Closed Sets in Topological Spaces

Closedness Is Defined by Complements

In Open Sets in Topological Spaces, openness was defined by membership in the chosen topology. Closedness is defined in relation to that same topology: a set is closed when its complement is open. This definition does not require a metric, and it makes clear that closedness, like openness, is a property of a set considered within a particular space.

Definition: Let $(X,\mathcal{T})$ be a topological space. A subset $F\subseteq X$ is closed in $X$ if its complement $X\setminus F$ is open in $X$.

The words “in $X$” matter. A subset may be closed in a subspace without being closed in the whole space. We will make that relationship precise below. First, the complement definition and the topology axioms give a useful collection of rules. The Closed-Set Characterization of a Topology established earlier in Topological Spaces says that closed sets include $\varnothing$ and $X$, are closed under arbitrary intersections, and are closed under finite unions. We will use those laws without proving them again.

The two closure rules have different scopes: arbitrary intersections of closed sets are closed, while only finite unions are guaranteed to be closed. This is the complement-side counterpart of the distinction between arbitrary unions and finite intersections of open sets. The difference follows from the corresponding De Morgan laws: the complement of an intersection is a union of complements, and the complement of a union is an intersection of complements.

Worked Example: Closed Sets in a Small Topology

Let $X=\{u,v,w\}$ and let $$ \mathcal{T}=\{\varnothing,\{u\},X\}. $$ A set is closed exactly when its complement belongs to $\mathcal{T}$. The complements of the three open sets are $$ X\setminus\varnothing=X,\qquad X\setminus\{u\}=\{v,w\},\qquad X\setminus X=\varnothing. $$ Thus the closed sets are $X$, $\{v,w\}$, and $\varnothing$. In particular, $\{u\}$ is open but not closed, since its complement $\{v,w\}$ is not open. Also, $\{v,w\}$ is closed but not open. A set need not be either both open and closed or neither: each property is determined by a separate membership test in the topology.

Closedness in a Subspace

Suppose $A\subseteq X$ is given the subspace topology. The open sets in $A$ are intersections $A\cap O$, where $O$ is open in $X$, as established in Open Sets in Topological Spaces. Taking complements within $A$, rather than within all of $X$, gives the corresponding description of closed sets in $A$.

Theorem: Let $A\subseteq X$, with the subspace topology, and let $F\subseteq A$. Then $F$ is closed in $A$ if and only if there is a set $C$ closed in $X$ such that $F=A\cap C$.

Proof. Suppose first that $F$ is closed in $A$. Then $A\setminus F$ is open in the subspace $A$. By the subspace characterization of open sets, there is an open set $O\subseteq X$ such that $$ A\setminus F=A\cap O. $$ Set $C=X\setminus O$. Since $O$ is open in $X$, $C$ is closed in $X$. Taking complements within $A$ in the displayed equality gives $$ F=A\setminus(A\cap O)=A\cap(X\setminus O)=A\cap C. $$ Thus $F$ has the required form.

Conversely, suppose $C$ is closed in $X$ and $F=A\cap C$. Then $$ A\setminus F=A\setminus(A\cap C)=A\cap(X\setminus C). $$ The set $X\setminus C$ is open in $X$, so $A\cap(X\setminus C)$ is open in the subspace $A$. Therefore $A\setminus F$ is open in $A$, which means that $F$ is closed in $A$. This proves both directions. $\square$

Worked Example: A Set Closed in a Subspace but Not in the Ambient Space

Give $\mathbb{R}$ its usual topology, let $A=(0,2)$, and consider $F=(0,1]$. The set $C=[0,1]$ is closed in $\mathbb{R}$, and $$ A\cap C=(0,2)\cap[0,1]=(0,1]=F. $$ The subspace theorem therefore shows that $F$ is closed in $A$. Equivalently, its complement in $A$ is $$ A\setminus F=(0,2)\setminus(0,1]=(1,2), $$ which is open in the subspace $A$.

But $F$ is not closed in $\mathbb{R}$. Its complement in $\mathbb{R}$ is $(-\infty,0]\cup(1,\infty)$, which is not open: no open interval around $0$ is contained in that complement, since every such interval contains positive numbers smaller than $1$. Relative closedness does not automatically imply ambient closedness. The theorem explains the distinction: the closed ambient set $[0,1]$ is first intersected with $A$, which removes the point $0$.

One useful consequence follows when $A$ itself is closed in $X$. If $F$ is closed in $A$, write $F=A\cap C$ for some closed $C\subseteq X$, as in the theorem. Both $A$ and $C$ are closed in $X$, and finite intersections of closed sets are closed. Hence $F$ is closed in $X$. Without the assumption that $A$ is closed, this conclusion can fail, as the example demonstrates.

