Closure and Interior Identities

Closure and Interior as Complementary Operations

Closure and interior describe opposite ways of approximating a set. The closure adds the points that cannot be separated from the set by an open neighborhood; the interior keeps the points surrounded by a neighborhood lying wholly inside the set. The two operations are linked by taking complements. That link lets us turn a question about closure into one about interior, or vice versa.

Recall that $x\in\overline{E}$ precisely when every open ball around $x$ intersects $E$, by the Neighborhood Characterization of Closure. Also, a point belongs to $\operatorname{int}(E)$ precisely when some open ball around it is contained in $E$. These descriptions suggest the relationship: a point fails to be in the closure of $E$ exactly when it has a neighborhood that avoids $E$, or, equivalently, lies inside the interior of the complement.

Theorem (Complement Identities for Closure and Interior): For every $E\subseteq\mathbb{R}$, $$ \mathbb{R}\setminus\overline{E} = \operatorname{int}(\mathbb{R}\setminus E) \qquad\text{and}\qquad \mathbb{R}\setminus\operatorname{int}(E) = \overline{\mathbb{R}\setminus E}. $$ Equivalently, $$ \overline{E} = \mathbb{R}\setminus\operatorname{int}(\mathbb{R}\setminus E) \qquad\text{and}\qquad \operatorname{int}(E) = \mathbb{R}\setminus\overline{\mathbb{R}\setminus E}. $$

Proof. We first prove $\mathbb{R}\setminus\overline{E}=\operatorname{int}(\mathbb{R}\setminus E)$. Let $x\in\mathbb{R}\setminus\overline{E}$. By the Neighborhood Characterization of Closure, there is an $r>0$ such that $B_r(x)\cap E=\varnothing$. This means $B_r(x)\subseteq\mathbb{R}\setminus E$, so $x$ is an interior point of $\mathbb{R}\setminus E$. Conversely, if $x\in\operatorname{int}(\mathbb{R}\setminus E)$, there is an $r>0$ with $B_r(x)\subseteq\mathbb{R}\setminus E$. Thus $B_r(x)\cap E=\varnothing$, which means $x\notin\overline{E}$. The first identity follows.

Apply that identity with $\mathbb{R}\setminus E$ in place of $E$. It gives $\mathbb{R}\setminus\overline{\mathbb{R}\setminus E}=\operatorname{int}(E)$. Taking complements of this identity also gives $\mathbb{R}\setminus\operatorname{int}(E)=\overline{\mathbb{R}\setminus E}$. These are the remaining forms. $\square$

Each equality has a direct neighborhood interpretation. A point is outside $\overline{E}$ if it has a neighborhood entirely in the complement. A point is outside $\operatorname{int}(E)$ if no neighborhood around it fits inside $E$; equivalently, every neighborhood intersects the complement, which places the point in the closure of the complement. Neither identity assumes that $E$ is open or closed.

Worked Example: Finding the Complementary Closure and Interior

Let $E=[-1,2)\cup\{4\}$. The interval $(-1,2)$ lies in the interior, and neither endpoint of that interval can be surrounded by an open interval contained in $E$. The isolated point $4$ is not interior either. Hence

\operatorname{int}(E)=(-1,2).

The closure includes the missing endpoint $2$, since every open interval around $2$ meets $[ -1,2)$, and it includes $4$, which is already in $E$. No other point is in the closure. Therefore

\overline{E}=[-1,2]\cup\{4\}.

The complement is $(-\infty,-1)\cup[2,4)\cup(4,\infty)$. Its interior is $(-\infty,-1)\cup(2,4)\cup(4,\infty)$: the points $-1$, $2$, and $4$ do not have open intervals contained in the complement. Its closure is $(-\infty,-1]\cup[2,\infty)$. Thus the complement identities give

\mathbb{R}\setminus\overline{E} =(-\infty,-1)\cup(2,4)\cup(4,\infty) =\operatorname{int}(\mathbb{R}\setminus E), $$ $$ \mathbb{R}\setminus\operatorname{int}(E) =(-\infty,-1]\cup[2,\infty) =\overline{\mathbb{R}\setminus E}.

