Closed Sets from Open Complements
Open sets are defined by neighborhoods of their points, while closed sets are defined by what happens outside them. This change of viewpoint gives a useful rule: complements exchange unions and intersections. Combined with the topology axioms for open sets, that rule tells us which combinations of closed sets remain closed. We will prove the general laws and then calculate several examples in the standard topology on \(\mathbb{R}\).
Throughout, \((X,\mathcal{T})\) is a topological space. Its topology \(\mathcal{T}\) contains \(\varnothing\) and \(X\), is closed under arbitrary unions, and is closed under finite intersections. A set \(U\subseteq X\) is open when \(U\in\mathcal{T}\).
For example, \(\varnothing\) and \(X\) are both open and closed in every topological space. They are open by the topology axioms. Also, \(X\setminus\varnothing=X\) and \(X\setminus X=\varnothing\) are open, so both sets are closed as well.
The word “finite” in the topology axiom includes intersections of one or more open sets; the intersection of zero sets is taken to be \(X\). For set operations on closed sets, we use the corresponding conventions: the intersection of an empty family of subsets of \(X\) is \(X\), and the union of an empty family is \(\varnothing\). These conventions ensure the laws below include their empty-family cases.
Complement Laws for Closed Sets
For any family of subsets \((A_\alpha)_{\alpha\in J}\) of \(X\), De Morgan’s laws say that a point fails to belong to an intersection exactly when it fails to belong to at least one member of the family. It fails to belong to a union exactly when it fails to belong to every member. In complement notation, these statements are:
These equalities follow directly by testing membership of any \(x\in X\). For an empty family, the first says \(X\setminus X=\varnothing\), and the second says \(X\setminus\varnothing=X\), in agreement with the conventions above.
Proof. Let \((C_\alpha)_{\alpha\in J}\) be any family of closed subsets of \(X\). By the definition of closed, \(X\setminus C_\alpha\) is open for every \(\alpha\in J\). De Morgan’s law gives \(X\setminus\bigcap_{\alpha\in J}C_\alpha=\bigcup_{\alpha\in J}(X\setminus C_\alpha)\). The right-hand side is open because arbitrary unions of open sets are open. Therefore the complement of \(\bigcap_{\alpha\in J}C_\alpha\) is open, so that intersection is closed. If \(J\) is empty, the intersection is \(X\), which is closed as established above.
Now let \(C_1,\ldots,C_n\) be a finite family of closed sets, where \(n\geq 1\). Each complement \(X\setminus C_i\) is open. De Morgan’s law gives \(X\setminus\bigcup_{i=1}^n C_i=\bigcap_{i=1}^n(X\setminus C_i)\). The right-hand side is open because finite intersections of open sets are open. Hence the union is closed. For \(n=0\), the union is \(\varnothing\), which is closed. A finite intersection is a special case of an arbitrary intersection, so it too is closed. These arguments do not establish that an arbitrary union of closed sets is closed; a counterexample will follow. \(\square\)
The proof uses exactly the relevant topology axioms: arbitrary unions of open sets and finite intersections of open sets are open. It does not require any separation property, metric, or special feature of the underlying space.
Closed Sets in the Standard Topology on \(\mathbb{R}\)
From here on, \(\mathbb{R}\) has its standard topology. In this topology, a set is open if every one of its points lies in an open interval contained in the set. In particular, open intervals and open rays are open. Their complements show that closed intervals and closed rays are closed. For instance, \(\mathbb{R}\setminus[a,b]=(-\infty,a)\cup(b,\infty)\) for \(a\leq b\), and both open rays on the right are open. Likewise, \(\mathbb{R}\setminus[a,\infty)=(-\infty,a)\), which is open.
Worked Example: An Infinite Intersection Shrinking to One Point
For each \(n\in\mathbb{N}\), let \(C_n=[-1/n,1/n]\). Compute \(\bigcap_{n\in\mathbb{N}}C_n\).
Each \(C_n\) is a closed interval: its complement is \((-\infty,-1/n)\cup(1/n,\infty)\), a union of open rays. The point \(0\) belongs to every \(C_n\), because \(-1/n\leq 0\leq 1/n\) for every positive integer \(n\).
Suppose \(x>0\). There is a positive integer \(n>1/x\), so \(1/n<x\). Thus \(x\notin[-1/n,1/n]\), and \(x\) is not in the intersection. If \(x<0\), choose a positive integer \(n>1/|x|\). Then \(1/n<|x|=-x\), so \(x<-1/n\) and again \(x\notin C_n\). Consequently no nonzero real number belongs to every \(C_n\), while \(0\) does.
Result: \(\bigcap_{n\in\mathbb{N}}[-1/n,1/n]=\{0\}\). This agrees with the theorem: each \(C_n\) is closed, and an arbitrary intersection of closed sets is closed. As a direct check, \(\mathbb{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)\) is open, so \(\{0\}\) is closed.
Worked Example: A Finite Union of Closed Intervals
Compute \(A\cup B\) and its complement in \(\mathbb{R}\), where \(A=[-3,-1]\) and \(B=[2,4]\).
The intervals do not overlap: every point of \(A\) is at most \(-1\), while every point of \(B\) is at least \(2\). Therefore the union is \([-3,-1]\cup[2,4]\). The endpoints \(-3,-1,2,4\) are included because each interval is closed.
A real number is outside this union precisely when it is less than \(-3\), strictly between \(-1\) and \(2\), or greater than \(4\). The endpoints themselves are not in the complement. Hence \(\mathbb{R}\setminus(A\cup B)=(-\infty,-3)\cup(-1,2)\cup(4,\infty)\). Each component is open, so their union is open. By the definition of closed, \(A\cup B\) is closed, as the finite-union law predicts.
Check: the points \(-3\) and \(-1\) belong to \(A\), and \(2\) and \(4\) belong to \(B\), so none of these four endpoints appears in the displayed complement. A point such as \(0\) lies strictly between \(-1\) and \(2\), while \(5\) is greater than \(4\); both are in the complement.
Worked Example: An Infinite Intersection That Is Empty
For \(n\in\mathbb{N}\), set \(F_n=[n,\infty)\). Find the intersection of all the \(F_n\).
Each \(F_n\) is a closed ray because its complement \((-\infty,n)\) is open. If a real number \(x\) belonged to every \(F_n\), then \(x\geq n\) for every positive integer \(n\). But the positive integers are not bounded above in \(\mathbb{R}\), so there is an \(n\in\mathbb{N}\) with \(n>x\). For this \(n\), \(x<n\), which means \(x\notin F_n\). Thus no real number belongs to every \(F_n\), and \(\bigcap_{n\in\mathbb{N}}F_n=\varnothing\).
Check: \(\varnothing\) is closed because its complement is \(\mathbb{R}\), which is open. The result is therefore consistent with the arbitrary-intersection law, including the possibility that the intersection contains no points.
Why Infinite Unions Need Care
The finite-union law cannot be extended to arbitrary unions. The distinction is not a technicality: even a countable union of closed subsets of the real line can fail to be closed. The next example verifies the failure directly from the open-set definition.
Worked Example: A Union of Closed Singletons That Is Not Closed
For each \(n\in\mathbb{N}\), let \(D_n=\{1/n\}\), and put \(D=\bigcup_{n\in\mathbb{N}}D_n=\{1/n:n\in\mathbb{N}\}\). Each \(D_n\) is closed: for any \(x\ne 1/n\), the open interval centered at \(x\) with radius \(|x-1/n|/2>0\) contains \(x\) and does not contain \(1/n\). Thus every point of \(\mathbb{R}\setminus D_n\) has an open interval contained in \(\mathbb{R}\setminus D_n\), proving that this complement is open.
However, \(D\) is not closed. The point \(0\) does not belong to \(D\), since \(1/n>0\) for every \(n\in\mathbb{N}\). If \(D\) were closed, its complement would be open and would contain \(0\). Openness would then give some \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq\mathbb{R}\setminus D\). Choose a positive integer \(n>1/\varepsilon\). Then \(0<1/n<\varepsilon\), so \(1/n\in(-\varepsilon,\varepsilon)\), but \(1/n\in D\). This contradicts the proposed containment. Therefore \(\mathbb{R}\setminus D\) is not open and \(D\) is not closed.
Check: the contradiction uses the strict inequality \(1/n<\varepsilon\), so the selected point lies inside the open interval. The example shows why the theorem asserts finite unions, not arbitrary unions: every individual \(D_n\) is closed, but their infinite union is not.
Common pitfall. A tempting but false argument is: “Every \(C_i\) is closed, so \(\bigcup_i C_i\) is closed.” It applies the finite-union law without checking that the family is finite. In the example above, each \(D_n\) is closed, but their countable union \(D\) is not. The correction is to take complements and ask which topology axiom applies: arbitrary intersections of open sets are not guaranteed to be open, whereas finite intersections are. Thus the complement argument proves the union law only for finite families.
| Operation on closed sets | Conclusion | Reason |
|---|---|---|
| Arbitrary intersection | Closed | Its complement is an arbitrary union of open sets. |
| Finite union | Closed | Its complement is a finite intersection of open sets. |
| Arbitrary union | May fail to be closed | Its complement is an arbitrary intersection of open sets, which need not be open. |
The complement viewpoint is useful because it translates a question about closed sets into one about open sets, where the topology axioms already provide the needed rules. When applying those rules, check whether the family is finite or arbitrary and include the empty-family conventions when they matter.
Check Your Understanding
For each question, identify the relevant complement and say whether the family is finite or arbitrary.
- Prove that the intersection of two closed sets is closed by using complements.
- For \(n\in\mathbb{N}\), compute \(\bigcap_{n\in\mathbb{N}}[n,\infty)\), and explain why each set in the family is closed.
- Compute \([-2,1]\cup[3,5]\) and write its complement as a union of open intervals or rays, with the correct endpoint strictness.
- In \(\mathbb{R}\), explain why \(\{2\}\) is closed by showing that its complement is open.
- Why does the complement argument for finite unions not prove that arbitrary unions of closed sets are closed? Give the counterexample from the tutorial.