Boundary Points

When Every Neighborhood Sees Both Sides

The previous tutorial described exterior points as points with a ball that misses a set. A point that is neither interior nor exterior has a different local behavior: however small a ball around it is, the ball still meets the set and also meets its complement. Such a point is called a boundary point. This neighborhood test lets us locate where a set meets the rest of the real line without requiring the point itself to belong to the set.

Definition (Boundary Point): Let $E\subseteq\mathbb{R}$. A point $x\in\mathbb{R}$ is a boundary point of $E$ if, for every $r>0$, $$ B_r(x)\cap E\neq\varnothing \qquad\text{and}\qquad B_r(x)\cap(\mathbb{R}\setminus E)\neq\varnothing. $$

The condition has two parts, and both matter. Every ball must meet $E$, so $x$ is in the closure of $E$. Every ball must also meet the complement, so $x$ is in the closure of the complement. A point outside $E$ can be a boundary point, as can a point inside $E$. Boundary status is determined by what happens arbitrarily close to the point, not just by whether the point belongs to the set.

A Characterization Using Closure

The Neighborhood Characterization of Closure, established earlier in this course, says that $x\in\overline{A}$ exactly when every ball around $x$ intersects $A$. Apply it once to $E$ and once to $\mathbb{R}\setminus E$. This gives a concise equivalent test for being a boundary point.

Theorem (Closure Characterization of Boundary Points): A point $x\in\mathbb{R}$ is a boundary point of $E\subseteq\mathbb{R}$ if and only if $$ x\in\overline{E}\cap\overline{\mathbb{R}\setminus E}. $$

Proof. Suppose $x$ is a boundary point of $E$. For every $r>0$, the ball $B_r(x)$ intersects $E$, so the Neighborhood Characterization of Closure gives $x\in\overline{E}$. The same definition says every such ball intersects $\mathbb{R}\setminus E$, so $x\in\overline{\mathbb{R}\setminus E}$. Thus $x$ belongs to their intersection.

Conversely, suppose $x\in\overline{E}\cap\overline{\mathbb{R}\setminus E}$. Membership in the first closure means that every ball $B_r(x)$, with $r>0$, intersects $E$. Membership in the second means that every such ball intersects $\mathbb{R}\setminus E$. These are precisely the two conditions in the definition of a boundary point. $\square$

The theorem can also be expressed using the interior. The Interior Points and the Closure of the Complement result from earlier in the course identifies an interior point of $E$ as a point not in $\overline{\mathbb{R}\setminus E}$. The closure characterization therefore says that a boundary point is in $\overline E$ but is not an interior point of $E$. In symbols, the boundary points are exactly the points in $\overline E\setminus\operatorname{int}(E)$.

Worked Example: The Boundary of a Half-Open Interval

Let $E=[-1,2)$. We claim that its boundary points are exactly $-1$ and $2$. Every ball around $-1$ contains $-1\in E$ and also contains points less than $-1$, which lie outside $E$. Every ball around $2$ contains $2\notin E$ and also contains points less than $2$ that lie in $E$. Thus both endpoints are boundary points.

If $-1<x<2$, choose $r=\frac12\min\{x+1,2-x\}>0$. Then $B_r(x)\subseteq[-1,2)$, so $x$ is an interior point, not a boundary point. If $x<-1$, choose $r=(-1-x)/2$. Every $y\in B_r(x)$ satisfies

y<x+r=\frac{x-1}{2}<-1,

so this ball misses $E$. If $x>2$, choose $r=(x-2)/2$. Every $y\in B_r(x)$ satisfies

y>x-r=\frac{x+2}{2}>2,

so this ball also misses $E$. These points are exterior, not boundary points. The two endpoints are therefore the only boundary points. Notice that $-1$ belongs to $E$, whereas $2$ does not; both nevertheless satisfy the same boundary test.

Boundary Points and Complements

The definition treats $E$ and its complement symmetrically: a ball must meet both. This gives another useful result. It is not necessary to recalculate boundary points from scratch after replacing a set by its complement.

Theorem (Boundary Points Are Unchanged by Taking Complements): A point $x$ is a boundary point of $E$ if and only if it is a boundary point of $\mathbb{R}\setminus E$.

Proof. Suppose $x$ is a boundary point of $E$. Every ball around $x$ intersects $E$ and $\mathbb{R}\setminus E$. For the complement, these are exactly the two sets that must be intersected, with their roles reversed. Therefore every ball around $x$ intersects both $\mathbb{R}\setminus E$ and its complement $E$, so $x$ is a boundary point of $\mathbb{R}\setminus E$. Reversing the roles of $E$ and $\mathbb{R}\setminus E$ proves the converse. $\square$

Worked Example: A Set with an Isolated Point

Consider $E=(0,1)\cup\{2\}$. The points $0$ and $1$ are boundary points: every ball around either endpoint meets $(0,1)$ and also contains points outside $E$. The point $2$ is also a boundary point. Every ball around $2$ meets $E$ because it contains $2$, and it meets the complement because it contains points different from $2$ that are not in $(0,1)$.

