Four Properties, Two Different Scales
Countability conditions can describe neighborhoods of individual points or the whole space. First countability is local: each point needs a countable collection of neighborhoods that refines every neighborhood of that point. Second countability is global: one countable collection must provide a base for the entire topology. Separability asks for a countable dense subset, while the Lindelof property asks whether open covers have countable subcovers. These conditions behave differently in the real line, discrete spaces and spaces with the cocountable topology.
First countability concerns a separate neighborhood collection at each point; second countability requires one collection for all points. One implication always holds: if \(\mathcal{B}\) is a countable base, then the members of \(\mathcal{B}\) containing \(x\) form a countable local base at \(x\). Indeed, for any open neighborhood \(U\) of \(x\), the base property gives \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). Thus second countability implies first countability.
The Usual Real Line
Worked Example: The Real Line Has All Four Properties
Give \(\mathbb{R}\) its standard topology. For each \(x\in\mathbb{R}\), consider the open intervals \((x-1/n,x+1/n)\), \(n\in\mathbb{N}\). They form a countable local base: if \(U\) is an open neighborhood of \(x\), there is an \(r>0\) such that \((x-r,x+r)\subseteq U\). Choose \(n\) with \(1/n<r\); then \((x-1/n,x+1/n)\subseteq U\). So \(\mathbb{R}\) is first countable.
The intervals \((a,b)\) with \(a,b\in\mathbb{Q}\) and \(a<b\) form a countable base. There are countably many such pairs because \(\mathbb{Q}\times\mathbb{Q}\) is countable. If \(x\in U\) and \(U\) is open, choose \(r>0\) with \((x-r,x+r)\subseteq U\). The density of \(\mathbb{Q}\) gives rationals \(a,b\) with \(x-r<a<x<b<x+r\). Hence \(x\in(a,b)\subseteq U\), proving the base property. Thus \(\mathbb{R}\) is second countable.
The set \(\mathbb{Q}\) is countable and dense in \(\mathbb{R}\): each nonempty open set contains an interval \((u,v)\) with \(u<v\), and that interval contains a rational. Therefore \(\mathbb{R}\) is separable.
Finally, let \(\mathcal{C}\) be an open cover of \(\mathbb{R}\). For every rational-endpoint interval \(B\) that is contained in some member of \(\mathcal{C}\), choose one such member \(C_B\). There are at most countably many choices. Given \(x\in\mathbb{R}\), some \(C\in\mathcal{C}\) contains \(x\); the base property supplies a rational-endpoint interval \(B\) with \(x\in B\subseteq C\). The chosen \(C_B\) also contains \(B\), and hence contains \(x\). Thus the chosen members cover \(\mathbb{R}\), proving it is Lindelof.
Check: the same rational-endpoint intervals establish second countability and, by selecting a cover member for each suitable basis element, the Lindelof property. The rational numbers themselves establish separability; these are different uses of countability.
Discrete Spaces: The Four Properties Agree on Size
Let \(X\) have the discrete topology \(\mathcal{P}(X)\), so every subset is open. In this topology, the only dense subset is \(X\) itself: if \(x\notin D\), the open set \(\{x\}\) misses \(D\). Likewise, the singleton open sets force any open cover to include a set containing each point. These observations make the size of \(X\) decisive.
Proof. For each \(x\in X\), the collection \(\{\{x\}\}\) is a finite local base, because \(\{x\}\) is open and is contained in every neighborhood of \(x\). Thus \(X\) is first countable. If \(X\) is countable, the collection \(\{\{x\}:x\in X\}\) is a countable base; \(X\) is also a countable dense subset of itself; and from any open cover one can choose, for each \(x\in X\), a cover member containing \(x\), obtaining at most countably many members that cover \(X\). Conversely, if \(X\) has a countable base \(\mathcal{B}\), then each \(x\in X\) belongs to some \(B\in\mathcal{B}\) with \(B\subseteq\{x\}\), so \(B=\{x\}\). Therefore \(x\mapsto\{x\}\) shows that \(X\) is countable. If \(X\) has a countable dense subset \(D\), the open set \(\{x\}\) for each \(x\in X\) must intersect \(D\), so \(x\in D\), and \(X=D\) is countable. If \(X\) is Lindelof, the open cover \(\{\{x\}:x\in X\}\) has a countable subcover, which is possible only if \(X\) is countable. These arguments include \(X=\varnothing\), for which all four properties hold. \(\square\)
Worked Example: Countable and Uncountable Discrete Spaces
Let \(X=\mathbb{N}\) with the discrete topology. The singleton sets \(\{n\}\), \(n\in\mathbb{N}\), are a countable base; \(\mathbb{N}\) is dense in itself; and any open cover has a subcover obtained by choosing one member containing each \(n\). The space is first countable by the local base \(\{\{n\}\}\), and it has the other three properties as well.
