Open Covers and the Lindelof Property
An open cover may have many more sets than are needed to cover a space. The Lindelof property asks whether, no matter how large an open cover is, some countable collection of its members still covers the whole space. Unlike compactness, it does not require a finite subcover. The distinction matters: an infinite space can require infinitely many members of a cover and still be Lindelof.
The family in the definition can be uncountable; it is the subfamily that must be at most countable. The definition also quantifies over every open cover: finding one cover with a countable subcover does not establish that a space is Lindelof.
Countable Spaces
Proof. Let \((X,\mathcal{T})\) be countable, and let \(\mathcal{U}\) be an open cover of \(X\). If \(X=\varnothing\), the empty subfamily covers \(X\). If \(X\) is finite and nonempty, for each \(x\in X\) choose a member \(U_x\in\mathcal{U}\) containing \(x\). The finitely many chosen sets cover \(X\).
Now suppose \(X\) is countably infinite and list its points as \(x_1,x_2,\ldots\). For each \(n\in\mathbb{N}\), choose \(U_n\in\mathcal{U}\) with \(x_n\in U_n\), which is possible because \(\mathcal{U}\) covers \(X\). This countable selection uses the usual countable choice principle of standard set theory. The family \(\{U_n:n\in\mathbb{N}\}\) is a subfamily of the original cover, and every point \(x_n\) belongs to its selected set \(U_n\). It therefore covers \(X\), proving that \(X\) is Lindelof. \(\square\)
Worked Example: A Countable Space That Is Not Compact
Give \(\mathbb{N}=\{1,2,\ldots\}\) the discrete topology, in which every subset is open. For each \(n\in\mathbb{N}\), let \(U_n=\{1,2,\ldots,n\}\). Each \(U_n\) is open, and the family \(\mathcal{U}=\{U_n:n\in\mathbb{N}\}\) covers \(\mathbb{N}\): the point \(k\) belongs to \(U_k\).
The whole family is a countable subcover. No finite subfamily covers \(\mathbb{N}\): if its largest index is \(m\), the union of its members is contained in \(\{1,\ldots,m\}\), so it misses \(m+1\). (The empty subfamily also does not cover the nonempty space.) Thus this cover has no finite subcover, but it does have a countable subcover.
Check: for any \(m\in\mathbb{N}\), \(U_m=\{1,\ldots,m\}\), and \(m+1\notin U_m\). On the other hand, every \(k\in\mathbb{N}\) belongs to \(U_k\). The cover therefore demonstrates both claims directly.
An Uncountable Discrete Space
In the discrete topology, every singleton is open. If the underlying set is uncountable, its family of singletons forms an open cover that cannot be reduced to a countable subcover: each selected singleton covers only its one point. Here is an explicit uncountable example.
Worked Example: The Power Set with the Discrete Topology
Let \(X=\mathcal{P}(\mathbb{N})\), the set of all subsets of \(\mathbb{N}\), with the discrete topology. First, \(X\) is uncountable. If its elements could be listed as \(A_1,A_2,\ldots\), form the subset \(D=\{n\in\mathbb{N}:n\notin A_n\}\). For each \(k\), membership of \(k\) in \(D\) is opposite to membership of \(k\) in \(A_k\). Hence \(D\ne A_k\) for every \(k\), contradicting that the list contains every subset of \(\mathbb{N}\).
Now consider the family \(\mathcal{U}=\{\{A\}:A\in X\}\). Every member is open because the topology is discrete, and the family covers \(X\) because each \(A\in X\) belongs to its singleton \(\{A\}\). If a subfamily of \(\mathcal{U}\) is countable, it contains at most countably many singletons and therefore covers at most countably many points of \(X\). Since \(X\) is uncountable, that subfamily cannot cover \(X\). This open cover has no countable subcover, so \(X\) is not Lindelof.
Check: selecting the singleton \(\{A\}\) covers \(A\) and no other point. Thus a subfamily indexed by a countable set covers at most countably many points, whereas the diagonal argument shows that \(X\) has more than countably many points.
