Small Sets and Complete Spaces
The \(L^p\) tutorials have used measure to quantify the size of sets and functions. Category provides a different way to describe how a set sits inside a topological space. A set can be dense and still be small in the category sense: the rationals are dense in the real line, yet they are a countable union of nowhere-dense sets. The key result is that this kind of smallness cannot cover a nonempty complete metric space.
We work in a metric space \((X,d)\). For \(x\in X\) and \(r>0\), write \(B(x,r)=\{y\in X:d(x,y)<r\}\). The closure \(\overline{A}\) and interior of a set \(A\subseteq X\) are taken in the metric space \(X\), unless another space is specified. In particular, whether a set is nowhere dense depends on its ambient space.
The closure in the definition matters. A nowhere-dense set need not itself be closed, but its closure must contain no nonempty open set. Every subset of a nowhere-dense set is nowhere dense, and every subset of a meagre set is meagre. A meagre set may nevertheless be dense. Residual and meagre are complementary notions: \(E\) is residual precisely when \(X\setminus E\) is meagre.
The Baire Category Theorem
Completeness is a powerful condition on a metric space: every Cauchy sequence has a limit in the space. The Baire Category Theorem connects this condition to category. Its proof constructs nested closed balls, each chosen to meet one more dense open set. Their radii shrink to zero, so completeness supplies a point that belongs to every one of those open sets.
Proof. Let \((X,d)\) be complete, and let \(G_1,G_2,\ldots\) be open dense subsets of \(X\). To prove that their intersection is dense, take an arbitrary nonempty open set \(U\subseteq X\). We will find a point in \(U\cap\bigcap_{n=1}^{\infty}G_n\).
Since \(G_1\) is dense, \(U\cap G_1\) is nonempty. It is also open. Choose \(x_1\in U\cap G_1\) and \(r_1>0\) small enough that the closed ball \(\overline{B}(x_1,r_1)=\{y:d(y,x_1)\leq r_1\}\) is contained in \(U\cap G_1\), with \(r_1\leq 2^{-1}\). Such a radius exists because \(U\cap G_1\) is open: some open ball around \(x_1\) is contained in it, and a smaller closed ball is then contained in that open ball.
Suppose \(x_n\) and \(r_n>0\) have been chosen. The open ball \(B(x_n,r_n)\) is nonempty, so density of \(G_{n+1}\) implies that \(B(x_n,r_n)\cap G_{n+1}\) is nonempty. This intersection is open. Choose \(x_{n+1}\) in it and choose \(r_{n+1}>0\), with \(r_{n+1}\leq 2^{-(n+1)}\), such that \(\overline{B}(x_{n+1},r_{n+1})\subseteq B(x_n,r_n)\cap G_{n+1}\). Thus the closed balls are nested, and for every \(n\), \(\overline{B}(x_n,r_n)\subseteq U\cap G_1\cap\cdots\cap G_n\).
For \(m,k\geq n\), both \(x_m\) and \(x_k\) lie in \(\overline{B}(x_n,r_n)\). Hence \(d(x_m,x_k)\leq d(x_m,x_n)+d(x_n,x_k)\leq 2r_n\). Since \(r_n\leq 2^{-n}\to0\), the sequence \((x_n)\) is Cauchy. Completeness gives a limit \(x\in X\). For each fixed \(n\), every \(x_m\) with \(m\geq n\) belongs to the closed set \(\overline{B}(x_n,r_n)\), so its limit \(x\) belongs to that closed ball as well. Consequently \(x\in U\) and \(x\in G_n\) for every \(n\). We have found a point of \(U\cap\bigcap_{n=1}^{\infty}G_n\). Because \(U\) was arbitrary, this intersection is dense.
