Boundary Identities

Boundaries Through Closure and Complement

Closure and interior identities provide ways to rewrite a set using neighborhoods and complements. Boundary identities bring both ideas together: a boundary point is approached both by points of the set and by points outside it. This description leads to an exact formula for the boundary, as well as useful rules for comparing the boundary of a set with the boundaries of sets built from it.

Recall the Closure Characterization of Boundary Points from “Boundary Points”: $x$ is a boundary point of $E$ precisely when every open ball centered at $x$ intersects both $E$ and $\mathbb{R}\setminus E$. Also, by the Neighborhood Characterization of Closure, $x\in\overline{E}$ precisely when every such ball intersects $E$. Applying that closure characterization to both sides of the complement gives the first identity.

Theorem (Boundary as the Intersection of Two Closures): For every $E\subseteq\mathbb{R}$, $$ \partial E = \overline{E}\cap\overline{\mathbb{R}\setminus E}. $$ Equivalently, a point lies on the boundary of $E$ exactly when every neighborhood meets both $E$ and its complement.

Proof. Let $x\in\mathbb{R}$. By the Closure Characterization of Boundary Points, $x\in\partial E$ if and only if every open ball centered at $x$ intersects $E$ and every such ball intersects $\mathbb{R}\setminus E$. The Neighborhood Characterization of Closure says that these two conditions are equivalent, respectively, to $x\in\overline{E}$ and $x\in\overline{\mathbb{R}\setminus E}$. Thus $x\in\partial E$ if and only if $x\in\overline{E}\cap\overline{\mathbb{R}\setminus E}$, proving the identity. $\square$

This identity is compatible with the Boundary Identity from “Boundary of a Set,” which gives $\partial E=\overline{E}\setminus\operatorname{int}(E)$. Indeed, the Complement Identities for Closure and Interior give $\overline{\mathbb{R}\setminus E}=\mathbb{R}\setminus\operatorname{int}(E)$. The two formulas express the same set in different ways: one emphasizes points approached from both sides, while the other removes the interior from the closure.

Worked Example: Using Both Closures to Find a Boundary

Let $E=[-3,-1)\cup(2,4)$. Its closure is $\overline{E}=[-3,-1]\cup[2,4]$. The complement is $(-\infty,-3)\cup[-1,2]\cup[4,\infty)$, whose closure is $(-\infty,-3]\cup[-1,2]\cup[4,\infty)$. Intersecting these closures gives

\partial E = \bigl([-3,-1]\cup[2,4]\bigr) \cap \bigl((-\infty,-3]\cup[-1,2]\cup[4,\infty)\bigr) = \{-3,-1,2,4\}.

Each listed point is approached by points of $E$ and by points of its complement. A point strictly inside either interval of $E$ is not in the closure of the complement, and a point strictly outside the displayed closed intervals is not in the closure of $E$. Neither can belong to the intersection.

Boundaries of Unions and Intersections

A boundary can arise only where membership in a set fails to be locally settled: every neighborhood contains points both in the set and outside it. If a point is not on the boundary of either $A$ or $B$, then each set has settled membership near that point. Taking a sufficiently small neighborhood makes both membership decisions hold at once. The union and intersection then also have settled membership there.

Theorem (Boundary of a Union and Intersection): For any $A,B\subseteq\mathbb{R}$, $$ \partial(A\cup B)\subseteq\partial A\cup\partial B \qquad\text{and}\qquad \partial(A\cap B)\subseteq\partial A\cup\partial B. $$

Proof. We prove both inclusions together. Suppose $x\notin\partial A\cup\partial B$. Then $x\notin\partial A$ and $x\notin\partial B$. By the Closure Characterization of Boundary Points, failure to be a boundary point means that some open ball about $x$ does not meet both a set and its complement. Thus there is a radius $r_A>0$ such that $B_{r_A}(x)$ is contained either in $A$ or in $\mathbb{R}\setminus A$. Likewise, there is an $r_B>0$ such that $B_{r_B}(x)$ is contained either in $B$ or in $\mathbb{R}\setminus B$.

Set $r=\min\{r_A,r_B\}$, which is positive. Since $B_r(x)\subseteq B_{r_A}(x)$ and $B_r(x)\subseteq B_{r_B}(x)$, membership in $A$ and membership in $B$ are each constant throughout $B_r(x)$: every point in this ball has the same membership status in $A$ as $x$, and the same status in $B$ as $x$. Therefore membership in $A\cup B$ is constant throughout $B_r(x)$, and membership in $A\cap B$ is also constant throughout $B_r(x)$. In either case, this ball cannot intersect both the set in question and its complement. Hence $x\notin\partial(A\cup B)$ and $x\notin\partial(A\cap B)$.

We have shown that every point outside $\partial A\cup\partial B$ is outside both $\partial(A\cup B)$ and $\partial(A\cap B)$. Taking the contrapositive gives both stated inclusions. $\square$

The theorem gives a useful restriction, not an equality in general. A point on the boundary of a component set might cease to be a boundary point after taking a union or intersection: another set may fill in the points on one side, or may remove the points on both sides. The following example shows both inclusions can be strict.

