Interior Points

When Does a Point Lie Inside a Set?

A point can belong to a set without lying securely inside it. For example, an endpoint of a closed interval belongs to the interval, but every open interval around that endpoint reaches outside the set. To distinguish such points from points surrounded by the set, we examine whether some open neighborhood of the point is contained entirely in the set.

We use the open balls on the real line, $B_r(x)=\{y\in\mathbb{R}:|y-x|<r\}$, where $r>0$. These are precisely the open intervals $(x-r,x+r)$. The definition below is pointwise: it asks whether there is at least one positive radius that works.

Definition (Interior Point): Let $E\subseteq\mathbb{R}$. A point $x\in\mathbb{R}$ is an interior point of $E$ if there exists $r>0$ such that $B_r(x)\subseteq E$. Equivalently, some open interval containing $x$ is contained in $E$.

The ball containment implies $x\in E$, because $x\in B_r(x)$. Thus an interior point must belong to the set. But membership alone does not suffice: a set may contain points that are on its edge or isolated from all its other points. The radius in the definition may depend on $x$; different interior points can require different radii.

Testing the Definition with Intervals

For an interval, the interior points are exactly those points that are not endpoints. The interval itself may include its endpoints, but no positive-radius ball centered at an endpoint can stay inside it. This gives a quick way to check the definition in familiar examples.

Worked Example: Interior Points of a Closed Interval

Let $E=[-3,2]$. If $-3<x<2$, both distances $x+3$ and $2-x$ are positive. Choose

r=\frac{1}{2}\min\{x+3,2-x\}.

Then $r>0$, and every $y$ with $|y-x|<r$ satisfies $-3<y<2$. Hence $B_r(x)\subseteq[-3,2]$, so every $x\in(-3,2)$ is an interior point.

The point $-3$ is not an interior point: for any $r>0$, the point $-3-r/2$ lies in $B_r(-3)$ but not in $[-3,2]$. Similarly, for any $r>0$, the point $2+r/2$ lies in $B_r(2)$ but not in $[-3,2]$. Therefore the interior points of $E$ are exactly $(-3,2)$. If the interval were the singleton $[-3,-3]$, there would be no interior points, since every ball around $-3$ contains points other than $-3$.

For an open interval $(a,b)$, each point already has a positive distance from both endpoints, so each point is interior. In contrast, in a half-open interval such as $[a,b)$, the included endpoint $a$ is not interior, while the points strictly between $a$ and $b$ are. It is the availability of a whole ball, not whether the set uses square or round brackets, that decides the question.

Worked Example: An Isolated Point Is Not Interior

Consider $E=\{5\}$. The point $5$ belongs to $E$, but for every $r>0$, the point $5+r/2$ satisfies $|(5+r/2)-5|=r/2<r$, so it lies in $B_r(5)$. It is not equal to $5$, so it does not belong to $E$. Thus no ball centered at $5$ is contained in $E$, and $5$ is not an interior point. Since it is the only point in $E$, the set has no interior points.

The same reasoning applies to any finite subset of $\mathbb{R}$. If $x$ is one of its points, a sufficiently small ball around $x$ contains no other points of that finite set, and every positive-radius ball contains points other than $x$. Consequently, no point of a finite set is interior.

A Useful Radius Estimate

The definition gives more than a yes-or-no test. If a ball around $x$ is contained in $E$, then points sufficiently close to $x$ also have their own balls contained in $E$. The radius available to a nearby point may be smaller, but it remains positive.

Theorem (Nearby Points of an Interior Point Are Interior): Suppose $B_r(x)\subseteq E$ for some $r>0$. If $y\in B_r(x)$, then $y$ is an interior point of $E$.

Proof. Since $y\in B_r(x)$, we have $|y-x|<r$. Define $\rho=r-|y-x|$, which is positive. If $z\in B_\rho(y)$, then the triangle inequality gives

|z-x|\leq |z-y|+|y-x|<\rho+|y-x|=r.

Therefore $z\in B_r(x)\subseteq E$. This holds for every $z\in B_\rho(y)$, so $B_\rho(y)\subseteq E$. By definition, $y$ is an interior point of $E$. $\square$

The estimate $\rho=r-|y-x|$ records how much room remains after moving the center from $x$ to $y$. As $y$ approaches the edge of $B_r(x)$, this guaranteed radius becomes smaller. It is still positive as long as $y$ remains strictly inside that ball.

Worked Example: Finding a Radius for a Nearby Point

Suppose $B_2(1)\subseteq E$, and take $y=2/3$. The distance from $y$ to the original center is

|y-1|=\left|\frac{2}{3}-1\right|=\frac{1}{3}<2.

The theorem permits the radius $\rho=2-\frac{1}{3}=\frac{5}{3}$. Indeed, if $|z-\frac{2}{3}|<\frac{5}{3}$, then

|z-1| \leq \left|z-\frac{2}{3}\right|+\left|\frac{2}{3}-1\right| <\frac{5}{3}+\frac{1}{3}=2.

Thus $z\in B_2(1)\subseteq E$ for every $z\in B_{5/3}(2/3)$, confirming that $2/3$ is an interior point of $E$. The conclusion uses only the original ball containment; it does not require a formula for the full set $E$.

Interior Points and the Complement

There is another way to decide whether $x$ is interior to $E$: check whether the complement can approach $x$. Recall the Neighborhood Characterization of Closure, established earlier in the course: a point belongs to the closure of a set exactly when every ball around it meets that set. Applying this to $\mathbb{R}\setminus E$ gives a useful pointwise characterization.

