Given a subset $A$ of a [topological space](/page/Topology) $X$, there are two fundamental questions one can ask about the relationship between $A$ and the ambient space. The first — which points definitely belong to $A$? — leads to the [interior](/page/Open%20Set). The second — which points are "infinitely close" to $A$, in the sense that every neighbourhood of the point meets $A$? — leads to the **closure**. The distinction matters because many natural constructions produce sets that are not [closed](/page/Closed%20Set). The open interval $(0,1) \subset \mathbb{R}$ misses its endpoints, yet sequences in $(0,1)$ can converge to $0$ or $1$. The [rationals](/page/Rational%20Numbers) $\mathbb{Q}$ form a countable set, yet every real number is the limit of a sequence of rationals. In each case, the original set $A$ is too small to capture the limiting behaviour of its own elements, and we need a systematic way to "complete" $A$ by adjoining the points that $A$ reaches toward but does not contain. The closure $\overline{A}$ is the result of that completion: the smallest closed set containing $A$, or equivalently, the set $A$ together with all its limit points. This concept is central to analysis, where density arguments, approximation theorems, and extension results all rest on understanding when $\overline{A} = X$. [example: A Set That Reaches Beyond Itself] Let $X = \mathbb{R}$ with the standard topology, and let $A = \{1/n : n \in \mathbb{N}\}$. Every element of $A$ is strictly positive, so $0 \notin A$. However, $0$ is "infinitely close" to $A$: for every $\varepsilon > 0$, the open interval $(-\varepsilon, \varepsilon)$ contains $1/n$ for all $n > 1/\varepsilon$, so every neighbourhood of $0$ meets $A$. On the other hand, no point $x > 1$ is close to $A$ in this sense. Choosing $\varepsilon = (x - 1)/2 > 0$, the interval $(x - \varepsilon, x + \varepsilon)$ lies entirely above $(1+x)/2 > 1$ and therefore misses $A$ entirely. The closure is $\overline{A} = A \cup \{0\} = \{0\} \cup \{1/n : n \in \mathbb{N}\}$. This set is closed: any convergent sequence in $\overline{A}$ either is eventually constant (and thus converges to a point $1/n$ or to $0$) or has infinitely many distinct terms, which forces convergence to $0$. [/example] ## Definition The closure can be characterised in several equivalent ways. The "top-down" definition, which is natural in the topological setting, describes the closure as the intersection of all closed sets containing $A$. The "bottom-up" definition builds the closure by adding limit points. The equivalence of these perspectives is one of the first results of the theory. [definition: Closure] Let $(X, \tau)$ be a [topological space](/page/Topology) and let $A \subset X$. The **closure** of $A$, denoted $\overline{A}$, is the intersection of all [closed sets](/page/Closed%20Set) containing $A$: \begin{align*} \overline{A} := \bigcap \{ F \subset X : F \text{ is closed and } A \subset F \}. \end{align*} [/definition] Since an arbitrary intersection of closed sets is closed, $\overline{A}$ is itself closed. Moreover, $A \subset \overline{A}$ because every closed set $F$ in the intersection contains $A$. If $C$ is any closed set with $A \subset C$, then $C$ appears in the intersection and so $\overline{A} \subset C$. This means $\overline{A}$ is the **smallest** closed set containing $A$ — it is the closed set obtained by adding to $A$ exactly the points that $A$ forces upon any closed superset, and nothing more. A set $A$ is closed if and only if $\overline{A} = A$. This is immediate: if $A$ is closed, then $A$ itself appears in the intersection, forcing $\overline{A} \subset A$; conversely, if $\overline{A} = A$, then $A$ equals a closed set. ## Equivalent Characterisations The intersection-of-closed-sets definition is clean and general, but it does not tell us how to decide whether a specific point belongs to $\overline{A}$. Testing membership by checking every closed superset of $A$ is impractical. We need pointwise criteria: conditions on a single point $x$ that determine whether $x \in \overline{A}$. There are two natural candidates. The first uses neighbourhoods: $x \in \overline{A}$ if and only if every open set containing $x$ meets $A$. The second uses limit points: $x \in \overline{A}$ if and only if $x$ is either in $A$ or is a limit point of $A$. In metric spaces, a third characterisation becomes available: $x \in \overline{A}$ if and only if some sequence in $A$ converges to $x$. [quotetheorem:1005] This characterisation reframes closure membership as a local condition: $x \in \overline{A}$ precisely when $A$ "intrudes" into every neighbourhood of $x$. Equivalently, $x \notin \overline{A}$ if and only if there exists an open set $U$ with $x \in U$ and $U \cap A = \varnothing$ — that is, $x$ can be separated from $A$ by an open set. The neighbourhood characterisation also reveals the relationship between closure and [interior](/page/Open%20Set). The complement $X \setminus \overline{A}$ consists of all points that have a neighbourhood disjoint from $A$, which is exactly the interior of $X \setminus A$: \begin{align*} X \setminus \overline{A} = (X \setminus A)^\circ, \qquad X \setminus A^\circ = \overline{X \setminus A}. \end{align*} These complement duality formulas mean that every result about closures has a dual statement about interiors, and vice versa. To state the second characterisation, we need the notion of a limit point. [definition: Limit Point] Let $(X, \tau)$ be a topological space and $A \subset X$. A point $x \in X$ is a **limit point** (or **accumulation point**) of $A$ if every open set $U$ containing $x$ satisfies $U \cap (A \setminus \{x\}) \neq \varnothing$. The set of all limit points of $A$ is denoted $A'$ and called the **derived set** of $A$. [/definition] The difference between the neighbourhood characterisation of closure and the definition of a limit point is the exclusion of $x$ itself: a limit point of $A$ must have every neighbourhood meeting $A$ in a point *other than $x$*. An isolated point of $A$ — a point $x \in A$ with some neighbourhood $U$ such that $U \cap A = \{x\}$ — belongs to $\overline{A}$ but is not a limit point of $A$. [quotetheorem:1006] This decomposition is often the most useful in practice: the closure consists of the original set together with whatever additional points are "forced" by the limit-taking process. In particular, $A$ is closed if and only if $A' \subset A$ — a closed set contains all its own limit points. ### Sequential Closure in Metric Spaces In a general topological space, sequences are not always sufficient to detect closure: there exist spaces in which a point belongs to $\overline{A}$ but no sequence in $A$ converges to it (the standard example is the ordinal space $[0, \omega_1)$, where $\omega_1$ is in the closure of the set of countable ordinals in $[0, \omega_1]$ but is not the limit of any sequence from that set). However, in [metric spaces](/page/Metric%20Space) — and more generally in first-countable spaces — sequences suffice. [quotetheorem:1007] The sequential characterisation is the workhorse of closure arguments in analysis. To prove that $x \in \overline{A}$, one constructs a sequence in $A$ converging to $x$. To prove that $x \notin \overline{A}$, one exhibits $\varepsilon > 0$ with $B(x, \varepsilon) \cap A = \varnothing$. In metric spaces, the closure also admits a distance characterisation. Define the distance from a point $x$ to a set $A$ by \begin{align*} \operatorname{dist}(x, A) := \inf_{a \in A} d(x, a). \end{align*} [quotetheorem:1008] The equivalence is immediate from the sequential characterisation: $\operatorname{dist}(x, A) = 0$ means that for every $n \in \mathbb{N}$, there exists $a_n \in A$ with $d(x, a_n) < 1/n$, which gives a sequence in $A$ converging to $x$. This characterisation is especially useful in [analysis](/page/Real%20Analysis) and [PDE theory](/page/Partial%20Differential%20Equation), where the function $x \mapsto \operatorname{dist}(x, A)$ is Lipschitz