Countability and Meagerness
A countable set can be spread throughout a space and still be small in the sense of Baire category. The key condition is that none of its points is isolated in the ambient metric space: then each individual point is nowhere dense, and a countable set is assembled from countably many nowhere-dense pieces. We will prove this claim and examine what it says about the Cantor set.
Let \((X,d)\) be a metric space, with its metric topology. A point \(x\in X\) is isolated if \(\{x\}\) is open in \(X\), or equivalently if some open ball about \(x\) contains no other point of \(X\). A metric space with no isolated points is often called perfect. If a subset of an ambient space is called a perfect set, the convention also requires that it be closed; the result below needs only the absence of isolated points in the space under consideration.
Meagerness is relative to the space: the same set can be meager in one space and not meager in another. A countable subset of a space with isolated points need not be meager. We first identify exactly why the no-isolated-points hypothesis matters.
When Is a Singleton Nowhere Dense?
Proof. First, \(\{x\}\) is closed. Indeed, for each \(y\ne x\), the open ball \(B(y,d(x,y)/2)\) does not contain \(x\), so it is contained in \(X\setminus\{x\}\). Thus \(X\setminus\{x\}\) is open. Consequently \(\overline{\{x\}}=\{x\}\). If \(x\) is not isolated, then \(\{x\}\) is not open. Since the only nonempty subset of \(\{x\}\) is \(\{x\}\) itself, its interior is empty, and \(\operatorname{int}\overline{\{x\}}=\varnothing\). Hence \(\{x\}\) is nowhere dense. Conversely, if \(x\) is isolated, then \(\{x\}\) is open and closed, so \(\operatorname{int}\overline{\{x\}}=\{x\}\ne\varnothing\). It is not nowhere dense. \(\square\)
Worked Example: A Countable Set in the Real Line
Consider \(A=\{1/n:n\in\mathbb{N}\}\) in \(\mathbb{R}\) with its standard topology. This set is countable: its points are listed by the positive integers \(n=1,2,3,\ldots\). The real line has no isolated points, because for any \(x\in\mathbb{R}\) and \(r>0\), the point \(x+r/2\) differs from \(x\) and belongs to \(B(x,r)=(x-r,x+r)\). The proposition therefore shows that each \(\{1/n\}\) is nowhere dense in \(\mathbb{R}\).
In particular, for each \(n\), \(\overline{\{1/n\}}=\{1/n\}\) and \(\operatorname{int}\{1/n\}=\varnothing\). Since \(A=\bigcup_{n\in\mathbb{N}}\{1/n\}\), this expresses \(A\) as a countable union of nowhere-dense sets. Thus \(A\) is meager in \(\mathbb{R}\). The argument concerns meagerness; it does not claim that the union has empty interior because the individual sets have empty interior alone. The definition of meager is about a countable union of nowhere-dense pieces.
Check: the first three pieces are \(\{1\}\), \(\{1/2\}\), and \(\{1/3\}\); each is closed with empty interior, and their union is exactly \(\{1,1/2,1/3\}\). The same description, with one singleton for each \(n\in\mathbb{N}\), gives all of \(A\).
The Countable-Set Theorem
Proof. If \(A=\varnothing\), it is meager because it is contained in the empty union of nowhere-dense sets. Otherwise, because \(A\) is countable, its points can be listed as \(a_1,a_2,\ldots\), with repetitions allowed if \(A\) is finite. The space has no isolated points, so the proposition shows that each singleton \(\{a_n\}\) is nowhere dense in \(X\). We have \(A=\bigcup_{n\in\mathbb{N}}\{a_n\}\). This is a countable union of nowhere-dense sets, so \(A\) is meager in \(X\). \(\square\)
The theorem applies to a subset \(A\) of \(X\), not just to \(X\) itself. Each singleton must be nowhere dense in the ambient space \(X\). This is why the hypothesis concerns isolated points of \(X\), rather than whether the points are isolated in \(A\).
Worked Example: A Countable Dense Subset of the Cantor Set
Construct the usual middle-thirds Cantor set as follows. Let \(C_0=[0,1]\), and at each stage remove the open middle third of every interval remaining from the previous stage. Write \(C_n\) for the union of the closed intervals remaining at stage \(n\), and let \(C=\bigcap_{n\ge 0}C_n\). At stage \(n\), the set \(C_n\) is a finite union of closed intervals of length \(3^{-n}\). Each \(C_n\) is closed in \(\mathbb{R}\), so \(C\) is closed in \(\mathbb{R}\), and we give it the subspace metric.
The space \(C\) has no isolated points. To check this, take \(x\in C\) and \(r>0\). Choose \(n\) with \(3^{-n}<r\), and choose a stage-\(n\) interval \(J\) containing \(x\). Both endpoints of \(J\) belong to \(C\): the construction removes open middle thirds and never removes an endpoint of a remaining interval. If \(x\) is one endpoint, take the other; if \(x\) is not an endpoint, take either endpoint. This gives \(y\in C\) with \(y\ne x\) and \(|y-x|\le 3^{-n}<r\). Hence every open ball in \(C\) about \(x\) contains another point.