Theorem (Interior Points and the Closure of the Complement): For every $E\subseteq\mathbb{R}$ and $x\in\mathbb{R}$, $$ x\text{ is an interior point of }E \quad\Longleftrightarrow\quad x\notin\overline{\mathbb{R}\setminus E}. $$

Proof. Suppose first that $x$ is an interior point of $E$. By definition, there is an $r>0$ such that $B_r(x)\subseteq E$. Thus $B_r(x)\cap(\mathbb{R}\setminus E)=\varnothing$. The Neighborhood Characterization of Closure implies that $x\notin\overline{\mathbb{R}\setminus E}$, because at least one ball around $x$ misses the complement.

Conversely, suppose $x\notin\overline{\mathbb{R}\setminus E}$. By the same characterization, there is an $r>0$ such that $B_r(x)\cap(\mathbb{R}\setminus E)=\varnothing$. Every point of $B_r(x)$ is therefore outside the complement, hence belongs to $E$. So $B_r(x)\subseteq E$, and $x$ is an interior point of $E$. Both implications are proved. $\square$

This criterion says that an interior point has a definite buffer from the complement: some entire ball around it avoids the complement. It is stronger than saying merely that $x\notin\mathbb{R}\setminus E$, which only says $x\in E$. A point can belong to $E$ while complement points occur arbitrarily close to it.

Worked Example: Interior of Two Closed Rays

Let

E=(-\infty,-1]\cup[3,\infty).

Every $x<-1$ is interior: choose $r=(-1-x)/2>0$. If $|y-x|<r$, then $y<x+r=(x-1)/2<-1$, so $y\in E$. Every $x>3$ is interior as well: with $r=(x-3)/2>0$, the condition $|y-x|<r$ gives $y>x-r=(x+3)/2>3$, so $y\in E$.

At $x=-1$, every ball contains points just to the right of $-1$, which are between $-1$ and $3$ and therefore outside $E$. At $x=3$, every ball contains points just to the left of $3$, also outside $E$. Points strictly between $-1$ and $3$ are not in $E$, so cannot be interior points. Hence the interior points are exactly

(-\infty,-1)\cup(3,\infty).

The complement is $(-1,3)$. Its closure is $[-1,3]$, so the complement-closure criterion gives $\mathbb{R}\setminus[-1,3]=(-\infty,-1)\cup(3,\infty)$, in agreement with the direct ball test.

Interior Points Are Not the Same as Limit Points

An interior point and a limit point are defined by different neighborhood tests. For an interior point, one ball must be contained in the set. For a limit point, every punctured ball must meet the set. A limit point need not belong to the set, and even when it does, it may fail to be interior. Conversely, an interior point is a limit point of the set, since every sufficiently small punctured ball around it contains points of the set other than the center. These distinctions prevent a common confusion between “points nearby” and “a whole neighborhood contained inside.”

Worked Example: An Interval Together with an Isolated Point

Let $E=(0,2)\cup\{5\}$. Every $x\in(0,2)$ is interior. For example, if $x\in(0,2)$, then $r=\frac12\min\{x,2-x\}>0$ satisfies $B_r(x)\subseteq(0,2)\subseteq E$.

Neither $0$ nor $2$ is interior: every ball around $0$ contains negative points, and every ball around $2$ contains points greater than $2$ and close enough to $2$ to be less than $5$; these points lie outside $E$. The point $5$ is also not interior, since for any $r>0$, $5+r/2\in B_r(5)$ but $5+r/2\notin E$. Thus the interior points of $E$ are exactly $(0,2)$.

The point $5$ is isolated in $E$, while $0$ and $2$ are limit points of $E$ that do not belong to $E$. None of these three points is interior. The example shows that membership, being a limit point, and being an interior point are distinct properties.

What the Definition Does—and Does Not—Say

The interior-point test is local: it asks what happens in some neighborhood of one specified point. It does not require the whole set to be an interval, bounded, or closed. The set can have several separated pieces, and a point is interior whenever at least one ball around that point stays within the set. Earlier in this course, the interior of a set was described as the largest open subset contained in it. The pointwise definition explains which points make up that open subset.

A common mistake is to choose a radius that is too large and then conclude that a point is not interior. Failure for one radius says nothing about whether a smaller positive radius works. The definition requires the existence of some suitable $r>0$, not that every radius work. A second mistake is to assume that every point of a set is interior. The endpoints of $[a,b]$ and the sole point of $\{c\}$ show why set membership alone cannot establish the required ball containment.

Start with the proposed point.
Check first whether it belongs to the set; a point outside the set cannot be interior.

Look for a positive radius.
Try to find $r>0$ such that every point less than distance $r$ from the proposed point remains in the set.

Test the complement if needed.
The point is interior exactly when some ball around it misses the complement, or equivalently when it is outside the closure of the complement.

Check Your Understanding

Use the pointwise definition and the results in this tutorial to answer the following questions.

State the definition of an interior point using an open ball.
Which points of the interval $[-3,2]$ are interior, and why are its endpoints excluded?
If $B_r(x)\subseteq E$ and $|y-x|<r$, what positive radius around $y$ is guaranteed to stay in $E$?
State the characterization of interior points using the closure of the complement.
Why is the point $5$ not interior to $(0,2)\cup\{5\}$?
Explain the difference between the neighborhood condition for an interior point and the one for a limit point.

What You'll Learn