From Nowhere Dense to Meager
A nowhere-dense set is small in a local sense: its closure contains no nonempty open set. But a set can be assembled from countably many nowhere-dense pieces, and the resulting union need not itself be nowhere dense. The notion of meagerness records this countable assembly. It is useful precisely because it describes a broader class of sets than nowhere-denseness alone.
Throughout, \(\mathbb{R}\) has its standard topology. In any topological space \((X,\mathcal{T})\), a set \(N\subseteq X\) is nowhere dense if \(\operatorname{int}\overline{N}=\varnothing\). Thus it is the closure of \(N\), not necessarily \(N\) itself, that must have empty interior.
Some texts define \(A\) to be meager when it is merely contained in a countable union of nowhere-dense sets. This gives the same class of sets. Indeed, if \(A\subseteq\bigcup_n N_n\), then \(A=\bigcup_n(A\cap N_n)\). Since \(A\cap N_n\subseteq N_n\), its closure is contained in \(\overline{N_n}\), so \(\operatorname{int}\overline{A\cap N_n}=\varnothing\). Each \(A\cap N_n\) is therefore nowhere dense, giving the equality form of the definition.
Basic Rules for Meager Sets
Proof. Suppose \(M=\bigcup_{n\in\mathbb{N}}N_n\), where each \(N_n\) is nowhere dense, and let \(A\subseteq M\). Then \(A=\bigcup_{n\in\mathbb{N}}(A\cap N_n)\). For each \(n\), the inclusion \(A\cap N_n\subseteq N_n\) implies \(\overline{A\cap N_n}\subseteq\overline{N_n}\), so \(\operatorname{int}\overline{A\cap N_n}=\varnothing\). Thus \(A\) is a countable union of nowhere-dense sets and is meager.
Now let \((M_j)_{j\in\mathbb{N}}\) be a sequence of meager sets. For every \(j\), choose a sequence \((N_{j,n})_{n\in\mathbb{N}}\) of nowhere-dense sets with \(M_j=\bigcup_{n\in\mathbb{N}}N_{j,n}\). The set of pairs \(\mathbb{N}\times\mathbb{N}\) can be enumerated as a sequence: list first the pairs with \(j+n=2\), then those with \(j+n=3\), and continue in order of increasing sum. Consequently, the sets \(N_{j,n}\) can also be listed as one sequence of nowhere-dense sets. Their union is \(\bigcup_{j\in\mathbb{N}}M_j=\bigcup_{j,n\in\mathbb{N}}N_{j,n}\), so this countable union is meager. \(\square\)
The argument only uses the definition of nowhere-denseness and the ability to enumerate pairs of positive integers. It does not require completeness, a metric, or a separation axiom.
Worked Example: A Sequence of Points
Let \(A=\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\). We show directly that \(A\) is meager. For each \(n\), take \(N_n=\{1/n\}\). A singleton \(\{x\}\) is closed in \(\mathbb{R}\), and its interior is empty: every open interval containing \(x\) also contains points other than \(x\). Hence \(\overline{N_n}=N_n\) and \(\operatorname{int}\overline{N_n}=\varnothing\) for every \(n\). Since \(A=\bigcup_{n\in\mathbb{N}}N_n\), the definition gives that \(A\) is meager.
Check: the points \(1/n\) are distinct, because \(1/n=1/m\) for positive integers \(m,n\) implies \(m=n\). Thus this really is a countable union of singleton sets; the argument does not depend on whether \(0\) is included in \(A\).
Worked Example: The Rationals Are Dense and Meager
In \(\mathbb{R}\), write \(\mathbb{Q}=\{q_n:n\in\mathbb{N}\}\), using an enumeration of the rationals. Such an enumeration can be obtained by listing reduced fractions \(p/q\), with \(p\in\mathbb{Z}\) and \(q\in\mathbb{N}\), in order of increasing \(|p|+q\); for each fixed value of \(|p|+q\), there are only finitely many pairs. Each singleton \(\{q_n\}\) is closed with empty interior, so it is nowhere dense. Therefore \(\mathbb{Q}=\bigcup_{n\in\mathbb{N}}\{q_n\}\) is meager.
