Closure as an Operation on Sets
Closure turns each subset of a topological space into a closed set. More importantly, it obeys a small collection of algebraic laws. These laws let us simplify expressions containing closures without recomputing each set from its neighborhoods. The four main properties are extensivity, monotonicity, idempotence, and preservation of finite unions.
Let \((X,\mathcal{T})\) be a topological space. A set is closed if its complement is open. For \(A\subseteq X\), define its closure to be the intersection of all closed subsets of \(X\) that contain \(A\). There is at least one such closed set, namely \(X\), so this intersection is defined. An intersection of closed sets is closed; hence \(\overline{A}\) is closed, contains \(A\), and is contained in every closed set containing \(A\).
- Extensive: \(A\subseteq\overline{A}\).
- Monotone: if \(A\subseteq B\), then \(\overline{A}\subseteq\overline{B}\).
- Idempotent: \(\overline{\overline{A}}=\overline{A}\).
- Finite-union preserving: \(\overline{A\cup B}=\overline{A}\cup\overline{B}\).
The empty set is also preserved: \(\overline{\varnothing}=\varnothing\), since \(\varnothing\) is itself closed. Thus the finite-union law includes unions of any finite number of sets, including a union with no sets, whose value is \(\varnothing\).
Why the Closure Laws Hold
Proof. Since \(\overline{A}\) is defined as an intersection of closed sets each containing \(A\), every point of \(A\) belongs to \(\overline{A}\). Therefore \(A\subseteq\overline{A}\), proving extensivity.
Suppose \(A\subseteq B\). The set \(\overline{B}\) is closed and contains \(B\), so it also contains \(A\). By the defining minimality of \(\overline{A}\), every closed set containing \(A\) contains \(\overline{A}\). Applying this to \(\overline{B}\) gives \(\overline{A}\subseteq\overline{B}\), proving monotonicity.
For idempotence, \(\overline{A}\) is closed, so its closure is itself: the smallest closed set containing \(\overline{A}\) is \(\overline{A}\). Hence \(\overline{\overline{A}}=\overline{A}\). This reasoning includes \(A=\varnothing\) and \(A=X\); neither requires a separate exception.
To prove the union law, first note that \(\overline{A}\cup\overline{B}\) is closed. Indeed, its complement is \((X\setminus\overline{A})\cap(X\setminus\overline{B})\), an intersection of two open sets, hence open. The set \(\overline{A}\cup\overline{B}\) contains \(A\cup B\). By minimality of closure, \(\overline{A\cup B}\subseteq\overline{A}\cup\overline{B}\). For the reverse inclusion, \(A\subseteq A\cup B\) and \(B\subseteq A\cup B\), so monotonicity gives \(\overline{A}\subseteq\overline{A\cup B}\) and \(\overline{B}\subseteq\overline{A\cup B}\). Taking their union proves \(\overline{A}\cup\overline{B}\subseteq\overline{A\cup B}\). The two inclusions establish equality.
Applying the binary union law repeatedly proves the law for any positive finite number of sets: at each step, take the union already formed as the first set and the next set as the second. For zero sets, the union is \(\varnothing\), and \(\overline{\varnothing}=\varnothing\). Thus the finite-union law holds in every case. \(\square\)
These laws also characterize the sets that are already closed. If \(A\) is closed, it is the smallest closed set containing itself, so \(\overline{A}=A\). Conversely, if \(\overline{A}=A\), then \(A\) is closed because \(\overline{A}\) is closed. This observation is useful when checking the result of a simplification.
Simplifying Nested Closures
The laws act like rewriting rules. A closure around another closure can be removed by idempotence. A closure around a finite union can be replaced by the union of the closures. If one set is contained in another, monotonicity gives a corresponding inclusion between their closures. These rules can be applied in succession, but each step must match a law: in particular, the finite-union law concerns unions, not intersections.
Worked Example: Intervals and an Isolated Point in \(\mathbb{R}\)
Use the standard topology on \(\mathbb{R}\). Let \(A=(0,1)\), \(B=(1,2)\), and \(C=\{4\}\). Simplify \(\overline{\overline{A}\cup(B\cup C)}\), and compute the resulting set.
