Boundary of a Set

From Boundary Points to the Boundary Set

The previous tutorial characterized boundary points one at a time: every ball around such a point meets both the set and its complement. We now collect all of those points into a single set, called the boundary of the set. This shift in viewpoint makes it possible to ask how the boundary itself behaves: whether it is closed, how it relates to the interior, and what an empty boundary tells us about the original set.

Definition (Boundary of a Set): Let $E\subseteq\mathbb{R}$. The boundary of $E$, denoted by $\partial E$, is the set of all boundary points of $E$: $$ \partial E=\{x\in\mathbb{R}: \text{every } B_r(x),\ r>0,\text{ meets both }E\text{ and }\mathbb{R}\setminus E\}. $$

The symbol $\partial E$ denotes a set, while saying that $x$ is a boundary point is a statement about an individual real number. The Closure Characterization of Boundary Points from the previous tutorial gives the set identity below. We will use it as a starting point, then establish consequences for the boundary as a whole.

Boundary Identity: For every $E\subseteq\mathbb{R}$, $$ \partial E=\overline{E}\cap\overline{\mathbb{R}\setminus E}. $$

The identity says that boundary points are exactly those points that lie in the closure of both sides. The previous tutorial also established that taking complements does not change boundary points. In the new notation, that result says $\partial E=\partial(\mathbb{R}\setminus E)$. Neither identity says that boundary points must belong to $E$: membership depends on whether $E$ is open, closed, or neither.

The Boundary in Terms of Closure and Interior

The Interior Points and the Closure of the Complement result from earlier in the course says that $x$ is an interior point of $E$ exactly when $x\notin\overline{\mathbb{R}\setminus E}$. Combining this with the Boundary Identity produces a useful formula: remove the interior of $E$ from its closure, and what remains is its boundary.

Theorem (Boundary as Closure Minus Interior): For every $E\subseteq\mathbb{R}$, $$ \partial E=\overline{E}\setminus\operatorname{int}(E). $$

Proof. By the Boundary Identity, $x\in\partial E$ if and only if $x\in\overline E$ and $x\in\overline{\mathbb{R}\setminus E}$. By the Interior Points and the Closure of the Complement result, the second condition is equivalent to $x\notin\operatorname{int}(E)$. Thus $x\in\partial E$ if and only if $x\in\overline E\setminus\operatorname{int}(E)$, which proves the set identity. $\square$

In particular, the boundary never overlaps the interior: $\partial E\cap\operatorname{int}(E)=\varnothing$. The formula also yields a disjoint decomposition of the closure. Since $\operatorname{int}(E)\subseteq\overline E$, every point of $\overline E$ either lies in the interior or lies in $\overline E\setminus\operatorname{int}(E)=\partial E$. Therefore

\overline E=\operatorname{int}(E)\cup\partial E, \qquad \operatorname{int}(E)\cap\partial E=\varnothing.

This decomposition is about the closure, not necessarily about $E$ itself. A point of $\partial E$ may or may not belong to $E$. For a closed set, boundary points belong to the set; for an open set, they do not, as established in the previous tutorial. The decomposition tells us that the closure consists of the points already safely inside $E$ together with the points at its boundary.

Worked Example: Two Intervals with Different Endpoint Membership

Let $E=[-3,-2]\cup(1,4)$. We claim that $\partial E=\{-3,-2,1,4\}$. At each of these four endpoints, every ball meets $E$ and its complement. For example, a ball around $-3$ contains $-3\in E$ and points less than $-3$ outside $E$; a ball around $1$ contains points just greater than $1$ in $E$ and $1\notin E$. The same reasoning applies at $-2$ and $4$, regardless of whether the endpoint belongs to $E$.

If $-3<x<-2$, choose $r=\frac12\min\{x+3,-2-x\}>0$. Then $B_r(x)\subseteq[-3,-2]$, so $x$ is interior, not boundary. If $1<x<4$, choose $r=\frac12\min\{x-1,4-x\}>0$; then $B_r(x)\subseteq(1,4)$, so these points are interior as well.

Every other point is exterior. More explicitly, if $x<-3$, take $r=(-3-x)/2$, which gives $x+r=(-3+x)/2<-3$, so $B_r(x)$ misses $E$. If $-2<x<1$, take $r=\frac12\min\{x+2,1-x\}>0$; that ball lies in the gap $(-2,1)$. If $x>4$, take $r=(x-4)/2$, which gives $x-r=(x+4)/2>4$. Thus no other points are boundary points. The example shows that the boundary records where the set ends, not which endpoints the set includes.

Every Boundary Is Closed

Although $E$ can be open, closed, or neither, its boundary is always closed. The reason is visible in the Boundary Identity: the boundary is the intersection of two closed sets. The closure of any set is closed, and arbitrary intersections of closed sets are closed, as established earlier in the course.

Theorem (The Boundary of Any Set Is Closed): For every $E\subseteq\mathbb{R}$, the set $\partial E$ is closed.

Proof. By the Boundary Identity, $$ \partial E=\overline E\cap\overline{\mathbb{R}\setminus E}. $$ Both $\overline E$ and $\overline{\mathbb{R}\setminus E}$ are closed, since each is the closure of a subset of $\mathbb R$. Their intersection is closed by the result that arbitrary intersections of closed sets are closed. Therefore $\partial E$ is closed. $\square$

The formula $\partial E=\overline E\setminus\operatorname{int}(E)$ gives another route to describing the boundary, but a difference of closed and open sets is not automatically the easiest way to establish closedness. The intersection formula is especially effective here because it immediately expresses the boundary using closed sets.

