Density and Separability
A subset can be small in cardinality yet still meet every open region of a space. The rational numbers in the usual real line provide the central example: they are countable, but every nonempty open interval contains a rational number. Separability asks whether a space has some countable subset with this density property. The answer depends on the topology, not just on the underlying set.
Throughout, a subset is called countable if it is finite or countably infinite. For \(A\subseteq X\), its closure \(\overline{A}\) is the smallest closed subset of \(X\) containing \(A\).
A useful way to test density is through open sets. A subset \(A\) is dense precisely when every nonempty open subset of \(X\) intersects \(A\). This criterion turns the definition involving closure into a direct test.
Proof. Suppose first that \(\overline{A}=X\). If a nonempty open set \(U\) did not intersect \(A\), then \(X\setminus U\) would be a closed set containing \(A\). The definition of closure would imply \(\overline{A}\subseteq X\setminus U\), contradicting \(\overline{A}=X\) and \(U\ne\varnothing\). Conversely, suppose every nonempty open set meets \(A\). If there were a point \(x\in X\setminus\overline{A}\), then \(X\setminus\overline{A}\) would be an open nonempty set. Since \(A\subseteq\overline{A}\), this open set would be disjoint from \(A\), a contradiction. Therefore \(\overline{A}=X\). \(\square\)
The Usual Real Line
In \(\mathbb{R}\) with its standard topology, the open intervals form a basis: every open neighborhood of a point contains an open interval around that point. Consequently, to prove a subset is dense, it suffices to show that every nonempty open interval intersects it.
Worked Example: The Rational Numbers Are Dense in \(\mathbb{R}\)
We show that \(\mathbb{Q}\) is countable and dense in \(\mathbb{R}\), so the usual real line is separable.
Countability. Every rational number has the form \(m/n\), where \(m\in\mathbb{Z}\) and \(n\in\mathbb{N}\). The pairs \((m,n)\) can be listed by increasing values of \(|m|+n\): for each fixed value there are only finitely many such pairs. This gives a countable list of pairs, and mapping each pair to \(m/n\) lists all rational numbers (with possible repetitions). Hence \(\mathbb{Q}\) is countable.
Density. Let \((a,b)\) be any nonempty open interval, so \(a<b\). By the Archimedean property, choose \(n\in\mathbb{N}\) such that \(n>1/(b-a)\). Let \(m=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer less than or equal to \(na\). Then \(m>na\), while \(m\le na+1\). Since \(1/n<b-a\), we obtain
Thus \(m/n\in\mathbb{Q}\cap(a,b)\). Every nonempty open interval meets \(\mathbb{Q}\), so the open-set test proves \(\overline{\mathbb{Q}}=\mathbb{R}\). Together with countability, this proves that \(\mathbb{R}\) is separable.
Check. The strict inequality \(a<m/n\) and the inequality \(m/n<b\) show that the rational lies inside the interval, not merely at an endpoint.
Density is always relative to the whole space under consideration. A subset may be dense in one space and fail to be dense in a larger one. In particular, \(\mathbb{Q}\cap(0,1)\) is not dense in \(\mathbb{R}\): the nonempty open interval \((2,3)\) is disjoint from it. It is, however, dense in the subspace \((0,1)\), where open sets are intersections of open subsets of \(\mathbb{R}\) with \((0,1)\). The distinction is the ambient space: the criterion tests every nonempty open set in that space.
Nor does countability alone imply density. For example, \(\mathbb{Z}\) is countable, but the nonempty open interval \((0,1)\) contains no integer. The open-set test therefore shows that \(\mathbb{Z}\) is not dense in \(\mathbb{R}\).
Discrete Spaces
A space \(X\) has the discrete topology when every subset of \(X\) is open. In this topology, every subset is also closed, because its complement is open. This makes density especially restrictive.
Proof. If \(A\) is dense, apply the open-set test to the open set \(\{x\}\) for each \(x\in X\). This set is nonempty, so \(\{x\}\cap A\ne\varnothing\), which forces \(x\in A\). Since this holds for every \(x\), \(A=X\). Conversely, \(X\) meets every nonempty open set, so \(X\) is dense in itself. Thus a discrete space has a countable dense subset exactly when \(X\) itself is countable. This includes the empty space: its underlying set is countable and is dense in itself. \(\square\)
Worked Example: Finite and Uncountable Discrete Spaces
Let \(X=\{a,b,c\}\) have the discrete topology. The subset \(A=\{a,b\}\) is not dense: the nonempty open set \(\{c\}\) misses \(A\). The full set \(X\) is dense, because every nonempty subset of \(X\) intersects \(X\). Since \(X\) is finite, it is a countable dense subset, and the space is separable.
