Closure (topology)

In topology, the closure of a subset $A$ of a topological space $X$ is the smallest closed set in $X$ containing $A$. It is commonly written $\overline{A}$, $\operatorname{cl}(A)$, or $\operatorname{cl}_X(A)$. Closure includes the original set together with all points that cannot be separated from it by an open neighborhood. The ambient space and its topology are essential: the same set can have different closures in different spaces. (maths.ed.ac.uk)

Definition and neighborhood characterization

Formally,

The family being intersected is nonempty because it contains $X$. Arbitrary intersections of closed sets are closed, so this expression is itself closed and establishes the existence of the smallest closed superset. In particular, $A$ is closed exactly when $\operatorname{cl}_X(A)=A$. (sites.math.northwestern.edu)

An equivalent characterization uses neighborhoods:

Thus, $x$ lies outside the closure precisely when some open set containing $x$ misses $A$. Points satisfying the neighborhood condition are called adherent points or closure points of $A$. (maths.ed.ac.uk)

Closure points should be distinguished from accumulation points, for which every neighborhood must meet $A\setminus\{x\}$. If $A'$ denotes the set of accumulation points, then

Consequently, an isolated point of $A$ belongs to its closure even though it is not an accumulation point. No Hausdorff assumption is needed for these characterizations. (sites.math.northwestern.edu)

Closure as an operator

Closure defines a map from the power set of $X$ to itself. Its fundamental properties are the Kuratowski closure axioms:

They express preservation of the empty set, extensivity, idempotence, and preservation of finite unions. Conversely, any operator on the subsets of $X$ satisfying these axioms determines a unique topology: its closed sets are exactly the subsets $F$ for which $\operatorname{cl}(F)=F$. Their complements are the open sets. (legacy-www.math.harvard.edu)

Closure is also monotone: $A\subseteq B$ implies $\overline{A}\subseteq\overline{B}$. However, it need not preserve intersections or infinite unions. For example, in the usual real line, $A=(0,1)$ and $B=(1,2)$ have empty intersection, while their closures intersect at $1$. Likewise, the union of the closed singletons $\{1/n\}$, for positive integers $n$, is not closed: its closure additionally contains $0$. These examples illustrate why the finite-union axiom cannot be extended to arbitrary unions. (fan.uni-wuppertal.de)

Metric spaces and convergence

In a metric space $(X,d)$, the neighborhood criterion becomes

where $B(x,\varepsilon)$ is an open ball. For nonempty $A$, this is equivalent to

The infimum need not be attained: a point outside $A$ may nevertheless have distance zero from it. (maths.ed.ac.uk)

In metric spaces, and more generally in every first-countable space, a point belongs to $\overline{A}$ exactly when it is the limit of a sequence of points of $A$. In the metric case, one can choose $a_n\in A\cap B(x,1/n)$; then $a_n\to x$. Points already in $A$ are covered by constant sequences. (math.ucla.edu)

In arbitrary topological spaces, sequential limits may capture only part of the closure. The unrestricted characterization uses a net, a generalization of a sequence indexed by a directed set: $x\in\overline{A}$ if and only if some net taking values in $A$ converges to $x$. (fan.uni-wuppertal.de)

Examples and dependence on topology

With the usual topology on the real numbers,

Both the rational numbers and the irrational numbers are therefore dense in $\mathbb R$. Generally, a subset is dense in $X$ exactly when its closure is $X$. (maths.ed.ac.uk)

In the discrete topology, every subset is closed and equals its closure. In the indiscrete topology, the only closed subsets are $\varnothing$ and $X$, so every nonempty subset has closure $X$. These cases show that closure describes topological proximity rather than necessarily numerical distance. (math.mit.edu)

If $A\subseteq Y\subseteq X$, with $Y$ carrying the subspace topology, then

For instance, $(0,1)$ has closure $[0,1]$ in $\mathbb R$, but closure $(0,1]$ in $Y=(0,2)$. (maths.ed.ac.uk)

Interior and continuous maps

Closure is dual to interior under complementation:

A continuous map $f:X\to Y$ satisfies

for every $A\subseteq X$; conversely, this condition characterizes continuity. Equality is not required. For example, the continuous map $f(x)=e^x$ sends the closed set $\mathbb R$ onto $(0,\infty)$, whose closure in $\mathbb R$ also contains $0$. (fan.uni-wuppertal.de)