Worked Example: Strict Inclusions for a Union and an Intersection

Take $A=[-2,0]$ and $B=[0,3]$. Their boundaries are $\partial A=\{-2,0\}$ and $\partial B=\{0,3\}$, so $\partial A\cup\partial B=\{-2,0,3\}$. Their union is $[-2,3]$, whose boundary consists only of its endpoints. Therefore

\partial(A\cup B)=\{-2,3\} \subsetneq \{-2,0,3\}=\partial A\cup\partial B.

Their intersection is $\{0\}$. A singleton has empty interior and is closed, so its boundary is itself. Consequently,

\partial(A\cap B)=\{0\} \subsetneq \{-2,0,3\}=\partial A\cup\partial B.

The point $0$ is a boundary point of each interval separately, but it is an interior point of their union: an open interval around $0$ lies inside $[-2,3]$. The endpoints $-2$ and $3$, on the other hand, are not in the intersection, so they cannot be boundary points of the singleton intersection.

By induction, the same reasoning gives, for any positive integer $m$ and sets $E_1,\ldots,E_m$, $\partial(\bigcup_{j=1}^{m}E_j)\subseteq\bigcup_{j=1}^{m}\partial E_j$ and $\partial(\bigcap_{j=1}^{m}E_j)\subseteq\bigcup_{j=1}^{m}\partial E_j$. The conclusion is still an inclusion: as the example demonstrates, some boundaries of the component sets may disappear after the operation.

Comparing a Set with Its Closure

Taking closure can add points to a set without changing the points it approaches. It can nevertheless change the boundary, because newly added points may fill portions of the original boundary. In general, the boundary of the closure is contained in the boundary of the original set.

Theorem (Boundary of the Closure): For every $E\subseteq\mathbb{R}$, $$ \partial\overline{E}\subseteq\partial E. $$

Proof. By the Boundary Identity, $\partial\overline{E}=\overline{\overline{E}}\setminus\operatorname{int}(\overline{E})$. Idempotence of closure gives $\overline{\overline{E}}=\overline{E}$, so $\partial\overline{E}=\overline{E}\setminus\operatorname{int}(\overline{E})$. The set $E$ is contained in $\overline{E}$; monotonicity of interior therefore gives $\operatorname{int}(E)\subseteq\operatorname{int}(\overline{E})$. It follows that \[ \overline{E}\setminus\operatorname{int}(\overline{E}) \subseteq \overline{E}\setminus\operatorname{int}(E) = \partial E, \] where the last equality is the Boundary Identity. This proves the theorem. $\square$

Worked Example: Closure Can Remove Boundary Points

Let $E=(0,1)\setminus\{1/2\}$. Every neighborhood of $1/2$ contains points of $E$, so $1/2\in\overline{E}$, even though $1/2\notin E$. The endpoints $0$ and $1$ are also in the closure, and no point outside $[0,1]$ is in it. Thus $\overline{E}=[0,1]$. The boundary of $E$ is $\{0,1/2,1\}$: the endpoints are approached from inside and outside $[0,1]$, and every neighborhood of $1/2$ meets both $E$ and its complement. But the boundary of $[0,1]$ is $\{0,1\}$. Hence

\partial\overline{E}=\{0,1\} \subsetneq \{0,1/2,1\}=\partial E.

Filling the missing point $1/2$ turns it into an interior point of the closure. This illustrates why the theorem asserts containment rather than equality.

Using Boundary Identities Carefully

The boundary identities serve different purposes. The formula $\partial E=\overline{E}\cap\overline{\mathbb{R}\setminus E}$ is an exact test: a boundary point must be approached from both the set and its complement. The union and intersection rules are containment results: a boundary point of the combined set must come from a boundary point of at least one component, but a component boundary need not survive the combination. Similarly, closing a set can remove boundary points but cannot create new ones outside its original boundary.

A common error is to replace one of these inclusions by equality without checking whether boundary points disappear. A sound calculation identifies which points are approached from each side, or uses the Boundary Identity $\partial E=\overline{E}\setminus\operatorname{int}(E)$ and computes both sets in that difference. When working with unions or intersections, it is also useful to check points shared by component sets: such points can change from boundary points to interior points, or remain as boundaries only for the combined set.

Choose a boundary formula.
Use the intersection of closures to track approach from both sides, or closure minus interior to compute the boundary directly.

For unions and intersections, start with containment.
Apply the inclusion into the union of component boundaries; do not assume every component boundary remains on the new boundary.

Check the points where membership changes.
Test endpoints, shared points, and removed points to determine whether the inclusion is strict.

Together, these identities turn boundary questions into questions about closure, interior, and local membership. They also make clear why a boundary can shrink when sets are combined or when closure fills a missing point: the boundary records where membership changes arbitrarily close to a point, not merely which points appeared as endpoints in an initial description.

Check Your Understanding

Use the boundary identities and proofs in this tutorial to answer the following questions.

Express $\partial E$ as an intersection involving the closures of $E$ and its complement.
Why does a point outside both $\partial A$ and $\partial B$ have a neighborhood on which membership in $A\cup B$ is constant?
State the inclusions for the boundary of a union and the boundary of an intersection.
For $A=[-2,0]$ and $B=[0,3]$, why is $0$ not a boundary point of $A\cup B$?
What containment relates $\partial\overline{E}$ and $\partial E$, and which interior inclusion helps prove it?

What You'll Learn