Complements of Finite Sets as Basic Open Sets
A basis is useful when its members are simple sets from which all open sets can be assembled. On an infinite set \(X\), a natural choice is to take as basic sets the complements of finite subsets. These sets are large: removing finitely many points never removes all of \(X\). We will check the basis criterion directly, then identify every open set generated by this collection.
Throughout, \(X\) is an infinite set. For each finite subset \(F\subseteq X\), write \(B_F=X\setminus F\), and let \(\mathcal{B}\) be the collection of all such sets. A basis for a topology on \(X\) is a collection of nonempty subsets that covers \(X\) and satisfies this local intersection condition: whenever \(x\in B_1\cap B_2\) for basis elements \(B_1,B_2\), there is a basis element \(B_3\) such that \(x\in B_3\subseteq B_1\cap B_2\).
The basis sets here are precisely the nonempty cofinite subsets. The set \(X\) itself is one of them, since \(X\setminus\varnothing=X\). Because \(X\) is infinite and \(F\) is finite, \(X\setminus F\ne\varnothing\) for every finite \(F\). These observations take care of the nonempty-set and covering requirements; the intersection condition is just as direct.
Verifying the Basis Criterion
Proof. Every \(B_F\in\mathcal{B}\) is nonempty, since otherwise \(X=F\) would be finite. Also \(X=B_{\varnothing}\in\mathcal{B}\), so every \(x\in X\) belongs to a basis element. Now take \(B_F,B_G\in\mathcal{B}\), and suppose \(x\in B_F\cap B_G\). By the definition of complement, \(B_F\cap B_G=(X\setminus F)\cap(X\setminus G)=X\setminus(F\cup G)\). The set \(F\cup G\) is finite, so \(X\setminus(F\cup G)\in\mathcal{B}\). It contains \(x\) and is equal to \(B_F\cap B_G\), so it is a suitable \(B_3\). Both basis conditions hold. \(\square\)
The construction gives more than a check of the criterion: it also gives an exact description of the topology. Recall that the topology generated by a basis consists of all unions of basis elements, including the empty union \(\varnothing\). We now show that these unions are exactly the cofinite subsets together with \(\varnothing\).
Proof. First consider a nonempty union of basis elements, written \(U=\bigcup_{i\in J}(X\setminus F_i)\), where \(J\ne\varnothing\) and each \(F_i\) is finite. An element \(x\in X\) is outside \(U\) exactly when it is outside every \(X\setminus F_i\). Thus \(X\setminus U=\bigcap_{i\in J}F_i\). Choose one index \(i_0\in J\). The intersection is a subset of \(F_{i_0}\), which is finite, so \(X\setminus U\) is finite. Therefore every nonempty union of basis elements is cofinite. The empty union is \(\varnothing\), so every set in the generated topology is either empty or cofinite.
Conversely, let \(U\subseteq X\) be cofinite. Then \(F=X\setminus U\) is finite, and \(U=X\setminus F=B_F\), which is itself a basis element and hence open in the generated topology. The empty set is also open, as the empty union. We have proved both inclusions, so the generated topology is exactly the cofinite topology. \(\square\)
In particular, on an infinite set every nonempty open set in this topology has finite complement. This is a strong restriction: an open set cannot omit infinitely many points. At the same time, the basis description makes a basic neighborhood of any point easy to specify: choose finitely many points to exclude, making sure the point of interest is not among them.
Worked Example: Basic Neighborhoods in the Cofinite Topology on \(\mathbb{N}\)
Let \(X=\mathbb{N}=\{1,2,3,\dots\}\), equipped with the topology generated by complements of finite subsets. Consider the basis elements \(B_{\{1,4\}}\) and \(B_{\{4,7\}}\). Written out, these are \(B_{\{1,4\}}=\mathbb{N}\setminus\{1,4\}\) and \(B_{\{4,7\}}=\mathbb{N}\setminus\{4,7\}\). Their intersection is \(\mathbb{N}\setminus\{1,4,7\}\), because a natural number must avoid all three excluded values \(1,4,7\) to belong to both sets.
For a point check, \(2\) is in both basis elements, and it is in \(\mathbb{N}\setminus\{1,4,7\}\). The point \(4\) is in neither, while \(1\) is excluded from the first and \(7\) from the second. The intersection is itself the basis element \(B_{\{1,4,7\}}\), as the criterion predicts.
For another neighborhood calculation, the basis element \(B_{\{2,5\}}=\mathbb{N}\setminus\{2,5\}\) contains \(8\), since \(8\notin\{2,5\}\). It does not contain \(2\) or \(5\), since each is explicitly excluded. Thus it is a basic open neighborhood of \(8\), but not of either excluded point.
