setminus \{0\}$ is not closed in $\mathbb{R}$. The failure occurs because the hyperbola "escapes to infinity" as $x \to 0$: the point $(0, y)$ is never in $H$ for any $y$, but points of $H$ accumulate toward the $y$-axis. If the domain were compact (say, $\pi_1$ restricted to a closed bounded subset of $\mathbb{R}^2$), the Closed Map Lemma would guarantee that the projection is closed. [/example] ### Quotient Maps and Closed Equivalence Relations Closed maps arise naturally in the theory of quotient spaces. If $\sim$ is an [equivalence relation](/page/Equivalence%20Relation) on $X$ and the quotient map $\pi: X \to X/{\sim}$ is a closed map, then the quotient space inherits many properties from $X$ (for instance, a quotient of a normal space by a closed equivalence relation is normal). The condition that $\pi$ is closed is equivalent to the **saturation** condition: for every closed set $F \subseteq X$, the saturation $\pi^{-1}(\pi(F)) = \{x \in X : x \sim y \text{ for some } y \in F\}$ is closed. This is a natural condition in many geometric constructions (collapsing a subspace to a point, identifying [boundary](/page/Boundary) points, forming orbit spaces). ## Closed Sets and Connectedness Closed sets interact with connectedness in a way that has no analogue for open sets: the closure of a connected set is always connected. This is a consequence of the general principle that "adding limit points cannot create gaps." The [closure of a connected set is connected](/theorems/297). More precisely, if $Y$ is a connected subspace of $X$, then $\overline{Y}$ is connected, and more generally any set $Z$ with $Y \subseteq Z \subseteq \overline{Y}$ is connected. The proof is by contradiction: if $Z = U \cup V$ were a disconnection (with $U, V$ disjoint, nonempty, and open in $Z$), then $Y = (U \cap Y) \cup (V \cap Y)$ would be a disconnection of $Y$, contradicting the connectedness of $Y$. The key point is that $Y$ is dense in $Z$ (since $Z \subseteq \overline{Y}$), so both $U$ and $V$ must meet $Y$. [example:Closure Preserving Connectedness] The graph $\Gamma = \{(x, \sin(1/x)) : x > 0\} \subset \mathbb{R}^2$ is connected (it is the continuous image of the connected set $(0, \infty)$). Its closure $\overline{\Gamma} = \Gamma \cup (\{0\} \times [-1, 1])$ — the **topologist's sine curve** — is therefore connected. The added segment $\{0\} \times [-1, 1]$ consists entirely of limit points of $\Gamma$: for any $y_0 \in [-1, 1]$, the oscillation of $\sin(1/x)$ near $x = 0$ guarantees points $(x_n, \sin(1/x_n))$ with $\sin(1/x_n) \to y_0$. [/example] The converse is false: if $\overline{A}$ is connected, $A$ need not be. For instance, $A = (0, 1) \cup (1, 2) \subset \mathbb{R}$ is disconnected ($1 \notin A$), but $\overline{A} = [0, 2]$ is connected. ## References - Munkres, J. R., *Topology* (2nd ed., 2000). - Willard, S., *General Topology* (1970). - Steen, L. A. and Seebach, J. A., *Counterexamples in Topology* (1978). - Kelley, J. L., *General Topology* (1955).