Which Points Have Room Inside a Set?
Closure identifies points that can be approached by points of a set. Interior asks a complementary question: which points have a whole neighborhood contained in the set? A point can belong to a set without having any such room. For example, an endpoint of a closed interval belongs to the interval, but every ball around that endpoint also contains points outside it.
Throughout, \((X,d)\) is a metric space and \(E\subseteq X\). The metric and the choice of \(X\) matter: balls are taken in the space under consideration. Thus the interior of a set viewed as a subspace can differ from its interior in a larger space.
The defining ball must be centered at the point being tested, and it must be contained in \(E\). It is not enough that the ball merely intersects \(E\), as it does for a closure point. Since \(x\in B_r(x)\), every interior point belongs to \(E\); consequently \(\operatorname{int}(E)\subseteq E\). If \(E=\varnothing\), its interior is empty, while \(\operatorname{int}(X)=X\), because every ball in \(X\) is contained in \(X\).
The Interior Is the Largest Open Part
The ball in the definition gives more than a test for individual points. If a ball around \(x\) lies in \(E\), then points sufficiently near \(x\) also have smaller balls lying in \(E\). This is why the collection of all interior points is itself open.
Proof. We already know that \(\operatorname{int}(E)\subseteq E\). To show that \(\operatorname{int}(E)\) is open, take any \(x\in\operatorname{int}(E)\). By definition, there is an \(r>0\) such that \(B_r(x)\subseteq E\). If \(z\in B_{r/2}(x)\), then for every \(w\in B_{r/2}(z)\), the triangle inequality gives $$ d(w,x)\leq d(w,z)+d(z,x)<\frac r2+\frac r2=r. $$ Thus \(w\in B_r(x)\subseteq E\), so \(B_{r/2}(z)\subseteq E\). This shows that \(z\in\operatorname{int}(E)\). Therefore \(B_{r/2}(x)\subseteq\operatorname{int}(E)\), and every point of \(\operatorname{int}(E)\) has a ball contained in it. By the definition of an open set, \(\operatorname{int}(E)\) is open.
Now let \(U\) be any open subset of \(X\) with \(U\subseteq E\). For each \(x\in U\), openness gives some \(r>0\) such that \(B_r(x)\subseteq U\). Since \(U\subseteq E\), this ball is contained in \(E\), so \(x\in\operatorname{int}(E)\). Hence \(U\subseteq\operatorname{int}(E)\). Together with the openness and containment already proved, this establishes the theorem. If \(U\) or \(E\) is empty, the same containment statements hold directly. \(\square\)
This result gives a useful way to identify an interior without testing every possible ball: find an open set contained in \(E\), then show that no larger open subset of \(X\) can be contained in \(E\). It also shows why the interior is not generally all of \(E\): points that cannot be included in any open subset of \(E\) are left out.
Worked Example: The Interior of a Closed Interval
Take \(E=[-1,3]\) in \(\mathbb{R}\) with its usual metric. If \(-1<x<3\), set \(r=\min\{x+1,3-x\}/2\). This is positive. For any \(y\in B_r(x)\), we have $$ y>x-r\geq -1+\frac{x+1}{2}>-1 \quad\text{and}\quad y<x+r\leq 3-\frac{3-x}{2}<3. $$ Thus \(B_r(x)\subseteq[-1,3]\), and every such \(x\) is an interior point.
At \(x=-1\), every ball contains points less than \(-1\): for any \(r>0\), the point \(-1-r/2\) lies in \(B_r(-1)\) but not in \(E\). At \(x=3\), the point \(3+r/2\) gives the corresponding obstruction. If \(x<-1\), then \(x\notin E\), so it cannot be an interior point; the same is true for \(x>3\), since every interior point belongs to \(E\). Therefore $$ \operatorname{int}([-1,3])=(-1,3). $$
Monotonicity and Finite Intersections
Interior behaves predictably when sets are enlarged or intersected. Enlarging a set cannot destroy a ball that was already contained in it. For intersections, a ball must fit inside both sets; if each set supplies a ball around the same point, the smaller of the two radii works for both.