continuous (with Lipschitz constant $1$) and provides a quantitative measure of how close a point is to $A$. ## The Closure Operator and the Kuratowski Axioms The passage from a set $A$ to its closure $\overline{A}$ defines a map on the power set of $X$. This map is not arbitrary — it satisfies four identities that completely characterise it. These identities, due to Kuratowski, show that the closure operation carries exactly the same information as the topology itself: given any map $\mathcal{P}(X) \to \mathcal{P}(X)$ satisfying these four axioms, there is a unique topology on $X$ for which that map is the closure operator. This is significant because it provides an alternative foundation for topology. Rather than starting with a collection $\tau$ of "open sets" satisfying the usual axioms (stability under arbitrary unions and finite intersections), one can start with a closure operator and derive the topology from it. The two approaches are equivalent, but the closure-operator perspective is sometimes more natural — particularly in contexts where the notion of "approaching a set" is more intuitive than the notion of "openness." [quotetheorem:1009] Several features of these axioms deserve attention. **Normalisation** says that the empty set has no limit points — nothing is close to nothing. This is the closure-operator counterpart of the topological axiom that $\varnothing$ is open. **Extensivity** says that adding limit points never shrinks a set. Together with idempotency, it means that closure is a "projection" onto the collection of closed sets: applying it once suffices, and applying it again changes nothing. **Idempotency** is the most substantive axiom. It asserts that $\overline{A}$ is closed — the limit points of $\overline{A}$ already belong to $\overline{A}$, so closing again adds nothing. Without idempotency, one might have to iterate the closure process transfinitely (and indeed, the analogous operation for convex hulls in infinite dimensions does require such iteration in some settings). **Preservation of binary unions** says that a point is close to $A \cup B$ if and only if it is close to $A$ or close to $B$. This is the property that ensures finite unions of closed sets are closed (since $\overline{A \cup B} = \overline{A} \cup \overline{B}$, a finite union of closed sets satisfies $\overline{F_1 \cup \cdots \cup F_n} = \overline{F_1} \cup \cdots \cup \overline{F_n} = F_1 \cup \cdots \cup F_n$, which is closed). The axiom does **not** extend to infinite unions: $\overline{\bigcup_{i=1}^\infty A_i}$ can be strictly larger than $\bigcup_{i=1}^\infty \overline{A_i}$. [example: Failure of Closure to Preserve Infinite Unions] Let $X = \mathbb{R}$ with the standard topology. For each $n \in \mathbb{N}$, define $A_n := \{1/n\}$. Each $A_n$ is a single point, hence closed: $\overline{A_n} = A_n$. Therefore: \begin{align*} \bigcup_{n=1}^\infty \overline{A_n} = \bigcup_{n=1}^\infty \{1/n\} = \{1/n : n \in \mathbb{N}\}. \end{align*} However, the closure of the union is: \begin{align*} \overline{\bigcup_{n=1}^\infty A_n} = \overline{\{1/n : n \in \mathbb{N}\}} = \{0\} \cup \{1/n : n \in \mathbb{N}\}, \end{align*} since $0$ is a limit point of $\{1/n : n \in \mathbb{N}\}$ (as computed in the opening example). The point $0$ belongs to the closure of the union but not to the union of the closures. [/example] ## Closure in Concrete Spaces The abstract definition of closure produces strikingly different behaviour depending on the topology of the ambient space. Three computations illustrate the range of possibilities: a familiar metric-space example where the closure is "almost" the original set, a case where closure is dramatic, and a case governed by a non-metric topology. ### Closure of Open Balls in Metric Spaces In $\mathbb{R}^n$ with the Euclidean metric, the closure of an open ball is the corresponding closed ball. This is the "expected" behaviour, but it relies on the geometry of the metric — it fails in general metric spaces. [example: Closure of