This calculation shows how a point can change roles under complementation: the point $2$ is missing from $E$ but belongs to $\overline{E}$, and so it is excluded from the interior of the complement.

Fixed-Point Tests for Open and Closed Sets

The complement identities also give tests for whether a set is already unchanged by one of the operations. A set is open when it equals its interior, and closed when it equals its closure. The identities express those tests using the closure or interior of the complementary set instead.

Theorem (Complementary Fixed-Point Tests): For every $E\subseteq\mathbb{R}$, $$ E\text{ is open} \quad\Longleftrightarrow\quad E=\mathbb{R}\setminus\overline{\mathbb{R}\setminus E}, $$ and $$ E\text{ is closed} \quad\Longleftrightarrow\quad E=\mathbb{R}\setminus\operatorname{int}(\mathbb{R}\setminus E). $$

Proof. By the Complement Identities for Closure and Interior, $\mathbb{R}\setminus\overline{\mathbb{R}\setminus E}=\operatorname{int}(E)$. Therefore the first equality in the theorem holds exactly when $E=\operatorname{int}(E)$, which is exactly when $E$ is open.

The other complement identity gives $\mathbb{R}\setminus\operatorname{int}(\mathbb{R}\setminus E)=\overline{E}$. Thus the second equality holds exactly when $E=\overline{E}$, which is exactly when $E$ is closed. This proves both equivalences. $\square$

These tests are especially useful when the complement has a simpler form than the original set. Instead of checking directly that every point of $E$ has a suitable neighborhood, one can study whether the complement’s closure leaves precisely $E$ behind. Likewise, closedness of $E$ can be checked by asking whether removing the interior of its complement leaves $E$.

Worked Example: Testing Openness and Closedness by Complements

Consider $U=(1,5)$. Its complement is $(-\infty,1]\cup[5,\infty)$, which is closed. The closure of that complement is itself, so

\mathbb{R}\setminus\overline{\mathbb{R}\setminus U} = \mathbb{R}\setminus\bigl((-\infty,1]\cup[5,\infty)\bigr) =(1,5)=U.

The fixed-point test confirms that $U$ is open. Now let $F=[1,5]$. Its complement is $(-\infty,1)\cup(5,\infty)$, whose interior is that same open set. Hence

\mathbb{R}\setminus\operatorname{int}(\mathbb{R}\setminus F) = \mathbb{R}\setminus\bigl((-\infty,1)\cup(5,\infty)\bigr) =[1,5]=F.

The second fixed-point test confirms that $F$ is closed. The tests are consistent with the endpoint behavior: the endpoints are absent from the open interval but retained by the closed interval.

Finite Laws and Their Limits

Closure and interior have familiar identities for finite unions and intersections. Earlier in this course, the Closure of a Finite Union theorem established that the closure of a finite union is the union of the closures. The Interior of a Finite Intersection theorem established the corresponding identity for the interior of a finite intersection. The complement identities explain why these two laws are paired: taking complements exchanges unions with intersections and exchanges closure with interior.

The other two combinations generally give inclusions rather than equalities. Earlier results give $\operatorname{int}(A)\cup\operatorname{int}(B)\subseteq\operatorname{int}(A\cup B)$ and $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$. The following examples show why replacing these inclusions by equalities would be unsafe.

Worked Example: Closure Does Not Always Distribute Over Intersection

Take $A=(-\infty,0)$ and $B=(0,\infty)$. They are disjoint, so $A\cap B=\varnothing$ and $\overline{A\cap B}=\varnothing$. Their closures are $\overline{A}=(-\infty,0]$ and $\overline{B}=[0,\infty)$, whose intersection is $\{0\}$. Therefore

\overline{A\cap B}=\varnothing \quad\text{but}\quad \overline{A}\cap\overline{B}=\{0\}.

