Interior and Closure Topologically

Two Ways to Approximate a Set

A topology tells us which subsets are open, but it is often useful to ask a more focused question: which points of a set can be surrounded by an open set that stays inside it, and which points can be approached by points of the set? The first question leads to the interior; the second leads to the closure.

The previous tutorial introduced neighborhoods and neighborhood bases. We will use neighborhoods to give a pointwise test for both operations. We will also use the result from Closed Sets in Topological Spaces that the closure of a set is the smallest closed set containing it. In this tutorial, “open” and “closed” always refer to the topology under discussion; no distance is needed.

Definition: Let $X$ be a topological space and $E\subseteq X$. The interior of $E$, denoted $\operatorname{int}(E)$ or $E^\circ$, is the union of all open sets contained in $E$. Equivalently, it is the largest open subset of $E$. The closure of $E$, denoted $\overline{E}$, is the intersection of all closed sets containing $E$. Equivalently, it is the smallest closed superset of $E$.

These definitions emphasize different directions. The interior stays within $E$, retaining only points that have enough room around them to remain in $E$. The closure contains $E$, adding any points that cannot be excluded from it by an open neighborhood. In particular, a point of the closure need not itself belong to $E$.

Neighborhood Tests

Theorem: Let $E\subseteq X$, where $X$ is a topological space, and let $x\in X$. Then $x\in\operatorname{int}(E)$ if and only if some open neighborhood of $x$ is contained in $E$. Also, $x\in\overline{E}$ if and only if every neighborhood of $x$ intersects $E$.

Proof. By definition, $\operatorname{int}(E)$ is a union of open subsets of $E$. If $x\in\operatorname{int}(E)$, then $x$ belongs to some open set $U\subseteq E$, so $U$ is an open neighborhood of $x$ contained in $E$. Conversely, if an open neighborhood $U$ of $x$ satisfies $U\subseteq E$, then $U$ is one of the open sets included in the union defining $\operatorname{int}(E)$. Thus $x\in\operatorname{int}(E)$.

For the closure test, first suppose $x\in\overline{E}$, and let $N$ be any neighborhood of $x$. There is an open set $U$ with $x\in U\subseteq N$. If $N\cap E=\varnothing$, then $U\cap E=\varnothing$, so the closed set $X\setminus U$ contains $E$ but does not contain $x$. This contradicts the fact that $\overline{E}$ is contained in every closed set containing $E$. Hence $N\cap E\neq\varnothing$.

Conversely, suppose every neighborhood of $x$ intersects $E$. If $x\notin\overline{E}$, then $X\setminus\overline{E}$ is an open neighborhood of $x$, because $\overline{E}$ is closed. It is disjoint from $E$, since $E\subseteq\overline{E}$. This contradicts the assumed intersection property. Therefore $x\in\overline{E}$. $\square$

The closure test can equivalently be phrased using only open neighborhoods: every open set containing $x$ must intersect $E$. Indeed, if every open neighborhood intersects $E$, then every neighborhood does as well, because every neighborhood contains an open neighborhood of its point. This pointwise description is particularly useful when a topology is specified by neighborhood bases.

Worked Example: Interior and Closure in the Real Line

Give $\mathbb{R}$ its usual topology and let $E=[0,1)$. A point $x$ belongs to the interior precisely when it has an open interval around it contained in $[0,1)$. Every point strictly between $0$ and $1$ has such an interval. The point $0$ does not: every open interval around $0$ contains negative numbers. The point $1$ is not in $E$. Therefore $$ \operatorname{int}(E)=(0,1). $$

Every point of $[0,1)$ belongs to the closure because $E$ is contained in its closure. The point $1$ also belongs: every open interval around $1$ contains points less than $1$ that lie in $E$. If $x<0$, an open interval around $x$ can be chosen disjoint from $E$; the same is true if $x>1$. Thus $$ \overline{E}=[0,1]. $$ This example shows both that the interior can omit points of the set and that the closure can contain points outside the set.

Basic Laws of Interior and Closure

Interior and closure respect inclusion: making a set larger cannot make its interior smaller or its closure smaller. Each operation is also idempotent: applying it twice has the same effect as applying it once. The finite union and intersection laws are slightly different for the two operations.

Theorem: Let $A,B\subseteq X$. Then $\operatorname{int}(A)\subseteq A$, $A\subseteq\overline{A}$, and both $\operatorname{int}(A)$ and $\overline{A}$ are unchanged by applying the same operation again. If $A\subseteq B$, then $\operatorname{int}(A)\subseteq\operatorname{int}(B)$ and $\overline{A}\subseteq\overline{B}$. Moreover, $$ \operatorname{int}(A\cap B)=\operatorname{int}(A)\cap\operatorname{int}(B), \qquad \overline{A\cup B}=\overline{A}\cup\overline{B}. $$

Proof. The interior is a union of open sets contained in $A$, so it is open and is a subset of $A$. Since $A$ itself is contained in its closure, $A\subseteq\overline{A}$. The interior is the largest open subset of $A$; hence the largest open subset of $\operatorname{int}(A)$ is $\operatorname{int}(A)$ itself. Thus $\operatorname{int}(\operatorname{int}(A))=\operatorname{int}(A)$. The closure is closed and contains $A$, so the smallest closed superset of $\overline{A}$ is $\overline{A}$ itself. Hence $\overline{\overline{A}}=\overline{A}$.

If $A\subseteq B$, every open subset of $A$ is an open subset of $B$. In particular, $\operatorname{int}(A)\subseteq\operatorname{int}(B)$. Also $\overline{B}$ is closed and contains $A$, so the smallest closed set containing $A$ is contained in $\overline{B}$. Therefore $\overline{A}\subseteq\overline{B}$.

