Rationals, Irrationals, and Category in the Real Line

Category in the Real Line

In the real line, a set can be small in the sense of category and still occur in every open interval. The rationals illustrate this: they are dense, yet they are meager. Their complement, the irrationals, is residual and dense. The point of the Baire category theorem here is not that the irrationals outnumber the rationals, but that a countable intersection of open dense sets in a complete metric space remains dense.

Throughout, \(\mathbb{R}\) has its standard topology, induced by the usual metric \(d(x,y)=|x-y|\). A set \(D\subseteq\mathbb{R}\) is dense if every nonempty open subset of \(\mathbb{R}\) intersects \(D\). Equivalently, \(\overline{D}=\mathbb{R}\). We will use open intervals as neighborhoods: every open neighborhood of a real number contains an open interval around that number.

These definitions describe different properties. Nowhere dense and meager refer to coverings by sets whose closures have empty interior. Residual means that the complement is meager; in an arbitrary space that alone need not make a set dense. The Baire theorem supplies the density conclusion in the complete-metric setting we use here.

Why Each Rational Can Be Covered by a Nowhere-Dense Set

Proof. To show that \(\{q\}\) is closed, take any \(x\in\mathbb{R}\setminus\{q\}\). Then \(|x-q|>0\), and the open interval \((x-|x-q|/2,x+|x-q|/2)\) contains \(x\) but not \(q\). Thus the complement of \(\{q\}\) is open, so \(\overline{\{q\}}=\{q\}\). To show that its interior is empty, suppose an open set \(U\) contains \(q\). There is some \(r>0\) for which \((q-r,q+r)\subseteq U\). This interval contains \(q+r/2\), which differs from \(q\). Hence \(U\) cannot be a subset of \(\{q\}\). No open neighborhood of \(q\) lies inside \(\{q\}\), so \(\operatorname{int}\{q\}=\varnothing\). Its closure therefore has empty interior, as required. \(\square\)

The Rationals Are Meager

To use the singleton result, we need a countable listing of the rationals. Every rational is \(m/n\) for an integer \(m\) and a positive integer \(n\). The pairs \((m,n)\in\mathbb{Z}\times\mathbb{N}\) can be listed in a sequence: list them in increasing order of \(|m|+n\), and within each fixed value of \(|m|+n\), list the finitely many possible pairs in a fixed order. Every pair occurs, because its value of \(|m|+n\) is a finite integer and each such group is finite. Applying \((m,n)\mapsto m/n\) gives a sequence \((q_n)_{n\in\mathbb{N}}\) whose terms include every rational. Repetitions do not affect the union of the singletons.

Proof. Use the sequence \((q_n)\) just described. Since every rational appears in it, \(\mathbb{Q}=\bigcup_{n\in\mathbb{N}}\{q_n\}\). By the proposition, each \(\{q_n\}\) is nowhere dense in \(\mathbb{R}\). The definition of meager is a countable union of nowhere-dense sets, so \(\mathbb{Q}\) is meager. \(\square\)

This proof uses the rationals’ countability and the nowhere-denseness of each singleton. In particular, it does not say that the rationals are absent from some interval. Indeed, every nonempty open interval contains rationals. The category property and density answer different questions.

The Irrationals Are Residual and Dense

For each \(q_n\in\mathbb{Q}\), the complement \(\mathbb{R}\setminus\{q_n\}\) is open because \(\{q_n\}\) is closed. It is also dense: any nonempty open interval contains at least two distinct points, so it cannot be contained in the singleton \(\{q_n\}\). Consequently the interval intersects \(\mathbb{R}\setminus\{q_n\}\).

The complete-metric Baire theorem, proved in earlier material, states that if a metric space is complete, then every countable intersection of open dense subsets is dense. The usual metric on \(\mathbb{R}\) is complete: every Cauchy sequence of real numbers converges to a real number. We can therefore apply the theorem to the open dense sets \(\mathbb{R}\setminus\{q_n\}\).

Proof. Since the sequence \((q_n)\) includes every rational, a real number belongs to every set \(\mathbb{R}\setminus\{q_n\}\) exactly when it is not rational. Therefore \(\mathbb{R}\setminus\mathbb{Q}=\bigcap_{n\in\mathbb{N}}(\mathbb{R}\setminus\{q_n\})\). Each set in this intersection is open and dense, as verified above. The intersection is a \(G_\delta\) by definition, and the complete-metric Baire theorem says that it is dense in \(\mathbb{R}\). Its complement is \(\mathbb{Q}\), which is meager by the preceding theorem. Thus the irrationals are residual as well as dense. \(\square\)

A Common Pitfall: Meager Does Not Mean Nondense

A tempting but incorrect conclusion is that a meager set must miss some nonempty open interval. The rationals disprove it: we proved that \(\mathbb{Q}\) is meager, and the interval argument above shows that every nonempty open interval contains a rational. Hence \(\mathbb{Q}\) is both meager and dense.

The erroneous reasoning often confuses “nowhere dense” with “meager.” Nowhere dense means that the closure has empty interior; meager allows a countable union of nowhere-dense sets. Neither definition says that the union misses an open set. In this example each rational is a nowhere-dense singleton, and the countable union of all those singletons is still dense.

Likewise, residual does not mean open or equal to the whole space. The irrationals are residual because their complement is meager, and they are dense here because the Baire theorem applies. They are not open: every open interval contains a rational, so no nonempty open interval can be contained in the irrationals. The key distinctions are summarized below.