Every $x\in(0,1)$ is an interior point: taking $r=\frac12\min\{x,1-x\}$ gives $B_r(x)\subseteq(0,1)$. Points outside $[0,1]\cup\{2\}$, other than the endpoint $2$ already considered, are exterior. In particular, if $1<x<2$, take $r=\frac12\min\{x-1,2-x\}$; this ball lies in $(1,2)$, so it misses $E$. If $x<0$ or $x>2$, a sufficiently small ball around $x$ also misses $E$. Thus the boundary points are exactly $0,1,2$.

The point $2$ highlights a distinction between a boundary point and a limit point of $E$. It is isolated in $E$: the ball $B_{1/2}(2)$ contains no point of $E$ other than $2$. Hence it is not a limit point of $E$, but it is a boundary point. The boundary test asks whether every ball meets both $E$ and its complement; it does not require other points of $E$ to approach $x$.

Boundary Points of Open and Closed Sets

For open and closed sets, the membership of boundary points has a particularly simple description. An open set cannot contain a boundary point: if $x$ belonged to the open set, some ball around $x$ would lie entirely inside it, contradicting the requirement that every ball meet the complement. A closed set has the corresponding property that it contains every point in its closure, including its boundary points.

Theorem (Boundary Points of Open and Closed Sets): If $U\subseteq\mathbb{R}$ is open, none of its boundary points belongs to $U$. If $F\subseteq\mathbb{R}$ is closed, every boundary point of $F$ belongs to $F$.

Proof. Let $x$ be a boundary point of an open set $U$. If $x\in U$, openness gives an $r>0$ such that $B_r(x)\subseteq U$. This ball does not intersect $\mathbb{R}\setminus U$, contrary to the definition of boundary point. Therefore $x\notin U$.

Now let $x$ be a boundary point of a closed set $F$. Every ball around $x$ intersects $F$, so the Neighborhood Characterization of Closure gives $x\in\overline F$. Since $F$ is closed, it contains its closure; hence $x\in F$. This proves both assertions. $\square$

Worked Example: Boundary Points of the Integers

Let $E=\mathbb{Z}$, the set of integers. Each integer $k$ is a boundary point. Given any $r>0$, the ball $B_r(k)$ contains $k\in E$, and it also contains a noninteger: for example, choose a positive integer $n$ large enough that $1/n<r$; then $k+1/n\in B_r(k)$ and $k+1/n\notin E$.

If $x\notin\mathbb{Z}$, there is an integer $k$ with $k<x<k+1$. Set $\delta=\min\{x-k,k+1-x\}>0$, and choose $r=\delta/2$. Every point in $B_r(x)$ lies strictly between $k$ and $k+1$, so the ball contains no integer. Thus $x$ is exterior to $\mathbb{Z}$, not a boundary point. The boundary points of the integers are exactly the integers.

This example also illustrates the open-and-closed result: $\mathbb{Z}$ is closed, so its boundary points belong to it, but its boundary points are not interior points. The closure characterization applies directly as well: each integer is in the closure of both $\mathbb{Z}$ and its complement.

What the Boundary Test Does—and Does Not—Say

A boundary point is neither required to belong to the set nor required to be a limit point of the set. For an open interval, its endpoints are boundary points outside the interval. For a singleton, its only point is a boundary point even though there are no distinct points of the singleton approaching it. The shared feature is that every ball, however small, contains at least one point of the set and at least one point outside it.

It is also useful to distinguish boundary points from points that are merely outside a set. If $x\notin E$, then $x$ might be exterior, as happens when some ball around $x$ misses $E$. Or it might be a boundary point, as happens at the excluded endpoint $2$ of $[-1,2)$. Being outside $E$ alone decides neither question; the behavior of all neighborhoods around $x$ decides it.

Check both intersections.
For a boundary point, every positive-radius ball must meet $E$ and must meet $\mathbb{R}\setminus E$.

Use closure when it simplifies the test.
The point must belong to both $\overline E$ and $\overline{\mathbb{R}\setminus E}$.

Do not infer membership from boundary status alone.
Boundary points of open sets lie outside the set; boundary points of closed sets lie inside it.

The central caution is not to check just one side. A point in $\overline E$ need not be a boundary point if it is an interior point of $E$, because a small ball may miss the complement. Similarly, a point in the closure of the complement need not be a boundary point if a ball misses $E$. Boundary status requires both closure conditions at once.

Check Your Understanding

Use the neighborhood definition and the results proved here to answer the following questions.

State the two conditions that every ball around a boundary point must satisfy.
How can boundary points be characterized using the closures of a set and its complement?
Why does taking the complement of a set leave its boundary points unchanged?
For $E=(0,1)\cup\{2\}$, why is $2$ a boundary point even though it is isolated in $E$?
What does the open-and-closed theorem say about whether boundary points belong to an open set or a closed set?

What You'll Learn