What Must Be Checked
To define a topology, it is not enough to name a collection of subsets that seems to have the right shape: the collection must satisfy the topology axioms. For the cofinite topology, the crucial distinction is that unions may have any number of terms, while intersections are required only for finite families. The proof uses complementary sets: the complement of a union is an intersection, and the complement of an intersection is a union.
Let \(X\) be an infinite set. A subset of \(X\) is cofinite when its complement in \(X\) is finite. We will check the topology axioms for the collection consisting of \(\varnothing\) and all cofinite subsets of \(X\). Including \(\varnothing\) explicitly matters: because \(X\) is infinite, its complement \(X\setminus\varnothing=X\) is not finite.
A Finite-Set Fact
The intersection argument will use the fact that a finite union of finite sets is finite. We record the fact with its finite-family case included.
Proof. For a family with zero members, its union is \(\varnothing\), a finite set. Suppose the claim holds for every family of \(n\) finite sets, and let \(F_1,\ldots,F_n,F_{n+1}\) be finite sets. By the induction hypothesis, \(F_1\cup\cdots\cup F_n\) is finite. Its union with the finite set \(F_{n+1}\) is finite, since it has at most \(|F_1\cup\cdots\cup F_n|+|F_{n+1}|\) elements. Induction proves the claim for every finite number of sets. \(\square\)
For two subsets \(A,B\subseteq X\), the complement identities we use are \(X\setminus(A\cup B)=(X\setminus A)\cap(X\setminus B)\) and \(X\setminus(A\cap B)=(X\setminus A)\cup(X\setminus B)\). The same identities hold for indexed families: the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. They follow by testing membership of an arbitrary point on each side.
The Topology Axioms
Proof. We verify each axiom. First, \(\varnothing\in\mathcal{T}_{\mathrm{cof}}\) by definition. Also \(X\setminus X=\varnothing\) is finite, so \(X\in\mathcal{T}_{\mathrm{cof}}\).
Next, let \((U_i)_{i\in J}\) be an arbitrary indexed family of members of \(\mathcal{T}_{\mathrm{cof}}\), where the index set \(J\) may be empty, finite, or infinite. If \(J=\varnothing\), its union is \(\varnothing\), which is open. If some \(U_j=X\), then \(\bigcup_{i\in J}U_i=X\), which is open. Now suppose \(J\ne\varnothing\) and no \(U_j\) equals \(X\). If every \(U_i=\varnothing\), their union is \(\varnothing\). Otherwise, choose an index \(j\) for which \(U_j\ne\varnothing\). Since \(U_j\) is open and is not \(\varnothing\), its complement \(X\setminus U_j\) is finite. The complement identity gives $$ X\setminus\bigcup_{i\in J}U_i=\bigcap_{i\in J}(X\setminus U_i)\subseteq X\setminus U_j. $$ A subset of a finite set is finite, so the complement of the union is finite. Therefore \(\bigcup_{i\in J}U_i\) is cofinite and belongs to \(\mathcal{T}_{\mathrm{cof}}\). This proves closure under arbitrary unions, including all the cases where the family is empty or contains \(\varnothing\).
Finally, let \(U_1,\ldots,U_n\) be a finite family of open sets. If \(n=0\), their intersection is \(X\), already shown to be open. Suppose \(n\ge1\). If one of the \(U_k\) is \(\varnothing\), then the intersection is \(\varnothing\), which is open. Otherwise every \(U_k\) is nonempty and cofinite, so each \(X\setminus U_k\) is finite. Then $$ X\setminus\bigcap_{k=1}^{n}U_k=\bigcup_{k=1}^{n}(X\setminus U_k). $$ The right-hand side is a finite union of finite sets, hence finite by the lemma. Thus \(\bigcap_{k=1}^{n}U_k\) is cofinite and open. All topology axioms hold. \(\square\)
The two closure checks use different amounts of information. In the union proof, when a nonempty open set occurs, the complement of the whole union is contained in that one finite complement. In the intersection proof, the complement is a union of complements, and finiteness is guaranteed because there are only finitely many of them. The proof does not claim that arbitrary intersections of open sets are open.
Worked Calculations
Worked Example: A Union and an Intersection in \(\mathbb{N}\)
Give \(\mathbb{N}=\{1,2,\dots\}\) the cofinite topology, and take \(U=\mathbb{N}\setminus\{1,4\}\) and \(V=\mathbb{N}\setminus\{2,4,7\}\). Both sets are open because their complements are finite.
For the union, an integer is missing from \(U\cup V\) exactly when it is missing from both \(U\) and \(V\). Therefore $$ \mathbb{N}\setminus(U\cup V) =\{1,4\}\cap\{2,4,7\} =\{4\}, $$ and \(U\cup V=\mathbb{N}\setminus\{4\}\), an open set.
