Points Between a Set and Its Complement
Interior identifies points with a ball contained in a set; closure identifies points whose every ball meets the set. The boundary brings these ideas together. A boundary point cannot be separated from the set by a ball, and it cannot be separated from the complement either. It may belong to the set, or it may lie outside it.
Throughout, \((X,d)\) is a metric space and \(E\subseteq X\). All closures, interiors, balls, and boundaries are taken in \(X\), unless a different ambient space is specified. This matters: the same subset can have different boundaries in different metric spaces.
The first formula says that a boundary point is in the closure of \(E\), but is not an interior point of \(E\). In particular, every boundary point is in \(\overline E\), but it need not be in \(E\) itself. The second formula expresses the same condition from both sides: a boundary point belongs to the closures of both \(E\) and its complement.
A Ball Test for Boundary Points
Recall that \(x\in\overline E\) exactly when every open ball centered at \(x\) intersects \(E\). Also, \(x\notin\operatorname{int}(E)\) exactly when no open ball centered at \(x\) is contained in \(E\). The latter condition says that every such ball contains at least one point outside \(E\). These observations give a direct test that does not require computing the closure first.
Proof. Suppose \(x\in\partial E\). By the definition of boundary, \(x\in\overline E\), so every ball \(B_r(x)\) meets \(E\). Also, \(x\notin\operatorname{int}(E)\). If some ball \(B_r(x)\) did not meet \(X\setminus E\), then \(B_r(x)\subseteq E\), making \(x\) an interior point. Thus every ball also meets \(X\setminus E\).
Conversely, suppose every ball centered at \(x\) meets both \(E\) and \(X\setminus E\). Since every such ball meets \(E\), we have \(x\in\overline E\). Since every such ball contains a point outside \(E\), no such ball is contained in \(E\), so \(x\notin\operatorname{int}(E)\). Therefore \(x\in\overline E\setminus\operatorname{int}(E)=\partial E\). This proves both directions. \(\square\)
The quantifier “for every \(r>0\)” is essential. A ball of one particular radius meeting both sets does not establish that its center is a boundary point; a smaller ball might lie entirely on one side. The criterion requires points of both sets to occur arbitrarily close to \(x\).
Worked Example: The Boundary of a Half-Open Interval
Take \(E=[-2,1)\) in \(\mathbb{R}\) with the usual metric. If \(-2<x<1\), choose \(r=\min\{x+2,1-x\}/2\). This radius is positive. Every \(y\in B_r(x)\) satisfies \[ y>x-r\geq -2+\frac{x+2}{2}>-2 \quad\text{and}\quad y<x+r\leq 1-\frac{1-x}{2}<1. \] Hence \(B_r(x)\subseteq[-2,1)\), so \(x\) is an interior point and not a boundary point.
At \(-2\), every ball meets \(E\), since it contains \(-2\), and meets the complement, since it contains \(-2-r/2<-2\). At \(1\), every ball meets \(E\), since \(1-r/2\in[-2,1)\) whenever \(0<r\leq 2\), and for larger \(r\) it contains points of \(E\) as well. Every ball around \(1\) also contains points greater than \(1\). Thus both endpoints satisfy the ball test.
If \(x<-2\), a ball of radius \((-2-x)/2\) around \(x\) misses \(E\); if \(x\geq1\), then \(x\notin\overline E\) when \(x>1\), while \(x=1\) has already been considered. Therefore \[ \partial[-2,1)=\{-2,1\}. \] The point \(1\) is a boundary point despite not belonging to \(E\).
The Boundary Is Closed and Complement-Invariant
The two-sided ball test also explains two general properties. A boundary does not distinguish a set from its complement: interchanging the two sets in the test changes nothing. Moreover, the boundary is closed. The closure formula gives a concise proof of both facts.
Proof. By definition, \(x\in\partial E\) exactly when \(x\in\overline E\) and \(x\notin\operatorname{int}(E)\). A point is not in the interior of \(E\) exactly when every ball around it meets \(X\setminus E\), which is exactly the condition \(x\in\overline{X\setminus E}\). Therefore \[ \partial E=\overline E\cap\overline{X\setminus E}. \] Interchanging \(E\) and \(X\setminus E\) leaves this intersection unchanged, so \(\partial E=\partial(X\setminus E)\).
