Two Different Ways to Describe a Space Countably
A countable dense subset uses points: it meets every nonempty open set. A countable basis uses open sets: every open set is a union of members of one countable collection. These properties are related, but they are not interchangeable in every topological space. Earlier in the course, “Second-Countable Spaces” established that every second-countable space is separable, and “Separable Spaces” established that every separable metric space is second-countable. Together, these results give an equivalence for metric spaces.
The metric hypothesis in the second implication matters. We will see a separable space with no countable basis. We will also prove two ways second-countability can be used: it passes to countable products of second-countable spaces and to targets of open continuous surjections. These results make the property useful beyond the metric setting.
Definitions and the Metric-Space Connection
The countability in these definitions applies to different collections. A countable basis may contain open sets of many different sizes and shapes; a countable dense subset need not itself be open or provide basis neighborhoods. In a metric space, the distance structure lets us turn points from a dense subset into a basis by using rational radii. That is why the result from “Separable Spaces” gives the reverse implication for metric spaces.
This theorem is a direct combination of the two results already proved earlier: every second-countable space is separable, and every separable metric space is second-countable. Neither implication should be transferred automatically to arbitrary topological spaces. The following example shows why the metric qualification cannot simply be omitted.
Worked Example: The Sorgenfrey Line Is Separable but Not Second-Countable
On \(\mathbb{R}\), consider the topology with basis consisting of intervals \([a,b)\), where \(a<b\). This is called the Sorgenfrey topology. The set \(\mathbb{Q}\) is countable and dense in this topology. Indeed, every nonempty open set contains a basic interval \([a,b)\), and there is a rational \(q\) with \(a<q<b\). Thus every nonempty open set meets \(\mathbb{Q}\).
Suppose, for contradiction, that \(\mathcal{B}\) were a countable basis for this topology. For each \(x\in\mathbb{R}\), the set \([x,x+1)\) is open and contains \(x\). The basis property gives some \(B_x\in\mathcal{B}\) such that \(x\in B_x\subseteq[x,x+1)\). Take the infimum of \(B_x\) in the usual order on \(\mathbb{R}\). Since \(x\in B_x\), its infimum is at most \(x\); since every point of \(B_x\) is at least \(x\), its infimum is at least \(x\). Therefore \(\inf B_x=x\).
If \(x\ne y\), then \(\inf B_x=x\) and \(\inf B_y=y\), so \(B_x\ne B_y\). Consequently, \(\mathcal{B}\) would contain a different member for every real number. That is impossible for a countable family. The Sorgenfrey line is therefore separable but not second-countable. This does not contradict the metric-space equivalence: the Sorgenfrey topology is not the usual metric topology on \(\mathbb{R}\).
Countable Products Preserve Second-Countability
A basic open set in a product topology restricts only finitely many coordinates. When there are countably many coordinates and each factor has a countable basis, there are still only countably many ways to make those finite restrictions. This gives a useful general theorem.
Proof. Let \(X_n\) be second-countable for each \(n\in\mathbb{N}\), and first suppose that every \(X_n\) is nonempty. For each \(n\), choose a countable basis \(\mathcal{B}_n\) for \(X_n\). Consider all product sets that restrict coordinates in a finite set \(F\subseteq\mathbb{N}\) and leave the other coordinates unrestricted:
These sets form a basis for the product topology. To check the basis refinement property, let \(U\) be open in the product and \(x\in U\). By the definition of the product topology, there is a basic product neighborhood restricting a finite set \(F\) of coordinates to open sets \(V_i\subseteq X_i\), with \(x_i\in V_i\), that is contained in \(U\). For each \(i\in F\), the basis property in \(X_i\) gives \(B_i\in\mathcal{B}_i\) with \(x_i\in B_i\subseteq V_i\). The resulting set \(C(F,(B_i)_{i\in F})\) contains \(x\) and is contained in \(U\). Thus the displayed family is a basis.
It remains to check countability. There are countably many finite subsets \(F\) of \(\mathbb{N}\). For a fixed finite \(F\), there are countably many choices of \(B_i\in\mathcal{B}_i\) for each \(i\in F\), since a finite product of countable sets is countable. The entire family is therefore a countable union of countable families, so it is countable. If some factor is empty, the product is empty and has the empty basis. This proves the result in all cases. \(\square\)
The theorem connects product topology directly to the metric-space equivalence, without requiring the product itself to have a metric. Since every second-countable space is separable, the countable product in the theorem is also separable. The following example verifies this conclusion concretely.
Worked Example: The Space of Binary Sequences
Give \(\{0,1\}\) the discrete topology and consider \(X=\{0,1\}^{\mathbb{N}}\) with the product topology. Each factor is second-countable: its two singletons form a finite basis. The countable-product theorem shows that \(X\) is second-countable.