The point $0$ can be approached from $A$ and from $B$, even though it is not in their intersection and the intersection has no points at all. Closure can preserve separate limiting behavior without creating a common point in the original sets.

Worked Example: Interior Does Not Always Distribute Over Union

Let $A=[-1,0]$ and $B=[0,1]$. Their interiors are $(-1,0)$ and $(0,1)$, so $\operatorname{int}(A)\cup\operatorname{int}(B)=(-1,0)\cup(0,1)$. But $A\cup B=[-1,1]$, whose interior is $(-1,1)$. Consequently,

\operatorname{int}(A)\cup\operatorname{int}(B) \subsetneq \operatorname{int}(A\cup B).

The shared point $0$ is not interior to either set separately: each one ends there. Once the sets are joined, however, points from both sides fill an entire neighborhood of $0$. The union has created an interior point that neither set had on its own.

There is a further limitation: a law valid for every finite family may fail for an infinite family. For example, the interior of a finite intersection of open intervals is governed by the finite-intersection identity. It does not follow that interior commutes with arbitrary intersections.

Worked Example: Interior and an Infinite Intersection

For each positive integer $n$, let $E_n=(-1/n,1/n)$. Every $E_n$ is open, so $\operatorname{int}(E_n)=E_n$. Their intersection is $\{0\}$: the point $0$ belongs to every interval, while any nonzero $x$ is excluded once $n>1/|x|$, since then $1/n<|x|$. A singleton contains no open interval, and therefore has empty interior. It follows that

\operatorname{int}\left(\bigcap_{n=1}^{\infty}E_n\right) =\operatorname{int}(\{0\}) =\varnothing, \qquad \bigcap_{n=1}^{\infty}\operatorname{int}(E_n) =\{0\}.

The distinction is between a neighborhood that works for each set separately and one neighborhood that works for the entire intersection. Here every $E_n$ is open, but the neighborhoods shrink toward $0$; no nonempty open interval survives inside their intersection.

Using the Identities Carefully

The complement identities are exact and hold for every subset of $\mathbb{R}$. By contrast, distribution rules must be checked against the number and type of set operations involved. Finite union and closure, and finite intersection and interior, have exact identities; the examples above show that the other pairings can be strict. Infinite families require particular care: openness of every set in a family does not guarantee that their intersection is open.

A reliable approach is to use complements when one operation is difficult to compute, then apply the exact dual identity. For instance, to find $\operatorname{int}(E)$, one may find the closure of $\mathbb{R}\setminus E$ and take its complement. This method is only as accurate as the closure calculation: omitting a limit point of the complement would incorrectly include that point in the interior of $E$.

Choose the easier side.
Decide whether the set or its complement has the simpler neighborhood behavior.

Apply the exact complement identity.
Use $\operatorname{int}(E)=\mathbb{R}\setminus\overline{\mathbb{R}\setminus E}$ or $\overline{E}=\mathbb{R}\setminus\operatorname{int}(\mathbb{R}\setminus E)$.

Check the operation’s scope.
Before distributing closure or interior over a union or intersection, verify that the relevant finite-family identity applies; do not assume it extends to arbitrary families.

The central lesson is that closure and interior are dual under complementation, but they do not distribute over every set operation in the same way. Keeping those two facts separate prevents a common error: turning a valid inclusion into an unsupported equality.

Check Your Understanding

Use the complement identities and the examples above to answer the following questions.

State the identity that expresses the complement of $\overline{E}$ as an interior.
Why does $x\notin\overline{E}$ imply that $x\in\operatorname{int}(\mathbb{R}\setminus E)$?
State the complementary fixed-point test for a set to be closed.
For $A=(-\infty,0)$ and $B=(0,\infty)$, compare $\overline{A\cap B}$ with $\overline{A}\cap\overline{B}$.
For $E_n=(-1/n,1/n)$, what are $\operatorname{int}(\bigcap_{n=1}^{\infty}E_n)$ and $\bigcap_{n=1}^{\infty}\operatorname{int}(E_n)$?