The Central Idea
Given a subset \(A\) of a topological space \(X\), we can classify points according to their local relationship with \(A\).
- The interior contains points that lie locally inside \(A\).
- The exterior contains points that lie locally outside \(A\).
- The boundary contains points whose neighborhoods meet both \(A\) and its complement.
- The closure contains \(A\) together with all points that cannot be separated from \(A\) by an open neighborhood.
These four constructions reveal the local structure of a set.
The Ambient Space Matters
Every statement about interior, exterior, closure, or boundary is made relative to an ambient topological space.
If
\[ A\subseteq X, \]then all neighborhoods, complements, and boundaries are interpreted inside \(X\).
Changing \(X\) can change all four constructions.
Interior Points
A point \(x\in A\) is an interior point of \(A\) if there is an open set \(U\) such that
\[ x\in U\subseteq A. \]In a metric space, this means that some open ball around \(x\) fits entirely inside \(A\):
\[ B_r(x)\subseteq A \]for some \(r>0\).
The Interior of a Set
The set of all interior points of \(A\) is called the interior of \(A\).
It is written
\[ \operatorname{int}(A) \]or sometimes
\[ A^\circ. \]Thus,
\[ \boxed{ \operatorname{int}(A) = \{x\in A:x\text{ has an open neighborhood contained in }A\} }. \]The Interior Is the Largest Open Set Inside \(A\)
The interior has an important structural interpretation:
\[ \boxed{\operatorname{int}(A)\text{ is the largest open subset of }A}. \]Every open set contained in \(A\) is contained in \(\operatorname{int}(A)\).
This makes the interior the complete open portion of the set.
Example: Interior of a Closed Interval
Let
\[ A=[0,5]\subseteq\mathbb{R}. \]Every point strictly between \(0\) and \(5\) has a small open interval contained inside \(A\).
The endpoints do not.
Therefore,
\[ \boxed{\operatorname{int}([0,5])=(0,5)}. \]Example: Interior of an Open Interval
Let
\[ A=(2,8). \]Every point of \(A\) is already an interior point.
Therefore,
\[ \boxed{\operatorname{int}((2,8))=(2,8)}. \]A set is open exactly when it equals its interior.
\[ \boxed{A\text{ is open}\iff \operatorname{int}(A)=A}. \]Exterior Points
A point \(x\) is an exterior point of \(A\) if it has an open neighborhood lying entirely outside \(A\).
Since the points outside \(A\) belong to the complement
\[ A^c=X\setminus A, \]the exterior of \(A\) is simply the interior of its complement.
\[ \boxed{\operatorname{ext}(A)=\operatorname{int}(A^c)}. \]The Exterior of a Set
The exterior of \(A\) is written
\[ \operatorname{ext}(A). \]It consists of points that have some neighborhood avoiding \(A\) completely.
In a metric space,
\[ x\in\operatorname{ext}(A) \]means that for some \(r>0\),
\[ B_r(x)\cap A=\varnothing. \]Example: Exterior of an Interval
Let
\[ A=[1,4]\subseteq\mathbb{R}. \]Points less than \(1\) and greater than \(4\) have neighborhoods that remain outside \(A\).
Therefore,
\[ \boxed{ \operatorname{ext}([1,4]) = (-\infty,1)\cup(4,\infty) }. \]Boundary Points
A point \(x\) is a boundary point of \(A\) if every open neighborhood of \(x\) intersects both \(A\) and its complement.
In a metric space, every ball around \(x\) satisfies
\[ B_r(x)\cap A\ne\varnothing \]and
\[ B_r(x)\cap A^c\ne\varnothing. \]The set of all boundary points is written
\[ \partial A. \]The Boundary Separates Local Inside from Local Outside
Interior points have neighborhoods contained in \(A\).
Exterior points have neighborhoods contained in \(A^c\).
Boundary points have neither kind of protected neighborhood.
Every neighborhood around a boundary point sees both sides.
Example: Boundary of an Interval
Let
\[ A=(0,1). \]The points \(0\) and \(1\) do not belong to \(A\), but every neighborhood around either point contains points from \(A\) and points outside \(A\).
