Exterior of a Set

From Interior to Exterior

The interior of a set collects points around which a whole open ball stays inside the set. The exterior asks for the corresponding points whose balls stay entirely outside it. This is not simply the set-theoretic complement: a point outside a set can still fail to be exterior if points of the set occur arbitrarily close to it. Endpoints of intervals and limit points of sequences are common examples of this distinction.

Definition (Exterior of a Set): Let $E\subseteq\mathbb{R}$. The exterior of $E$, denoted $\operatorname{ext}(E)$, is the set of all $x\in\mathbb{R}$ for which there exists $r>0$ such that $B_r(x)\cap E=\varnothing$. Equivalently, $$ \operatorname{ext}(E)=\{x\in\mathbb{R}:\text{there exists }r>0\text{ such that }B_r(x)\subseteq\mathbb{R}\setminus E\}. $$

Recall that $B_r(x)=\{y\in\mathbb{R}:|y-x|<r\}$. The equivalence in the definition follows because a ball misses $E$ exactly when each of its points lies in $\mathbb{R}\setminus E$. In particular, every exterior point lies outside $E$, since $x\in B_r(x)$. But being outside $E$ alone is not enough: there must be a positive-radius ball around the point that misses $E$.

Exterior, Interior, and Closure

The definition makes the exterior the interior of the complement. The Neighborhood Characterization of Closure, established earlier in this course, gives a second useful description: a point is outside $\overline{E}$ exactly when some ball around it misses $E$. Together these observations give an identity that often turns an exterior calculation into a closure calculation.

Theorem (Exterior as the Complement of the Closure): For every $E\subseteq\mathbb{R}$, $$ \operatorname{ext}(E) = \operatorname{int}(\mathbb{R}\setminus E) = \mathbb{R}\setminus\overline{E}. $$

Proof. First, $x\in\operatorname{ext}(E)$ if and only if there is an $r>0$ such that $B_r(x)\cap E=\varnothing$. This is equivalent to $B_r(x)\subseteq\mathbb{R}\setminus E$, which, by the definition of interior, is equivalent to $x\in\operatorname{int}(\mathbb{R}\setminus E)$.

For the other equality, the Neighborhood Characterization of Closure says that $x\in\overline{E}$ if and only if every ball around $x$ intersects $E$. Negating this statement, $x\notin\overline{E}$ if and only if there exists $r>0$ such that $B_r(x)\cap E=\varnothing$. The right-hand condition is precisely $x\in\operatorname{ext}(E)$. Therefore $\operatorname{ext}(E)=\mathbb{R}\setminus\overline{E}$, as required. $\square$

This identity also shows that the exterior is open: the closure $\overline{E}$ is closed, so its complement is open. Equivalently, the exterior is the interior of a set, and the Interior of a Set result established earlier shows that every such interior is open. The exterior and the interior of $E$ are disjoint, but they need not cover the real line. Points in neither are on the boundary, a topic developed in the next tutorial.

Worked Example: Exterior of a Union of Two Rays

Let $E=(-\infty,-2]\cup[3,\infty)$. Its complement is $(-2,3)$, but the exterior is not all of that complement: the endpoints $-2$ and $3$ do not have balls disjoint from $E$. Using the closure description, $E$ is closed, so $\overline{E}=E$. Thus

\operatorname{ext}(E) =\mathbb{R}\setminus\bigl((-\infty,-2]\cup[3,\infty)\bigr) =(-2,3).

To check the endpoint issue directly, every ball around $-2$ contains $-2\in E$, and every ball around $3$ contains $3\in E$. Neither endpoint is exterior. In contrast, if $-2<x<3$, the positive radius $r=\frac12\min\{x+2,3-x\}$ gives $B_r(x)\subseteq(-2,3)$, so $x$ is exterior.

Worked Example: Exterior of a Finite Set

Take $F=\{-3,2,6\}$. A finite set is closed, so $\overline{F}=F$. The theorem gives

\operatorname{ext}(F)=\mathbb{R}\setminus\{-3,2,6\} =(-\infty,-3)\cup(-3,2)\cup(2,6)\cup(6,\infty).

For example, $x=4$ is exterior: the radius $r=1$ gives $B_1(4)=(3,5)$, which contains none of $-3,2,6$. The point $2$, although it is approached by other real numbers outside $F$, is not exterior because every ball around $2$ contains $2$ itself, which belongs to $F$. More generally, every point of $F$ fails the exterior test for this same reason.

Exteriors of Finite Unions

A ball that misses a union must miss every set making up that union. For finitely many sets, the separate balls in this observation can be replaced by one common ball: take the smallest of the finitely many positive radii. This gives an exact formula for the exterior of a finite union.

Theorem (Exterior of a Finite Union): Let $E_1,\ldots,E_m\subseteq\mathbb{R}$, where $m$ is a positive integer. Then $$ \operatorname{ext}\left(\bigcup_{j=1}^{m}E_j\right) = \bigcap_{j=1}^{m}\operatorname{ext}(E_j). $$

Proof. Suppose first that $x\in\operatorname{ext}\left(\bigcup_{j=1}^{m}E_j\right)$. There is an $r>0$ such that $B_r(x)$ misses $\bigcup_{j=1}^{m}E_j$. Since each $E_j$ is contained in that union, $B_r(x)\cap E_j=\varnothing$ for every $j$. Thus $x\in\operatorname{ext}(E_j)$ for every $j$, so $x\in\bigcap_{j=1}^{m}\operatorname{ext}(E_j)$.

