Complementation Exchanges Closure and Interior
Closure and interior are related, but they are not interchangeable operations. A complement changes the relationship between them: the closure of a complement is the complement of an interior, and the interior of a complement is the complement of a closure. Keeping the complement in the right place is essential, especially when a set has endpoints that belong to one version of the set but not another.
Throughout, \((X,\mathcal{T})\) is a topological space and \(A\subseteq X\). A set is open if it belongs to \(\mathcal{T}\), and it is closed if its complement in \(X\) is open. The closure \(\overline{A}\) is the smallest closed subset of \(X\) containing \(A\); the interior \(\operatorname{int}A\) is the largest open subset of \(X\) contained in \(A\). In particular, \(\overline{A}\) is closed and contains \(A\), while \(\operatorname{int}A\) is open and contained in \(A\).
These identities describe a duality: to pass from the closure operation to the interior operation, complement the set before applying one operation and complement the result afterward. The formulas do not say that closure equals interior, or that either operation can be omitted.
Proof from the Definitions
We first prove the closure identity directly. A useful observation is that \(X\setminus\operatorname{int}A\) is closed, since \(\operatorname{int}A\) is open. It also contains \(X\setminus A\): if \(x\notin A\), then \(x\notin\operatorname{int}A\), because \(\operatorname{int}A\subseteq A\). Thus the smallest-closed-superset property gives \(\overline{X\setminus A}\subseteq X\setminus\operatorname{int}A\).
Proof. We prove the reverse inclusion as well. Suppose \(x\notin\overline{X\setminus A}\). Since \(\overline{X\setminus A}\) is closed, \(U=X\setminus\overline{X\setminus A}\) is open and contains \(x\). The closure contains the set whose closure is being taken, so \(X\setminus A\subseteq\overline{X\setminus A}\). Therefore \(U\cap(X\setminus A)=\varnothing\), which means \(U\subseteq A\). Since \(U\) is open and contains \(x\), the definition of interior gives \(x\in\operatorname{int}A\). We have shown that every point outside \(\overline{X\setminus A}\) is outside \(X\setminus\operatorname{int}A\), or equivalently that \(X\setminus\operatorname{int}A\subseteq\overline{X\setminus A}\). Combining the two inclusions proves the first identity.
For the second identity, apply the first one to \(X\setminus A\) in place of \(A\). Since \(X\setminus(X\setminus A)=A\), it gives \(\overline{A}=X\setminus\operatorname{int}(X\setminus A)\). Taking complements of both sides yields \(X\setminus\overline{A}=\operatorname{int}(X\setminus A)\), as required. These arguments also cover \(A=\varnothing\), \(A=X\), and the empty space: any pointwise statements in an empty space are vacuous. \(\square\)
The proof uses a basic feature of complementation: if \(C\subseteq D\), then \(X\setminus D\subseteq X\setminus C\). This reversal of inclusion explains why a largest-open-subset description turns into a smallest-closed-superset description when complements are taken.
Worked Example: An Interval with an Included Endpoint
Let \(X=\mathbb{R}\) with its standard topology and \(A=(0,1]\). We compute the interior and closure and then check both identities, paying particular attention to the point \(1\).
Interior. The interval \((0,1)\) is open and is contained in \(A\). Neither endpoint can belong to an open subset of \(A\): every open neighborhood of \(0\) contains negative numbers, and every open neighborhood of \(1\) contains numbers greater than \(1\). Thus \(\operatorname{int}A=(0,1)\).
Closure. The set \([0,1]\) is closed in \(\mathbb{R}\), as its complement \((-\infty,0)\cup(1,\infty)\) is open, and it contains \(A\). Any closed set \(F\) containing \(A\) must also contain \(0\): if \(0\notin F\), the open set \(\mathbb{R}\setminus F\) contains \(0\), so it contains some interval \((-\varepsilon,\varepsilon)\), which intersects \((0,1]\). This contradicts \(A\subseteq F\). Since \(1\in A\), every such \(F\) contains \(1\) as well. Hence the smallest closed superset is \(\overline{A}=[0,1]\).
Now \(X\setminus A=(-\infty,0]\cup(1,\infty)\). Its closure is \((-\infty,0]\cup[1,\infty)\): this set is closed because its complement \((0,1)\) is open, and any closed superset of \(X\setminus A\) must contain \(1\), by the same neighborhood argument used for \(0\) above. Meanwhile \(X\setminus\operatorname{int}A=X\setminus(0,1)=(-\infty,0]\cup[1,\infty)\). In particular, \(1\) is included in this complement, not excluded.
