From Nowhere Dense to Meager
A nowhere dense set cannot fill any open region, even after taking its closure. The previous tutorial showed that finite unions of nowhere dense sets retain this property, while a countable union need not: the rational numbers are a countable union of singletons, yet are dense in \(\mathbb{R}\). The term meager (also called of first category) is designed to describe countable unions of nowhere dense sets without requiring the union itself to be nowhere dense.
As before, closures and interiors are taken in the specified ambient space \(X\). This convention is essential: a set may be meager in one space and not meager in another. A set contained in a meager set is also meager, so the definition can be stated using either a countable union or a subset of one.
Some texts define a meager set as a countable union of nowhere dense sets rather than as a subset of such a union. These definitions agree. If \(A\subseteq\bigcup_n N_n\), then \(A=\bigcup_n(A\cap N_n)\). Each subset of a nowhere dense set is nowhere dense: its closure is contained in the closure of the larger set, so its closure has empty interior. Thus \(A\) itself is a countable union of nowhere dense sets.
The word “countable” includes finite collections, since a finite list can be extended by empty sets. A nowhere dense set is therefore meager. Countable unions of meager sets are meager: gather the countably many nowhere dense sets covering each set into one countable list. These closure properties make meager sets a useful way to organize topologically small pieces, even when their union is dense.
Closed Nowhere Dense Covers
In practice, it is often convenient to cover a meager set by closed nowhere dense sets. Taking closures supplies this form of cover without changing the essential smallness: by definition, the closure of a nowhere dense set has empty interior.
Proof. Suppose \(A\) is meager. Choose nowhere dense sets \(N_n\) such that \(A\subseteq\bigcup_{n=1}^{\infty}N_n\), and set \(F_n=\overline{N_n}^{\,X}\). Each \(F_n\) is closed. Since \(N_n\) is nowhere dense, \(\operatorname{int}_X(\overline{N_n}^{\,X})=\varnothing\), so \(F_n\) is nowhere dense as well. Also \(A\subseteq\bigcup_{n=1}^{\infty}F_n\).
Conversely, suppose \(A\subseteq\bigcup_{n=1}^{\infty}F_n\), where every \(F_n\) is closed and nowhere dense. Each \(F_n\) is in particular nowhere dense, so this is a countable nowhere dense cover of \(A\). By the definition, \(A\) is meager. \(\square\)
This characterization gives a useful description of the complement of a meager set. If \(F_n\) are closed nowhere dense sets covering \(A\), then \(X\setminus\bigcup_n F_n=\bigcap_n(X\setminus F_n)\) is a countable intersection of open dense sets, and it is contained in \(X\setminus A\). Such an intersection is called a dense \(G_\delta\) when it is dense. The Baire Category Theorem supplies that density in Baire spaces; outside that setting, the intersection need not be dense.
Worked Example: The Dyadic Rationals
Let \(D=\{k/2^n:k\in\mathbb{Z},\ n\in\mathbb{N}\}\subseteq\mathbb{R}\). This set is countable, since it is a countable union, over \(n\), of countable sets of integer multiples of \(2^{-n}\). Every singleton \(\{d\}\) is nowhere dense in \(\mathbb{R}\): it is closed, and it contains no nonempty open interval. Therefore \(D\), a countable union of nowhere dense singletons, is meager.
In fact, \(D\) is dense. Let \((a,b)\) be any nonempty open interval. Choose \(n\) so large that \(2^n(b-a)>2\), and let \(k=\lfloor 2^n a\rfloor+1\). Then \(k>2^n a\), while \(k\leq 2^n a+1<2^n b\). Dividing by the positive number \(2^n\) gives \(a<k/2^n<b\), with \(k/2^n\in D\). Thus every nonempty open interval meets \(D\). This example shows that a meager set need not be nowhere dense and need not even be small in the sense of being nondense.
Its complement \(\mathbb{R}\setminus D\) is comeager by definition. It is also dense: every nonempty open interval contains a real number not in \(D\), since an interval is uncountable whereas \(D\) is countable. In this case both the meager set and its comeager complement are dense.
