One Countable Family for the Whole Topology
First-countability asks for a countable local base at each point. The countable family can depend on the point, so this condition does not necessarily give one countable collection that describes open sets throughout the space. Second-countability makes that stronger, global demand: a single countable basis must generate every open set.
This distinction matters because a global basis can be used across the whole space at once. Besides giving a countable local base at every point, it allows us to reduce arbitrary open covers to countable ones. We will prove these consequences and see how second-countability behaves under passage to subspaces.
Definition and Immediate Consequence
The equivalence in the definition is the basis property established earlier in this course. The members of \(\mathcal{B}\) need not be neighborhoods of every point; rather, those that contain a particular point supply local neighborhoods there.
Proof. Let \(\mathcal{B}\) be a countable basis for \(X\), and fix \(x\in X\). Define \(\mathcal{B}_x=\{B\in\mathcal{B}:x\in B\}\). This collection is countable because it is a subcollection of a countable family. Each of its members is an open neighborhood of \(x\). If \(U\) is any open neighborhood of \(x\), the basis property gives \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Then \(B\in\mathcal{B}_x\), so \(\mathcal{B}_x\) is a local base at \(x\). Since \(x\) was arbitrary, \(X\) is first-countable. \(\square\)
The converse need not hold: having a countable local base at each point does not require the local bases at all points to fit into one countable family. The theorem gives a direct way to use results about first-countable spaces, including the sequential characterization of closure and the sequential criterion for continuity from “First-Countable Spaces.”
Worked Example: A Countable Basis for the Real Line
Consider \(\mathbb{R}\) with its usual topology. Let \(\mathcal{B}=\{(p,q):p,q\in\mathbb{Q},\ p<q\}\). The rationals are countable, so the collection of ordered pairs \((p,q)\in\mathbb{Q}\times\mathbb{Q}\) is countable; hence \(\mathcal{B}\) is countable.
To check the basis property, take an open set \(U\subseteq\mathbb{R}\) and \(x\in U\). There is \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). By the density of \(\mathbb{Q}\) in \(\mathbb{R}\), choose rationals \(p,q\) satisfying \(x-\varepsilon<p<x<q<x+\varepsilon\). Then \(x\in(p,q)\subseteq U\). Thus \(\mathcal{B}\) is a countable basis, and \(\mathbb{R}\) is second-countable.
This example also illustrates why basis members are useful: the endpoints can be selected from a fixed countable set, even though the open set \(U\) and the point \(x\) were arbitrary.
Second-Countability Passes to Subspaces
The subspace topology can be described using intersections with open sets of the original space. Intersecting a countable basis with the subspace therefore gives a countable basis there as well.
Proof. Let \(\mathcal{B}\) be a countable basis for \(X\), and let \(A\subseteq X\). By the basis theorem for subspaces, the collection \(\mathcal{B}_A=\{A\cap B:B\in\mathcal{B}\}\) is a basis for the subspace topology on \(A\). It is countable because it is the image of the countable collection \(\mathcal{B}\) under the operation \(B\mapsto A\cap B\). Thus \(A\) is second-countable. This argument also covers \(A=\varnothing\), whose subspace topology has a finite, and therefore countable, basis. \(\square\)
Worked Example: A Subspace of the Real Line
Let \(A=[-2,2]\subseteq\mathbb{R}\). The family \(\{A\cap(p,q):p,q\in\mathbb{Q},\ p<q\}\) is countable, since it is obtained from the countable rational-interval basis by taking intersections with \(A\). By the subspace basis theorem, it is a basis for \(A\).
For example, a neighborhood of the endpoint \(-2\) in \(A\) can be \(A\cap(-3,-1)= [-2,-1)\). This is a subspace-open set even though it is not open in \(\mathbb{R}\). To find a basis member inside it that contains \(-2\), take rational \(p,q\) with \(p<-2<q<-1\), such as \(p=-3\) and \(q=-3/2\). Then \(A\cap(p,q)=[-2,-3/2)\subseteq[-2,-1)\). The intersections account for the subspace’s neighborhoods, including those at its endpoints.
Open Covers and the Lindelöf Property
A second important consequence is that open covers can be reduced to countable subcovers. A space is called Lindelöf if every open cover has a countable subcover. Compactness requires a finite subcover; the Lindelöf property weakens “finite” to “countable.”
Proof. Let \(\mathcal{B}\) be a countable basis for \(X\), and let \(\mathcal{U}\) be an open cover of \(X\). Consider the subcollection \(\mathcal{B}'=\{B\in\mathcal{B}: B\subseteq U\text{ for some }U\in\mathcal{U}\}\). It is countable because it is a subcollection of \(\mathcal{B}\).
