One Countable Family for All Open Sets
To describe a topology, one could list every open set. A basis gives a more economical description: it is a family of open sets from which all the others can be assembled using unions. When that family is countable, the topology has a countable global description, even if the space itself is infinite.
Throughout, \((X,\mathcal{T})\) is a topological space. A family \(\mathcal{B}\) of subsets of \(X\) is a basis for \(\mathcal{T}\) if every member of \(\mathcal{B}\) is open and every open set is a union of members of \(\mathcal{B}\). A basis is countable if it is finite or can be listed in a sequence. Having a countable basis is also called second countability.
- For every \(x\in X\), there is \(B\in\mathcal{B}\) with \(x\in B\).
- If \(B_1,B_2\in\mathcal{B}\) and \(x\in B_1\cap B_2\), then there is \(B_3\in\mathcal{B}\) such that \(x\in B_3\subseteq B_1\cap B_2\).
Proof. Suppose first that \(\mathcal{B}\) is a basis for a topology. Since \(X\) is open and is a union of basis members, each \(x\in X\) belongs to some \(B\in\mathcal{B}\). If \(x\in B_1\cap B_2\), that intersection is open. Express it as a union of basis members; one of those members contains \(x\), and is contained in \(B_1\cap B_2\). Thus both conditions hold.
Conversely, suppose the two conditions hold, and let \(\mathcal{T}_{\mathcal{B}}\) be the collection of all unions of members of \(\mathcal{B}\), including the empty union \(\varnothing\). The first condition gives \(X=\bigcup_{B\in\mathcal{B}}B\), so \(X\in\mathcal{T}_{\mathcal{B}}\). Arbitrary unions of members of \(\mathcal{T}_{\mathcal{B}}\) are still unions of members of \(\mathcal{B}\). To check finite intersections, it is enough to check two such unions, say \(U=\bigcup_i B_i\) and \(V=\bigcup_j C_j\). If \(x\in U\cap V\), then \(x\in B_i\cap C_j\) for some \(i,j\). By the second condition, some \(D\in\mathcal{B}\) satisfies \(x\in D\subseteq B_i\cap C_j\subseteq U\cap V\). Doing this for every point shows \(U\cap V\) is a union of basis members; the empty intersection of two sets is also a union, namely \(\varnothing\). Hence \(\mathcal{T}_{\mathcal{B}}\) is a topology.
Finally, suppose the members of \(\mathcal{B}\) are open in an existing topology \(\mathcal{T}\), and every point of every \(U\in\mathcal{T}\) lies in some \(B\in\mathcal{B}\) contained in \(U\). Then \(U\) is exactly the union of the basis members contained in it. Conversely, if every open set is such a union, each of its points belongs to one of the sets in that union. This proves the stated characterization for a given topology. If \(X=\varnothing\), the empty family satisfies both conditions and generates the topology \(\{\varnothing\}\), so the criterion includes the empty space. \(\square\)
Worked Examples
Worked Example: The Usual Topology on \(\mathbb{R}\)
Give a countable basis for \(\mathbb{R}\) with its standard topology. Consider \(\mathcal{B}=\{(p,q):p,q\in\mathbb{Q},\ p<q\}\), where \(\mathbb{Q}\) is the set of rational numbers.
Countability. Every rational number can be written as \(m/n\) with \(m\in\mathbb{Z}\) and \(n\in\mathbb{N}\). These integer pairs can be listed by increasing \(|m|+n\); for each fixed value there are only finitely many pairs. Thus \(\mathbb{Q}\) is countable, and listing pairs of rationals by increasing positions in their lists shows that \(\mathbb{Q}\times\mathbb{Q}\), and hence \(\mathcal{B}\), is countable.
Covering points. For any \(x\in\mathbb{R}\), density of \(\mathbb{Q}\) gives rationals \(p<x<q\). Thus \(x\in(p,q)\). For completeness, density follows by choosing \(n\in\mathbb{N}\) with \(n(b-a)>1\) when \(a<b\), and setting \(m=\lfloor na\rfloor+1\). Then \(na<m\le na+1<nb\), so \(a<m/n<b\). Applying this to intervals on either side of \(x\) gives the required \(p\) and \(q\).
Intersection condition. Suppose \(x\in(p_1,q_1)\cap(p_2,q_2)\). Let \(a=\max(p_1,p_2)\) and \(b=\min(q_1,q_2)\). Then \(a<x<b\). By density choose rationals \(p,q\) with \(a<p<x<q<b\). Consequently \(x\in(p,q)\subseteq(p_1,q_1)\cap(p_2,q_2)\). Each member of \(\mathcal{B}\) is open in the standard topology, so the basis criterion shows that \(\mathcal{B}\) is a countable basis.