For the intersection, an integer is missing from \(U\cap V\) when it is missing from at least one of \(U,V\). Hence $$ \mathbb{N}\setminus(U\cap V) =\{1,4\}\cup\{2,4,7\} =\{1,2,4,7\}, $$ so \(U\cap V=\mathbb{N}\setminus\{1,2,4,7\}\), also open. Check: the two computed complements have respectively \(1\) and \(4\) elements, so both are finite, as required by the definition.
Worked Example: An Infinite Union of Cofinite Sets
For each \(n\in\mathbb{N}\), let \(U_n=\mathbb{N}\setminus\{1,n+1\}\). Each \(U_n\) is open because its complement \(\{1,n+1\}\) has at most two elements.
Using complements, $$ \mathbb{N}\setminus\bigcup_{n\in\mathbb{N}}U_n =\bigcap_{n\in\mathbb{N}}\{1,n+1\}. $$ The number \(1\) belongs to every set in this intersection. No number \(m\ge2\) does: choosing \(n=m\) gives the set \(\{1,m+1\}\), which does not contain \(m\). Thus the intersection of the complements is exactly \(\{1\}\), and \(\bigcup_{n\in\mathbb{N}}U_n=\mathbb{N}\setminus\{1\}\). Its complement is finite, so the infinite union is open.
Check: \(1\) is absent from every \(U_n\). For any \(m\ge2\), the set \(U_m=\mathbb{N}\setminus\{1,m+1\}\) contains \(m\), because \(m\ne1\) and \(m\ne m+1\). Hence the union contains every number other than \(1\), exactly as computed.
Worked Example: An Infinite Intersection Need Not Be Open
For \(n\in\mathbb{N}\), define \(V_n=\mathbb{N}\setminus\{2n\}\). Each \(V_n\) is open: its complement consists of the single number \(2n\). A number belongs to every \(V_n\) precisely when it is not equal to \(2n\) for any positive integer \(n\). Therefore $$ \bigcap_{n\in\mathbb{N}}V_n=\{1,3,5,\dots\}, \qquad \mathbb{N}\setminus\bigcap_{n\in\mathbb{N}}V_n=\{2,4,6,\dots\}. $$ The complement is infinite, so the intersection is not cofinite. It is not \(\varnothing\), either, and consequently it is not in \(\mathcal{T}_{\mathrm{cof}}\).
Check: every odd positive integer avoids every excluded value \(2n\), while every even positive integer equals \(2n\) for some \(n\in\mathbb{N}\) and is therefore excluded from \(V_n\). This verifies the intersection exactly. There is no conflict with the topology axiom: it requires finite intersections, not intersections of arbitrary families.
A Common Pitfall: Reversing the Complement Rule
For the family \(U_n\) in the second example, a wrong calculation might read \(\mathbb{N}\setminus\bigcup_n U_n=\bigcup_n(\mathbb{N}\setminus U_n)=\bigcup_n\{1,n+1\}=\mathbb{N}\), and then conclude that the union of the \(U_n\) is empty. The first equality is wrong: complements turn unions into intersections, not unions. The correct calculation is \(\mathbb{N}\setminus\bigcup_n U_n=\bigcap_n\{1,n+1\}=\{1\}\), so the union is \(\mathbb{N}\setminus\{1\}\).
The topology proof depends on preserving that distinction. For an arbitrary union of open sets, the relevant complement is an arbitrary intersection; it is enough that this intersection is contained in the finite complement of one nonempty open member. For a finite intersection of open sets, the relevant complement is a finite union of finite sets. The finiteness in this second step would not follow for an arbitrary family.
Check Your Understanding
For each question, use the definition of the cofinite topology and state whether the relevant complement is finite.
- On an infinite set \(X\), why is \(X\) cofinite and why is \(\varnothing\) included separately in \(\mathcal{T}_{\mathrm{cof}}\)?
- Let \(X=\mathbb{N}\), \(A=\mathbb{N}\setminus\{3,8\}\), and \(B=\mathbb{N}\setminus\{3,5\}\). Compute \(A\cup B\) and \(A\cap B\) by first computing their complements.
- In the arbitrary-union proof, what changes if the family has no members? What if one member equals \(X\)?
- Explain why a finite intersection containing \(\varnothing\) is open, without using the cofinite-complement calculation.
- For \(W_n=\mathbb{N}\setminus\{n\}\), compute \(\bigcap_{n\in\mathbb{N}}W_n\). Is it open in the cofinite topology on \(\mathbb{N}\)?