For each \(B\in\mathcal{B}'\), choose one set \(U_B\in\mathcal{U}\) such that \(B\subseteq U_B\). The resulting family \(\{U_B:B\in\mathcal{B}'\}\) is countable. We verify that it covers \(X\). If \(x\in X\), choose \(U\in\mathcal{U}\) with \(x\in U\). The set \(U\) is open, so the basis property supplies \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). This \(B\) belongs to \(\mathcal{B}'\), and \(x\in B\subseteq U_B\). Hence \(x\) lies in a member of the selected family. Since every \(x\in X\) is covered, the family is a countable subcover. If \(X=\varnothing\), the empty subfamily is already a subcover. Therefore \(X\) is Lindelöf. \(\square\)
Worked Example: Reducing a Cover of the Real Line
For each \(n\geq1\), let \(U_n=(-n,n)\). The family \(\{U_n:n\geq1\}\) covers \(\mathbb{R}\): given \(x\in\mathbb{R}\), choose an integer \(n>|x|\), which gives \(x\in(-n,n)\). This particular cover is already countable, and it has no finite subcover because any finite collection of these intervals is contained in \((-N,N)\) for its largest index \(N\), leaving points outside that interval uncovered.
The Lindelöf theorem is more general than this example: it applies to every open cover of \(\mathbb{R}\), not just this one. Since the rational intervals form a countable basis, any open cover of \(\mathbb{R}\) has a countable subcover, even if the original cover is uncountable. This conclusion does not say that the subcover must be finite.
A Countable Basis Also Gives a Countable Dense Set
A subset \(D\) of \(X\) is dense in \(X\) if \(\overline{D}=X\). Equivalently, every nonempty open subset of \(X\) meets \(D\), by the neighborhood characterization of closure. A space with a countable dense subset is called separable. This gives a further connection to the next topic in the series.
Proof. Let \(\mathcal{B}\) be a countable basis for \(X\). For each nonempty \(B\in\mathcal{B}\), choose a point \(x_B\in B\), and let \(D=\{x_B:B\in\mathcal{B},\ B\ne\varnothing\}\). The set \(D\) is countable. If \(U\) is a nonempty open set, choose \(x\in U\). The basis property gives \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Since \(B\) is nonempty, its chosen point \(x_B\) belongs to \(D\cap U\). Thus every nonempty open set meets \(D\), so \(D\) is dense. If \(X=\varnothing\), take \(D=\varnothing\), which is dense in \(X\). Therefore \(X\) is separable. \(\square\)
Recognizing Spaces That Are Not Second-Countable
A useful test is to ask what a basis must contain in order to distinguish points and neighborhoods. In a discrete space every subset is open, so a basis must be able to refine each singleton \(\{x\}\). This places a restriction on how many points the space can have if the basis is countable.
Worked Example: An Uncountable Discrete Space
Let \(X\) be uncountable with the discrete topology. Suppose \(\mathcal{B}\) were a basis for this topology. Fix \(x\in X\). Since \(\{x\}\) is open and \(x\in\{x\}\), the basis property requires some \(B_x\in\mathcal{B}\) such that \(x\in B_x\subseteq\{x\}\). These inclusions force \(B_x=\{x\}\).
Consequently, every singleton \(\{x\}\) must occur as a member of \(\mathcal{B}\). Distinct points give distinct singleton sets, so \(\mathcal{B}\) must be uncountable. This contradicts the supposition that it was countable. Thus an uncountable discrete space is not second-countable. In contrast, a countable discrete space is second-countable: its collection of all singletons is a countable basis.
The contrast shows that second-countability is a restriction on how many basic open sets are needed, not simply a statement that the underlying set is countable. The real line is uncountable but second-countable; an uncountable discrete space cannot be second-countable.
What the Countable Basis Provides
A countable basis is one global collection of open sets that can refine every open neighborhood. This makes second-countability stronger than first-countability, and it has several useful consequences: second-countable spaces are first-countable, their subspaces remain second-countable, every open cover has a countable subcover, and every such space has a countable dense subset.
A common pitfall is to confuse a countable local base at each point with a countable basis for the whole space. First-countability is local; second-countability requires one countable family to serve everywhere. Another is to confuse the Lindelöf conclusion with compactness: a countable subcover need not be finite, as the cover of \(\mathbb{R}\) by \((-n,n)\) demonstrates.
Check Your Understanding
Use the definition and proved results to answer the following questions.
- What refinement property must a basis satisfy for each point in an open set?
- How does a countable basis produce a countable local base at a fixed point?
- Why does intersecting a countable basis with a subspace give a countable family?
- In the Lindelöf proof, why does every point belong to one of the selected cover sets?
- Why must every singleton occur in a basis for an uncountable discrete space?
- Does second-countability imply that every open cover has a finite subcover, or only a countable one?