Therefore,
\[ \boxed{\partial(0,1)=\{0,1\}}. \]Boundary Points Need Not Belong to the Set
The boundary of a set is not necessarily contained in the set.
For example,
\[ \partial(0,1)=\{0,1\}, \]but
\[ 0\notin(0,1) \]and
\[ 1\notin(0,1). \]Boundary membership and set membership are separate questions.
Boundary Points Can Belong to the Set
For the closed interval
\[ [0,1], \]the boundary is still
\[ \{0,1\}. \]This time both boundary points are members of the set.
\[ \boxed{\partial[0,1]=\{0,1\}}. \]Open and Closed Intervals Can Have the Same Boundary
Compare
\[ (0,1) \]and
\[ [0,1]. \]Their interiors differ only in whether the endpoints were included in the original set, but their boundaries are identical:
\[ \partial(0,1)=\partial[0,1]=\{0,1\}. \]Boundary describes local transition behavior, not merely membership.
The Closure of a Set
The closure of \(A\) is the smallest closed set containing \(A\).
It is written
\[ \overline{A} \]or
\[ \operatorname{cl}(A). \]It contains all points of \(A\) together with all points that are forced by the limiting behavior of \(A\).
Neighborhood Definition of Closure
A point \(x\) belongs to the closure of \(A\) if every open neighborhood of \(x\) intersects \(A\).
Thus,
\[ \boxed{ x\in\overline{A} \iff U\cap A\ne\varnothing \text{ for every neighborhood }U\text{ of }x }. \]In a metric space, equivalently,
\[ B_r(x)\cap A\ne\varnothing \]for every \(r>0\).
The Closure Contains the Set
Every point of \(A\) automatically belongs to \(\overline{A}\).
Therefore,
\[ \boxed{A\subseteq\overline{A}}. \]The closure may equal \(A\), or it may add new boundary or limit points.
Example: Closure of an Open Interval
Let
\[ A=(2,6). \]Every neighborhood of \(2\) intersects \(A\), and every neighborhood of \(6\) intersects \(A\).
Therefore both endpoints belong to the closure:
\[ \boxed{\overline{(2,6)}=[2,6]}. \]Example: Closure of a Closed Interval
Let
\[ A=[2,6]. \]This set already contains all of its boundary and limit points.
Therefore,
\[ \boxed{\overline{[2,6]}=[2,6]}. \]A set is closed exactly when it equals its closure.
\[ \boxed{A\text{ is closed}\iff\overline{A}=A}. \]Closure and Limit Points
In many familiar spaces, the closure can be understood as
\[ \boxed{ \overline{A} = A\cup \{\text{limit points of }A\} }. \]This description is especially useful in metric spaces.
Example: A Sequence Set
Consider
\[ A= \left\{ 1,\frac12,\frac13,\frac14,\ldots \right\}. \]The point \(0\) is not in \(A\), but every neighborhood of \(0\) contains terms \(1/n\) for sufficiently large \(n\).
Therefore,
\[ \boxed{ \overline{A} = A\cup\{0\} }. \]Closure and Boundary
The closure of a set consists of its interior together with its boundary.
\[ \boxed{ \overline{A} = \operatorname{int}(A)\cup\partial A }. \]These two parts are disjoint:
\[ \operatorname{int}(A)\cap\partial A=\varnothing. \]The Boundary from Closure and Interior
Because the closure consists of interior points plus boundary points,
\[ \boxed{ \partial A = \overline{A}\setminus\operatorname{int}(A) }. \]This is often a convenient computational formula.
The Boundary from Two Closures
A boundary point is simultaneously close to \(A\) and to its complement.
Therefore,
\[ \boxed{ \partial A = \overline{A}\cap\overline{A^c} }. \]This formula captures the idea that boundary points cannot be separated locally from either side.
The Boundary Is Shared by a Set and Its Complement
Since the same transition separates \(A\) from \(A^c\),
\[ \boxed{\partial A=\partial(A^c)}. \]The boundary does not belong uniquely to one side.
Interior and Exterior Are Disjoint
No point can have one neighborhood contained in \(A\) and another neighborhood contained in \(A^c\) while belonging locally to both sides.