Now let \(Y\) be an uncountable set with the discrete topology. Every point still has the one-element local base \(\{\{y\}\}\), so \(Y\) is first countable. But \(\{\{y\}:y\in Y\}\) is an open cover with no countable subcover, and a dense subset must contain every \(y\). Also, any base must contain each singleton: a base element \(B\) with \(y\in B\subseteq\{y\}\) must equal \(\{y\}\). Thus \(Y\) is not Lindelof, separable or second countable.
Check: the contrast is not about neighborhoods at a point; those are simple in both spaces. It is the uncountable number of points that prevents the three global properties.
Uncountable Cocountable Spaces
Let \(X\) be uncountable and give it the cocountable topology: the open sets are \(\varnothing\) and all sets \(U\subseteq X\) for which \(X\setminus U\) is countable. A nonempty open set therefore omits at most countably many points.
Proof. Let \(\mathcal{C}\) be an open cover of \(X\). Since \(X\) is nonempty, choose a nonempty \(U_0\in\mathcal{C}\). The set \(C=X\setminus U_0\) is countable. For each \(x\in C\), choose a member \(U_x\in\mathcal{C}\) containing \(x\). Then \(U_0\) together with the at most countably many sets \(U_x\) covers \(X\), proving the Lindelof property.
If \(D\subseteq X\) is countable, then \(X\setminus D\) is a nonempty open set disjoint from \(D\), so \(D\) is not dense. Hence no countable subset is dense, and \(X\) is not separable.
Suppose for contradiction that \(x\in X\) has a countable local base \(\mathcal{U}_x\). The set \(C=\bigcup_{V\in\mathcal{U}_x}(X\setminus V)\) is countable, since it is a countable union of countable sets. Choose \(y\in X\setminus(C\cup\{x\})\), possible because \(X\) is uncountable. The set \(U=X\setminus\{y\}\) is an open neighborhood of \(x\). The local base property gives \(V\in\mathcal{U}_x\) with \(V\subseteq U\), so \(y\notin V\). This implies \(y\in X\setminus V\subseteq C\), contradicting the choice of \(y\). Thus no point has a countable local base, and the space is not first countable. Since second countability implies first countability, it is not second countable either. \(\square\)
Worked Example: Testing a Cocountable Space
Take \(X=\mathbb{R}\) and give it the cocountable topology, not its standard topology. Let \(D=\mathbb{Q}\). Since \(\mathbb{Q}\) is countable, \(U=\mathbb{R}\setminus\mathbb{Q}\) is open and nonempty, and \(U\cap D=\varnothing\). Therefore \(D\) is not dense in this topology, even though it is dense in the standard topology.
For contrast, let \(E=\mathbb{R}\setminus\mathbb{Q}\), which is uncountable. If \(U\) is any nonempty cocountable-open set, write \(U=\mathbb{R}\setminus C\) for a countable \(C\). If \(U\cap E\) were empty, then \(E\subseteq C\), contradicting that \(E\) is uncountable. Thus every nonempty open set meets \(E\), so \(E\) is dense.
Check: this verifies the strict distinction: a countable complement defines an open set, and a countable subset cannot meet every nonempty open set. An uncountable subset does meet every such set.
Comparison and a Common Pitfall
| Space | First countable | Second countable | Separable | Lindelof |
|---|---|---|---|---|
| \(\mathbb{R}\), standard topology | Yes | Yes | Yes | Yes |
| Countable discrete space | Yes | Yes | Yes | Yes |
| Uncountable discrete space | Yes | No | No | No |
| Uncountable cocountable space | No | No | No | Yes |
Common pitfall. It is incorrect to infer that a space with simple neighborhoods must have a countable base. In an uncountable discrete space, every point has the especially simple local base \(\{\{x\}\}\), but a base for the whole space must contain every singleton and is therefore uncountable. Conversely, the uncountable cocountable space is Lindelof without being first countable: its open covers can be reduced using one open set and its countable complement, while no point has a countable local base. The four properties should be checked from their definitions rather than treated as interchangeable.
Check Your Understanding
Use the definitions to justify each conclusion, paying attention to which topology is being used.
- Which of the four properties hold for an uncountable set with the discrete topology?
- In an uncountable cocountable space, why is the complement of a countable subset an open set that obstructs separability?
- Explain why a countable base would give a countable local base at each point.
- Give an example of a space that is Lindelof but not first countable, and verify both properties.
- Why does \(\mathbb{Q}\) being dense in the standard real line not imply that it is dense in the cocountable topology on \(\mathbb{R}\)?