Computing from the Open-Set List
In a finite topological space, the topology is often given by listing every open set. That list contains all the information needed to calculate the interior, closure, exterior, and boundary of any subset. The key is to use the definitions systematically: collect open subsets for the interior, and use closed supersets or neighborhoods for the closure.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). A set is closed if its complement in \(X\) is open. The interior of \(A\) is the largest open subset of \(A\); the closure of \(A\) is the smallest closed set containing \(A\). The exterior of \(A\) is the interior of its complement, and the boundary consists of points whose every open neighborhood meets both \(A\) and \(X\setminus A\).
For a finite open-set list, the interior calculation is direct: retain each listed open set that is a subset of \(A\), then take their union. For closure, first take complements of the listed open sets to obtain the closed sets, retain the closed sets that contain \(A\), then intersect them. The empty set and \(X\) are always open and closed, respectively, so they must be included in the relevant lists.
Proof. A point \(x\) is outside \(\overline{A}\) exactly when some open neighborhood of \(x\) misses \(A\): if such a neighborhood is \(U\), then the closed set \(X\setminus U\) contains \(A\) and misses \(x\); conversely, if \(x\notin\overline{A}\), the complement of the closed set \(\overline{A}\) is an open neighborhood of \(x\) missing \(A\). An open neighborhood misses \(A\) exactly when it is contained in \(X\setminus A\). Thus \(x\notin\overline{A}\) exactly when \(x\in\operatorname{int}(X\setminus A)\), proving the first identity. A point is not in the boundary precisely when it has a neighborhood contained in \(A\), or a neighborhood contained in \(X\setminus A\). These alternatives describe \(\operatorname{int}A\) and \(\operatorname{int}(X\setminus A)\); using the first identity, the points in neither interior are exactly \(\overline{A}\setminus\operatorname{int}A\), and also exactly \(\overline{A}\cap\overline{X\setminus A}\). \(\square\)
Discrete and Indiscrete Topologies
The discrete topology on \(X\) is \(\mathcal{P}(X)\), the collection of all subsets of \(X\). The indiscrete topology is \(\{\varnothing,X\}\). These two cases give useful reference points for computations in more complicated finite spaces.
Worked Example: A Discrete Space
Let \(X=\{a,b,c\}\) with the discrete topology, and take \(A=\{a,c\}\). Since every subset is open, \(A\) itself is open, and every subset of \(A\) is among the open sets used to form its interior. Their union is \(A\), so \(\operatorname{int}A=\{a,c\}\).
Every subset is also closed, because its complement is open. In particular, \(A\) is a closed set containing itself. Intersecting all closed supersets of \(A\) gives \(A\), since \(A\) is one of those supersets and every one contains it. Hence \(\overline{A}=\{a,c\}\). The complement is \(X\setminus A=\{b\}\), which is open, so the exterior is \(\operatorname{int}(X\setminus A)=\{b\}\). Therefore \(\partial A=\overline{A}\setminus\operatorname{int}A=\varnothing\).
Check: Each point has the open neighborhood consisting of that point alone. For \(a,c\), this neighborhood is contained in \(A\); for \(b\), it is contained in \(X\setminus A\). No point has every neighborhood meeting both sides, confirming the empty boundary.
Worked Example: An Indiscrete Space
Let \(X=\{p,q,r\}\) have the indiscrete topology \(\{\varnothing,X\}\), and let \(A=\{p\}\). The only open set contained in \(A\) is \(\varnothing\), so \(\operatorname{int}A=\varnothing\). The only closed sets are also \(\varnothing\) and \(X\). Of these, only \(X\) contains \(A\), so \(\overline{A}=X\). The complement \(X\setminus A=\{q,r\}\) is nonempty and proper, and its only open subset is again \(\varnothing\). Thus its interior, the exterior of \(A\), is empty. It follows that \(\partial A=X\).
Check: The only open neighborhood of any point is \(X\). It meets \(A\) at \(p\) and meets \(X\setminus A\) at \(q\) and \(r\). Therefore every point lies in the boundary, as calculated.
The empty and whole-set cases are different: \(\operatorname{int}\varnothing=\overline{\varnothing}=\varnothing\), while \(\operatorname{int}X=\overline{X}=X\). For the closure of \(\varnothing\), every closed set contains \(\varnothing\), including \(\varnothing\) itself, so the intersection of all closed supersets is \(\varnothing\). The exterior of \(\varnothing\) is \(X\), and the exterior of \(X\) is \(\varnothing\); both boundaries are empty.
A Finite Non-Hausdorff Example
In a finite topology, not every subset need be open or closed. To calculate reliably, write down the open sets and their complements before deciding whether a particular subset contributes to an interior or a closure.
Worked Example: A Three-Point Topology
Let \(X=\{a,b,c\}\) and \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\). This is a topology: it contains \(\varnothing\) and \(X\); intersections of listed sets are listed; and every union of listed sets is listed. Its closed sets, obtained by taking complements, are \(X,\{b,c\},\{c\},\varnothing\).