Proof. Suppose \(x\in\operatorname{int}(E)\). There is an \(r>0\) with \(B_r(x)\subseteq E\). Since \(E\subseteq F\), we also have \(B_r(x)\subseteq F\), so \(x\in\operatorname{int}(F)\). This proves monotonicity.
For the intersection identity, first suppose \(x\in\operatorname{int}(E\cap F)\). Some ball \(B_r(x)\) is contained in \(E\cap F\), hence in each of \(E\) and \(F\). Therefore \(x\in\operatorname{int}(E)\cap\operatorname{int}(F)\).
Conversely, suppose \(x\in\operatorname{int}(E)\cap\operatorname{int}(F)\). Choose \(r,s>0\) such that \(B_r(x)\subseteq E\) and \(B_s(x)\subseteq F\). With \(t=\min\{r,s\}>0\), we have \(B_t(x)\subseteq B_r(x)\) and \(B_t(x)\subseteq B_s(x)\). Consequently \(B_t(x)\subseteq E\cap F\), so \(x\in\operatorname{int}(E\cap F)\). The two inclusions prove the identity, including when either set is empty. \(\square\)
Applying the two-set identity repeatedly gives the corresponding equality for every finite intersection. The same equality does not hold for arbitrary intersections. For example, in \(\mathbb{R}\), let \(U_n=(-1/n,1/n)\) for each positive integer \(n\). Every \(U_n\) is open, so $$ \bigcap_{n=1}^{\infty}\operatorname{int}(U_n) =\bigcap_{n=1}^{\infty}U_n =\{0\}. $$ But \(\operatorname{int}(\{0\})=\varnothing\), since every ball around \(0\) contains nonzero real numbers. Thus the interior of the infinite intersection is empty, although the intersection of the individual interiors is not.
Worked Example: A Union of Two Closed Intervals
In \(\mathbb{R}\), let \(E=[-3,-1]\cup[2,4]\). Each point in \((-3,-1)\cup(2,4)\) has a sufficiently small ball contained in its interval, and therefore in \(E\). For instance, at a point \(x\in(-3,-1)\), choose \(r=\min\{x+3,-1-x\}/2\); then \(B_r(x)\subseteq(-3,-1)\). For \(x\in(2,4)\), take \(r=\min\{x-2,4-x\}/2\), which gives \(B_r(x)\subseteq(2,4)\).
The four endpoints are not interior points. At \(-3\), every ball contains a point less than \(-3\); at \(-1\), every ball contains a point strictly between \(-1\) and \(2\). At \(2\), every ball contains a point strictly between \(-1\) and \(2\), and at \(4\), every ball contains a point greater than \(4\). Each indicated point lies outside \(E\). Points not in \(E\) cannot be interior points because the interior is contained in the set. Hence $$ \operatorname{int}\bigl([-3,-1]\cup[2,4]\bigr) =(-3,-1)\cup(2,4). $$
Interior and Distance from the Complement
The distance-to-a-set construction used for closure also gives a numerical test for interior. Here the relevant set is the complement \(X\setminus E\): an interior point has a positive-sized ball that avoids this complement. When the complement is nonempty, this is equivalent to having positive distance from it.
Proof. Suppose \(x\in\operatorname{int}(E)\). There is an \(r>0\) such that \(B_r(x)\subseteq E\). Every \(y\in X\setminus E\) must satisfy \(d(x,y)\geq r\), since otherwise \(y\in B_r(x)\). Taking the infimum over all such \(y\) gives \(d(x,X\setminus E)\geq r>0\).
Conversely, suppose \(d(x,X\setminus E)=a>0\). For every \(y\in X\setminus E\), the definition of an infimum gives \(d(x,y)\geq a\). Therefore no such \(y\) belongs to \(B_{a/2}(x)\), because \(d(x,y)<a/2\) would contradict \(d(x,y)\geq a\). It follows that \(B_{a/2}(x)\subseteq E\), so \(x\in\operatorname{int}(E)\). \(\square\)
The hypothesis that the complement is nonempty makes the distance in this statement an ordinary finite infimum. If \(E=X\), then every point is interior directly from the definition, and no distance-to-the-empty-set convention is needed.