Therefore,
\[ \boxed{ \operatorname{int}(A) \cap \operatorname{ext}(A) = \varnothing }. \]Interior, Exterior, and Boundary Partition the Space
Every point of the ambient space \(X\) belongs to exactly one of three regions:
- the interior of \(A\);
- the exterior of \(A\);
- the boundary of \(A\).
Thus,
\[ \boxed{ X = \operatorname{int}(A) \cup \partial A \cup \operatorname{ext}(A) }. \]These three sets are pairwise disjoint.
The Complement of the Closure
A point lies outside the closure of \(A\) exactly when it has a neighborhood avoiding \(A\).
That is precisely the definition of exterior.
\[ \boxed{ X\setminus\overline{A} = \operatorname{ext}(A) }. \]Equivalently,
\[ \boxed{ \overline{A} = X\setminus\operatorname{ext}(A) }. \]The Complement of the Interior
The complement of the interior of \(A\) is the closure of the complement:
\[ \boxed{ X\setminus\operatorname{int}(A) = \overline{A^c} }. \]Likewise,
\[ \boxed{ X\setminus\overline{A} = \operatorname{int}(A^c) }. \]These are topological versions of complement duality.
Interior of a Complement
Since exterior is defined through the complement,
\[ \boxed{ \operatorname{ext}(A) = \operatorname{int}(A^c) }. \]Similarly,
\[ \operatorname{ext}(A^c) = \operatorname{int}(A). \]Taking complements swaps the roles of inside and outside while preserving the boundary.
Interior of the Empty Set
The empty set contains no points, so it contains no interior points.
\[ \boxed{ \operatorname{int}(\varnothing)=\varnothing }. \]Its closure is also empty:
\[ \boxed{ \overline{\varnothing}=\varnothing }. \]Its boundary is therefore empty as well.
Interior of the Whole Space
The whole space \(X\) is open in itself.
Therefore,
\[ \boxed{\operatorname{int}(X)=X}. \]Since \(X\) is also closed,
\[ \boxed{\overline{X}=X}. \]Its boundary is empty:
\[ \boxed{\partial X=\varnothing}. \]Interior and Closure Are Idempotent
Once the interior has been taken, taking it again changes nothing:
\[ \boxed{ \operatorname{int}(\operatorname{int}(A)) = \operatorname{int}(A) }. \]Likewise,
\[ \boxed{ \overline{\overline{A}} = \overline{A} }. \]These operations stabilize after one application.
Interior Is Monotone
If
\[ A\subseteq B, \]then every open set contained in \(A\) is also contained in \(B\).
Therefore,
\[ \boxed{ \operatorname{int}(A) \subseteq \operatorname{int}(B) }. \]Closure Is Monotone
If
\[ A\subseteq B, \]then every point forced into the closure of \(A\) is also forced into the closure of \(B\).
Thus,
\[ \boxed{ \overline{A} \subseteq \overline{B} }. \]Interior of an Intersection
Interior behaves particularly well with finite intersections.
\[ \boxed{ \operatorname{int}(A\cap B) = \operatorname{int}(A)\cap\operatorname{int}(B) }. \]A point is safely inside both sets exactly when it is safely inside their intersection.
Interior of a Union
We always have
\[ \operatorname{int}(A)\cup\operatorname{int}(B) \subseteq \operatorname{int}(A\cup B). \]Equality need not hold.
Two sets can fit together so that their union gains interior points that neither set had separately.
Closure of a Union
Closure behaves particularly well with finite unions:
\[ \boxed{ \overline{A\cup B} = \overline{A}\cup\overline{B} }. \]Every point adherent to either set is adherent to their union.
Closure of an Intersection
We always have
\[ \overline{A\cap B} \subseteq \overline{A}\cap\overline{B}. \]Equality need not hold.
A point may lie arbitrarily close to both sets without lying arbitrarily close to points belonging to their intersection.
Example in the Plane: Open Disk
Let
\[ A= \{(x,y):x^2+y^2<1\}. \]This is the open unit disk.
