From Local Basis Elements to a Countable Subcover
An open cover may have many more members than are needed to cover a space. The useful feature of a countable basis is that it supplies only countably many local open sets to consider, even when the original cover is uncountable. By keeping basis elements that fit inside cover members, then assigning a cover member to each retained element, we obtain a countable subcover.
Let \((X,\mathcal{T})\) be a topological space. A family \(\mathcal{B}\subseteq\mathcal{T}\) is a basis for \(\mathcal{T}\) if, whenever \(U\in\mathcal{T}\) and \(x\in U\), there is \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). The space is second countable if it has a countable basis. An open cover of \(X\) is a family \(\mathcal{U}\) of open subsets of \(X\) whose union is \(X\). A subcover is a subfamily of \(\mathcal{U}\) that still has union \(X\). A space is Lindelof if every open cover has a countable subcover.
The key is to ask which basis elements are small enough to fit inside at least one member of the given cover. There may be many cover members containing a particular basis element; the proof needs just one of them.
The Countable Refinement
Proof. The family \(\mathcal{B}_{\mathcal{U}}\) is a subfamily of the countable family \(\mathcal{B}\), so it is countable. Every member of \(\mathcal{B}_{\mathcal{U}}\) is open because every basis element is open. Its definition also says that each such member is contained in some member of \(\mathcal{U}\). It remains to show that these basis elements cover \(X\). If \(x\in X\), the cover property gives some \(U\in\mathcal{U}\) with \(x\in U\). By the basis condition, there is \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Thus \(B\in\mathcal{B}_{\mathcal{U}}\), and \(x\in B\). This works for every \(x\in X\); if \(X=\varnothing\), the covering condition is vacuous. \(\square\)
This lemma constructs a countable open refinement: its members are open, they cover the space, and each lies inside some original cover member. A refinement need not itself be a subfamily of the original cover. The next step transfers the countability from the refinement back to the given cover.
Proof. Let \(\mathcal{B}\) be a countable basis and \(\mathcal{U}\) any open cover of \(X\). Form \(\mathcal{B}_{\mathcal{U}}\) as in the lemma. For each \(B\in\mathcal{B}_{\mathcal{U}}\), choose a set \(U_B\in\mathcal{U}\) such that \(B\subseteq U_B\). The selected family \(\{U_B:B\in\mathcal{B}_{\mathcal{U}}\}\) is countable, since it is the image of a countable family. It is a subfamily of \(\mathcal{U}\). For any \(x\in X\), the lemma gives \(B\in\mathcal{B}_{\mathcal{U}}\) with \(x\in B\). Then \(x\in B\subseteq U_B\), so \(x\) belongs to a selected cover member. Therefore the selected family covers \(X\), and is the required countable subcover. If \(X=\varnothing\), the empty family is already a countable subcover. The selections here use the usual countable choice implicit in standard set-theoretic foundations for topology. \(\square\)
The mechanism does not require that \(X\) be Hausdorff, or satisfy any other separation axiom. Only the basis condition and countability of the basis enter the proof.
Worked Example: An Uncountable Cover of the Real Line
Give \(\mathbb{R}\) its standard topology and consider the cover indexed by all real numbers: \[ \mathcal{U}=\{U_a:a\in\mathbb{R}\},\qquad U_a=(a-1,a+1). \] Each \(U_a\) is open. The family covers \(\mathbb{R}\), since for any \(x\in\mathbb{R}\), choosing \(a=x\) gives \(x\in(x-1,x+1)\). Its members are distinct: if \(U_a=U_b\), their left endpoints satisfy \(a-1=b-1\), so \(a=b\). Thus this is an uncountable cover.
Use the countable basis consisting of intervals \((p,q)\) with rational endpoints \(p<q\). The theorem guarantees a countable subcover. In this case we can exhibit one directly: take \(U_n=(n-1,n+1)\) for \(n\in\mathbb{Z}\). For any \(x\in\mathbb{R}\), choose an integer \(n\) with \(n\leq x<n+1\). Then \(n-1<x\), and \(x<n+1\), so \(x\in(n-1,n+1)=U_n\). Hence these integer-indexed members cover \(\mathbb{R}\).
