What the Boundary Records
The boundary of a set records where points can be approached both from the set and from outside it. It is a closed set, even when the original set is neither open nor closed. Its structure also gives simple tests: a set has empty boundary exactly when it is both open and closed, and for open or closed sets one part of the general boundary formula simplifies.
Throughout, \((X,\mathcal{T})\) is a topological space and \(A\subseteq X\). The closure \(\overline{A}\) is the smallest closed set containing \(A\); the interior \(\operatorname{int}A\) is the largest open set contained in \(A\). A set is clopen if it is both open and closed. A point belongs to \(\overline{A}\) exactly when every open neighborhood of it meets \(A\), and it belongs to \(\operatorname{int}A\) exactly when it has an open neighborhood contained in \(A\).
To verify the equivalent formula, a point belongs to \(\overline{A}\cap\overline{X\setminus A}\) exactly when every open neighborhood meets \(A\) and every open neighborhood meets \(X\setminus A\). That is precisely the defining condition for membership in \(\partial A\). Another useful form is \(\partial A=\overline{A}\setminus\operatorname{int}A\). Indeed, a point is outside \(\operatorname{int}A\) exactly when every open neighborhood fails to be contained in \(A\), or equivalently meets \(X\setminus A\). Thus belonging to \(\overline{A}\) and being outside \(\operatorname{int}A\) means meeting both sides in every neighborhood.
Every Boundary Is Closed
Proof. From the equivalent formula, \(\partial A=\overline{A}\cap\overline{X\setminus A}\). A closure is closed, so both sets in this intersection are closed. An intersection of closed sets is closed: its complement is the union of their open complements, which is open. Therefore \(\partial A\) is closed. This argument also covers \(A=\varnothing\) and \(A=X\), since the closure of the empty set is empty and the closure of \(X\) is \(X\). \(\square\)
The closedness belongs to the boundary, not necessarily to the set whose boundary is being taken. The formula also identifies what can lie on the boundary: a point in the interior of \(A\) cannot be there, since it has a neighborhood contained in \(A\); a point in the exterior cannot be there, since it has a neighborhood missing \(A\). Every other point is on the boundary.
Worked Example: An Open Interval in the Real Line
Use the standard topology on \(\mathbb{R}\), and let \(A=(0,2)\). Since \(A\) is open, \(\operatorname{int}A=A\). Its closure is \([0,2]\): every neighborhood of a point in \([0,2]\) meets \((0,2)\), while a point outside \([0,2]\) has a small open interval around it disjoint from \((0,2)\). Hence \(\partial A=\overline{A}\setminus\operatorname{int}A=[0,2]\setminus(0,2)=\{0,2\}\).
At \(0\), every interval \((-\varepsilon,\varepsilon)\), with \(\varepsilon>0\), contains a point of \(A\), for example \(\min(\varepsilon,1)\!/2\), and contains points outside \(A\), such as \(-\varepsilon/2\). At \(2\), every neighborhood also meets both sets: it contains \(2-\min(\varepsilon,1)\!/2\in A\) and \(2+\varepsilon/2\notin A\). If \(x\in(0,2)\), an interval of radius smaller than \(\min(x,2-x)\) around \(x\) lies entirely in \(A\), so \(x\notin\partial A\). Points outside \([0,2]\) have neighborhoods missing \(A\), so they are not boundary points either.
Check: \(\{0,2\}\) is closed in \(\mathbb{R}\), since its complement \((-\infty,0)\cup(0,2)\cup(2,\infty)\) is open. The computed boundary therefore agrees with the general theorem.
When Is the Boundary Empty?
Proof. Suppose first that \(\partial A=\varnothing\). The formula \(\partial A=\overline{A}\setminus\operatorname{int}A\) then gives \(\overline{A}\subseteq\operatorname{int}A\). Since \(\operatorname{int}A\subseteq A\subseteq\overline{A}\), all three sets are equal. In particular \(A=\operatorname{int}A\), so \(A\) is open, and \(A=\overline{A}\), so \(A\) is closed. Conversely, suppose \(A\) is both open and closed. Then \(\operatorname{int}A=A\) and \(\overline{A}=A\), so \(\partial A=\overline{A}\setminus\operatorname{int}A=A\setminus A=\varnothing\). The argument includes \(A=\varnothing\) and \(A=X\), which are clopen in every topology. \(\square\)
This equivalence concerns the boundary in the ambient space \(X\). It is not enough that a set be open or closed relative to some smaller subspace; the topology used must be specified. The criterion is especially useful when the boundary is easier to characterize than the open and closed sets themselves, or when one wants to know whether a subset separates into two topologically distinct parts.
Worked Example: A Nontrivial Boundary in the Sierpinski Topology
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and take \(A=\{a\}\). The closed sets are the complements of the open sets, namely \(\varnothing,\{b\},X\). The only closed set containing \(A\) is \(X\), so \(\overline{A}=X\). The set \(A\) is open, hence \(\operatorname{int}A=A\). Therefore \(\partial A=\overline{A}\setminus\operatorname{int}A=X\setminus\{a\}=\{b\}\).
