From Open-Set Axioms to Closed-Set Rules
The topology axioms are stated using open sets, but many arguments are expressed using closed sets. The link is complementation: taking the complement reverses unions and intersections. This converts the axioms into useful rules for closed sets, while also showing exactly where a finiteness condition is essential.
Let \((X,\mathcal{T})\) be a topological space. Its topology \(\mathcal{T}\) contains \(\varnothing\) and \(X\), is closed under arbitrary unions, and is closed under finite intersections. A set \(C\subseteq X\) is closed when \(X\setminus C\) is open. For a family \((A_i)_{i\in I}\), the index set \(I\) may be finite, infinite, or empty. The empty union is \(\varnothing\), and the empty intersection is \(X\).
Proof. First let \((C_i)_{i\in I}\) be any family of closed subsets of \(X\). For each \(i\), the set \(X\setminus C_i\) is open. De Morgan’s law gives \[ X\setminus\bigcap_{i\in I}C_i=\bigcup_{i\in I}(X\setminus C_i). \] The right-hand side is open by the arbitrary-union axiom. Therefore the complement of \(\bigcap_{i\in I}C_i\) is open, so \(\bigcap_{i\in I}C_i\) is closed. If \(I=\varnothing\), the intersection is \(X\), which is closed because its complement \(\varnothing\) is open.
Now let \(C_1,\ldots,C_m\) be a finite family of closed sets, with \(m\geq 1\). Each \(X\setminus C_j\) is open, and the finite-intersection axiom says that \(\bigcap_{j=1}^{m}(X\setminus C_j)\) is open. Since \[ X\setminus\bigcup_{j=1}^{m}C_j=\bigcap_{j=1}^{m}(X\setminus C_j), \] the complement of the union is open, so the union is closed. If \(m=0\), the union is \(\varnothing\), which is closed because its complement \(X\) is open. This proves both claims, with the empty-family cases included. \(\square\)
Why the Two Rules Have Different Scope
The first rule permits any number of closed sets in an intersection because its complement is a union, and the topology axioms allow unions of any size. The second rule permits only finitely many closed sets in a union because its complement is an intersection, and the axioms guarantee openness only for finite intersections.
The same complement calculations also state the open-set rules in closed-set language. Arbitrary unions of open sets are open, and finite intersections of open sets are open. Neither rule can simply be enlarged to arbitrary intersections of open sets or arbitrary unions of closed sets. The following examples make the distinction concrete.
Worked Example: Translating a Finite Union
Suppose \(C_1,C_2,C_3\) are closed subsets of an arbitrary topological space \(X\). Compute the complement of their union and use the axioms to determine whether the union is closed.
Since each \(C_j\) is closed, its complement \(U_j=X\setminus C_j\) is open, for \(j=1,2,3\). De Morgan’s law gives \[ X\setminus(C_1\cup C_2\cup C_3) =(X\setminus C_1)\cap(X\setminus C_2)\cap(X\setminus C_3) =U_1\cap U_2\cap U_3. \] The topology axiom for finite intersections makes \(U_1\cap U_2\cap U_3\) open. Thus \(C_1\cup C_2\cup C_3\) is closed.
Check: Taking the complement of the result gives \(X\setminus(U_1\cap U_2\cap U_3)=C_1\cup C_2\cup C_3\), as required. The reasoning uses the finite-intersection axiom exactly three times’ worth of sets; it does not establish the corresponding claim for infinitely many closed sets.
A Concrete Topology to Test the Rules
To see why arbitrary intersections require care, we use a specific infinite space. Let \(X=\{0\}\cup\mathbb{N}\). For each integer \(n\geq 2\), define \[ U_n=\{0\}\cup\{k\in\mathbb{N}:k\geq n\}, \] and let \(\mathcal{T}=\{\varnothing,X\}\cup\{U_n:n\geq 2\}\).
Here is a check that these sets form a topology. A union of a family containing \(X\) is \(X\), and a union of only empty sets is \(\varnothing\). For a nonempty family of sets \(U_n\), the set of its indices has a least integer \(r\), and the union is \(U_r\). For finite intersections, any intersection involving \(\varnothing\) is empty; an intersection involving \(X\) leaves the other sets unchanged; and the intersection of \(U_{n_1},\ldots,U_{n_m}\) is \(U_{\max(n_1,\ldots,n_m)}\). The empty finite intersection is \(X\). Thus \(\mathcal{T}\) contains \(\varnothing,X\) and satisfies both union and finite-intersection axioms.