The set $\operatorname{int}(A)\cap\operatorname{int}(B)$ is open and is contained in $A\cap B$, so it is contained in $\operatorname{int}(A\cap B)$. In the other direction, $\operatorname{int}(A\cap B)$ is an open subset of both $A$ and $B$. It is therefore contained in both interiors, and hence in their intersection. This proves the interior identity.

For the closure identity, monotonicity gives $\overline{A}\cup\overline{B}\subseteq\overline{A\cup B}$. The set $\overline{A}\cup\overline{B}$ is closed, because a finite union of closed sets is closed, and it contains $A\cup B$. By the smallest-closed-superset property, $\overline{A\cup B}\subseteq\overline{A}\cup\overline{B}$. The two inclusions prove the equality. $\square$

Notice the asymmetry: interior always distributes over finite intersections, while closure distributes over finite unions. The corresponding reverse equalities need not hold. For example, in $\mathbb{R}$, let $A=(-\infty,0]$ and $B=[0,\infty)$. Then $A\cup B=\mathbb{R}$, so $\operatorname{int}(A\cup B)=\mathbb{R}$, while $\operatorname{int}(A)\cup\operatorname{int}(B)=(-\infty,0)\cup(0,\infty)$, which omits $0$. Thus the interior of a union can be larger than the union of the interiors.

Worked Example: A Finite Topology

Let $X=\{a,b,c\}$ have topology $$ \mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}. $$ The closed sets, obtained by taking complements of the open sets, are $\varnothing$, $\{b,c\}$, $\{c\}$, and $X$. Take $E=\{b\}$. The only open set containing $b$ is $\{a,b\}$, and it is not contained in $E$. There is no nonempty open subset of $E$, so $$ \operatorname{int}(E)=\varnothing. $$

To find the closure, inspect the closed supersets of $E$. These are $\{b,c\}$ and $X$, whose intersection is $\{b,c\}$. Thus $\overline{E}=\{b,c\}$. The neighborhood test gives the same answer: every neighborhood of $c$ contains an open set containing $c$, and the only such open set is $X$, which intersects $E$. By contrast, $a$ has the open neighborhood $\{a\}$, which does not intersect $E$, so $a\notin\overline{E}$.

Complementarity and Topological Duality

Interior and closure are related by complements. A point fails to be interior to $E$ exactly when every open neighborhood around it reaches outside $E$. That is precisely the condition for the point to lie in the closure of the complement.

Theorem: For every $E\subseteq X$, $$ X\setminus\operatorname{int}(E)=\overline{X\setminus E} \qquad\text{and}\qquad X\setminus\overline{E}=\operatorname{int}(X\setminus E). $$

Proof. Fix $x\in X$. The neighborhood test says that $x\notin\operatorname{int}(E)$ if and only if there is no open neighborhood of $x$ contained in $E$. Equivalently, every open neighborhood of $x$ contains a point outside $E$. By the closure test applied to $X\setminus E$, this is equivalent to $x\in\overline{X\setminus E}$. Thus the first identity holds. Apply that identity with $X\setminus E$ in place of $E$: $$ X\setminus\operatorname{int}(X\setminus E)=\overline{E}. $$ Taking complements of both sides gives the second identity. $\square$

This duality lets us transfer facts about one operation to the other. For instance, the closure of a set is closed, so its complement is open; the second identity says this complement is exactly the interior of the original complement. The identities also provide a practical check on calculations: finding the closure of $E$ is equivalent to finding the complement of the interior of $X\setminus E$.

Worked Example: The Cofinite Topology

Let $X$ be an infinite set with the cofinite topology: the open sets are $\varnothing$ and all subsets whose complements in $X$ are finite. Choose an infinite subset $E\subseteq X$ whose complement is also infinite. No nonempty open set can be contained in $E$, because every nonempty open set has finite complement, whereas $X\setminus E$ is infinite. Hence $\operatorname{int}(E)=\varnothing$.

Every nonempty open set intersects $E$. Indeed, if $U$ is nonempty and open, then $X\setminus U$ is finite; an infinite $E$ cannot be contained in that finite complement, so $U\cap E\neq\varnothing$. The closure test now gives $\overline{E}=X$. This example illustrates how the topology, rather than any underlying distance, determines which sets are interior or closure points.

Why These Operations Matter

Interior and closure turn local information into global sets. The interior collects all points with a sufficiently small open neighborhood lying inside the set; the closure collects all points whose every neighborhood meets it. The neighborhood tests make both ideas applicable in an arbitrary topological space, while the algebraic laws allow calculations to be simplified before examining individual points.

A common mistake is to assume that the closure consists only of points in the set. The closure always contains the set, but it can be strictly larger, as with $[0,1)$ in the real line or the infinite subset in the cofinite example. Another pitfall is to assume that closure distributes over arbitrary unions. The theorem above proves the identity for two sets, and therefore for any finite union by induction; an infinite union can behave differently. Keeping track of which operation distributes over which finite set operation prevents many incorrect simplifications.

Check Your Understanding

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

Give the neighborhood test for membership in the interior and the neighborhood test for membership in the closure.
Why is the interior of a set contained in the set, while the closure contains the set?
State the finite intersection law for interior and the finite union law for closure.
Explain why the interior of a union need not equal the union of the interiors.
How can the closure of a set be found from the interior of its complement?
In the cofinite topology on an infinite set, why does every infinite subset have closure equal to the whole space?