One Topology, Two Different Countability Properties
A space is Lindelof if every open cover has a countable subcover. It is separable if it has a countable dense subset. These properties concern different tasks: Lindelofness asks how many open sets suffice to cover the space, while separability asks how many points suffice to meet every nonempty open set. An uncountable set with the cocountable topology provides a direct example of a space that is Lindelof but not separable.
Throughout, “countable” includes finite sets. Let \(X\) be an uncountable set. Its cocountable topology consists of the empty set and all subsets whose complements in \(X\) are countable. We first verify that this definition really gives a topology.
Proof that this is a topology. The sets \(\varnothing\) and \(X\) belong to \(\mathcal{T}\), since \(X\setminus X=\varnothing\). For any family of open sets, if every member is empty then its union is empty. Otherwise choose a nonempty member \(U\); the complement of the union is a subset of \(X\setminus U\), so it is countable. Thus the union is open. For two open sets \(U,V\), if either is empty then \(U\cap V=\varnothing\). Otherwise \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\), a union of two countable sets, hence countable. So \(U\cap V\) is open. These are the topology axioms. \(\square\)
Here we use the standard set-theoretic fact that a finite union of countable sets is countable. The same fact for a countable union is used below; in the usual set theory with countable choice, a countable union of countable sets is countable.
Why Every Open Cover Has a Countable Subcover
Proof. Let \(\mathcal{U}\) be an open cover of \(X\). Since \(X\) is nonempty, some member \(U_0\in\mathcal{U}\) contains a point of \(X\), so \(U_0\ne\varnothing\). By the definition of the topology, \(C=X\setminus U_0\) is countable. If \(C=\varnothing\), then \(U_0=X\) and the one-set family \(\{U_0\}\) is a subcover. Otherwise, for each \(x\in C\), choose a member \(U_x\in\mathcal{U}\) with \(x\in U_x\), which exists because \(\mathcal{U}\) covers \(X\). The family consisting of \(U_0\) and all the \(U_x\) is countable because \(C\) is countable. It covers \(U_0\) and every point of \(C\), hence all of \(X\). This selection uses countable choice for the countable set \(C\), as available in the usual set theory. Therefore every open cover has a countable subcover. \(\square\)
The argument depends on one key feature: a single nonempty open set leaves at most countably many points uncovered. Those remaining points can be handled one at a time using cover members. No countable base is needed, and the chosen subcover need not be finite.
Worked Example: The Cocountable Topology on \(\mathbb{R}\)
Give \(\mathbb{R}\) the cocountable topology, not its standard topology. Consider the open cover \(\mathcal{U}=\{U_n:n\in\mathbb{N}\}\), where \(U_n=\mathbb{R}\setminus\{n,n+1,n+2,\ldots\}\).
Each set is open. Its complement \(\{n,n+1,n+2,\ldots\}\) is countable, so \(U_n\) is open in the cocountable topology.
The family covers \(\mathbb{R}\). If \(x\) is not a positive integer, then \(x\) belongs to every \(U_n\), since no positive integer tail contains \(x\). If \(x=k\in\mathbb{N}\), then \(k\notin U_n\) when \(n\le k\), but \(k\in U_{k+1}\), because \(k\notin\{k+1,k+2,\ldots\}\). Thus every real number belongs to at least one member of the cover.
No finite subfamily covers. In a nonempty finite selection of indices, let \(N\) be the largest index selected. The sets increase with their indices: if \(n\le N\), then \(\{N,N+1,\ldots\}\subseteq\{n,n+1,\ldots\}\), so \(U_n\subseteq U_N\). The finite union is therefore \(U_N\). It misses every integer \(k\ge N\), so it is not all of \(\mathbb{R}\). The empty selection also fails to cover \(\mathbb{R}\).
Check: the full family is a countable cover, while every finite subfamily misses infinitely many positive integers. This verifies that this particular cover has no finite subcover, even though the space is Lindelof.
Exactly Which Subsets Are Dense?
The open-set test for density is especially useful here. In any topological space, \(D\) is dense if and only if every nonempty open set meets \(D\). Indeed, the neighborhood characterization of closure says that \(x\in\overline{D}\) exactly when every open neighborhood of \(x\) meets \(D\). If \(\overline{D}=X\), every nonempty open set \(U\) contains some \(x\), and then \(U\) is a neighborhood of \(x\) that must meet \(D\). Conversely, if every nonempty open set meets \(D\), every neighborhood of every point meets \(D\), giving \(\overline{D}=X\).
Proof. Suppose first that \(D\) is countable. Then \(X\setminus D\) is open by the definition of the cocountable topology, and it is nonempty because \(X\) is uncountable whereas \(D\) is countable. This nonempty open set is disjoint from \(D\), so the open-set test shows that \(D\) is not dense.