A basis member specifies the values of a sequence at finitely many coordinates and places no restrictions on the other coordinates. For example, one such member is \(\{x\in X:x_2=1,\ x_5=0\}\). There are countably many finite sets of coordinates and finitely many possible specifications on each one, which is the countability mechanism in the theorem.
There is also a direct countable dense subset. Let \(D\) be the set of binary sequences that are eventually zero. For each finite set of coordinates at which the sequence may be \(1\), there are finitely many choices, so \(D\), a countable union of finite families, is countable. Any nonempty basic open set specifies values at only finitely many coordinates. Choose the sequence that matches those specifications and is zero everywhere else. It belongs to \(D\) and to that basic open set. Since every nonempty open set contains a basic open set, every nonempty open set meets \(D\). Thus \(D\) is dense in \(X\).
Open Continuous Surjections Preserve Second-Countability
A continuous map pulls open sets back to open sets. An open map sends open sets forward to open sets. When a map has both properties and is onto, a countable basis in its domain produces a countable basis in its target.
Proof. Let \(\mathcal{B}\) be a countable basis for \(X\), and form the family \(\mathcal{C}=\{f[B]:B\in\mathcal{B}\}\). Because \(f\) is open, every member of \(\mathcal{C}\) is open in \(Y\). The family is countable because it is the image of the countable family \(\mathcal{B}\).
We verify the basis property. Let \(V\) be open in \(Y\) and take \(y\in V\). Surjectivity gives \(x\in X\) with \(f(x)=y\). Continuity makes \(f^{-1}(V)\) open in \(X\), and \(x\in f^{-1}(V)\). Thus some \(B\in\mathcal{B}\) satisfies \(x\in B\subseteq f^{-1}(V)\). Applying \(f\) gives \(y\in f[B]\subseteq V\), and \(f[B]\in\mathcal{C}\). Every point of every open set in \(Y\) is therefore contained in a member of \(\mathcal{C}\) lying inside that open set. Hence \(\mathcal{C}\) is a countable basis for \(Y\). If \(X\) is empty, surjectivity forces \(Y\) to be empty, which is second-countable. \(\square\)
Worked Example: A Projection onto the Real Line
Let \(\pi:\mathbb{R}^2\to\mathbb{R}\) be the projection \(\pi(x,y)=x\), using the usual topologies. It is surjective. It is continuous because the inverse image of an open set \(V\subseteq\mathbb{R}\) is \(V\times\mathbb{R}\), which is open in the product topology.
It is also open. A basic open rectangle \(I\times J\), where \(I\) and \(J\) are open intervals, has image \(I\) under \(\pi\). Every open set in \(\mathbb{R}^2\) is a union of basic open rectangles, and the image of a union is the union of the images. Thus the image of every open set is open. Since \(\mathbb{R}^2\) is second-countable, the open-surjection theorem applies and yields that \(\mathbb{R}\) is second-countable. Here the conclusion is consistent with the familiar countable basis of rational-endpoint intervals.
What the Comparison Does—and Does Not—Say
| Setting or construction | Conclusion |
|---|---|
| Any topological space | Second-countability implies separability, by the earlier theorem. |
| Metric space | Separability and second-countability are equivalent. |
| Countable product of second-countable spaces | The product is second-countable, and hence separable. |
| Open continuous surjection from a second-countable space | The target is second-countable, and hence separable. |
| General topological space | Separability alone does not imply second-countability. |
The main pitfall is to treat a countable dense set as if it were a countable basis. Density asks whether each nonempty open set contains a point from a specified set; a basis must instead refine every open neighborhood at every point. The Sorgenfrey example shows these demands can differ. When a metric is available, rational-radius balls bridge the gap, but without a metric that construction is unavailable.
Second-countability is especially useful when a space is built from simpler spaces or obtained as the target of a well-behaved map. Countable products preserve it because only finitely many coordinates are restricted in a basic neighborhood. Open continuous surjections preserve it because basis neighborhoods can be carried forward while continuity ensures they remain within a chosen target neighborhood. These are distinct mechanisms, and each depends on the stated hypotheses.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Which collection is countable in the definition of second-countability, and which collection is countable in the definition of separability?
- What role does the metric play in the equivalence between separability and second-countability?
- Why are there only countably many basic product sets when there are countably many factors with countable bases?
- In the binary-sequence example, why does every nonempty open set meet the eventually-zero sequences?
- Which two mapping hypotheses are used to show that a countable basis in \(X\) gives a countable basis in \(Y\)?
- How can the Sorgenfrey line be separable while failing to be second-countable?