Worked Example: Intersecting Closed Sets in the Example Space
For \(n\geq2\), the complement of \(U_n\) is \[ C_n=X\setminus U_n=\{1,2,\ldots,n-1\}. \] Each \(C_n\) is closed because its complement \(U_n\) is open. Consider \(C_2\cap C_3\cap C_4\). The sets are \(C_2=\{1\}\), \(C_3=\{1,2\}\), and \(C_4=\{1,2,3\}\), so \[ C_2\cap C_3\cap C_4=\{1\}. \] The result is \(C_2\), hence closed.
Now take the intersection of the whole family \((C_n)_{n\geq2}\). The point \(1\) belongs to every \(C_n\), since \(1\leq n-1\) for every \(n\geq2\). The point \(2\) does not belong to \(C_2=\{1\}\), and any integer \(k\geq2\) also fails to belong to \(C_2\). The point \(0\) belongs to none of the \(C_n\). Therefore \[ \bigcap_{n\geq2}C_n=\{1\}=C_2, \] which is closed. This computation illustrates that an intersection of infinitely many closed sets may be nonempty, as well as the general theorem that it is closed.
Check: The complement of the computed intersection is \(X\setminus\{1\}=U_2\), which is open. This verifies closedness directly from the definition.
Worked Example: An Arbitrary Intersection of Open Sets
In the same space, every \(U_n\) is open. Compute their intersection, using the strict membership test \(k\in U_n\) exactly when \(k\geq n\) for a positive integer \(k\).
The point \(0\) belongs to every \(U_n\) by definition. Fix any positive integer \(k\). For \(n=k+1\), we have \(n\geq2\) and \(k<n\), so \(k\notin U_{k+1}\). Thus no positive integer belongs to every \(U_n\), and \[ \bigcap_{n\geq2}U_n=\{0\}. \] This intersection is not open: the only nonempty open sets in \(\mathcal{T}\) are \(X\) and the sets \(U_n\), and each contains positive integers, whereas \(\{0\}\) contains none.
The same example shows that arbitrary unions of closed sets need not be closed. Since \(C_n=X\setminus U_n=\{1,\ldots,n-1\}\), every \(C_n\) is closed. Every positive integer \(k\) belongs to \(C_{k+1}\), and \(0\) belongs to no \(C_n\). Hence \[ \bigcup_{n\geq2}C_n=\mathbb{N}. \] Its complement in \(X\) is \(\{0\}\), which is not open by the calculation above. Therefore \(\mathbb{N}\) is not closed.
Check: The complement of \(\bigcup_{n\geq2}C_n\) is indeed \(\{0\}\), the same nonopen set obtained by intersecting the \(U_n\). This agrees with De Morgan’s law and confirms that the arbitrary-union claim for closed sets fails here.
Common Pitfall: Dropping “Finite”
A common incorrect argument says: “Each \(U_n\) is open, so their intersection is open.” The premise is true in the example above, but the conclusion is false: the intersection is \(\{0\}\), which is not open in the specified topology. The correction is to check which axiom applies. The topology axioms guarantee finite intersections of open sets, not arbitrary ones.
There is no contradiction with the closed-set theorem. The complement of an arbitrary intersection of closed sets is an arbitrary union of open sets, which the axioms allow. By contrast, the complement of an arbitrary union of closed sets is an arbitrary intersection of open sets, which the axioms do not guarantee. The distinction is about what follows from the axioms in every topological space; some particular spaces may have stronger closure properties.
Check Your Understanding
For each question, identify the relevant topology axiom and, when useful, take complements.
- Explain why an arbitrary intersection of closed sets is closed, including when the family is empty.
- Explain why a union of zero closed sets is closed, and why a finite nonempty union of closed sets is closed.
- In the example space, compute \(C_3\cup C_5\) and verify that it is closed by taking its complement.
- In the example space, does \(\bigcap_{n\geq2}U_n\) contain the positive integer \(4\)? Give the index \(n\) that decides the membership test.
- Which open-set axiom would be needed to deduce that an arbitrary union of closed sets is closed?