Conversely, suppose that \(D\) is uncountable. Let \(U\) be any nonempty open set. Its complement \(C=X\setminus U\) is countable. If \(U\cap D=\varnothing\), then \(D\subseteq C\), which would make \(D\) countable. This contradicts the assumption that \(D\) is uncountable. Hence \(U\cap D\ne\varnothing\) for every nonempty open set \(U\), so \(D\) is dense. \(\square\)
This characterizes density completely: in this topology, no finite or countably infinite subset is dense, while every uncountable subset is dense. In particular, the whole space \(X\) is uncountable and dense in itself, but it has no countable dense subset. Thus it is not separable.
Worked Example: Countable and Uncountable Subsets of \(\mathbb{R}\)
Again give \(X=\mathbb{R}\) the cocountable topology. We compare \(D=\mathbb{Q}\) with \(E=\mathbb{R}\setminus\mathbb{Q}\). The density conclusion depends on this topology; it is not a calculation in the standard topology on \(\mathbb{R}\).
The rationals are not dense here. The set \(\mathbb{Q}\) is countable, so \(U=\mathbb{R}\setminus\mathbb{Q}\) is open. It is nonempty because \(\mathbb{R}\) is uncountable and \(\mathbb{Q}\) is countable. Also \(U\cap\mathbb{Q}=\varnothing\). Therefore \(\mathbb{Q}\) is not dense in the cocountable topology.
The irrationals are dense here. The set \(E\) is uncountable: if it were countable, then \(\mathbb{R}=\mathbb{Q}\cup E\) would be countable, contradicting that \(\mathbb{R}\) is uncountable. By the theorem, \(E\) is dense. More explicitly, for any nonempty open set \(U=\mathbb{R}\setminus C\), the set \(C\cup\mathbb{Q}\) is countable. It cannot contain all of \(\mathbb{R}\), so \(\mathbb{R}\setminus(C\cup\mathbb{Q})=U\cap E\ne\varnothing\).
Check: the two conclusions follow from the same criterion: \(\mathbb{Q}\) is countable, whereas \(E\) is uncountable. The open set \(\mathbb{R}\setminus\mathbb{Q}\) also directly witnesses why \(\mathbb{Q}\) fails to be dense.
Worked Example: A Countable Dense Subset Cannot Be Recovered from a Cover
Let \(X\) be any uncountable set with the cocountable topology, and let \(D\subseteq X\) be countable. Consider the single open set \(U=X\setminus D\).
The complement \(X\setminus U=D\) is countable, so \(U\) is open. It is nonempty because \(X\) is uncountable and \(D\) is countable. The one-set family \(\{U\}\) covers \(U\), but it does not cover \(X\): every point of \(D\) lies outside \(U\). Also \(U\cap D=\varnothing\), so \(D\) fails the open-set test for density.
This computation illustrates the different roles of points and open sets. Lindelofness allows a countable subfamily to cover the space when the starting family covers all of \(X\). It does not say that a countable set of points meets every nonempty open set. Here the open complement of any countable set gives an explicit nonempty open set that the set misses.
A Common Mistake: Confusing Lindelofness with Separability
A tempting but incorrect argument says: “Every nonempty open set is uncountable, so every countable set must meet every nonempty open set.” The conclusion does not follow. An uncountable set can be disjoint from a particular countable set. For example, in the cocountable topology on \(\mathbb{R}\), the open set \(\mathbb{R}\setminus\mathbb{Q}\) is uncountable but is disjoint from \(\mathbb{Q}\). Thus \(\mathbb{Q}\) is not dense.
The corrected density test asks whether the candidate set meets each nonempty open set. For a countable candidate \(D\), the open set \(\mathbb{R}\setminus D\) is nonempty and misses \(D\), so \(D\) cannot be dense. By contrast, Lindelofness starts with a family already covering \(X\) and selects countably many members of that family. These are distinct requirements, so being Lindelof does not imply being separable.
| Property | What is counted? | Cocountable topology on uncountable \(X\) |
|---|---|---|
| Lindelof | Open sets in a subcover of an open cover | Every open cover has a countable subcover |
| Separable | Points in a dense subset | No countable subset is dense |
| Dense subset | Points that meet every nonempty open set | Exactly the uncountable subsets |
Check Your Understanding
Use the definition of the cocountable topology and the open-set test for density.
- On an uncountable set \(X\), why is \(X\setminus A\) open whenever \(A\subseteq X\) is countable?
- Show directly that a countable subset \(D\subseteq X\) is not dense by giving a nonempty open set disjoint from it.
- Prove that every uncountable subset of \(X\) is dense.
- In the cover \(U_n=\mathbb{R}\setminus\{n,n+1,\ldots\}\), which member contains a given positive integer \(k\) for the first time?
- In one sentence, distinguish the countable selection made in the Lindelof proof from the countable set ruled out by the density theorem.