The Cocountable Topology: Lindelof But Not Second Countable

The Cocountable Topology

A space can have a strong countability property for open covers without having a countable basis. The cocountable topology gives a direct example: on an uncountable set, every open cover has a countable subcover, but no countable family can describe all the local open sets as a basis. The distinction is between selecting open sets to cover the space and finding basis elements inside every open neighborhood.

Let \(X\) be an uncountable set. A subset \(U\subseteq X\) is called cocountable if \(X\setminus U\) is countable. We equip \(X\) with the cocountable topology: its open sets are \(\varnothing\) and all cocountable subsets of \(X\). Here “countable” includes finite sets.

First, the stated collection really is a topology. It contains \(\varnothing\) and \(X\), since \(X\setminus X=\varnothing\) is countable. If \(U\) and \(V\) are cocountable, then \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\) is countable, so their intersection is cocountable. For any family of open sets, if all of them are empty, their union is empty. Otherwise, choose one nonempty member \(U\) of the family. The complement of the union is contained in \(X\setminus U\), which is countable, so the union is cocountable. Thus arbitrary unions and finite intersections are open, as required.

Why the Space Is Lindelof

Proof. Let \(\mathcal{U}\) be an open cover of \(X\). Since \(X\) is nonempty and \(\mathcal{U}\) covers \(X\), some member \(U_0\in\mathcal{U}\) is nonempty. By the definition of the topology, \(C=X\setminus U_0\) is countable. If \(C=\varnothing\), then \(U_0=X\) is already a finite subcover. Otherwise, list the points of \(C\) as \(c_1,c_2,\ldots\), stopping if \(C\) is finite. For each listed point \(c_n\), choose \(U_n\in\mathcal{U}\) with \(c_n\in U_n\); these choices use only countable choice. The family consisting of \(U_0\) and the chosen \(U_n\)'s is countable. It covers \(U_0\), and it covers every point of \(C\) by construction, so it covers \(X\). Therefore \(\mathcal{U}\) has a countable subcover. \(\square\)

The argument does not require a countable subcover to be finite. The remainder \(C\) may have infinitely many points, and an open cover can require infinitely many of its members to cover those points. What matters is that after choosing one nonempty open set, only countably many points remain to be covered.

Countable Intersections and the Basis Obstruction

A useful feature of this topology is that a countable intersection of nonempty open sets is still nonempty. This fact is not the Lindelof property; it will instead prevent a countable family from being a basis.

Proof. Since each \(U_n\) is nonempty and open, \(C_n=X\setminus U_n\) is countable. The complement of the intersection is \(X\setminus\bigcap_{n\in\mathbb{N}}U_n=\bigcup_{n\in\mathbb{N}}C_n\), which is countable because a countable union of countable sets is countable. If the intersection were empty, this union would equal \(X\), making \(X\) countable. That contradicts the assumption that \(X\) is uncountable. Therefore the intersection is nonempty. \(\square\)

Proof. Suppose, to the contrary, that \(\mathcal{B}\) is a countable basis. It cannot consist only of empty sets, since a basis must provide a basis element containing each point of \(X\). List its nonempty members as a finite or countable family. The intersection of these nonempty basis elements is nonempty: in the countably infinite case this follows from the proposition, and in the finite case it follows by the same complement argument, since a finite union of countable sets is countable. Choose \(x\) in this intersection.

Because \(X\) is uncountable, there is \(y\in X\) with \(y\ne x\). The set \(X\setminus\{x\}\) is open, since its complement is a singleton. It contains \(y\), so the basis property gives \(B\in\mathcal{B}\) with \(y\in B\subseteq X\setminus\{x\}\). In particular, \(B\) is nonempty. But \(x\) belongs to every nonempty member of \(\mathcal{B}\), including \(B\). This contradicts \(B\subseteq X\setminus\{x\}\). Thus no countable basis exists. \(\square\)

A Common Pitfall and a Countable Comparison

Common pitfall. One might guess that the sets with finite complement form a countable basis. There are two problems with this guess. First, the family of all finite-complement sets is not countable when \(X\) is uncountable: it includes \(X\setminus\{x\}\) for every \(x\in X\), and these are distinct sets. Second, even if one tries to select a countable subfamily of them, that subfamily cannot be a basis. If its members are \(X\setminus F_n\), where each \(F_n\) is finite, then \(\bigcup_n F_n\) is countable. Choose \(x\in X\setminus\bigcup_n F_n\), possible because \(X\) is uncountable. Every selected set \(X\setminus F_n\) contains \(x\). For any \(y\ne x\), none of those sets can both contain \(y\) and be contained in \(X\setminus\{x\}\). So the basis condition fails at the open neighborhood \(X\setminus\{x\}\) of \(y\).

This also illustrates why Lindelofness does not give second countability. Lindelofness selects countably many members from a cover that is already given. A basis must meet a different demand: for every open neighborhood and every point in it, some basis element must fit inside that neighborhood.

The uncountability assumption is essential for the failure of second countability:

The key contrast is now explicit. On an uncountable cocountable space, the countable complement of one nonempty open set makes open covers manageable, yielding the Lindelof property. At the same time, countably many nonempty open sets have a common point, which obstructs any countable basis. These conclusions come from the same definition but address distinct ways of organizing open sets.