Let \(E\) be the set of all endpoints of intervals appearing at any finite stage. Each stage has finitely many such endpoints, so \(E\) is a countable union of finite sets and is countable. Every endpoint belongs to \(C\), as noted above. Furthermore, \(E\) is dense in \(C\): given \(x\in C\) and \(r>0\), choose \(n\) and an interval \(J\) as before. An endpoint \(e\) of \(J\) belongs to \(E\), and \(|e-x|\le 3^{-n}<r\). Thus every open neighborhood of \(x\) in \(C\) meets \(E\).
The countable-set theorem now shows that \(E\) is meager in \(C\), even though \(E\) is dense in \(C\). The point of the example is not that density fails, but that density and meagerness describe different properties.
Worked Example: The Cantor Set Is Not Countable
The previous example established that \(C\) is a nonempty closed subset of \(\mathbb{R}\) with no isolated points. It is nonempty because \(0\) remains at every stage and hence belongs to \(C\). We now use completeness to distinguish the whole space \(C\) from its countable dense subset \(E\).
The real line with its usual metric is complete, and \(C\) is closed in \(\mathbb{R}\), so \(C\) is complete in its subspace metric. Indeed, any Cauchy sequence in \(C\) converges in \(\mathbb{R}\); since \(C\) is closed, its limit belongs to \(C\). The complete-metric Baire theorem, proved in the earlier material on Baire spaces, says that every nonempty complete metric space is a Baire space. In particular, such a space is not meager in itself: if it were a countable union of nowhere-dense sets, the complements of their closures would be open dense sets with empty intersection, contrary to the Baire theorem.
If \(C\) were countable, the countable-set theorem applied with \(X=C\) and \(A=C\) would make \(C\) meager in itself. This contradicts the Baire theorem, since \(C\) is nonempty and complete. Therefore \(C\) is uncountable. The same reasoning explains the role of completeness: the theorem about countable subsets only gives meagerness; the Baire theorem supplies the contradiction that proves uncountability.
Other Spaces and the Role of the Hypothesis
Worked Example: The Rational Numbers as a Metric Space
Give \(\mathbb{Q}\) the metric inherited from \(\mathbb{R}\). It has no isolated points: if \(q\in\mathbb{Q}\) and \(r>0\), choose \(m\in\mathbb{N}\) with \(1/m<r\). Then \(q+1/m\) is rational, differs from \(q\), and lies in \(B(q,r)\). Also, \(\mathbb{Q}\) is countable. Applying the theorem with \(X=A=\mathbb{Q}\), we find that \(\mathbb{Q}\) is meager in itself.
This does not contradict the complete-metric Baire theorem, which requires completeness, not just the absence of isolated points. In the subspace metric, \(\mathbb{Q}\) is not complete: rational sequences can converge in \(\mathbb{R}\) to an irrational number, which is not in \(\mathbb{Q}\). Thus the example separates two facts: no isolated points ensure that countable subsets are meager, while completeness is an additional hypothesis in the Baire conclusion.
The no-isolated-points condition cannot simply be dropped. For example, let \(X=\mathbb{N}\) have the discrete metric \(d(m,n)=1\) for \(m\ne n\). Every subset of \(X\) is open and closed. If \(N\subseteq X\) is nowhere dense, then \(\operatorname{int}\overline{N}=N\), so \(N=\varnothing\). A countable union of nowhere-dense subsets is therefore empty, and the nonempty countable set \(X\) is not meager in itself. Its points are isolated, exactly the case excluded by the theorem.
Common pitfall. It is incorrect to infer that a dense set cannot be meager. In the Cantor-set example, \(E\) is dense in \(C\) because every relative open neighborhood in \(C\) meets \(E\); yet \(E\) is meager in \(C\) because it is countable and \(C\) has no isolated points. Density tests whether every open neighborhood meets a set. Meagerness tests whether the set is contained in a countable union of nowhere-dense sets. These tests are compatible.
Why the Result Matters
In Baire category, a set is meager if it is contained in a countable union of nowhere-dense sets. The countable-set theorem gives a practical way to identify such sets in spaces without isolated points: it is enough to check countability. In a nonempty complete metric space, the Baire theorem then ensures that the whole space is not meager in itself. Together, the results show why a countable subset of a nonempty complete metric space with no isolated points cannot be the entire space, as the Cantor-set example illustrates.
| Question | Test or conclusion | Hypothesis |
|---|---|---|
| Is \(\{x\}\) nowhere dense? | Yes exactly when \(x\) is not isolated | \(X\) is a metric space |
| Is a countable subset meager? | Yes | \(X\) is a metric space with no isolated points |
| Can a dense set be meager? | Yes; the countable endpoint set \(E\) is dense and meager in \(C\) | Density and meagerness are distinct properties |
| Can a countable space without isolated points be meager in itself? | Yes; \(\mathbb{Q}\) with its usual metric is an example | Completeness is not assumed in the countable-set theorem |
Check Your Understanding
For each question, specify the space in which closure, interior, or meagerness is being considered.
- Why is a singleton in a metric space closed, and what additional condition makes it nowhere dense?
- Show that every finite subset of a metric space with no isolated points is nowhere dense.
- Can a countable dense subset of a metric space with no isolated points be meager? Give the Cantor-set example.
- Why does the countable-set theorem not imply that \(\mathbb{Q}\) is uncountable?
- In a discrete metric space, which subsets are nowhere dense, and why does the countable-set theorem not apply to its points?