Interior of a Set

From Interior Points to the Interior of a Set

The previous tutorial asked whether one specified point has a ball contained in a set. We can collect all points that pass that test to form a new set: the interior. This changes the question from “Is $x$ inside $E$?” to “Which points of the real line have some room around them that stays inside $E$?” The resulting set can be empty, can equal $E$, or can be a proper subset of $E$.

Definition (Interior of a Set): Let $E\subseteq\mathbb{R}$. The interior of $E$, denoted $\operatorname{int}(E)$, is the set of all interior points of $E$: $$ \operatorname{int}(E)=\{x\in\mathbb{R}:\text{there exists }r>0\text{ such that }B_r(x)\subseteq E\}. $$

Recall that $B_r(x)=\{y\in\mathbb{R}:|y-x|<r\}$, where $r>0$. The definition immediately implies $\operatorname{int}(E)\subseteq E$: if $B_r(x)\subseteq E$, then $x\in B_r(x)$, so $x\in E$. It also implies that points outside $E$ cannot enter its interior. But a point in $E$ may still fail to be in $\operatorname{int}(E)$, as happens at the endpoint of a closed interval.

Earlier, in “Arbitrary Unions of Open Sets,” we established that $\operatorname{int}(E)$ is the largest open subset of $E$. That result gives a useful set-level viewpoint: to identify the interior, find the open portion of $E$ and exclude points for which no whole neighborhood fits inside $E$. In particular, $\operatorname{int}(\varnothing)=\varnothing$ and $\operatorname{int}(\mathbb{R})=\mathbb{R}$.

Calculating Interiors Directly

For intervals, the calculation usually comes down to whether a point has positive distance from the endpoints that restrict it. Endpoints included in the original set still fail the ball test if every ball around them reaches beyond the set. Open rays have no finite endpoint on one side, but their finite endpoint is excluded from the interior for the same reason.

Worked Example: Interior of a Bounded Interval

Let $E=[-4,3]$. If $-4<x<3$, both $x+4$ and $3-x$ are positive. Set

r=\frac12\min\{x+4,3-x\}.

Then $r>0$. If $y\in B_r(x)$, then $|y-x|<r$, so $y>x-r\geq -4$ and $y<x+r\leq 3$. Thus $B_r(x)\subseteq[-4,3]$, and $x\in\operatorname{int}(E)$.

Neither endpoint is interior. For any $r>0$, the point $-4-r/2$ lies in $B_r(-4)$ but not in $E$, and $3+r/2$ lies in $B_r(3)$ but not in $E$. Therefore

\operatorname{int}([-4,3])=(-4,3).

The open ball test explains why the answer does not retain the closed interval’s endpoints, even though both endpoints belong to the original set.

The interior need not be nonempty. For example, if a set consists only of isolated points, each ball around any one of them also contains points not in the set. More generally, to show that a point is not interior, it is enough to find a point outside the set in every ball around it. The outside point may depend on the radius.

Worked Example: A Set with an Isolated Point and an Interval

Consider $E=(2,5)\cup\{8\}$. Each $x\in(2,5)$ is interior: the radius

r=\frac12\min\{x-2,5-x\}

is positive, and $B_r(x)\subseteq(2,5)\subseteq E$. The point $8$ is not interior. For every $r>0$, the point $8+r/2$ lies in $B_r(8)$, but it is neither in $(2,5)$ nor equal to $8$. Hence it is outside $E$. We conclude that

\operatorname{int}(E)=(2,5).

This example separates three ideas: $8$ belongs to the set, but it is not an interior point; the points in $(2,5)$ belong to the set and are interior; points outside $E$ are not interior either.

Interiors and Unions

If an open ball around $x$ is contained in one set $E$, it is also contained in any larger set that contains $E$. This is the pointwise reason behind the Monotonicity and Idempotence of the Interior result established earlier: enlarging a set cannot remove its interior points. For unions, there is a related inclusion that is useful even when the pieces are not open.

Theorem (Interior of a Union Contains the Union of the Interiors): For any family of sets $\{E_\alpha:\alpha\in A\}$ in $\mathbb{R}$, $$ \bigcup_{\alpha\in A}\operatorname{int}(E_\alpha) \subseteq \operatorname{int}\left(\bigcup_{\alpha\in A}E_\alpha\right). $$

Proof. If the indexing set $A$ is empty, both unions are empty, so the inclusion holds. Otherwise, take $x\in\bigcup_{\alpha\in A}\operatorname{int}(E_\alpha)$. By the definition of union, there is some $\alpha_0\in A$ such that $x\in\operatorname{int}(E_{\alpha_0})$. By the definition of interior, there is an $r>0$ with $B_r(x)\subseteq E_{\alpha_0}$. Since $E_{\alpha_0}\subseteq\bigcup_{\alpha\in A}E_\alpha$, we have $$ B_r(x)\subseteq\bigcup_{\alpha\in A}E_\alpha. $$ Thus $x\in\operatorname{int}\left(\bigcup_{\alpha\in A}E_\alpha\right)$. Every point in the left-hand side belongs to the right-hand side, proving the inclusion. $\square$

The reverse inclusion need not hold: points from different sets can fit together to form a neighborhood even though neither set provides a neighborhood by itself. Thus one must not replace the inclusion in the theorem with an equality without checking additional hypotheses.