The Cocountable Topology
The cocountable topology reverses the behavior of the discrete example. On an uncountable set, its open sets are the empty set and those whose complements are countable. A nonempty open set can omit only countably many points, which lets one member of an open cover take care of all but countably many points at once.
Worked Example: An Uncountable Cocountable Space
Let \(X=\mathcal{P}(\mathbb{N})\), which is uncountable by the diagonal argument above, and give it the cocountable topology. We first check that this is a topology. The sets \(\varnothing\) and \(X\) belong to it. For an arbitrary union, if every set in the union is empty, the union is empty. Otherwise choose a nonempty member \(U_0\). The complement of the union is contained in \(X\setminus U_0\), which is countable, so the union is open. For a finite intersection of nonempty open sets \(U_1,\ldots,U_m\), its complement is \(\bigcup_{j=1}^m(X\setminus U_j)\), a finite union of countable sets and hence countable. If any set in the intersection is empty, the intersection is open as well. These checks verify the topology axioms.
Let \(\mathcal{U}\) be any open cover of \(X\). Since \(X\ne\varnothing\), at least one member \(U_0\in\mathcal{U}\) is nonempty. Because \(U_0\) is open, \(C=X\setminus U_0\) is countable. If \(C=\varnothing\), then \(U_0=X\) is already a one-set subcover.
If \(C\ne\varnothing\), list its points as \(c_1,c_2,\ldots\) when it is infinite, or as a finite list when it is finite. For each listed point \(c_n\), choose \(U_n\in\mathcal{U}\) with \(c_n\in U_n\); this again uses only countable choice. The selected family consisting of \(U_0\) and the sets \(U_n\) is at most countable. Every point in \(U_0\) is covered by \(U_0\), and every point outside \(U_0\) belongs to \(C\), so is covered by its chosen \(U_n\). Thus the selected family is a subcover. Since the original cover was arbitrary, this uncountable space is Lindelof.
Check: the points not covered by \(U_0\) are exactly in \(C=X\setminus U_0\). The construction selects a cover member for each such point; therefore no point of \(X\) is left uncovered, and only countably many members are selected.
Compactness and a Common Pitfall
Compactness requires every open cover to have a finite subcover. A finite family is at most countable, so the definition immediately gives one implication.
Proof. Let \(X\) be compact and let \(\mathcal{U}\) be any open cover of \(X\). By compactness, \(\mathcal{U}\) has a finite subcover. A finite subfamily is at most countable, so \(\mathcal{U}\) has a countable subcover. This holds for every open cover, and therefore \(X\) is Lindelof. \(\square\)
The converse fails: the cover of discrete \(\mathbb{N}\) by \(U_n=\{1,\ldots,n\}\) has a countable subcover but no finite subcover. Consequently, this space is Lindelof by the theorem on countable spaces, but it is not compact.
Common pitfall. A wrong conclusion is: “The cover \(\{U_n:n\in\mathbb{N}\}\) has no finite subcover, so \(\mathbb{N}\) is not Lindelof.” The calculation only rules out a finite subcover: for finitely many indices with maximum \(m\), the union misses \(m+1\). It does not rule out a countable subcover. In fact, the entire family is countable and covers every \(k\in\mathbb{N}\), since \(k\in U_k\). The correction is to check the definition’s requirement—an at most countable subcover—not to impose compactness’s stronger finite requirement.
The examples show why the number of points alone does not determine whether a space is Lindelof. Countable spaces are Lindelof regardless of their topology; an uncountable discrete space fails the property; and an uncountable space with the cocountable topology satisfies it. The decisive question is how many members of a given open cover are needed to cover the points.
Check Your Understanding
For each question, use the open-cover definition and state whether finite or countably infinite subcovers are being considered.
- Why does every countable topological space have a countable subcover for each open cover?
- In an uncountable discrete space, why does the cover by all singletons have no countable subcover?
- In the cocountable topology, why does choosing one nonempty member of a cover leave at most countably many points to cover?
- Does the cover \(U_n=\{1,\ldots,n\}\) of discrete \(\mathbb{N}\) have a finite subcover? Does it have a countable subcover?
- Which implication holds between compactness and the Lindelof property, and why does the example on \(\mathbb{N}\) show that the converse fails?