Worked Example: Interior from Distance to the Complement
Let \(E=(-\infty,0]\cup[2,\infty)\) in \(\mathbb{R}\). Its complement is \((0,2)\). If \(x<0\), then every \(y\in(0,2)\) satisfies \(|x-y|>-x\), while values of \(y\) can be chosen arbitrarily close to \(0\). Hence \(d(x,(0,2))=-x>0\). If \(x>2\), values of \(y\) can be chosen arbitrarily close to \(2\), and every \(y\in(0,2)\) satisfies \(|x-y|>x-2\). Thus \(d(x,(0,2))=x-2>0\).
At \(x=0\), choosing \(y=1/n\) for integers \(n\geq1\) gives \(d(0,(0,2))=0\): the distances \(1/n\) approach zero, and distances are nonnegative. At \(x=2\), choosing \(y=2-1/n\) for integers \(n\geq1\) likewise gives \(d(2,(0,2))=0\). Points strictly between \(0\) and \(2\) are not in \(E\), so they cannot be interior points. The distance characterization therefore yields $$ \operatorname{int}(E)=(-\infty,0)\cup(2,\infty). $$ The endpoints belong to \(E\), but their distance from the complement is zero, which explains why they are excluded from its interior.
The Ambient Space Matters
An interior is always relative to the metric space in which its balls are formed. The metric does not have to change for the answer to change: restricting the space changes which points are available to lie outside a ball-contained set. The Open Sets in a Restricted Metric result from earlier in this course formalizes this relationship between the ambient and subspace viewpoints.
Worked Example: The Same Interval in Two Metric Spaces
Let \(A=[0,1]\), first considered as a subset of \(\mathbb{R}\), and then considered as the entire metric space \(X=A\) with the restricted usual metric. In \(\mathbb{R}\), the interior is \((0,1)\): each \(x\in(0,1)\) has a ball of radius \(\min\{x,1-x\}/2\) contained in \([0,1]\), while every ball around \(0\) contains negative numbers and every ball around \(1\) contains numbers greater than \(1\). Thus neither endpoint is an interior point in \(\mathbb{R}\).
Now take \(X=A\). For every \(x\in X\) and every \(r>0\), the ball \(B_r^X(x)\) consists only of points of \(X=[0,1]\); in particular, \(B_r^X(x)\subseteq A\). Every point is therefore an interior point relative to \(X\), and $$ \operatorname{int}_{\mathbb{R}}([0,1])=(0,1), \qquad \operatorname{int}_{[0,1]}([0,1])=[0,1]. $$ The difference comes from the ambient space, not from a change in the formula for the metric.
Using the Interior Carefully
For a direct calculation, the defining ball condition is often the clearest tool: show that each proposed interior point has a ball inside the set, and rule out every other point. The largest-open-subset theorem is useful when a candidate open set is apparent. Monotonicity and the finite-intersection identity simplify set operations, while distance to the complement can turn the ball condition into a calculation.
Two distinctions prevent common errors. First, belonging to \(E\) does not imply being an interior point; endpoints in the interval examples belong to the sets but admit no ball contained in them. Second, “interior” is not an absolute property of a subset independent of context. Always identify the ambient space and the metric before deciding which balls are being used. In particular, do not apply the finite-intersection identity to an infinite intersection without checking it: the intervals \((-1/n,1/n)\) show that equality can fail.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why must every interior point of \(E\) belong to \(E\)?
- How does the proof that the interior is open use the triangle inequality?
- If \(E\subseteq F\), why must \(\operatorname{int}(E)\subseteq\operatorname{int}(F)\)?
- Why does the finite-intersection identity not automatically extend to arbitrary intersections?
- What does positive distance from \(X\setminus E\) tell you about a ball around the point?
- Why can the same set have different interiors when viewed in two different metric spaces?