The neighborhood definition checks the result directly. The only open neighborhood of \(b\) is \(X\), which meets \(A\) at \(a\) and \(X\setminus A\) at \(b\); thus \(b\in\partial A\). The open neighborhood \(\{a\}\) of \(a\) misses \(X\setminus A\), so \(a\notin\partial A\). Since \(X\) has no other points, \(\partial A=\{b\}\). The singleton \(\{b\}\) is closed in this topology, as its complement \(\{a\}\) is open. Also, \(A\) is not closed, so the empty-boundary criterion predicts that its boundary is nonempty.
Formulas for Open and Closed Sets
Proof. If \(A\) is open, then \(\operatorname{int}A=A\). Substitution into \(\partial A=\overline{A}\setminus\operatorname{int}A\) gives \(\partial A=\overline{A}\setminus A\). If \(A\) is closed, then \(\overline{A}=A\). Substituting this into the same formula gives \(\partial A=A\setminus\operatorname{int}A\). Both conclusions follow directly from the definition of interior and closure; neither requires \(A\) to be both open and closed. \(\square\)
For an open set, the boundary consists of the points added when taking its closure. For a closed set, the boundary consists of its points that are not interior points. These descriptions are compatible: if a set is clopen, both formulas give the empty set. They also make explicit that a boundary need not be contained in the original set: the boundary of an open set can lie entirely outside it.
Worked Example: A Closed Interval in the Real Line
Again use the standard topology on \(\mathbb{R}\), now with \(C=[0,2]\). This set is closed because its complement \((-\infty,0)\cup(2,\infty)\) is open. Its interior is \((0,2)\): each point strictly between \(0\) and \(2\) has a small interval contained in \(C\), while every neighborhood of either endpoint contains points outside \(C\). The closed-set formula gives \(\partial C=C\setminus\operatorname{int}C=[0,2]\setminus(0,2)=\{0,2\}\).
To check using neighborhoods, every neighborhood of \(0\) or \(2\) meets \(C\) and its complement, so both endpoints are boundary points. A point \(x\in(0,2)\) has a neighborhood contained in \(C\), and a point outside \([0,2]\) has a neighborhood disjoint from \(C\). Neither kind of point is on the boundary. Thus there are no omitted boundary points.
Check: \(C\setminus\operatorname{int}C\) removes exactly the interior points and leaves the two endpoints. The answer matches the direct neighborhood test and is closed.
Worked Example: Empty Boundary in Discrete and Indiscrete Spaces
Let \(X=\{p,q\}\). First give \(X\) the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). For \(A=\{p\}\), the set \(A\) is both open and closed, so \(\overline{A}=A\) and \(\operatorname{int}A=A\). Therefore \(\partial A=A\setminus A=\varnothing\). Directly, each point has its singleton as an open neighborhood; \(\{p\}\) has a neighborhood missing the complement, and \(\{q\}\) has a neighborhood missing \(A\). Neither point is boundary. In fact every subset of a discrete space is clopen, so every subset has empty boundary.
Now give the same underlying set the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\), and again take \(A=\{p\}\). The only closed sets are also \(\varnothing,X\). Thus the smallest closed superset of \(A\) is \(X\), so \(\overline{A}=X\). The only open subset of \(A\) is \(\varnothing\), so \(\operatorname{int}A=\varnothing\). It follows that \(\partial A=X\setminus\varnothing=X\). The neighborhood check gives the same result: the only open neighborhood of either point is \(X\), and \(X\) meets both \(A=\{p\}\) and \(X\setminus A=\{q\}\).
Check: in the discrete case, \(\partial\{p\}=\varnothing\) because \(\{p\}\) is clopen. In the indiscrete case, \(\{p\}\) is neither open nor closed, and its boundary is nonempty. These calculations illustrate both directions of the empty-boundary criterion.
A Common Pitfall
A frequent error is to conclude that the boundary of an open set is contained in the set. For \(A=(0,2)\subseteq\mathbb{R}\), the incorrect calculation might be \(\partial A=\overline{A}\cap A=[0,2]\cap(0,2)=(0,2)\). This uses the wrong operation: boundary points must be approachable from both \(A\) and its complement, so the formula is \(\overline{A}\cap\overline{X\setminus A}\), not \(\overline{A}\cap A\). Equivalently, because \(A\) is open, \(\partial A=\overline{A}\setminus A=[0,2]\setminus(0,2)=\{0,2\}\). The neighborhood test confirms the correction: points inside \(A\) have neighborhoods contained in \(A\), while the endpoints have every neighborhood meeting both sides.
The main structural facts can be kept together: boundaries are always closed; an empty boundary is equivalent to being clopen; and openness or closedness allows one to replace the general formula by a single set difference. In applying these statements, first identify the topology and the relevant hypothesis—open, closed, or neither—then use the corresponding formula and, when needed, confirm the result from neighborhoods.
Check Your Understanding
For each question, specify the topology and justify the conclusion using the boundary formulas or the neighborhood definition.
- In the standard topology on \(\mathbb{R}\), find the boundary of the open set \((1,4)\).
- Prove directly from the definition that \(\partial\varnothing=\varnothing\) and \(\partial X=\varnothing\).
- Suppose \(\partial A=\varnothing\). Explain why \(\overline{A}=\operatorname{int}A\), and deduce that \(A\) is clopen.
- In the indiscrete topology on a two-point set \(X=\{p,q\}\), compute the boundary of \(\{q\}\).
- Give an example of a closed set whose boundary is nonempty, and verify the boundary using both the closed-set formula and neighborhoods.