Closure from Closed Supersets
A set can fail to be closed because it is missing points that every closed set containing it must include. One way to identify precisely which points are forced is to collect all closed sets containing the original set and intersect them. This construction works in every topological space, and its result is the smallest closed set that still contains the original set.
Let \((X,\mathcal{T})\) be a topological space. A set \(C\subseteq X\) is closed when \(X\setminus C\) is open. For any \(A\subseteq X\), there is at least one closed set containing \(A\): the whole space \(X\) is closed, since its complement \(\varnothing\) is open. Thus the collection of closed supersets of \(A\) is never empty.
The phrase “all closed supersets” matters: a point belongs to \(\overline{A}\) exactly when it belongs to every closed set containing \(A\). The next result establishes that this intersection itself is closed and gives the precise minimality property.
Proof. Write \(\mathcal{C}=\{C\subseteq X:C\text{ is closed and }A\subseteq C\}\), so \(\overline{A}=\bigcap_{C\in\mathcal{C}}C\). Each \(C\in\mathcal{C}\) contains \(A\), so every point of \(A\) belongs to every such \(C\); hence \(A\subseteq\overline{A}\). The complement of the intersection is \(X\setminus\overline{A}=\bigcup_{C\in\mathcal{C}}(X\setminus C)\). Every \(X\setminus C\) in this union is open, and an arbitrary union of open sets is open. Therefore \(X\setminus\overline{A}\) is open, so \(\overline{A}\) is closed. If \(F\) is any closed set containing \(A\), then \(F\in\mathcal{C}\), and an intersection is contained in each set being intersected; thus \(\overline{A}\subseteq F\). Finally, suppose \(D\) is another closed set containing \(A\) and contained in every closed set containing \(A\). Applying this property to \(F=\overline{A}\) gives \(D\subseteq\overline{A}\). Applying the already established minimality of \(\overline{A}\) to the closed superset \(D\) gives \(\overline{A}\subseteq D\). Hence \(D=\overline{A}\), proving uniqueness. \(\square\)
This proof uses only the topology axioms that arbitrary unions of open sets are open and that the whole space is open. It does not require a metric, a notion of distance, or any separation property.
First Consequences of the Definition
Proof. Suppose first that \(F\) is closed and \(A\subseteq F\). It is then one of the closed supersets in the definition, so the smallest-closed-superset theorem gives \(\overline{A}\subseteq F\). Conversely, \(A\subseteq\overline{A}\) by that theorem. Therefore, if \(\overline{A}\subseteq F\), then \(A\subseteq F\). For the empty set, \(\varnothing\) is itself closed and contains \(\varnothing\), so the smallest closed superset of \(\varnothing\) is contained in \(\varnothing\); since it also contains \(\varnothing\), it equals \(\varnothing\). For \(A=X\), any subset of \(X\) containing \(X\) must be \(X\), so the only closed superset of \(X\) is \(X\), and \(\overline{X}=X\). \(\square\)
The proposition gives a practical way to prove an answer: exhibit a closed set \(F\) containing \(A\), then show that every closed set containing \(A\) must also contain all of \(F\). The first fact gives \(\overline{A}\subseteq F\); the second gives the reverse inclusion.
Computing Closures in the Real Line
In the following examples, \(\mathbb{R}\) has its standard topology, whose basic open sets are open intervals. To show a proposed set is closed, we verify that its complement is open. To show that points are required in every closed superset, we use the fact that the complement of a closed set is open.
Worked Example: The Closure of an Open Interval
Find the closure of \(A=(0,1)\) in \(\mathbb{R}\).
The candidate is \([0,1]\). First, \(A\subseteq[0,1]\). The complement is \(\mathbb{R}\setminus[0,1]=(-\infty,0)\cup(1,\infty)\), which is open: it is a union of open intervals, since each of its points lies in an open interval contained in the same ray. Thus \([0,1]\) is closed, and the smallest-closed-superset theorem gives \(\overline{A}\subseteq[0,1]\).
For the reverse inclusion, let \(C\) be any closed set with \((0,1)\subseteq C\). Every point of \((0,1)\) is already in \(C\). Suppose \(0\notin C\). Then \(X\setminus C\), here \(\mathbb{R}\setminus C\), is an open set containing \(0\). Consequently there is some \(\delta>0\) such that \((-\delta,\delta)\subseteq\mathbb{R}\setminus C\). The number \(t=\min(\delta/2,1/2)\) satisfies \(0<t<1\) and \(t<\delta\). Hence \(t\in(0,1)\subseteq C\), but also \(t\in(-\delta,\delta)\subseteq\mathbb{R}\setminus C\), a contradiction. Therefore \(0\in C\).
Suppose instead that \(1\notin C\). Openness of \(\mathbb{R}\setminus C\) gives some \(\eta>0\) such that \((1-\eta,1+\eta)\subseteq\mathbb{R}\setminus C\). Set \(s=1-\min(\eta/2,1/2)\). Then \(0<s<1\) and \(1-\eta<s<1+\eta\). This puts \(s\) both in \(C\), because \(s\in(0,1)\), and in \(\mathbb{R}\setminus C\), a contradiction. Thus \(1\in C\) as well.
We have proved that every closed superset \(C\) of \(A\) contains \([0,1]\), so \([0,1]\subseteq\overline{A}\). Combining this with the earlier inclusion gives \(\overline{(0,1)}=[0,1]\).