Conversely, suppose $x\in\bigcap_{j=1}^{m}\operatorname{ext}(E_j)$. For each $j$, there is an $r_j>0$ such that $B_{r_j}(x)\cap E_j=\varnothing$. Because there are only finitely many radii, their minimum $r=\min\{r_1,\ldots,r_m\}$ exists and is positive. The containment $B_r(x)\subseteq B_{r_j}(x)$ holds for every $j$, so $B_r(x)$ misses every $E_j$. It therefore misses their union, and $x\in\operatorname{ext}\left(\bigcup_{j=1}^{m}E_j\right)$. Both inclusions hold, proving the identity. $\square$

The finite nature of the collection matters. For an arbitrary family, a point may have a ball avoiding each individual set, while no single positive radius works for all the sets together. One always has the inclusion

\operatorname{ext}\left(\bigcup_{\alpha\in A}E_\alpha\right) \subseteq \bigcap_{\alpha\in A}\operatorname{ext}(E_\alpha).

Indeed, a ball avoiding the union avoids each member of the family. The reverse inclusion can fail because the individual radii may become arbitrarily small. The following example verifies that failure, including the additional closure point that is easy to overlook.

Worked Example: A Countable Union with an Additional Closure Point

For each positive integer $n$, let $E_n=\{1/n\}$, and put $S=\bigcup_{n=1}^{\infty}E_n=\{1/n:n\geq1\}$. We first determine its closure. Every point $1/k$ belongs to $\overline{S}$ because it belongs to $S$. Also $0\in\overline{S}$: given $r>0$, choose a positive integer $n$ large enough that $1/n<r$. Then $1/n\in S\cap B_r(0)$.

Now let $x\notin S\cup\{0\}$. Since $1/n\to0$, choose a positive integer $N$ such that $1/n<|x|/2$ whenever $n\geq N$. For such $n$, the reverse triangle inequality gives

|x-1/n|\geq |x|-1/n>|x|/2.

There are only finitely many terms $1/n$ with $1\leq n<N$, and none equals $x$. If this finite collection is nonempty, the minimum of the positive distances $|x-1/n|$ over those indices is positive. Choose a positive radius smaller than that minimum and smaller than $|x|/2$. If the finite collection is empty, choose any positive radius smaller than $|x|/2$. In either case, the resulting ball around $x$ misses every $1/n$: it misses the finite initial collection by the choice of radius, and it misses the tail by the displayed inequality. Thus $x\notin\overline{S}$, proving

\overline{S}=S\cup\{0\}.

Each singleton $E_n$ is closed, so $\operatorname{ext}(E_n)=\mathbb{R}\setminus\{1/n\}$. Consequently,

\bigcap_{n=1}^{\infty}\operatorname{ext}(E_n) =\mathbb{R}\setminus S, \qquad \operatorname{ext}\left(\bigcup_{n=1}^{\infty}E_n\right) =\mathbb{R}\setminus(S\cup\{0\}).

The two sets differ at $0$: it belongs to every individual exterior because $0\neq1/n$ for every positive integer $n$, but it is not exterior to the union because points $1/n$ lie arbitrarily close to it. This confirms that the finite-union identity cannot be extended to arbitrary unions.

Using the Exterior Carefully

The formula $\operatorname{ext}(E)=\mathbb{R}\setminus\overline{E}$ is often the quickest way to calculate an exterior. It also gives a useful order rule: if $A\subseteq B$, then $\overline{A}\subseteq\overline{B}$, so $\operatorname{ext}(B)\subseteq\operatorname{ext}(A)$. Enlarging a set can only remove exterior points, not create new ones. This reversal of inclusion is consistent with the neighborhood definition: a ball that avoids the larger set necessarily avoids the smaller one.

A common pitfall is to identify the exterior with $\mathbb{R}\setminus E$. The latter includes every point not in $E$, even points that are limits of points in $E$. The sequence example shows the difference: $0\notin S$, yet $0\notin\operatorname{ext}(S)$. Another pitfall is to assume that the finite-union identity holds for infinitely many sets. Each set can be avoided by a ball around the same point, but the radii may shrink toward zero, leaving no ball that avoids the entire union.

Test for a neighborhood outside the set.
To prove that $x$ is exterior, find a positive radius $r$ such that $B_r(x)\cap E=\varnothing$.

Use the closure when convenient.
The exterior is $\mathbb{R}\setminus\overline{E}$, so points in the closure are not exterior even if they do not belong to $E$.

Check how many sets are involved.
Exteriors distribute over finite unions as an equality. For arbitrary unions, the exterior of the union is contained in the intersection of the individual exteriors, but equality can fail.

Check Your Understanding

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

Define $\operatorname{ext}(E)$ using open balls. Why must every exterior point lie outside $E$?
State the relationship between the exterior, the interior of the complement, and the closure of a set.
What is the exterior of $(-\infty,-2]\cup[3,\infty)$, and why are its endpoints not exterior?
Why does the formula for the exterior of a finite union use the minimum of finitely many radii?
For $E_n=\{1/n\}$, why does $0$ belong to every $\operatorname{ext}(E_n)$ but not to $\operatorname{ext}(\bigcup_{n=1}^{\infty}E_n)$?

What You'll Learn