the Open Ball] Let $X = \mathbb{R}^n$ with the Euclidean metric $d(x,y) = |x - y|$. Fix $x_0 \in \mathbb{R}^n$ and $r > 0$, and let $A = B(x_0, r) = \{x \in \mathbb{R}^n : |x - x_0| < r\}$. We claim $\overline{B(x_0, r)} = \overline{B}(x_0, r) = \{x \in \mathbb{R}^n : |x - x_0| \le r\}$. The inclusion $\overline{B(x_0, r)} \subset \overline{B}(x_0, r)$ follows because $\overline{B}(x_0, r)$ is closed and contains $B(x_0, r)$, so it contains the smallest closed superset. For the reverse inclusion, let $x$ satisfy $|x - x_0| = r$ (a boundary point). Define the sequence $(a_n)_{n=1}^\infty$ by: \begin{align*} a_n := x_0 + \left(1 - \frac{1}{n}\right)(x - x_0). \end{align*} Then $|a_n - x_0| = (1 - 1/n) r < r$, so $a_n \in B(x_0, r)$. Moreover: \begin{align*} |a_n - x| = \left|\left(1 - \frac{1}{n}\right)(x - x_0) - (x - x_0)\right| = \frac{1}{n}|x - x_0| = \frac{r}{n} \to 0. \end{align*} So $a_n \to x$, which proves $x \in \overline{B(x_0, r)}$ by the sequential characterisation. [/example] This argument depends on the "convexity" of the Euclidean metric: we can interpolate between $x_0$ and a boundary point $x$ and stay inside the open ball. In a discrete metric space $d(x,y) = 1$ for $x \neq y$, the open ball $B(x_0, 1) = \{x_0\}$ is already closed (every singleton is clopen in the discrete topology), so $\overline{B(x_0, 1)} = \{x_0\}$, while $\overline{B}(x_0, 1) = X$. The closure of the open ball can be a strict subset of the closed ball. ### Closure of the Rationals The rationals $\mathbb{Q}$ are a countable set, yet they are dense in $\mathbb{R}$: $\overline{\mathbb{Q}} = \mathbb{R}$. This is a far more dramatic closure than the previous example — the closure is the entire ambient space. [example: Density of the Rationals] Let $X = \mathbb{R}$ with the standard topology, and let $A = \mathbb{Q}$. We show that $\overline{\mathbb{Q}} = \mathbb{R}$ by verifying the neighbourhood criterion: for every $x \in \mathbb{R}$ and every $\varepsilon > 0$, the interval $(x - \varepsilon, x + \varepsilon)$ contains a rational number. Fix $x \in \mathbb{R}$ and $\varepsilon > 0$. Choose $n \in \mathbb{N}$ with $n > 1/\varepsilon$ (possible by the Archimedean property of $\mathbb{R}$). Consider the set $\{m/n : m \in \mathbb{Z}\}$ of rational numbers with denominator $n$. Consecutive elements $m/n$ and $(m+1)/n$ are spaced $1/n < \varepsilon$ apart. Since the real line is covered by the intervals $[m/n, (m+1)/n]$ as $m$ ranges over $\mathbb{Z}$, there exists $m_0 \in \mathbb{Z}$ with $m_0/n \le x < (m_0 + 1)/n$. Then: \begin{align*} |x - m_0/n| < 1/n < \varepsilon, \end{align*} so $m_0/n \in (x - \varepsilon, x + \varepsilon) \cap \mathbb{Q}$. Since $x$ and $\varepsilon$ were arbitrary, every open set meeting $\mathbb{R}$ contains a rational, and $\overline{\mathbb{Q}} = \mathbb{R}$. [/example] The same argument, applied to the irrationals $\mathbb{R} \setminus \mathbb{Q}$, shows that $\overline{\mathbb{R} \setminus \mathbb{Q}} = \mathbb{R}$ as well: between any two rationals lies an irrational (for instance, $a + (b-a)/\sqrt{2}$ for $a < b$ rational). So $\mathbb{R}$ contains two disjoint dense subsets — one countable, one uncountable. This phenomenon is impossible in the discrete topology, where every set is closed and no proper subset is dense. ### Closure in the Cofinite Topology [example: Closure in the Cofinite Topology] Let $X = \mathbb{R}$ with the cofinite topology: the closed sets are $\mathbb{R}$ and the finite subsets of $\mathbb{R}$. Let $A \subset \mathbb{R}$ be any infinite subset (for instance, $A = \mathbb{N}$). The closure $\overline{A}$ is the intersection of all closed sets containing $A$. A closed set $F$ in the cofinite topology is either finite or all of $\mathbb{R}$. Since $A$ is infinite, no finite set contains $A$, so the only closed set containing $A$ is $\mathbb{R}$ itself. Therefore $\overline{A} = \mathbb{R}$. Conversely, if $A = \{p\}$ is a single point, then $\{p\}$ is itself finite, hence closed in the cofinite topology, so $\overline{\{p\}} = \{p\}$. This shows that in the cofinite topology, every infinite set is dense. The topology is too coarse to distinguish infinite sets from the whole space. [/example] ## Dense Subsets and Approximation The most important instance of closure in analysis is the case $\overline{A} = X$: the set $A$ is [dense](/page/Dense%20Subset) in $X$. Density is the topological condition that makes approximation arguments possible — it guarantees that every point of $X$ can be reached as a limit of elements of $A$. [definition: Dense Subset] Let $(X, \tau)$ be a topological space. A subset $A \subset X$ is **dense** in $X$ if $\overline{A} = X$, or equivalently, if every nonempty open set $U \in \tau$ satisfies $U \cap A \neq \varnothing$. [/definition] Density is the bridge between "easy" and "hard" objects. If a property $P$ holds for all elements of a dense subset $A$ and is preserved under limits (in an appropriate sense), then $P$ holds for all elements of $X$. The strength of this strategy depends on three ingredients: 1. **The dense subset must be tractable.** The elements of $A$ should be objects for which verification is straightforward — smooth functions, polynomials, rational numbers, simple functions. 2. **The property must be stable.** The property $P$ must pass to limits in the topology of $X$. An inequality that holds for smooth functions passes to $W^{1,p}$ limits if both sides are continuous with respect to the [Sobolev norm](/page/Sobolev%20Space). 3. **The topology must be appropriate.** Density depends on the topology. The polynomials are dense in $C([0,1])$ with the supremum norm ([Weierstrass Approximation Theorem](/theorems/480)) but not in $C^1([0,1])$ with the $C^1$ norm (approximating a function in $C^1$ requires approximating its derivative as well). [example: Density Depends on the Topology] Let $A = C^\infty([0,1])$, the smooth functions on $[0,1]$. **In $L^2([0,1])$:** The set $A$ is dense. Given $f \in L^2([0,1])$ and $\varepsilon > 0$, mollification produces $f_\varepsilon \in C^\infty$ with $\|f - f_\varepsilon\|_{L^2} < \varepsilon$ (after extending $f$ by zero and restricting the mollification to $[0,1]$). **In $L^\infty([0,1])$:** The set $A$ is **not** dense. Every element of $\overline{A}$ in the $L^\infty$ norm is a continuous function (since the uniform limit of continuous functions is continuous). The function $\mathbb{1}_{[0, 1/2]}$ belongs to $L^\infty([0,1])$ but not to $\overline{A}$, because $\|\mathbb{1}_{[0,1/2]} - g\|_{L^\infty} \ge 1/2$ for any continuous $g$ (the function $g$ must take values near $1$ and near $0$ on opposite sides of $x = 1/2$, but by [continuity](/page/Continuity) it passes through $1/2$, while $\mathbb{1}_{[0,1/2]}$ jumps). The same set of functions can be dense or non-dense depending on the norm, which is why density results always specify the topology. [/example] The concept of density also leads to the notion of [separability](/page/Separability): a topological space is separable if it contains a countable dense subset. The real line $\mathbb{R}$ is separable (with $\mathbb{Q}$ as a countable dense subset), and so are the Lebesgue spaces $L^p(\mathbb{R}^n)$ for $1 \le p < \infty$. The space $L^\infty(\mathbb{R}^n)$ is not separable. Separability has deep consequences for the structure of [Banach spaces](/page/Banach%20Space) — for instance, a Banach space is separable if and only if its closed unit ball is metrizable in the [weak topology](/page/Weak%20Topology). ## Closure and Continuity The closure operator interacts tightly with [continuous maps](/page/Continuity). In fact, the closure characterisation of continuity is often more useful than the standard open-preimage definition, because it gives direct control over limit processes: a continuous function cannot "tear apart" a set from its closure. [quotetheorem:1010] Condition (2) is the "closure" formulation: applying $f$ to the closure of $A$ lands inside the closure of the image. Equivalently, if $x$ is a limit point of $A$, then $f(x)$ is in the closure of $f(A)$ — limits are not "lost" under continuous maps. The inclusion in condition (2) cannot be replaced by equality. A continuous map can collapse a set, making $f(\overline{A})$ strictly smaller than $\overline{f(A)}$. [example: Strict Inclusion in the Closure-Image Relationship] Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \sin(x)$, and let $A = (0, \pi)$. Then: \begin{align*} f(A) &= f((0, \pi)) = (0, 1], \\ \overline{A} &= [0, \pi], \\ f(\overline{A}) &= f([0, \pi]) = [0, 1], \\ \overline{f(A)} &= \overline{(0, 1]} = [0, 1]. \end{align*} In this case $f(\overline{A}) = \overline{f(A)} = [0,1]$, so equality holds. Now consider $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = 1/(1+x^2)$, and let $A = (0, \infty)$. Then: \begin{align*} g(A) &= g((0, \infty)) = (0, 1), \\ \overline{A} &= [0, \infty), \\ g(\overline{A}) &= g([0, \infty)) = (0, 1], \\ \overline{g(A)} &= \overline{(0, 1)} = [0, 1]. \end{align*} Here $g(\overline{A}) = (0, 1] \subsetneq [0, 1] = \overline{g(A)}$. The value $0$ belongs to $\overline{g(A)}$ because $g(n) = 1/(1+n^2) \to 0$, but $0 \notin g(\overline{A})$ because $g(x) > 0$ for all $x \in [0, \infty)$. The inclusion is strict. [/example] Condition (3) — that preimages of closed sets are closed — is the [closed-set characterisation of continuity](/page/Continuity), dual to the standard open-set characterisation. It is often the most convenient formulation for proving continuity in practice, because verifying that a preimage is closed (e.g., by showing it contains its limit points) can be more direct than checking the open-set condition. ## Closure and Subspaces When $A$ is a subset of a topological space $X$ and we equip $A$ with the [subspace topology](/page/Topology), the notion of closure becomes relative. The closure of a subset $B \subset A$ in the subspace topology need not agree with its closure in $X$. [quotetheorem:1011] This formula follows from the definition of the subspace topology: the closed sets of $A$ are exactly the sets of the form $F \cap A$ where $F$ is closed in $X$. The smallest such set containing $B$ is $\overline{B} \cap A$. [example: Relative vs. Absolute Closure] Let $X = \mathbb{R}$, $A = (0, 2)$ with the subspace topology, and $B = (0, 1)$. In $X = \mathbb{R}$: $\overline{B} = [0, 1]$. In $A = (0, 2)$: $\overline{B}^A = \overline{B} \cap A = [0, 1] \cap (0, 2) = (0, 1]$. The point $0$ belongs to $\overline{B}$ (the closure in $\mathbb{R}$) but not to $\overline{B}^A$ (the closure in $(0,2)$). This is because $0 \notin A$, so $0$ is not a candidate for membership in the relative closure. In the subspace $(0, 2)$, the set $(0, 1]$ is closed: its complement in $A$ is $(1, 2)$, which is open in the subspace topology. [/example] ## Closure and the Boundary The [boundary](/page/Boundary) $\partial A$ of a set $A$ consists of the points that lie on the "edge" — belonging to the closure of both $A$ and its complement. The closure, interior, and boundary of a set are related by clean decomposition formulas. [definition: Boundary] Let $(X, \tau)$ be a topological space and $A \subset X$. The **boundary** of $A$, denoted $\partial A$, is: \begin{align*} \partial A := \overline{A} \cap \overline{X \setminus A}. \end{align*} Equivalently, $\partial A = \overline{A} \setminus A^\circ$, where $A^\circ$ denotes the interior of $A$. [/definition] The boundary collects the points where $A$ and its complement are both "present" in every neighbourhood. The decomposition of $\overline{A}$ into interior and boundary is: \begin{align*} \overline{A} = A^\circ \cup \partial A, \end{align*} and this union is disjoint ($A^\circ \cap \partial A = \varnothing$, since interior points have a neighbourhood contained in $A$, while boundary points have every neighbourhood meeting $X \setminus A$). [quotetheorem:1012] This trichotomy makes precise the intuitive picture of a set having an inside, an edge, and an outside. The closure captures the first two pieces: $\overline{A} = A^\circ \cup \partial A$. [example: Interior, Boundary, and Closure in $\mathbb{R}$] Let $A = [0, 1) \cup \{2\} \subset \mathbb{R}$. **Interior:** A point $x$ is in $A^\circ$ if some open interval $(x - \varepsilon, x + \varepsilon) \subset A$. For $x \in (0, 1)$, taking $\varepsilon = \min(x, 1-x)$ works. The point $x = 0$ fails: every interval $(-\varepsilon, \varepsilon)$ contains negative numbers not in $A$. The point $x = 2$ fails: every interval $(2-\varepsilon, 2+\varepsilon)$ contains points like $2 + \varepsilon/2 \notin A$. So $A^\circ = (0, 1)$. **Closure:** By the sequential characterisation, $0 \in \overline{A}$ (the sequence $1/n \to 0$) and $1 \in \overline{A}$ (the sequence $1 - 1/n \to 1$). The point $2 \in A \subset \overline{A}$. No other points are limits of sequences in $A$: for $x > 2$ or $1 < x < 2$ or $x < 0$, the ball $B(x, \varepsilon)$ for sufficiently small $\varepsilon$ misses $A$. So $\overline{A} = [0, 1] \cup \{2\}$. **Boundary:** $\partial A = \overline{A} \setminus A^\circ = ([0, 1] \cup \{2\}) \setminus (0,1) = \{0, 1, 2\}$. **Verification of trichotomy:** $A^\circ = (0,1)$, $\partial A = \{0, 1, 2\}$, and $(X \setminus A)^\circ = (-\infty, 0) \cup (1, 2) \cup (2, \infty)$. Their union is $\mathbb{R}$, and they are pairwise disjoint. [/example] ## Closure and Connectedness A fundamental property of the closure operation is that it preserves [connectedness](/page/Connectedness): if $A$ is a connected subset of a topological space, then $\overline{A}$ is connected as well. This is not a generic property of "enlargement" operations — arbitrary supersets of connected sets need not be connected — but closure has the special property that the new points it adds cannot create a separation. [quotetheorem:297] The "more general" statement is the key: any set squeezed between a connected set and its closure inherits connectedness. This is because a separation of $B$ would restrict to a separation of $A$ (since $A$ is dense in $B$ and open sets that partition $B$ must partition $A$), contradicting the connectedness of $A$. This result has a useful converse for detecting connected components. If $C$ is a connected component of $X$ (a maximal connected subset), then $C$ is closed. Indeed, $\overline{C}$ is connected and contains $C$, so by maximality $\overline{C} = C$. Connected components are therefore always closed — but they need not be open (consider the rationals $\mathbb{Q} \subset \mathbb{R}$, whose connected components are singletons, which are closed but not open in $\mathbb{Q}$). ## Working with Closures: Standard Arguments This section collects the techniques that arise repeatedly in closure-based arguments throughout analysis and topology. Each technique is a pattern that the reader will encounter in proofs across many areas of mathematics. ### The Density Argument The most pervasive closure-based technique in analysis is the **density argument**: prove a result for a dense subset, then extend to the full space by passing to limits. The general structure is as follows. **Goal:** Show that a property $P$ holds for all $x \in X$. **Strategy:** 1. Identify a dense subset $D \subset X$ (so $\overline{D} = X$) for which $P$ is easy to verify. 2. Prove $P$ for all $d \in D$. 3. For arbitrary $x \in X$, choose a sequence (or net) $d_n \in D$ with $d_n \to x$. 4. Show that $P$ passes to the limit: if $P(d_n)$ holds for all $n$, then $P(x)$ holds. The critical step is (4), which requires the property $P$ to be "closed" — stable under the relevant notion of convergence. If $P$ is an inequality of the form $\|Tx\| \le C\|x\|$ where $T$ is a continuous linear operator, then $P$ passes to limits automatically. If $P$ is the validity of an identity, both sides must be continuous in $x$. [example: Extending a Bounded Linear Operator by Density] Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space), let $D \subset X$ be a dense subspace, and let $Y$ be a [Banach space](/page/Banach%20Space). Suppose $T_0: D \to Y$ is a bounded linear operator with $\|T_0 d\|_Y \le M\|d\|_X$ for all $d \in D$. **Claim:** $T_0$ extends uniquely to a bounded linear operator $T: X \to Y$ with $\|T\| = \|T_0\|$. **Construction.