Interior as the Largest Open Subset

The Interior as an Open Approximation

The closure of a set is the smallest closed set that contains it. The interior gives a complementary kind of approximation: it retains as much of the set as possible while requiring the result to be open. This viewpoint is useful when a set has many points but contains no interval, or when endpoints belong to a set without being surrounded by nearby points of that set.

Recall the Theorem (The Interior Is the Largest Open Subset), established earlier in this course. For any $E\subseteq\mathbb{R}$, $\operatorname{int}(E)$ is open, is contained in $E$, and contains every open subset of $E$. In particular, if $U$ is open and $U\subseteq E$, then $U\subseteq\operatorname{int}(E)$. We use that result here as a tool; the focus is on ways to apply the largest-open-subset property.

Recall (Largest Open Subset): The interior $\operatorname{int}(E)$ is open, satisfies $\operatorname{int}(E)\subseteq E$, and contains every open set that is a subset of $E$.

“Largest” means largest with respect to set inclusion. It does not mean that the interior has the greatest number of points among subsets of $E$. Rather, every open subset of $E$ must fit inside it. Consequently, a practical way to identify an interior is to find open pieces contained in the set and collect all such pieces.

Interior as a Union of Open Intervals

Open intervals are the basic building blocks of open subsets of the real line. The earlier Theorem (Open Sets as Unions of Open Intervals) says that every open set in $\mathbb{R}$ is a union of open intervals. Applying that fact to the interior gives a useful interval-by-interval description.

Theorem (Interval Representation of the Interior): For every $E\subseteq\mathbb{R}$, $$ \operatorname{int}(E) = \bigcup\{(a,b):a,b\in\mathbb{R},\ a<b,\ (a,b)\subseteq E\}. $$ The union is empty if no nonempty open interval is contained in $E$.

Proof. Let $V$ denote the union on the right-hand side. Every interval in this union is contained in $E$, so $V\subseteq E$. Each interval is open, and arbitrary unions of open sets are open. Thus $V$ is an open subset of $E$. By the Largest Open Subset property, $V\subseteq\operatorname{int}(E)$.

For the reverse inclusion, $\operatorname{int}(E)$ is open. By the Theorem (Open Sets as Unions of Open Intervals), it is a union of open intervals contained in $\operatorname{int}(E)$. Since $\operatorname{int}(E)\subseteq E$, each of those intervals is also contained in $E$, and so appears among the intervals used to form $V$. Therefore $\operatorname{int}(E)\subseteq V$. The two inclusions prove the equality. $\square$

This representation turns the task of finding an interior into a search for entire intervals lying inside the original set. A point that belongs to $E$ contributes to the interior only if it lies in at least one such interval. Being in $E$ alone is not enough.

Worked Example: An Interval with Included and Excluded Endpoints

Let $E=[-3,2)$. Every point $x$ with $-3<x<2$ lies in an open interval contained in $E$. For instance, choose $r=\frac12\min\{x+3,2-x\}$. Both quantities in the minimum are positive, so $r>0$, and $(x-r,x+r)\subseteq[-3,2)$. Hence $(-3,2)\subseteq\operatorname{int}(E)$.

Neither endpoint belongs to the interior. The point $-3$ is not in any open interval contained in $E$, since every open interval containing $-3$ also contains points less than $-3$. Those points are not in $E$. The point $2$ is not in $E$ at all. Since the interior is contained in $E$, it cannot contain $2$. Thus

\operatorname{int}([-3,2))=(-3,2).

The interval representation explains why endpoint inclusion in $E$ does not necessarily persist in its interior: the endpoint must be part of an open interval lying wholly inside $E$.

Worked Example: Isolated Points Do Not Enlarge the Interior

Consider $E=\{0\}\cup(2,4)\cup\{7\}$. The open interval $(2,4)$ is contained in $E$, so the Largest Open Subset property gives $(2,4)\subseteq\operatorname{int}(E)$.

No open interval containing $0$ can be contained in $E$. Indeed, every such interval contains points different from $0$ and sufficiently close to $0$; these points are not in $E$. The same reasoning applies at $7$: every open interval containing $7$ contains points other than $7$ and sufficiently close to it, none of which belong to $E$. Finally, if a point lies outside $(2,4)$ and is not one of the two isolated points, it cannot be in the interior: it either lies outside $E$, or is an endpoint of $(2,4)$ and every interval around it includes points outside $E$. Therefore

\operatorname{int}(E)=(2,4).

Adding isolated points can enlarge a set without enlarging its interior. The interior records the open pieces that the set contains, not every point that the set happens to include.

Recognizing an Empty Interior

The interval representation also gives a useful test for when the interior is empty. A set has empty interior precisely when it contains no nonempty open interval. Equivalently, every nonempty open interval contains at least one point outside the set. This second form can be more convenient: instead of trying to list all open subsets of $E$, test whether the complement reaches into every interval.