First take \(A=\{b\}\). Among the open sets, only \(\varnothing\) is contained in \(\{b\}\), so \(\operatorname{int}A=\varnothing\). The closed supersets of \(\{b\}\) are \(X\) and \(\{b,c\}\); their intersection is \(\{b,c\}\). Thus \(\overline{A}=\{b,c\}\). The complement \(\{a,c\}\) contains the open set \(\{a\}\), but neither \(\{a,b\}\) nor \(X\) is contained in it. Hence \(\operatorname{int}(X\setminus A)=\{a\}\), the exterior is \(\{a\}\), and the boundary is \(\{b,c\}\).
Neighborhood check: The open neighborhoods of \(b\) are \(\{a,b\}\) and \(X\); both meet \(A\) and its complement. The only open neighborhood of \(c\) is \(X\), which also meets both. The point \(a\) has the neighborhood \(\{a\}\), which misses \(A\). So the closure is \(\{b,c\}\), and exactly \(b,c\) are boundary points.
This space is not Hausdorff. Any open neighborhood of \(c\) must be \(X\), since no other listed open set contains \(c\). Any open neighborhood of \(b\) is either \(\{a,b\}\) or \(X\), and both intersect \(X\). Thus \(b\) and \(c\) have no disjoint open neighborhoods.
For completeness, the table gives all four quantities for every subset. Each entry follows by retaining the listed open subsets of \(A\), intersecting the listed closed supersets of \(A\), and using the exterior and boundary formulas above.
| \(A\) | \(\operatorname{int}A\) | \(\overline{A}\) | Exterior | \(\partial A\) |
|---|---|---|---|---|
| \(\varnothing\) | \(\varnothing\) | \(\varnothing\) | \(X\) | \(\varnothing\) |
| \(\{a\}\) | \(\{a\}\) | \(X\) | \(\varnothing\) | \(\{b,c\}\) |
| \(\{b\}\) | \(\varnothing\) | \(\{b,c\}\) | \(\{a\}\) | \(\{b,c\}\) |
| \(\{c\}\) | \(\varnothing\) | \(\{c\}\) | \(\{a,b\}\) | \(\{c\}\) |
| \(\{a,b\}\) | \(\{a,b\}\) | \(X\) | \(\varnothing\) | \(\{c\}\) |
| \(\{a,c\}\) | \(\{a\}\) | \(X\) | \(\varnothing\) | \(\{b,c\}\) |
| \(\{b,c\}\) | \(\varnothing\) | \(\{b,c\}\) | \(\{a\}\) | \(\{b,c\}\) |
| \(X\) | \(X\) | \(X\) | \(\varnothing\) | \(\varnothing\) |
For instance, for \(A=\{c\}\), the closed supersets are \(\{c\},\{b,c\},X\), whose intersection is \(\{c\}\). The open subsets of its complement \(\{a,b\}\) are \(\varnothing,\{a\},\{a,b\}\), whose union is \(\{a,b\}\). The boundary is therefore \(\{c\}\setminus\varnothing=\{c\}\), matching the table.
A Reliable Procedure and a Common Pitfall
For a finite space, a compact calculation can be organized as follows. First list the topology and obtain the closed sets by complementation. Next, for the chosen \(A\), union all listed open sets contained in \(A\) to find \(\operatorname{int}A\). Intersect all closed sets containing \(A\) to find \(\overline{A}\). Finally calculate \(\operatorname{int}(X\setminus A)\) and \(\overline{A}\setminus\operatorname{int}A\). The open-neighborhood test provides an independent check: a point lies in \(\overline{A}\) exactly when every open neighborhood of it intersects \(A\).
Common pitfall. A mistaken closure calculation sometimes says that the only closed superset of \(\varnothing\) that matters is \(X\), and therefore concludes \(\overline{\varnothing}=X\). This omits the closed superset \(\varnothing\) itself. The family of all closed supersets of \(\varnothing\) contains \(\varnothing\), and the intersection of a family containing \(\varnothing\) is \(\varnothing\). Thus \(\overline{\varnothing}=\varnothing\), in every topology.
A second error is to treat a subset as open merely because it appears as a subset of an open set. In the three-point example, \(\{b\}\subseteq\{a,b\}\), but \(\{b\}\) is not in the open-set list. It cannot be used as an open subset when forming an interior. The interior of \(\{b\}\) is consequently \(\varnothing\), not \(\{b\}\).
Check Your Understanding
Use the open-set list in each question; distinguish being a subset of an open set from being open itself.
- In the indiscrete topology on \(X=\{u,v\}\), find the interior, closure, exterior, and boundary of \(\{u\}\).
- In the three-point topology above, verify the table entries for \(A=\{a,c\}\) by listing its open subsets and the closed sets that contain it.
- In a discrete topology, what are the interior and closure of an arbitrary subset \(A\subseteq X\), and why?
- Explain why \(\overline{\varnothing}=\varnothing\) using the intersection of closed supersets, rather than using a neighborhood test.
- In the three-point topology above, use the open neighborhoods of \(a\) to check whether \(a\) lies in the closure of \(\{b\}\).