Its interior is itself:
\[ \operatorname{int}(A)=A. \]Its boundary is the unit circle:
\[ \partial A= \{(x,y):x^2+y^2=1\}. \]Its closure is the closed unit disk:
\[ \overline{A} = \{(x,y):x^2+y^2\le1\}. \]Example in the Plane: Closed Disk
Let
\[ F= \{(x,y):x^2+y^2\le1\}. \]Then
\[ \operatorname{int}(F) = \{(x,y):x^2+y^2<1\}, \] \[ \partial F = \{(x,y):x^2+y^2=1\}, \]and
\[ \boxed{\overline{F}=F}. \]Example: A Circle Has Empty Interior in the Plane
Consider the unit circle
\[ S^1= \{(x,y):x^2+y^2=1\}. \]As a subset of \(\mathbb{R}^2\), every open disk around every point of the circle contains points not on the circle.
Therefore,
\[ \boxed{\operatorname{int}_{\mathbb{R}^2}(S^1)=\varnothing}. \]The circle is closed, so
\[ \overline{S^1}=S^1. \]Hence,
\[ \boxed{\partial S^1=S^1}. \]Relative Interior Changes with the Ambient Space
The previous example changes if the circle itself is taken as the ambient space.
If
\[ X=S^1, \]then \(S^1\) is the whole space.
Therefore,
\[ \operatorname{int}_{S^1}(S^1)=S^1. \]The same underlying set has empty interior in \(\mathbb{R}^2\) but full interior in itself.
Subspace Example on an Interval
Let
\[ X=[0,3] \]with the topology inherited from \(\mathbb{R}\), and let
\[ A=[0,1). \]Relative to \(X\), the point \(0\) is an interior point of \(A\).
Indeed,
\[ [0,1)=X\cap(-1,1). \]Therefore,
\[ \operatorname{int}_X(A)=[0,1). \]Relative to \(\mathbb{R}\), however,
\[ \operatorname{int}_{\mathbb{R}}(A)=(0,1). \]Dense Sets and Closure
A subset \(A\) of \(X\) is dense in \(X\) when
\[ \overline{A}=X. \]This means every nonempty open region of \(X\) contains points of \(A\).
Dense sets may still omit many points.
Example: The Rational Numbers Are Dense
In the usual topology on \(\mathbb{R}\),
\[ \mathbb{Q} \]is dense because every open interval contains rational numbers.
Therefore,
\[ \boxed{\overline{\mathbb{Q}}=\mathbb{R}}. \]Yet
\[ \operatorname{int}(\mathbb{Q})=\varnothing, \]because every open interval also contains irrational numbers.
The Irrational Numbers Are Also Dense
Let
\[ \mathbb{R}\setminus\mathbb{Q} \]denote the irrational numbers.
Every open interval contains irrational numbers, so
\[ \boxed{ \overline{\mathbb{R}\setminus\mathbb{Q}} = \mathbb{R} }. \]Thus two disjoint sets can both be dense in the same space.
Boundary of a Dense Set with Empty Interior
If
\[ \overline{A}=X \]and
\[ \operatorname{int}(A)=\varnothing, \]then
\[ \partial A = \overline{A}\setminus\operatorname{int}(A) = X. \]Therefore the rational numbers satisfy
\[ \boxed{\partial\mathbb{Q}=\mathbb{R}}. \]Isolated Points
A point \(x\in A\) is an isolated point of \(A\) if some neighborhood of \(x\) contains no other points of \(A\).
Isolated points can belong to the closure even when they are not limit points.
This is why the closure is more accurately described as
\[ A\cup\{\text{limit points of }A\} \]rather than merely as the set of limit points.
Example: The Integers
In \(\mathbb{R}\),
\[ \mathbb{Z} \]has no interior points.
Each integer is isolated from the others, but \(\mathbb{Z}\) is closed.
Therefore,
\[ \operatorname{int}(\mathbb{Z})=\varnothing, \] \[ \overline{\mathbb{Z}}=\mathbb{Z}, \]and
\[ \boxed{\partial\mathbb{Z}=\mathbb{Z}}. \]Open Sets Have No Boundary Points Inside Their Exterior
If \(A\) is open, then
\[ A=\operatorname{int}(A). \]Consequently,
\[ \partial A=\overline{A}\setminus A. \]For an open set, the boundary consists precisely of the extra points added by taking the closure.