Worked Example: The Interior of a Union Can Be Larger

Let $E=(-\infty,0]\cup[1,\infty)$ and $F=[0,1]$. The interval calculation gives $$ \operatorname{int}(E)=(-\infty,0)\cup(1,\infty), \qquad \operatorname{int}(F)=(0,1). $$

Their interiors together omit $0$ and $1$: $$ \operatorname{int}(E)\cup\operatorname{int}(F) =(-\infty,0)\cup(0,1)\cup(1,\infty). $$

However, $E\cup F=\mathbb{R}$: points at most $0$ or at least $1$ belong to $E$, while points between $0$ and $1$ belong to $F$. Therefore $\operatorname{int}(E\cup F)=\mathbb{R}$. In particular, both $0$ and $1$ belong to the interior of the union even though neither belongs to the union of the individual interiors. This verifies strict inclusion in the theorem for these sets.

Interior of a Set Difference

A more precise formula is available when points are removed from a set. To remain interior after removing $F$, a point must first be interior to $E$. It must also have some neighborhood that avoids $F$. The Neighborhood Characterization of Closure, established earlier, says that $x\notin\overline{F}$ exactly when some ball around $x$ misses $F$. Combining these two neighborhood requirements gives an exact identity.

Theorem (Interior of a Set Difference): For any $E,F\subseteq\mathbb{R}$, $$ \operatorname{int}(E\setminus F) = \operatorname{int}(E)\setminus\overline{F}. $$

Proof. Suppose first that $x\in\operatorname{int}(E\setminus F)$. There is an $r>0$ such that $B_r(x)\subseteq E\setminus F$. Since $E\setminus F\subseteq E$, the same ball lies in $E$, so $x\in\operatorname{int}(E)$. Also $B_r(x)\cap F=\varnothing$, because no point of $E\setminus F$ belongs to $F$. The Neighborhood Characterization of Closure therefore gives $x\notin\overline{F}$. Hence $x\in\operatorname{int}(E)\setminus\overline{F}$.

Conversely, suppose $x\in\operatorname{int}(E)\setminus\overline{F}$. There are positive radii $r_1,r_2$ such that $B_{r_1}(x)\subseteq E$ and $B_{r_2}(x)\cap F=\varnothing$. Let $r=\min\{r_1,r_2\}$, which is positive. Then $B_r(x)\subseteq E$, and $B_r(x)\cap F=\varnothing$. Thus every point in $B_r(x)$ belongs to $E$ and not to $F$, giving $B_r(x)\subseteq E\setminus F$. Therefore $x\in\operatorname{int}(E\setminus F)$. Both inclusions hold, proving the identity. $\square$

Worked Example: Removing a Closed Interval

Take $E=[-2,5]$ and $F=[1,3]$. Then $E\setminus F=[-2,1)\cup(3,5]$. Directly, the endpoints $-2$ and $5$ cannot be interior, and neither can $1$ or $3$: every ball around $1$ or $3$ reaches into the removed interval. Points strictly between $-2$ and $1$, and strictly between $3$ and $5$, do have balls contained in $E\setminus F$. Hence $$ \operatorname{int}(E\setminus F)=(-2,1)\cup(3,5). $$

The identity gives the same answer. We have $\operatorname{int}(E)=(-2,5)$ and $\overline{F}=[1,3]$. Thus $$ \operatorname{int}(E)\setminus\overline{F} =(-2,5)\setminus[1,3] =(-2,1)\cup(3,5). $$ The closure of the removed set matters: its endpoints must also be excluded from the interior, even though the endpoints themselves are removed from $E\setminus F$ as part of $F$.

Worked Example: Removing One Point from an Open Interval

Let $E=(-3,3)$ and $F=\{0\}$. The set difference is $(-3,0)\cup(0,3)$, which is already open, so its interior is the set itself. Using the identity, $\operatorname{int}(E)=(-3,3)$ and $\overline{\{0\}}=\{0\}$. Therefore $$ \operatorname{int}(E\setminus F) =(-3,3)\setminus\{0\} =(-3,0)\cup(0,3). $$

The formula also helps when the removed set is not closed: it is its closure, not just the set itself, that identifies points with removed points arbitrarily near. The neighborhood must avoid all of $F$, which is impossible at any point of $\overline{F}$.

Using the Interior Without Overstating It

The interior is a way to retain precisely those points around which a whole open neighborhood fits. It can be computed directly from intervals, or through identities involving unions and set differences. The union theorem always gives an inclusion, but the example shows why equality can fail: separate pieces may jointly fill in a neighborhood. By contrast, the set-difference identity is an equality because it accounts for both requirements at once—room inside $E$ and a neighborhood avoiding $F$.

A common error is to treat “belongs to the set” as equivalent to “belongs to the interior.” The endpoints in the interval examples disprove that equivalence. Another is to assume that every operation on sets passes through the interior unchanged. The union inclusion is safe, but equality requires justification. For set differences, forgetting $\overline{F}$ can miss points where removed points occur arbitrarily close, even if the point itself is not in $F$.

Identify candidate points.
Interior points must belong to the original set; points outside it can be discarded immediately.

Find a ball or rule one out.
To prove a point is interior, exhibit a positive radius whose ball stays inside. To disprove it, find an outside point in every ball.

Use the set operation carefully.
For unions, the union of the interiors is always contained in the interior of the union, but may be smaller. For a difference, use $\operatorname{int}(E\setminus F)=\operatorname{int}(E)\setminus\overline{F}$.

Check Your Understanding

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

Give the definition of $\operatorname{int}(E)$ using open balls, and explain why $\operatorname{int}(E)\subseteq E$.
What is the interior of $[-4,3]$, and how can a point in either endpoint’s ball show that endpoint is not interior?
State the inclusion relating the interior of a union to the union of the interiors. Why can equality fail?
For $E=[-2,5]$ and $F=[1,3]$, use the set-difference identity to compute $\operatorname{int}(E\setminus F)$.
Why does the formula for $\operatorname{int}(E\setminus F)$ use $\overline{F}$ rather than just $F$?
What two neighborhood conditions must a point satisfy to be interior to $E\setminus F$?

What You'll Learn