In the standard topology, \(\overline{(0,1)}=[0,1]\): every neighborhood of an endpoint meets \((0,1)\), every point of \((0,1)\) belongs to the interval, and each point outside \([0,1]\) has an open interval around it disjoint from \((0,1)\). The same reasoning gives \(\overline{B}=[1,2]\). The singleton \(C=\{4\}\) is closed, so \(\overline{C}=\{4\}\).
Apply the binary union law first to the inner union, then to the outer one, and use idempotence on \(A\):
Check: \([0,2]\cup\{4\}\) is closed: its complement is \((-\infty,0)\cup(2,4)\cup(4,\infty)\), a union of open intervals. It contains the original set \(\overline{A}\cup(B\cup C)=[0,1]\cup(1,2)\cup\{4\}\). Its points in \([0,2]\) are exactly the points forced by closing the intervals, and the point \(4\) is already a closed singleton. Thus the computed set agrees with the closure laws.
Worked Example: Closure in the Sierpinski Topology
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Simplify \(\overline{\overline{\{a\}}\cup(\{b\}\cup\overline{\varnothing})}\).
The closed sets are the complements of the open sets: they are \(\varnothing\), \(\{b\}\), and \(X\). The only closed set containing \(\{a\}\) is \(X\), so \(\overline{\{a\}}=X\). The set \(\{b\}\) is closed, so \(\overline{\{b\}}=\{b\}\). Also \(\overline{\varnothing}=\varnothing\).
Substitution gives the full calculation:
The operator laws give the same answer without evaluating the whole inner set at once: idempotence and finite-union preservation reduce the expression to \(\overline{\{a\}}\cup\overline{\{b\}}\cup\overline{\varnothing}=X\cup\{b\}\cup\varnothing=X\). Check: the original inner set is \(X\), which is closed, so its closure is \(X\), as computed.
Worked Example: A Finite-Complement Topology
Let \(X=\mathbb{Z}\) with the finite-complement topology: the open sets are \(\varnothing\) and all subsets whose complement in \(X\) is finite. Let \(E\) be the set of even integers and \(F=\{1,2\}\). Simplify \(\overline{\overline{E}\cup F}\).
The closed sets are \(X\) and the finite subsets of \(X\). The set \(E\) is infinite. Any closed set containing \(E\) cannot be finite, so the only closed set containing \(E\) is \(X\). Therefore \(\overline{E}=X\). The set \(F\) is finite and closed, so \(\overline{F}=F\).
Using idempotence and the union law,
Check: the inner set \(\overline{E}\cup F\) is \(X\cup\{1,2\}=X\), and \(X\) is closed, so its closure is \(X\). The result also illustrates why the topology matters: closure is determined by the closed sets of the specified topology, not by the usual distance on the integers.
A Common Pitfall: Do Not Replace Unions by Intersections
The finite-union law does not say that closure distributes over every set operation. A common incorrect calculation is to assume \(\overline{A\cap B}=\overline{A}\cap\overline{B}\). In \(\mathbb{R}\) with its standard topology, take \(A=(0,1)\) and \(B=(1,2)\). Their intersection is empty, whereas their closures both contain \(1\):
Thus the proposed equality would claim \(\varnothing=\{1\}\), which is false. The correction is to use only a law whose hypotheses match the expression: closure preserves finite unions, while no corresponding equality for intersections follows from the closure-operator laws. The examples above show the practical benefit of checking each algebraic step rather than manipulating closure symbols as if they obeyed every familiar distributive rule.
Check Your Understanding
For each question, identify which closure law applies and state any topology being used.
- Prove that if \(A\subseteq B\), then \(\overline{A}\subseteq\overline{B}\), using the definition of closure as the smallest closed superset.
- Simplify \(\overline{\overline{A}\cup\overline{B}}\) for arbitrary subsets \(A,B\) of a topological space.
- In the standard topology on \(\mathbb{R}\), compute \(\overline{(2,5)\cup\{8\}}\).
- In the finite-complement topology on \(\mathbb{Z}\), compute the closure of the set of odd integers and justify the answer from the closed sets.
- For \(A=(0,1)\) and \(B=(1,2)\) in the standard topology on \(\mathbb{R}\), compare \(\overline{A\cap B}\) with \(\overline{A}\cap\overline{B}\). What does this show about the finite-union law?