Comeager Sets and the Baire Property
A set is comeager when everything outside it is meager. The definition concerns the complement, not the set’s own closure or interior. In particular, a comeager set need not be open or closed. In earlier tutorials, sets whose complements are meager were also discussed under the name residual; in a Baire space, residual sets are dense.
Worked Example: The Integers and Their Complement
The integers \(\mathbb{Z}\) are closed in \(\mathbb{R}\). For example, if \(x\notin\mathbb{Z}\), choose an integer \(m\) with \(m<x<m+1\). The distance from \(x\) to the nearer of \(m\) and \(m+1\) is positive, so a sufficiently small open interval around \(x\) contains no integers. Thus the complement of \(\mathbb{Z}\) is open, proving that \(\mathbb{Z}\) is closed.
The integers have empty interior: every nonempty open interval contains nonintegers. Because they are closed, their closure is themselves, and hence \(\mathbb{Z}\) is nowhere dense. It follows that \(\mathbb{Z}\) is meager, and \(\mathbb{R}\setminus\mathbb{Z}\) is comeager. The complement is dense, since every nonempty open interval contains a noninteger. Here the meager set is itself nowhere dense, unlike the dense dyadic rationals in the preceding example.
The relationship between comeagerness and density depends on the ambient space. The Baire Category Theorem says that in a Baire space no nonempty open set is contained in a meager set. Consequently, every comeager subset of a Baire space is dense. Indeed, if a comeager set \(C\) were not dense, some nonempty open set would be disjoint from \(C\); that open set would be contained in \(X\setminus C\), which is meager, contradicting the Baire property. This is the same open-set obstruction used in the Baire Category Theorem and its corollaries.
Worked Example: A Space That Is Meager in Itself
Now take \(X=\mathbb{Q}\) with the metric inherited from \(\mathbb{R}\). Every singleton \(\{q\}\) is closed in \(X\). It has empty interior in \(X\), because every relatively open neighborhood of a rational number contains other rational numbers. Therefore each singleton is nowhere dense in \(X\).
Since \(\mathbb{Q}\) is countable, it is a countable union of these relatively nowhere dense singletons. Thus the whole space \(X=\mathbb{Q}\) is meager in itself. Its complement in \(X\) is empty, so the empty set is comeager in \(X\). But the empty set is not dense in the nonempty space \(\mathbb{Q}\). There is no contradiction with the Baire Category Theorem: \(\mathbb{Q}\) is not a Baire space. This example makes clear why “comeager implies dense” requires a Baire-space hypothesis.
Why Category Is Not Measure
Meagerness is a topological notion: it is defined using closure, interior, and countable unions of nowhere dense sets. It does not mean “small in length,” “probability zero,” or “few points.” The dyadic rationals are meager and dense, so meager sets can occur in every open interval. Conversely, a set can be nowhere dense and still have positive Lebesgue measure; category and measure answer different questions.
This distinction matters when a theorem gives a category conclusion. A comeager set in a Baire space is dense, but density alone does not say that it has full measure. Likewise, a measure-zero set can be dense. When working with both topological and measure-theoretic notions of smallness, check which definition a result uses rather than transferring conclusions from one setting to the other.
The practical pattern is to identify nowhere dense pieces, combine them into a meager exceptional set, and then study its complement. In a Baire space, that complement is dense; in a complete metric space, the Baire Category Theorem guarantees the needed Baire property. A conclusion that holds outside a meager set is therefore a topological “generic” conclusion, not a claim that the exceptional set is empty.
Check Your Understanding
Use the definitions, examples, and Baire Category Theorem to answer the following questions.
- Why can a countable union of nowhere dense sets be dense without being nowhere dense?
- How does taking closures turn a nowhere dense cover into a closed nowhere dense cover?
- What is the difference between a meager set and a comeager set?
- Why does the comeager set \(\mathbb{R}\setminus D\) in the dyadic-rational example remain dense?
- Why does the empty set being comeager in \(\mathbb{Q}\) not contradict the Baire Category Theorem?
- Why should a category statement not automatically be interpreted as a measure statement?