To see the equivalent formulation, suppose \(A_n\) is nowhere dense for each \(n\). Then \(G_n=X\setminus\overline{A_n}\) is open and dense. The theorem says \(\bigcap_n G_n\) is dense, so it is nonempty in every nonempty open set. This intersection is disjoint from \(\bigcup_n A_n\), which therefore cannot contain any nonempty open set. Thus the union has empty interior. Conversely, if every countable union of nowhere-dense sets has empty interior, the complements of the closures of any countable family of nowhere-dense sets have dense intersection; this gives the dense-intersection formulation. \(\square\)
Worked Example: The Rationals Are Dense but Meagre
In \(\mathbb{R}\) with its usual metric, each singleton \(\{q\}\) is closed. It has empty interior: every open interval around \(q\) contains points other than \(q\). Thus each singleton is nowhere dense. The rationals are countable, so they can be listed as \(q_1,q_2,\ldots\), and \(\mathbb{Q}=\bigcup_{n=1}^{\infty}\{q_n\}\). It follows that \(\mathbb{Q}\) is meagre in \(\mathbb{R}\).
The rationals are also dense: every nonempty open interval contains a rational number. There is no conflict between density and meagreness. Density says that every open interval meets \(\mathbb{Q}\); meagreness says that \(\mathbb{Q}\) is a countable union of sets whose closures contain no open interval.
Since \(\mathbb{R}\) is complete, the Baire Category Theorem applies. The complement \(\mathbb{R}\setminus\mathbb{Q}\), the irrationals, is a countable intersection of open dense sets: \(\mathbb{R}\setminus\mathbb{Q}=\bigcap_{n=1}^{\infty}(\mathbb{R}\setminus\{q_n\})\). It is therefore dense. In fact, the theorem shows that this intersection meets every nonempty open interval.
Reading Category in the Ambient Space
The same subset can have different category properties in different spaces. For example, \(\mathbb{Q}\) is meagre in \(\mathbb{R}\), as just shown. But if the ambient space is \(\mathbb{Q}\) with its subspace metric, then \(\mathbb{Q}\) is the whole space, and its interior in that space is \(\mathbb{Q}\), not empty. It is not nowhere dense in \(\mathbb{Q}\). Whenever category is used, specify the space in which closure and interior are being taken.
Worked Example: A Closed Nowhere-Dense Set in the Plane
Consider the horizontal axis \(L=\{(x,0):x\in\mathbb{R}\}\) in \(\mathbb{R}^2\) with the Euclidean metric. The set is closed: if \((x_n,0)\in L\) converges to \((x,y)\), then the second coordinates converge to \(y\), while they are all zero, so \(y=0\) and \((x,y)\in L\).
The interior of \(L\) is empty. Indeed, given any \((x,0)\in L\) and any radius \(r>0\), the point \((x,r/2)\) lies within distance \(r\) of \((x,0)\) but does not lie in \(L\). Thus no open ball centered on \(L\) is contained in \(L\). Since \(L\) is closed, \(\overline{L}=L\), and its closure has empty interior. Hence \(L\) is nowhere dense in \(\mathbb{R}^2\). Its complement is open and dense: it is open because \(L\) is closed, and every open ball contains a point off the horizontal axis.
This example illustrates the definition directly. A set can contain infinitely many points and extend without bound while still being nowhere dense. The relevant question is not how many points it has or how far it stretches, but whether its closure contains an open region.
A Complete Space Without Isolated Points Is Uncountable
The Baire Category Theorem has a useful consequence about the size of a space. A point \(x\) is isolated if \(\{x\}\) is open in the space, or equivalently if some open ball around \(x\) contains no other point of the space. In a space without isolated points, each singleton is nowhere dense. Completeness then prevents a countable list of such singletons from covering the space.
Proof. Let \(X\) be a nonempty complete metric space with no isolated points. In a metric space, each singleton \(\{x\}\) is closed. Since \(x\) is not isolated, \(\{x\}\) has empty interior, so it is nowhere dense in \(X\).