Closed Sets Contain Their Boundary
If \(A\) is closed, then
\[ \overline{A}=A. \]Therefore,
\[ \partial A = A\setminus\operatorname{int}(A). \]So every boundary point of a closed set belongs to the set.
A Set Is Open Exactly When It Avoids Its Boundary
A set \(A\) is open if and only if
\[ A\cap\partial A=\varnothing. \]In that case, every point of \(A\) lies in its interior.
A Set Is Closed Exactly When It Contains Its Boundary
A set \(A\) is closed if and only if
\[ \boxed{\partial A\subseteq A}. \]This gives another useful test for closedness.
Common Error: Treating Interior as "Everything Except Endpoints"
That shortcut works for some intervals but not in general.
A set may have infinitely many boundary points, no obvious endpoints, or empty interior.
Interior must be determined from neighborhoods.
Common Error: Assuming the Boundary Must Be Inside the Set
Boundary points may belong to \(A\), belong to \(A^c\), or include examples of both.
For
\[ (0,1), \]neither boundary point belongs to the set.
For
\[ [0,1], \]both do.
Common Error: Assuming Closure Means "Add the Endpoints"
Adding endpoints is only a special interval example.
In general, closure adds every point whose every neighborhood intersects the set.
A dense set such as \(\mathbb{Q}\) has
\[ \overline{\mathbb{Q}}=\mathbb{R}, \]so its closure adds far more than two endpoints.
Common Error: Confusing Exterior with Complement
The exterior is not always the entire complement.
Instead,
\[ \operatorname{ext}(A)=\operatorname{int}(A^c). \]Boundary points may lie in the complement without belonging to the exterior.
For example, if
\[ A=(0,1), \]then \(0\in A^c\), but \(0\notin\operatorname{ext}(A)\) because every neighborhood of \(0\) meets \(A\).
Common Error: Forgetting the Ambient Space
The same set can have different interiors and boundaries in different ambient spaces.
Always identify the space first.
For example,
\[ [0,1] \]has interior \((0,1)\) in \(\mathbb{R}\), but has interior \([0,1]\) when \([0,1]\) itself is the entire space.
A Reliable Classification Procedure
- Identify the ambient topological space \(X\).
- Choose a point and examine its neighborhoods.
- If some neighborhood lies inside \(A\), classify the point as interior.
- If some neighborhood lies inside \(A^c\), classify the point as exterior.
- If every neighborhood meets both \(A\) and \(A^c\), classify the point as boundary.
- Collect all points whose every neighborhood meets \(A\) to obtain the closure.
A Fast Interval Procedure on \(\mathbb{R}\)
For ordinary intervals on the real line:
- Remove exposed endpoints to find the interior.
- Include limiting endpoints to find the closure.
- The transition points form the boundary.
- The remaining open regions form the exterior.
This shortcut is useful for intervals, but the neighborhood definitions are the general method.
Example: Analyze \([1,4)\)
Let
\[ A=[1,4). \]The interior is
\[ \boxed{\operatorname{int}(A)=(1,4)}. \]The closure is
\[ \boxed{\overline{A}=[1,4]}. \]The boundary is
\[ \boxed{\partial A=\{1,4\}}. \]The exterior is
\[ \boxed{ \operatorname{ext}(A) = (-\infty,1)\cup(4,\infty) }. \]Example: Analyze a Finite Set
Let
\[ A=\{1,3,5\}\subseteq\mathbb{R}. \]No point of \(A\) contains an open interval lying entirely in \(A\), so
\[ \operatorname{int}(A)=\varnothing. \]The set is closed, so
\[ \overline{A}=A. \]Therefore,
\[ \boxed{\partial A=A}. \]Its exterior is
\[ \boxed{\operatorname{ext}(A)=\mathbb{R}\setminus A}. \]Example: Analyze the Rational Numbers
Let
\[ A=\mathbb{Q}\subseteq\mathbb{R}. \]Every open interval contains both rational and irrational numbers.