By the Closure Is the Smallest Closed Superset result from earlier in this course, both \(\overline E\) and \(\overline{X\setminus E}\) are closed. Their intersection is closed, so the displayed identity shows that \(\partial E\) is closed. This also covers \(E=\varnothing\) and \(E=X\): in either case the boundary is empty. \(\square\)
The identity \(\partial E=\overline E\setminus\operatorname{int}(E)\) is useful when the closure and interior are already known. The intersection identity is often more useful when the set and its complement are easy to examine separately. Either way, the ambient space remains part of the calculation.
Worked Example: The Boundary of the Integers
Let \(E=\mathbb{Z}\) in \(\mathbb{R}\). Each integer \(n\) is in \(\overline E\), since every ball around \(n\) contains \(n\). If \(x\notin\mathbb{Z}\), choose an integer \(k\) such that \(k<x<k+1\). The positive number \(r=\min\{x-k,k+1-x\}/2\) gives a ball \(B_r(x)\) containing no integers. Thus \(x\notin\overline{\mathbb{Z}}\), and consequently \(\overline{\mathbb{Z}}=\mathbb{Z}\).
No integer is interior. Given \(n\in\mathbb{Z}\) and \(r>0\), let \(t=\min\{r/2,1/2\}\). Then \(0<t<r\), and \(n+t\) is not an integer: it lies strictly between \(n\) and \(n+1\). Therefore \(B_r(n)\) contains a point outside \(\mathbb Z\). Hence \(\operatorname{int}(\mathbb Z)=\varnothing\), and \[ \partial\mathbb{Z} =\overline{\mathbb{Z}}\setminus\operatorname{int}(\mathbb{Z}) =\mathbb{Z}. \]
This example shows that a boundary need not be just a few endpoints. Every integer is a boundary point, even though \(\mathbb{Z}\) has no interior points in \(\mathbb{R}\).
Boundaries Depend on the Metric Space
The Open Sets in a Restricted Metric result from earlier in this course explains why calculations can change when the ambient space is restricted. Since closure and interior depend on the open balls available in that space, the boundary does too. A set that has points on both sides in a larger space may have no outside points at all when it is the entire space.
Worked Example: An Interval as a Subset and as a Space
Let \(A=(0,1)\). First view \(A\) as a subset of \(\mathbb{R}\). Its closure in \(\mathbb{R}\) is \([0,1]\), and its interior in \(\mathbb{R}\) is \((0,1)\). Therefore \[ \partial_{\mathbb{R}}A=[0,1]\setminus(0,1)=\{0,1\}. \] Indeed, every ball around \(0\) or \(1\) meets both \(A\) and its complement in \(\mathbb{R}\).
Now take \(X=(0,1)\) itself as the ambient metric space, and consider \(E=X\). Every ball in \(X\) consists entirely of points of \(X\). Thus \(E=X\) is both closed and open relative to \(X\): its closure and interior in \(X\) are both \(X\). It follows that \[ \partial_X X=X\setminus X=\varnothing. \] The two answers differ because the complement used in the boundary test differs: in the subspace \(X\), there are no points outside \(E=X\).
Interpreting the Boundary Correctly
The boundary can be viewed as the interface between a set and its complement, but “interface” should not be taken to mean that the boundary must be outside the set. For a closed interval, its endpoints belong to the interval; for a half-open interval, one boundary point belongs and one does not. The general definition permits both possibilities.
A useful consequence of the closure and interior characterizations is that, when both \(E\) and \(X\setminus E\) are nonempty, a point is on the boundary precisely when its distance from each of these sets is zero. Indeed, the Distance Characterization of Closure says that distance zero is equivalent to belonging to the closure, and the Distance Characterization of Interior says that positive distance from the complement is equivalent to being interior. At a boundary point, neither side can be avoided by choosing a sufficiently small ball.
In calculations, check both sides of the boundary test. Showing that every ball meets \(E\) establishes only that the point is in \(\overline E\); it does not show that the point is on the boundary. The ball must also meet \(X\setminus E\) at every radius. Conversely, a point outside \(E\) can still be a boundary point, as \(1\) is for \([-2,1)\). Always specify the ambient metric space before applying these tests.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What must every ball around a boundary point meet?
- Why can a point outside \(E\) still belong to \(\partial E\)?
- How does the intersection formula show that \(\partial E=\partial(X\setminus E)\)?
- Why is the boundary closed in every metric space?
- What is the boundary of \(\mathbb{Z}\) in \(\mathbb{R}\), and why?
- Why does \((0,1)\) have a different boundary in \(\mathbb{R}\) than it does when treated as the whole ambient space?