Suppose, for contradiction, that \(X\) is countable. List its points as \(x_1,x_2,\ldots\), repeating points if \(X\) is finite. Then \(X=\bigcup_{n=1}^{\infty}\{x_n\}\) is a countable union of nowhere-dense sets. By the Baire Category Theorem, this union has empty interior. But \(X\), as a subset of itself, is open in itself and has interior \(X\), which is nonempty. This is a contradiction. Therefore \(X\) is uncountable. \(\square\)
Worked Example: An Uncountable Space of Binary Sequences
Let \(X=\{0,1\}^{\mathbb{N}}\), the set of all sequences \(x=(x_1,x_2,\ldots)\) with each \(x_k\) equal to zero or one. Define \(d(x,y)=\sum_{k=1}^{\infty}2^{-k}|x_k-y_k|\). The sum is finite because \(|x_k-y_k|\leq1\) and \(\sum_{k=1}^{\infty}2^{-k}=1\). This is a metric: it is nonnegative and symmetric, it is zero only when every coordinate agrees, and its triangle inequality follows by applying \(|x_k-z_k|\leq |x_k-y_k|+|y_k-z_k|\) in each coordinate and summing.
We verify completeness. Let \((x^{(j)})_{j\geq1}\) be Cauchy in this metric. Fix a coordinate \(k\). If two sequences disagree at coordinate \(k\), their distance is at least \(2^{-k}\). Since \((x^{(j)})\) is Cauchy, there is an index after which the distance between any two terms is less than \(2^{-k}\). Therefore their \(k\)th coordinates must all agree after that index. Define \(x_k\) to be this eventual value, for each \(k\), to obtain \(x\in X\).
To show \(x^{(j)}\to x\), fix \(\varepsilon>0\) and choose \(N\) with \(2^{-N}<\varepsilon\). Each of the first \(N\) coordinates is eventually constant, so for all sufficiently large \(j\), \(x^{(j)}_k=x_k\) for \(1\leq k\leq N\). For such \(j\), \(d(x^{(j)},x)\leq\sum_{k=N+1}^{\infty}2^{-k}=2^{-N}<\varepsilon\). Thus \(X\) is complete.
Finally, \(X\) has no isolated points. Given \(x\in X\) and \(r>0\), choose \(k\) so large that \(2^{-k}<r\), and form \(y\) by changing only the \(k\)th coordinate of \(x\). Then \(y\ne x\) and \(d(x,y)=2^{-k}<r\). Every open ball around \(x\) therefore contains another point. The theorem just proved shows that \(X\) is uncountable.
What Category Does—and Does Not—Say
A meagre set is small in the category sense, not necessarily in measure, cardinality, or geometric extent. The rationals provide a useful warning: they are dense and countable, but meagre in \(\mathbb{R}\). Conversely, a category statement alone does not give a numerical measure estimate. Measure and category answer different questions, and neither notion should be silently substituted for the other.
The role of completeness is equally important. The Baire Category Theorem does not say that every metric space has the dense-intersection property. Its proof depends on finding the limit of the Cauchy sequence of nested-ball centers inside the space. Incomplete spaces may fail the conclusion; for instance, in \(\mathbb{Q}\) with the usual metric, the singletons are nowhere dense in \(\mathbb{Q}\), and their countable union is all of \(\mathbb{Q}\). Thus \(\mathbb{Q}\) is not a Baire space.
A practical way to use the theorem is to express a desired set as a countable intersection of open dense sets. The theorem then guarantees that the desired set is dense in a complete metric space. Equivalently, to prove that a set cannot contain an open region, it can be enough to show it is a countable union of nowhere-dense sets. In either form, keep track of the ambient space and use completeness only where it has been established.
Check Your Understanding
Use the definitions and the complete-metric-space result to answer the following questions.
- Why does a set being nowhere dense depend on the ambient space in which its closure and interior are taken?
- How can the rationals be both dense and meagre in the real line?
- In the nested-ball proof, where is completeness used, and why must the limit lie in every chosen closed ball?
- Why does a complete metric space with no isolated points have uncountably many points?
- What fails when the Baire Category Theorem is applied to the rational numbers with their usual metric?