** For $x \in X$, choose $(d_n)_{n=1}^\infty$ in $D$ with $d_n \to x$. The sequence $(T_0 d_n)$ is Cauchy in $Y$: \begin{align*} \|T_0 d_n - T_0 d_m\|_Y = \|T_0(d_n - d_m)\|_Y \le M\|d_n - d_m\|_X \to 0. \end{align*} Since $Y$ is complete, $T_0 d_n \to y$ for some $y \in Y$. Define $Tx := y$. To verify well-definedness, suppose $(d_n')$ is another sequence in $D$ with $d_n' \to x$. Then: \begin{align*} \|T_0 d_n - T_0 d_n'\|_Y \le M\|d_n - d_n'\|_X \le M(\|d_n - x\|_X + \|x - d_n'\|_X) \to 0, \end{align*} so both sequences converge to the same limit. Linearity of $T$ follows from linearity of $T_0$ and uniqueness of limits. The bound $\|Tx\|_Y \le M\|x\|_X$ passes to the limit: \begin{align*} \|Tx\|_Y = \lim_{n \to \infty} \|T_0 d_n\|_Y \le \lim_{n \to \infty} M\|d_n\|_X = M\|x\|_X. \end{align*} This extension argument — sometimes called the BLT (Bounded Linear Transformation) theorem — is the prototype for all density-based constructions. The completeness of $Y$ is essential: without it, the Cauchy sequence $(T_0 d_n)$ might not converge, and the extension would fail. [/example] ### The Closure Containment Argument To prove that a closed set $F$ contains a set $A$, it suffices to show that $F$ contains a dense subset of $A$. Formally: if $D \subset A$ with $\overline{D} \supset A$ and $D \subset F$, then $A \subset \overline{D} \subset F$ (since $F$ is closed, $\overline{D} \subset F$). This technique appears frequently in [functional analysis](/page/Functional%20Analysis). For instance, to show that the range of an operator contains a given subspace, one often shows that the range contains a dense subset and then concludes by closure. ### The "Separate and Close" Argument Many proofs in topology reduce to the following pattern: to show that two sets $A$ and $B$ have a specific relationship, separate them using open sets, then use the characterisation of closure in terms of open sets to derive the conclusion. [example: Disjoint Closures in Hausdorff Spaces] Let $(X, \tau)$ be a [Hausdorff space](/page/Hausdorff%20Space) (any two distinct points can be separated by disjoint open sets). Let $x, y \in X$ with $x \neq y$. Choose disjoint open sets $U, V \in \tau$ with $x \in U$ and $y \in V$. Since $U \cap V = \varnothing$ and $x \in U$, the point $x$ does not belong to $V$, so $V \cap \{x\} = \varnothing$. But $V$ is an open neighbourhood of $y$ with $V \cap \{x\} = \varnothing$. By the neighbourhood characterisation of closure, $y \notin \overline{\{x\}}$. Therefore, in a Hausdorff space, $\overline{\{x\}} = \{x\}$ for every $x$ — singletons are closed. This is a necessary condition for the Hausdorff property and is in fact equivalent to it when the space is $T_1$. [/example] ### Sequential Closure Arguments in Metric Spaces In [metric spaces](/page/Metric%20Space), the sequential characterisation of closure converts topological closure arguments into explicit sequence constructions. The pattern is: **To prove $x \in \overline{A}$:** Construct a sequence $(a_n)$ in $A$ with $d(a_n, x) \to 0$, often by choosing $a_n \in A \cap B(x, 1/n)$. **To prove $\overline{A} \subset B$:** Take an arbitrary $x \in \overline{A}$, choose a sequence $(a_n)$ in $A$ with $a_n \to x$, and show $x \in B$ using properties of the sequence and the set $B$. **To prove $A$ is closed:** Show that whenever $(a_n)$ is a sequence in $A$ with $a_n \to x$, we have $x \in A$. This sequential approach is often simpler than the neighbourhood approach, because sequences can be manipulated with standard limit theorems (algebraic operations on limits, comparison, squeeze). ## References - Munkres, J. R., *Topology* (2nd ed., 2000). - Willard, S., *General Topology* (1970). - Kelley, J. L., *General Topology* (1955). - Rudin, W., *Principles of Mathematical Analysis* (3rd ed., 1976). - Kuratowski, C., *Topologie I* (1933).