The Closure of a Set

A set that is not closed can still be contained in many closed sets. Their intersection gives a natural candidate for the smallest closed set containing it. This construction is available in every topological space, even when there is no distance or notion of convergence to use.

Definition: For $E\subseteq X$, the closure of $E$ in $X$, denoted $\overline{E}$, is the intersection of all closed subsets of $X$ that contain $E$: $$ \overline{E}=\bigcap\{F\subseteq X:E\subseteq F\text{ and }F\text{ is closed in }X\}. $$

The family being intersected is nonempty because $X$ is closed and contains $E$. The arbitrary-intersection law for closed sets therefore ensures that $\overline{E}$ is closed. The next theorem spells out the most important consequence of the definition: the closure is not merely a closed set containing $E$; it is contained in every such set.

Theorem: For every $E\subseteq X$, the closure $\overline{E}$ is closed, contains $E$, and is contained in every closed set that contains $E$. In particular, $E$ is closed if and only if $\overline{E}=E$.

Proof. The family of closed supersets of $E$ includes $X$, so its intersection is defined. Since an arbitrary intersection of closed sets is closed, $\overline{E}$ is closed. Every member of the family contains $E$, so every point of $E$ belongs to their intersection; hence $E\subseteq\overline{E}$.

Now let $C$ be any closed set containing $E$. It is one of the sets in the defining family, and an intersection is contained in each member of the family. Thus $\overline{E}\subseteq C$. This proves the smallest-closed-superset property.

If $E$ is closed, then $E$ itself is a closed superset of $E$, so the property just proved gives $\overline{E}\subseteq E$. Together with $E\subseteq\overline{E}$, this yields $\overline{E}=E$. Conversely, if $\overline{E}=E$, then $E$ is closed because $\overline{E}$ is closed. $\square$

Worked Example: The Closure of an Open Interval

In the usual topology on $\mathbb{R}$, take $E=(2,5)$. The interval $[2,5]$ is closed and contains $E$, so the smallest-closed-superset property gives $$ \overline{E}\subseteq[2,5]. $$ To show that no smaller closed set can contain $E$, let $C$ be any closed subset of $\mathbb{R}$ containing $(2,5)$. If $2\notin C$, then the open set $\mathbb{R}\setminus C$ contains $2$. It must therefore contain some open interval $(2-r,2+r)$ for $r>0$. But that interval contains points in $(2,5)$, contradicting $(2,5)\subseteq C$. Hence $2\in C$. The same argument at $5$ shows that $5\in C$. Since $C$ already contains $(2,5)$, it contains $[2,5]$.

Every closed set containing $E$ therefore contains $[2,5]$, while $[2,5]$ is itself one such closed set. Consequently, $$ \overline{(2,5)}=[2,5]. $$ The endpoints are included not because they belong to $E$, but because every closed superset of $E$ must contain them.

Infinite Unions and a Common Pitfall

It is tempting to extend the finite-union law and assume that a union of any number of closed sets is closed. The topology axioms do not guarantee this. The following example shows exactly where an infinite union can fail: its complement is an infinite intersection of open sets, and such an intersection need not be open.

Worked Example: A Countable Union of Closed Sets That Is Not Closed

In the usual topology on $\mathbb{R}$, every singleton $\{1/n\}$ is closed. Consider their union $$ E=\bigcup_{n=1}^{\infty}\left\{\frac{1}{n}\right\} =\left\{1,\frac12,\frac13,\ldots\right\}. $$ The point $0$ does not belong to $E$, but every open interval around $0$ contains some $1/n$: given $r>0$, choose a positive integer $n>1/r$, so $0<1/n<r$. Thus no open interval around $0$ lies in $\mathbb{R}\setminus E$. The complement of $E$ is not open, so $E$ is not closed. In particular, an infinite union of closed sets need not be closed, even though each set in the union is closed.

This example also illustrates why the closure construction is useful. Since every closed set containing $E$ must contain $0$, we have $0\in\overline{E}$. More generally, closure identifies precisely the smallest closed set that contains a given set; it does not assert that the original set was closed.

Keep the relative space in view whenever you test closedness. In $X$, the relevant complement is $X\setminus F$; in a subspace $A$, it is $A\setminus F$. The subspace theorem translates the latter test into an ambient description, and the closure theorem gives a reliable way to enlarge any set to a closed one without adding points unnecessarily.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What condition on the complement defines a subset as closed in a topological space?
State the theorem characterizing a set closed in a subspace $A\subseteq X$ using a closed set in $X$.
Why can $(0,1]$ be closed in the subspace $(0,2)$ without being closed in $\mathbb{R}$?
How is the closure of $E\subseteq X$ constructed, and what makes it the smallest closed set containing $E$?
Why does the union of the sets $\{1/n\}$ fail to be closed in the usual topology on $\mathbb{R}$?