Theorem (Empty Interior and Interval Intersection): For $E\subseteq\mathbb{R}$, the following are equivalent:

$\operatorname{int}(E)=\varnothing$.
Every nonempty open interval $(a,b)$, where $a<b$, intersects $\mathbb{R}\setminus E$.

Proof. Suppose first that $\operatorname{int}(E)=\varnothing$. Let $(a,b)$ be any nonempty open interval. If it did not intersect $\mathbb{R}\setminus E$, then $(a,b)\subseteq E$. Since $(a,b)$ is open and contained in $E$, the Largest Open Subset property would give $(a,b)\subseteq\operatorname{int}(E)$, contradicting that the interior is empty. Thus every such interval intersects the complement.

Conversely, suppose every nonempty open interval intersects $\mathbb{R}\setminus E$. If $\operatorname{int}(E)$ were nonempty, choose $x\in\operatorname{int}(E)$. The interior is open, so there is some $r>0$ such that $(x-r,x+r)\subseteq\operatorname{int}(E)\subseteq E$. This is a nonempty open interval that does not intersect $\mathbb{R}\setminus E$, contrary to the assumption. Hence $\operatorname{int}(E)=\varnothing$. $\square$

This criterion is a way to prove an interior is empty without determining the entire interior. It is enough to show that every open interval contains a point missing from $E$. The criterion concerns intervals of positive length: testing only individual points, or only intervals with a specified center, would not establish the required condition.

Worked Example: The Integers Have Empty Interior

Let $E=\mathbb{Z}$. We verify that every nonempty open interval intersects $\mathbb{R}\setminus\mathbb{Z}$. Given $a<b$, choose a positive integer $n$ so large that $n(b-a)>2$. There is an integer $k$ satisfying $\lfloor na\rfloor+1\leq k\leq\lfloor na\rfloor+2$; choose the smallest integer strictly greater than $na$, so $na<k\leq na+1$. Set $x=k/n$. Then $x>a$, and $x\leq a+1/n<b$, since $1/n<(b-a)/2<b-a$. Thus $x\in(a,b)$.

If $x$ is not an integer, it is already in $\mathbb{R}\setminus\mathbb{Z}$. If $x$ is an integer, choose a positive integer $m$ large enough that $1/m<b-x$, and put $y=x+1/(2m)$. Then $x<y<b$, and $y$ is not an integer because $0<y-x=1/(2m)<1$. In either case the interval contains a noninteger. The Empty Interior and Interval Intersection theorem therefore gives

\operatorname{int}(\mathbb{Z})=\varnothing.

The key point is not merely that the integers are missing many real numbers. It is that no open interval can fit wholly inside the integers.

Using Maximality Without Overclaiming

The largest-open-subset property has a simple proof strategy. If an open set $U$ is contained in $E$, then $U\subseteq\operatorname{int}(E)$. This can identify a lower bound for the interior. To show that a proposed open set $U$ is exactly the interior, one must also rule out any additional interior points—for example, by showing that no open interval around a point outside $U$ stays inside $E$.

A common mistake is to treat the interior as though it were the set of all points that have any nearby members of $E$. The definition requires a whole neighborhood around the point to be contained in $E$. For example, an endpoint of an interval may have points of $E$ arbitrarily close on one side, but every open interval around that endpoint extends to the other side as well. Such an endpoint need not be an interior point.

The two results in this tutorial give complementary ways to work with the interior:

To identify which points belong to the interior, look for open intervals contained in the set.
To prove that the interior is empty, show that every nonempty open interval meets the complement.
To prove a proposed open set lies in the interior, use the Largest Open Subset property.
To prove equality with a proposed interior, check both containment directions.

Find open pieces inside the set.
Identify open intervals that are wholly contained in $E$; their union is contained in $\operatorname{int}(E)$.

Check for additional points.
For points not in the proposed union, test whether every open interval around them reaches outside $E$.

For an empty interior, test every interval.
Show that each nonempty open interval contains a point of $\mathbb{R}\setminus E$, then apply the Empty Interior and Interval Intersection theorem.

Check Your Understanding

Use the largest-open-subset property and the interval criteria developed here to answer the following questions.

What does “largest” mean in the phrase “largest open subset,” and why does it not refer to the number of points?
How does the interval representation of $\operatorname{int}(E)$ identify which open intervals contribute to the interior?
Why does an endpoint of $[-3,2)$ fail to belong to its interior?
State the interval-intersection criterion equivalent to $\operatorname{int}(E)=\varnothing$.
What additional step is needed after showing that an open set $U$ is contained in $E$, if the goal is to prove $U=\operatorname{int}(E)$?

Interior as the Largest Open Subset

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