Check on a particular open set. For the open interval \((0,1)\), its rational-endpoint basis members contained in it are precisely the intervals \((p,q)\) with \(0\le p<q\le1\). Every \(x\in(0,1)\) lies in one of them by rational density. Their union is therefore \((0,1)\), as required.
Worked Example: A Countable Discrete Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(\mathbb{N})\). Take \(\mathcal{B}=\{\{n\}:n\in\mathbb{N}\}\).
The family is countable because its members are indexed by \(\mathbb{N}\). Every \(n\in X\) belongs to \(\{n\}\). If \(n\in\{i\}\cap\{j\}\), then \(i=j=n\), and \(\{n\}\) itself is a basis member contained in the intersection. The criterion applies. Moreover, for every \(A\subseteq\mathbb{N}\), \(A=\bigcup_{n\in A}\{n\}\), including \(A=\varnothing\) as the empty union. Thus these singletons generate every open set of the discrete topology.
Check. The set \(\{2,5\}\) is represented as \(\{2\}\cup\{5\}\); the intersection \(\{2,5\}\cap\{5,7\}=\{5\}\) is itself a basis member. This verifies the union and intersection behavior directly.
Worked Example: The Cofinite Topology on a Finite Set
Let \(X=\{a,b,c,d\}\) and give it the cofinite topology: a set is open if it is empty or its complement in \(X\) is finite. Since \(X\) is finite, every subset has finite complement. Therefore every subset is open and this topology is the discrete topology. Take \(\mathcal{B}=\{\{a\},\{b\},\{c\},\{d\}\}\).
This is a finite, hence countable, family. Each point is in its singleton. Two distinct singleton basis members have empty intersection, so there is no point to test in that case. If two singleton members have a point in common, they are the same singleton, which is a basis member contained in their intersection. The basis criterion is satisfied. Every open set is the union of the singletons of its elements; for example, \(\{a,c\}=\{a\}\cup\{c\}\), and \(\varnothing\) is the empty union. There are \(2^4=16\) subsets of \(X\), all generated this way.
Global Bases, Local Bases, and a Pitfall
A countable basis is a single countable family that works across the whole space. A countable local base, by contrast, is a family of neighborhoods chosen at one specified point. These are different definitions: checking a family at one point does not show that it generates the open sets everywhere. The basis criterion makes the global requirement explicit: it asks for coverage at every point and a suitable basis member inside each relevant intersection.
Common mistake. One might claim that all open intervals \((a,b)\) form a countable basis for \(\mathbb{R}\), on the grounds that an interval is determined by its two endpoints and “pairs of endpoints can be counted.” The calculation is invalid: the endpoints range over \(\mathbb{R}\), not over a countable set. In fact, the intervals \((x-1,x+1)\), as \(x\) ranges over \(\mathbb{R}\), are all distinct: equality of two such intervals forces equality of their left endpoints, hence of their centers. So that family is uncountable. The correction is to restrict endpoints to \(\mathbb{Q}\), which is countable and still provides intervals small enough to fit around every point inside every open set, as the real-line example showed.
Countability also depends on the space and its topology, not just on whether one can describe individual open sets. For contrast, let \(X=\mathbb{R}\) have the cofinite topology. This space has no countable basis. Suppose \(\mathcal{B}\) were a countable basis and fix \(x\in X\). For every \(y\ne x\), the set \(X\setminus\{y\}\) is open and contains \(x\). The basis property therefore supplies \(B_y\in\mathcal{B}\) with \(x\in B_y\subseteq X\setminus\{y\}\). In particular, \(y\notin B_y\). Every nonempty open set in this cofinite topology has finite complement, so each \(X\setminus B\), for \(B\in\mathcal{B}\) containing \(x\), is finite. There are at most countably many such \(B\); the union of their finite complements is countable. But that union must contain every \(y\ne x\), since \(y\notin B_y\). This would make \(\mathbb{R}\setminus\{x\}\) countable, a contradiction. Thus the finiteness of the set in the finite cofinite example matters.
The practical test is straightforward: propose a family of open sets, check that it covers every point, then check that intersections can be refined around each point. If the family is countable, those checks establish a countable basis; a list of open sets that merely looks manageable does not.
Check Your Understanding
For each question, identify the topology and use the basis criterion where appropriate.
- Why do rational-endpoint intervals cover every point of every open interval in the usual topology on \(\mathbb{R}\)?
- Verify the intersection condition for the family of singleton subsets of a countable discrete space.
- On \(X=\{u,v,w\}\) with the cofinite topology, list a finite basis and express \(\{u,w\}\) as a union of its members.
- Explain why a family that is a local base at one point has not, from that fact alone, been shown to be a basis for the whole topology.
- In the cofinite topology on an uncountable set, which step in the contradiction for a proposed countable basis uses uncountability?