At the same time, \(\mathbb{Q}\) is dense in \(\mathbb{R}\). To verify this, take any nonempty open interval \((a,b)\). Choose \(n\in\mathbb{N}\) so large that \(n(b-a)>1\), and set \(k=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer no larger than \(na\). Then \(na<k\leq na+1<nb\). Thus \(a<k/n<b\), and \(k/n\) is rational. Every nonempty open interval therefore meets \(\mathbb{Q}\).
This example shows that dense and meager are not opposites: a set can meet every nonempty open interval while still being a countable union of nowhere-dense sets.
Worked Example: An Uncountable Meager Set
Let \(C_0=[0,1]\), and construct \(C_{n+1}\) from \(C_n\) by removing the open middle third of each interval component. The standard Cantor set is \(C=\bigcap_{n=0}^{\infty}C_n\). We verify that it is nowhere dense, and hence meager.
Each \(C_n\) is a finite union of closed intervals, so it is closed. Therefore \(C\), their intersection, is closed. At stage \(n\geq 1\), \(C_n\) is a union of finitely many disjoint closed intervals, each of length \(3^{-n}\). If \(C\) contained a nonempty open interval \(J=(a,b)\), then \(J\subseteq C_n\) for every \(n\). An interval contained in this finite union must lie in one of its component intervals; otherwise it would cross a removed gap. Hence \(b-a\leq 3^{-n}\) for every \(n\). Choosing \(n\) with \(3^{-n}<b-a\) gives a contradiction. Thus \(C\) has empty interior. Since \(C\) is closed, \(\operatorname{int}\overline{C}=\operatorname{int}C=\varnothing\), so \(C\) is nowhere dense and is meager.
In fact, \(C\) is uncountable. For each sequence \((\varepsilon_n)_{n\in\mathbb{N}}\) with \(\varepsilon_n\in\{0,1\}\), form \(x=\sum_{n=1}^{\infty}2\varepsilon_n/3^n\). The partial choices of left or right thirds show that \(x\) lies in the closed interval of \(C_m\) specified by its first \(m\) choices, for every \(m\); hence \(x\in C\). If two binary sequences first differ at index \(k\), the difference contributed at that index has magnitude \(2/3^k\), while all later terms together can change the value by at most \(\sum_{n>k}2/3^n=1/3^k\). Their values are therefore distinct. There are uncountably many binary sequences: any proposed list misses the sequence whose \(n\)-th digit differs from the \(n\)-th digit of the \(n\)-th listed sequence. This gives an injection of an uncountable set into \(C\). Thus meagerness does not imply countability.
Check: \(C\) is meager for the direct reason that it is itself nowhere dense; it is also a countable union of nowhere-dense sets by writing \(C=C\cup\varnothing\cup\varnothing\cup\cdots\).
A Common Mistake and Why Category Matters
Common mistake. It is incorrect to conclude that a countable union of nowhere-dense sets is itself nowhere dense. For a concrete counterexample, write \(\mathbb{Q}=\bigcup_n\{q_n\}\). Every set \(\{q_n\}\) is nowhere dense, but the union is dense in \(\mathbb{R}\). Its closure is \(\mathbb{R}\), and \(\operatorname{int}\overline{\mathbb{Q}}=\mathbb{R}\ne\varnothing\), so \(\mathbb{Q}\) is not nowhere dense. The corrected conclusion is that \(\mathbb{Q}\) is meager: being a countable union of nowhere-dense sets is exactly what meagerness asserts.
Meagerness is a way of describing a set assembled from pieces whose closures contain no open region. The examples show why it is not simply another name for “small”: a meager set can be dense, as \(\mathbb{Q}\) is, and uncountable, as the Cantor set is. The countable-union rule is the basic bookkeeping fact that lets the notion apply to complicated sets built in stages.
Check Your Understanding
For each question, identify the ambient space and use the definition of nowhere-dense or meager sets.
- Why is every singleton in \(\mathbb{R}\) nowhere dense in the standard topology?
- Show that every countable subset of \(\mathbb{R}\) is meager.
- If \(A\subseteq M\) and \(M\) is meager, explain why \(A\) is meager.
- Why does the density of \(\mathbb{Q}\) not prevent it from being meager?
- Can a meager subset of \(\mathbb{R}\) be uncountable? Give the example established above and state why it qualifies.