Worked Example: The Boundary of a Finite Set

Let $F=\{-1,1\}$. We claim that $\partial F=F$. First, each $a\in F$ is a boundary point. Every ball around $a$ meets $F$ because it contains $a$; it also contains points not in $F$. For instance, given $r>0$, choose $t$ with $0<t<r$ and $t\neq 2$. Then $a+t\in B_r(a)$ and $a+t\notin F$.

Now take $x\notin F$. Both $|x+1|$ and $|x-1|$ are positive, so $$ r=\frac12\min\{|x+1|,|x-1|\}>0. $$ If $y\in B_r(x)$, then $|y-x|<r$. The triangle inequality gives $$ |y+1|\geq |x+1|-|y-x|>|x+1|-r\geq r>0 $$ and $$ |y-1|\geq |x-1|-|y-x|>|x-1|-r\geq r>0. $$ Thus $y\neq-1$ and $y\neq1$, so $B_r(x)\cap F=\varnothing$. Hence $x$ is not a boundary point. This proves $\partial F=F$. The boundary is closed here, and the interior is empty: a ball around any point contains points other than the two elements of $F$.

What an Empty Boundary Tells Us

A set has empty boundary precisely when it has no points at which every ball meets both the set and its complement. This happens exactly when the set and its complement are both open—that is, when the set is both open and closed. The next result makes this connection precise.

Theorem (Empty Boundary and Open-and-Closed Sets): A set $E\subseteq\mathbb{R}$ has $\partial E=\varnothing$ if and only if $E$ is both open and closed.

Proof. Suppose first that $\partial E=\varnothing$. Let $x\in E$. Every ball around $x$ meets $E$, since it contains $x$. Because $x$ is not a boundary point, it cannot be true that every ball around $x$ also meets $\mathbb{R}\setminus E$. Hence some ball around $x$ misses the complement and is contained in $E$. This holds for every $x\in E$, so $E$ is open.

Now let $x\in\mathbb{R}\setminus E$. Every ball around $x$ meets the complement, since it contains $x$. Again, $x\notin\partial E$ means that some ball around $x$ misses $E$. Thus the complement is open. By the Complement Characterization of Closedness established earlier, $E$ is closed.

Conversely, suppose $E$ is both open and closed. If $x\in E$, openness gives a ball around $x$ contained in $E$, so that ball misses the complement. If $x\notin E$, closedness implies the complement is open, so a ball around $x$ is contained in the complement and misses $E$. In either case, $x$ is not a boundary point. Therefore $\partial E=\varnothing$. $\square$

The earlier theorem that the only subsets of $\mathbb R$ that are both open and closed are $\varnothing$ and $\mathbb R$ now has an equivalent boundary formulation: these are the only sets with empty boundary. This conclusion is special to the real line and should not be assumed for arbitrary spaces.

Worked Example: The Boundary of Two Rays

Let $E=(-\infty,0)\cup[2,\infty)$. We claim that $\partial E=\{0,2\}$. For every $r>0$, the ball $B_r(0)$ contains points less than $0$ in $E$ and points greater than $0$ but less than $2$ outside $E$. The ball $B_r(2)$ contains $2\in E$ and points less than $2$ but greater than $0$ outside $E$. Thus $0$ and $2$ are boundary points.

If $x<0$, choose $r=-x/2>0$. Every $y\in B_r(x)$ satisfies $y<x+r=x/2<0$, so the ball lies in $E$, and $x$ is interior. If $x>2$, choose $r=(x-2)/2$; then every $y\in B_r(x)$ satisfies $y>x-r=(x+2)/2>2$, so $x$ is interior too.

For $0<x<2$, take $r=\frac12\min\{x,2-x\}>0$. Each $y\in B_r(x)$ satisfies $0<y<2$, so the ball misses $E$. These points are exterior. This exhausts the real line and proves the claim. Here $2$ belongs to $E$ but $0$ does not; $E$ is neither open nor closed.

Using the Boundary Set in Practice

When computing $\partial E$, it is often helpful to decide first which points are interior and which are exterior, then identify the points left over. Equivalently, use $\partial E=\overline E\setminus\operatorname{int}(E)$ or $\partial E=\overline E\cap\overline{\mathbb{R}\setminus E}$, depending on which description is easier to calculate. The intersection identity is particularly useful for proving that a boundary is closed; the closure-minus-interior identity is particularly useful for locating it.

One common pitfall is to confuse the boundary with the set itself or with the closure. A boundary may contain points outside $E$, as at the excluded endpoint of an interval, and a boundary may equal $E$, as for a finite set. The boundary is specifically the part of the closure that is not interior. It marks the points where no neighborhood can be confined entirely to either side.

Find the interior.
Identify points with a ball contained in $E$; these are not boundary points.

Find the closure.
Include points every ball around which meets $E$.

Subtract the interior from the closure.
The remaining points form $\partial E=\overline E\setminus\operatorname{int}(E)$.

Check Your Understanding

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

What does the notation $\partial E$ represent, and how is it described using the closures of $E$ and its complement?
State the formula for the boundary using the closure and interior of $E$.
Why is the boundary of every subset of $\mathbb R$ closed?
How does the closure of $E$ decompose into its interior and boundary?
What is the relationship between having an empty boundary and being both open and closed?
For $E=(-\infty,0)\cup[2,\infty)$, which points form the boundary, and why are the points strictly between $0$ and $2$ not boundary points?

What You'll Learn