For the other identity, \(\operatorname{int}(X\setminus A)=(-\infty,0)\cup(1,\infty)\): this is an open subset of \(X\setminus A\), while neither \(0\) nor \(1\) has an open neighborhood contained in \(X\setminus A\). Also \(X\setminus\overline{A}=X\setminus[0,1]=(-\infty,0)\cup(1,\infty)\). Both identities hold.
Check: the computed sets differ exactly at the endpoints: \(X\setminus\operatorname{int}A\) includes \(0\) and \(1\), whereas \(X\setminus\overline{A}\) includes neither. This agrees with the two different formulas.
Worked Example: Every Subset of a Discrete Space
Let \(X=\{p,q,r\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. Every subset is also closed, because its complement is another subset and hence is open. For any \(A\subseteq X\), the largest open subset contained in \(A\) is \(A\), and the smallest closed superset containing \(A\) is also \(A\). Therefore \(\operatorname{int}A=A=\overline{A}\).
For a specific calculation, take \(A=\{p,r\}\). Then \(X\setminus A=\{q\}\), and \(\overline{X\setminus A}=\{q\}=X\setminus\operatorname{int}A\). Also, \(\operatorname{int}(X\setminus A)=\{q\}=X\setminus\overline{A}\). The equalities hold because both operations leave every subset unchanged in this topology.
Check: the complement of \(\{p,r\}\) in \(X=\{p,q,r\}\) is \(\{q\}\), and the complement of \(\{q\}\) is \(\{p,r\}\). Thus the set calculations use complements relative to \(X\), as the identities require.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Its closed sets are the complements of its open sets, namely \(\varnothing,\{b\},X\). This example shows that the identities do not depend on a metric or on the usual behavior of intervals.
Take \(A=\{a\}\). The largest open set contained in \(A\) is \(\{a\}\), so \(\operatorname{int}A=\{a\}\). The only closed set containing \(\{a\}\) is \(X\), so \(\overline{A}=X\). Since \(X\setminus A=\{b\}\), the only closed set containing \(X\setminus A\) that is smallest is \(\{b\}\), and the only open subset of \(\{b\}\) is \(\varnothing\). Therefore \(\overline{X\setminus A}=\{b\}=X\setminus\operatorname{int}A\), and \(\operatorname{int}(X\setminus A)=\varnothing=X\setminus\overline{A}\).
Check: the complement of \(\operatorname{int}A=\{a\}\) is \(\{b\}\), while the complement of \(\overline{A}=X\) is \(\varnothing\). These are exactly the closure of \(\{b\}\) and the interior of \(\{b\}\), respectively.
Why the Placement of Complements Matters
A common mistake is to treat the two identities as though they said that closure and interior are interchangeable, or to drop an endpoint while taking a complement. For \(A=(0,1]\subseteq\mathbb{R}\), the incorrect calculation \(X\setminus\operatorname{int}A=(-\infty,0)\cup(1,\infty)\) forgets that the complement of \((0,1)\) includes both endpoints. The correct calculation is \(X\setminus\operatorname{int}A=(-\infty,0]\cup[1,\infty)\). In particular, \(1\) belongs to this set because \(1\notin\operatorname{int}A\). The identity then says this is \(\overline{X\setminus A}\), which also contains \(1\) as a limit point of \((1,\infty)\).
It is also incorrect to replace \(X\setminus\overline{A}\) by \(X\setminus\operatorname{int}A\): the former is \((-\infty,0)\cup(1,\infty)\), while the latter includes \(0\) and \(1\). The safe procedure is to write the full identity first, then calculate the interior or closure, and only then take the complement relative to the stated space.
Check Your Understanding
For each question, take complements relative to the specified space.
- In \(\mathbb{R}\) with its standard topology, let \(A=[0,1)\). Find \(\operatorname{int}A\) and \(\overline{A}\), then use both identities to compute \(\overline{\mathbb{R}\setminus A}\) and \(\operatorname{int}(\mathbb{R}\setminus A)\).
- In a discrete topology, explain why both identities hold for every subset \(A\), including \(A=\varnothing\) and \(A=X\).
- In the Sierpinski space from the example, find the interior and closure of \(\{b\}\), and verify both complement identities for this set.
- State which endpoint belongs to \(X\setminus\operatorname{int}(0,1]\) and explain why.
- Starting from \(\overline{X\setminus A}=X\setminus\operatorname{int}A\), show how to obtain the other identity by replacing \(A\) with \(X\setminus A\) and taking complements.