Now let \(Y\) be an uncountable set with the discrete topology. If \(D\subseteq Y\) is countable, choose \(y\in Y\setminus D\), which exists because \(Y\) is uncountable. The open set \(\{y\}\) is nonempty and disjoint from \(D\), so \(D\) is not dense. The only dense subset is \(Y\), which is uncountable. Therefore \(Y\) is not separable.
Check. In both cases the test uses a singleton, which is open specifically because the topology is discrete.
The Cocountable Topology
Let \(X\) be an uncountable set. The cocountable topology on \(X\) consists of \(\varnothing\) and all subsets whose complements in \(X\) are countable. A nonempty open set can omit countably many points, but it cannot omit an uncountable subset.
Proof. First suppose \(A\) is uncountable. Let \(U\) be any nonempty open subset of \(X\). By the definition of the topology, \(X\setminus U\) is countable. If \(U\cap A=\varnothing\), then \(A\subseteq X\setminus U\), which would make \(A\) countable. This contradicts the assumption that \(A\) is uncountable. Thus every nonempty open set meets \(A\), and the open-set test shows \(A\) is dense.
Now suppose \(A\) is countable. Because \(X\) is uncountable, \(X\setminus A\ne\varnothing\). Its complement \(A\) is countable, so \(X\setminus A\) is a nonempty open set in the cocountable topology. It is disjoint from \(A\), and the open-set test shows that \(A\) is not dense. We have proved both directions. Every countable subset therefore fails to be dense, so the space is not separable. \(\square\)
Worked Example: Dense Subsets in a Cocountable Space
Let \(X\) be an uncountable set with the cocountable topology, and let \(A\subseteq X\). We test two choices.
First choice: \(A\) is countable. The set \(U=X\setminus A\) is nonempty, because \(X\) is uncountable and \(A\) is countable. It is open, since its complement \(A\) is countable. Also, \(U\cap A=\varnothing\). Thus \(A\) is not dense.
Second choice: \(A\) is uncountable. Let \(U\) be a nonempty open set. Its complement \(X\setminus U\) is countable. If \(U\cap A\) were empty, then every element of \(A\) would belong to \(X\setminus U\), so \(A\) would be countable. This contradicts the choice of \(A\). Therefore \(U\cap A\ne\varnothing\) for every such \(U\), and \(A\) is dense.
For a concrete instance, take \(A=X\setminus\{x_0\}\) for some \(x_0\in X\). This is uncountable: if it were countable, then \(X=(X\setminus\{x_0\})\cup\{x_0\}\) would be countable. Hence \(A\) is dense by the theorem. In contrast, the singleton \(\{x_0\}\) is countable and is not dense. The open set \(X\setminus\{x_0\}\) is nonempty and misses it.
Check. The test uses nonempty open sets throughout. In the singleton case, \(X\setminus\{x_0\}\ne\varnothing\) follows from the assumption that \(X\) is uncountable.
What Separability Does—and Does Not—Say
Separability is an existence property: it asks whether at least one countable dense subset exists. It does not say that every countable subset is dense, as the example of \(\mathbb{Z}\subseteq\mathbb{R}\) shows. Nor does it say that every dense subset is countable: in the usual real line, \(\mathbb{R}\) itself is dense and uncountable. In the cocountable topology on an uncountable set, the situation is reversed in a useful way: every uncountable subset is dense, but no countable subset is dense.
The open-set test is the practical tool in all three settings. In \(\mathbb{R}\), it reduces density to meeting every open interval. In a discrete space, singleton open sets force a dense set to contain every point. In a cocountable space, the size of the complement of an open set determines whether it can avoid a proposed subset. The same definition thus leads to different conclusions because the available open sets differ.
Common pitfall. A countable set is not automatically dense just because it is dense in some region. For example, \(\mathbb{Q}\cap(0,1)\) does not meet \((2,3)\), so it is not dense in \(\mathbb{R}\), even though it is dense in the subspace \((0,1)\). Always identify the space first, then test the nonempty open sets of that space.
Check Your Understanding
For each question, identify the ambient topology before applying the open-set test.
- Why does the existence of a countable dense subset in \(\mathbb{R}\) imply separability?
- In a discrete space \(X\), can a proper subset of \(X\) be dense? Explain using an open set.
- Is \(\mathbb{Z}\) dense in \(\mathbb{R}\) with the standard topology? Give a nonempty open set that decides the question.
- In an uncountable space with the cocountable topology, why does every uncountable subset meet every nonempty open set?
- Is \(\mathbb{Q}\cap(0,1)\) dense in \(\mathbb{R}\)? Is it dense in the subspace \((0,1)\)? Give the relevant open-set tests.