Therefore,
\[ \operatorname{int}(\mathbb{Q})=\varnothing, \] \[ \operatorname{ext}(\mathbb{Q})=\varnothing, \] \[ \overline{\mathbb{Q}}=\mathbb{R}, \]and
\[ \boxed{\partial\mathbb{Q}=\mathbb{R}}. \]Practice 1: Interior
Find
\[ \operatorname{int}([-4,7]). \]Practice 2: Closure
Find
\[ \overline{(-2,5)}. \]Practice 3: Boundary
Find
\[ \partial[-2,5). \]Practice 4: Exterior
Find the exterior of
\[ A=[0,3] \]in \(\mathbb{R}\).
Practice 5: Finite Set
Let
\[ A=\{-1,2,6\}. \]Find its interior, closure, and boundary in \(\mathbb{R}\).
Practice 6: Sequence Set
Let
\[ A= \left\{ 1,\frac12,\frac13,\ldots \right\}. \]Find \(\overline{A}\).
Practice 7: Open Disk
Let
\[ A=\{(x,y):x^2+y^2<9\}. \]Find its boundary and closure in \(\mathbb{R}^2\).
Practice 8: Whole Space
Let \(A=X\).
Find
\[ \operatorname{int}(A),\qquad \operatorname{ext}(A),\qquad \partial A,\qquad \overline{A}. \]Practice 9: Complement Relationship
Complete the identity:
\[ \operatorname{ext}(A)=\boxed{\phantom{\operatorname{int}(A^c)}}. \]Practice 10: Partition of the Space
Complete:
\[ X= \operatorname{int}(A) \cup \boxed{\phantom{\partial A}} \cup \operatorname{ext}(A). \]Practice Answers
Practice 1
\[ \boxed{\operatorname{int}([-4,7])=(-4,7)}. \]Practice 2
\[ \boxed{\overline{(-2,5)}=[-2,5]}. \]Practice 3
Both endpoints are boundary points:
\[ \boxed{\partial[-2,5)=\{-2,5\}}. \]Practice 4
\[ \boxed{ \operatorname{ext}([0,3]) = (-\infty,0)\cup(3,\infty) }. \]Practice 5
A finite subset of \(\mathbb{R}\) has empty interior and is closed.
\[ \boxed{\operatorname{int}(A)=\varnothing}, \] \[ \boxed{\overline{A}=A}, \] \[ \boxed{\partial A=A}. \]Practice 6
The point \(0\) is the missing limit point.
\[ \boxed{ \overline{A} = A\cup\{0\} }. \]Practice 7
The boundary is the circle of radius \(3\):
\[ \boxed{ \partial A= \{(x,y):x^2+y^2=9\} }. \]The closure includes that circle:
\[ \boxed{ \overline{A} = \{(x,y):x^2+y^2\le9\} }. \]Practice 8
\[ \boxed{\operatorname{int}(X)=X}, \] \[ \boxed{\operatorname{ext}(X)=\varnothing}, \] \[ \boxed{\partial X=\varnothing}, \] \[ \boxed{\overline{X}=X}. \]Practice 9
\[ \boxed{ \operatorname{ext}(A)=\operatorname{int}(A^c) }. \]Practice 10
\[ \boxed{ X= \operatorname{int}(A) \cup \partial A \cup \operatorname{ext}(A) }. \]What to Remember
- The interior consists of points with neighborhoods contained entirely in the set.
- The exterior is the interior of the complement.
- Boundary points have every neighborhood meeting both the set and its complement.
- The closure consists of points whose every neighborhood meets the set.
- The interior is the largest open subset of a set.
- The closure is the smallest closed set containing a set.
- \(\overline{A}=\operatorname{int}(A)\cup\partial A\).
- \(\partial A=\overline{A}\setminus\operatorname{int}(A)\).
- \(\partial A=\partial(A^c)\).
- Interior, boundary, and exterior partition the ambient space.
The key structural picture is:
\[ \boxed{ \text{inside} \quad|\quad \text{boundary} \quad|\quad \text{outside} }. \]Topology makes each part precise through neighborhoods rather than visual appearance alone.
Where We Go Next
Interior, exterior, closure, and boundary describe how a single set sits inside its surrounding space.
The next stages of topology use these ideas to study neighborhoods, limit points, connectedness, compactness, continuity, and qualitative structure. Those concepts shift attention from individual sets toward the behavior of spaces and functions as a whole.