Check: for \(x=2.4\), take \(n=2\), since \(2\leq2.4<3\). The selected interval is \(U_2=(1,3)\), and \(1<2.4<3\). The direct subcover is countable because \(\mathbb{Z}\) is countable, while the original cover has one distinct member for each real \(a\).
Worked Example: The Selection Procedure on a Discrete Space
Let \(X=\mathbb{N}\) with the discrete topology, in which every subset is open. The singleton family \(\mathcal{B}=\{\{n\}:n\in\mathbb{N}\}\) is a countable basis: if \(x\in U\subseteq X\), then \(\{x\}\in\mathcal{B}\) and \(x\in\{x\}\subseteq U\).
Consider the open cover \(U_k=\{1,2,\ldots,k\}\), for \(k\in\mathbb{N}\). It covers \(X\) because \(n\in U_n\) for every \(n\in\mathbb{N}\). For the basis element \(\{n\}\), the cover members containing it are exactly \(U_k\) with \(k\geq n\): the condition \(\{n\}\subseteq U_k\) holds precisely when \(n\leq k\). Choose the least such index, \(k=n\). The selection procedure therefore assigns \(\{n\}\) to \(U_n\), and its selected family is \(\{U_n:n\in\mathbb{N}\}\), a countable subcover.
This particular cover has no finite subcover. If a finite selection has largest index \(m\), then each selected set is contained in \(U_m=\{1,\ldots,m\}\), so the point \(m+1\) is missed. Thus “countable subcover” cannot in general be strengthened to “finite subcover.”
Worked Example: A Finite Space and a Countable Bound
Let \(X=\{p,q\}\) have the discrete topology and basis \(\mathcal{B}=\{\{p\},\{q\}\}\). For any open cover \(\mathcal{U}\), the basis elements that fit inside some cover member form a subfamily of these two singletons. For each such singleton, choose one cover member containing it. The selected cover family has at most two members.
To check coverage, \(p\in X\) belongs to some \(U\in\mathcal{U}\); then \(\{p\}\subseteq U\), so the singleton is retained and its chosen cover member contains \(p\). The same argument applies to \(q\). Hence the selected family covers \(X\). If the cover contains \(X\) itself, one cover member suffices; otherwise, the two singletons can be covered by at most two chosen members. This example shows the selection argument directly, including the possibility that different basis elements are assigned to the same cover member.
A Common Mistake and the Point of the Argument
Common pitfall. A countable basis does not mean that there are only countably many open sets, and it does not mean that every open cover is already countable. In the real-line example, the family \(\{(a-1,a+1):a\in\mathbb{R}\}\) is an uncountable open cover, despite the rational-endpoint intervals giving a countable basis. The incorrect inference “countable basis, therefore countable cover” confuses the basis with the cover.
The correction is to use the basis as an intermediate family. For every point, first find a basis element that contains the point and lies inside some cover member. There are only countably many possible basis elements to retain. Then choose one original cover member for each retained element. This proves the existence of a countable subcover; it does not assert that every member of the original cover must be selected, or that the selected subcover is unique.
The practical value is that a countable basis reduces a covering question with potentially many open sets to a countable list of local choices. The proof applies to arbitrary open covers, not merely covers by basis elements, and it makes no assumption about how the cover is indexed. When applying it, keep the containment direction straight: the retained basis element must satisfy \(B\subseteq U\), so the chosen cover member really does cover every point of \(B\).
Check Your Understanding
Answer using the basis condition and the selection argument.
- In the countable-refinement lemma, why does a point \(x\in X\) belong to some retained basis element?
- Why is the family of chosen cover members countable even if the original cover is uncountable?
- In the discrete-space example, which indices \(k\) satisfy \(\{n\}\subseteq U_k\), and why is the least such index \(n\)?
- Does the theorem require \(X\) to be Hausdorff? Identify the hypotheses actually used in the proof.
- Explain why a countable basis does not imply that every open cover is countable.