Worked Example: A Finite Set
Find the closure of \(A=\{-1,2\}\) in \(\mathbb{R}\).
The set \(A\) contains itself and is closed. To verify closedness directly, take any \(x\in\mathbb{R}\setminus A\). Both \(|x+1|\) and \(|x-2|\) are positive, so \(r=\min(|x+1|,|x-2|)/2>0\). Every \(y\in(x-r,x+r)\) satisfies \(|y-x|<r\). If \(y=-1\), then \(|x+1|=|x-y|<r\leq |x+1|/2\), which is impossible. If \(y=2\), then \(|x-2|=|x-y|<r\leq |x-2|/2\), also impossible. Thus \((x-r,x+r)\subseteq\mathbb{R}\setminus A\). Every point of the complement has an open interval around it still in the complement, so the complement is open and \(A\) is closed.
Since \(A\) is a closed set containing itself, the smallest closed superset \(\overline{A}\) is contained in \(A\). The definition also gives \(A\subseteq\overline{A}\). Therefore \(\overline{\{-1,2\}}=\{-1,2\}\).
Check: The result is closed, as verified above, and contains the original set. No smaller set can contain both \(-1\) and \(2\), so it is indeed the smallest closed superset.
Worked Example: A Reciprocal Sequence and Its Limit
Let \(A=\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\), where \(\mathbb{N}=\{1,2,\dots\}\). Find \(\overline{A}\).
Consider \(F=\{0\}\cup A\). We first verify that \(F\) is closed. If \(x<0\), the interval \((x-|x|/2,x+|x|/2)\) lies below \(0\), so it misses \(F\). If \(x>1\), the interval \((x-(x-1)/2,x+(x-1)/2)\) lies above \(1\), and it too misses \(F\). These intervals show that every such \(x\) has an open neighborhood in \(\mathbb{R}\setminus F\).
Now let \(0<x<1\) and suppose \(x\notin A\). There is an integer \(n\geq1\) such that \(n<1/x<n+1\): choose the integer part of \(1/x\), which is not an integer because \(x\notin A\). Taking reciprocals of positive quantities reverses the inequalities, giving \(1/(n+1)<x<1/n\). Set \(r=\min(x-1/(n+1),1/n-x)/2\), which is positive. Then \((x-r,x+r)\) lies strictly between \(1/(n+1)\) and \(1/n\). No number \(1/k\), for \(k\in\mathbb{N}\), lies strictly between those two consecutive reciprocal values: if \(k\leq n\), then \(1/k\geq1/n\); if \(k\geq n+1\), then \(1/k\leq1/(n+1)\). Also \(0\) is not in this interval. Thus this open neighborhood misses \(F\). The points not in \(F\) are exactly the points covered by these cases, so \(\mathbb{R}\setminus F\) is open and \(F\) is closed.
Since \(A\subseteq F\), the minimality theorem gives \(\overline{A}\subseteq F\). To prove the other inclusion, let \(C\) be any closed set containing \(A\). It already contains every point \(1/n\). If \(0\notin C\), openness of \(\mathbb{R}\setminus C\) provides a \(\delta>0\) such that \((-\delta,\delta)\subseteq\mathbb{R}\setminus C\). Choose an integer \(n>1/\delta\), possible because the positive integers are unbounded. Then \(0<1/n<\delta\), so \(1/n\) belongs both to \(A\subseteq C\) and to \((-\delta,\delta)\subseteq\mathbb{R}\setminus C\), a contradiction. Hence \(0\in C\), and \(F\subseteq C\) for every closed superset \(C\) of \(A\). By the definition of closure, \(F\subseteq\overline{A}\).
Combining the inclusions yields \(\overline{A}=\{0\}\cup\{1/n:n\in\mathbb{N}\}\). The verification of closedness of \(F\) and the argument that every closed superset contains \(F\) check both requirements of the construction.
A Common Pitfall: Taking a Union
The closure is an intersection of closed supersets, not their union. For example, in \(\mathbb{R}\), the whole space \(\mathbb{R}\) is itself a closed superset of \(A=(0,1)\). Therefore the union of all closed supersets of \(A\) contains \(\mathbb{R}\), so that union is \(\mathbb{R}\), not the closure \([0,1]\). The intersection has the right meaning: it retains exactly the points that every closed superset is forced to contain.
When computing a closure from the definition, two checks prevent common errors:
- Show that the proposed answer is a closed set containing the original set.
- Show that every closed set containing the original set also contains the proposed answer.
The first check proves the closure is contained in the candidate; the second proves the candidate is contained in the closure. Both are needed to establish equality.
Check Your Understanding
Use the definition as an intersection of closed supersets, and justify both inclusions whenever you propose a closure.
- In \(\mathbb{R}\), find the closure of \((2,5)\).
- In \(\mathbb{R}\), find the closure of \(\{0,3,7\}\). What fact about finite subsets of \(\mathbb{R}\) can you use?
- Explain why \(\overline{\varnothing}=\varnothing\) using the definition, not a neighborhood test.
- Suppose \(F\) is closed and \(A\subseteq F\). Prove directly from the smallest-closed-superset property that \(\overline{A}\subseteq F\).
- For \(A=(0,1)\), what is the union of all closed supersets of \(A\), and why is it not the closure?