Check: The complement of \(\mathbb{N}\setminus\{1,4,7\}\) in \(\mathbb{N}\) is exactly \(\{1,4,7\}\), a finite set. So the intersection computed above is cofinite and open, in agreement with the theorem.
Worked Example: A Union of Basis Elements in \(\mathbb{R}\)
In this example, equip \(X=\mathbb{R}\) with the cofinite topology, not its standard topology. Compute the union \(B_{\{1,4\}}\cup B_{\{2,4,9\}}\).
The two basis sets are \(\mathbb{R}\setminus\{1,4\}\) and \(\mathbb{R}\setminus\{2,4,9\}\). A real number is outside their union exactly when it is outside both sets, which means it belongs to both excluded finite sets. Their shared elements are obtained by comparing \(\{1,4\}\) with \(\{2,4,9\}\): the only shared element is \(4\). Hence \[ B_{\{1,4\}}\cup B_{\{2,4,9\}}=\mathbb{R}\setminus\{4\}. \]
For instance, \(1\) is missing from \(B_{\{1,4\}}\), but it belongs to \(B_{\{2,4,9\}}\), so it belongs to the union. The point \(4\) is excluded from both sets, so it is not in the union. Every real number other than \(4\) belongs to at least one of the two sets: a number other than \(1\) or \(4\) belongs to the first, and \(1\) belongs to the second. This verifies the computed union point by point.
Check: The complement of the union is \(\{4\}\), which is finite. Therefore the answer is cofinite and open in the specified topology.
Worked Example: A Subset That Is Not Open
Let \(X=\mathbb{N}\) with the cofinite topology, and let \(A=\{2,4,6,\dots\}\), the set of positive even natural numbers. The complement \(X\setminus A=\{1,3,5,\dots\}\) is infinite: it contains the distinct odd numbers \(2n-1\) for every \(n\in\mathbb{N}\). Thus \(A\) is not cofinite.
The open-set description shows that \(A\) is not open. It is not empty, and if it were open its complement would have to be finite; its complement is infinite. In contrast, \(X\setminus\{2,5\}\) is open, because its complement is the finite set \(\{2,5\}\). It is also a basis element, not merely a union of several basis elements.
Check: The first set omits infinitely many points of \(X\), while the second omits exactly the two points \(2\) and \(5\). Only the second meets the cofinite condition.
A Common Intersection Error
Common pitfall. When intersecting complements, it is easy to intersect the excluded sets instead of taking their union. For example, one might incorrectly calculate \((X\setminus\{1,4\})\cap(X\setminus\{4,7\})=X\setminus\{4\}\). That answer is wrong: it includes \(1\), even though \(1\notin X\setminus\{1,4\}\); it also includes \(7\), even though \(7\notin X\setminus\{4,7\}\). The correct calculation is \[ (X\setminus\{1,4\})\cap(X\setminus\{4,7\}) =X\setminus\bigl(\{1,4\}\cup\{4,7\}\bigr) =X\setminus\{1,4,7\}. \]
The set identity used here follows directly from membership: \(x\) is in the left side exactly when \(x\notin\{1,4\}\) and \(x\notin\{4,7\}\), which is exactly when \(x\notin\{1,4,7\}\). Keeping track of the excluded points is often the quickest way to avoid reversing a union and an intersection.
What the Basis Description Provides
The basis description is useful in two complementary ways. To construct an open set, take a union of complements of finite sets. To test whether a set is open, it is enough to check whether it is empty or its complement is finite. The proof above establishes that these are not merely convenient tests: they characterize precisely the topology generated by the basis.
The assumption that \(X\) is infinite also matters. If \(X\) were finite, the complement of a finite subset could be empty, and every subset of \(X\) would have finite complement. The collection of all complements of finite subsets would then be the collection of all subsets, giving the discrete topology rather than the nontrivial cofinite topology described here. On an infinite set, by contrast, every basis element is nonempty, and subsets with infinite complement are excluded from the topology unless they are empty.
Check Your Understanding
Use the basis and the open-set characterization to answer the following.
- For an infinite set \(X\), compute \((X\setminus F)\cap(X\setminus G)\) in terms of \(F\) and \(G\), and explain why the result is a basis element.
- In the cofinite topology on \(\mathbb{Z}\), is \(\mathbb{Z}\setminus\{-2,0,3\}\) open? Give the reason.
- In the cofinite topology on \(\mathbb{N}\), is the set of odd positive integers open? Explain using its complement.
- Prove directly that every nonempty union of basis elements has finite complement.
- Why does the collection of complements of finite subsets give the discrete topology when \(X\) is finite?