Rational Interval Bases for Borel Sets

Why Rational Intervals Suffice

Open subsets of the real line can have complicated shapes: they may be unbounded or have infinitely many separate components. Yet to generate the Borel sigma-algebra, it is enough to use intervals whose endpoints are rational. The reason is twofold: rational numbers are dense, so these intervals can fit inside any open neighborhood, and there are only countably many such intervals. Countability matters because a sigma-algebra is closed under countable unions, not arbitrary unions.

We work with the real line \(\mathbb{R}\), its usual topology, and its Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\), the sigma-algebra generated by the open subsets of \(\mathbb{R}\). Thus \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) is the measurable space in view; no measure is needed here. Define the family of rational-endpoint open intervals by

We will prove that \(\mathcal{I}_{\mathbb{Q}}\) is countable, that it forms a basis for the usual topology, and that \(\sigma(\mathcal{I}_{\mathbb{Q}})=\mathcal{B}_{\mathbb{R}}\). Here, saying it is a basis means that for every open set \(U\subseteq\mathbb{R}\) and every \(x\in U\), there is an interval \(I\in\mathcal{I}_{\mathbb{Q}}\) such that \(x\in I\subseteq U\).

Density and Countability

Proof. Let \(a<b\), so \(b-a>0\). By the Archimedean property, choose a positive integer \(n\) such that \(n(b-a)>1\), or equivalently \(nb-na>1\). Choose the integer \(k=\lfloor na\rfloor+1\). The definition of the floor gives \(\lfloor na\rfloor\le na<\lfloor na\rfloor+1\), and hence \(na<k\le na+1\). Since \(nb>na+1\), we obtain \(na<k<nb\). Dividing by \(n>0\) gives \(a<k/n<b\), and \(k/n\) is rational. \(\square\)

This density statement can be applied independently on either side of a point: whenever \(a<x<b\), there are rationals \(p,q\) with \(a<p<x<q<b\). To see this, apply the lemma to \((a,x)\) to find \(p\), and to \((x,b)\) to find \(q\). In particular \(p<q\), so \((p,q)\) is one of the intervals in \(\mathcal{I}_{\mathbb{Q}}\).

Proof. The integers \(\mathbb{Z}\) can be listed in a sequence, for example by listing \(0,1,-1,2,-2,\ldots\), so \(\mathbb{Z}\) is countable. The positive integers are countable as well. Pairs \((m,n)\in\mathbb{Z}\times\{1,2,\ldots\}\) can be listed by diagonals: list pairs in order of increasing \(|m|+n\), and within each diagonal use the fixed order on integers. Each diagonal contains only finitely many pairs, and every pair occurs on some diagonal. Thus this product is countable. The map \((m,n)\mapsto m/n\) has image \(\mathbb{Q}\), so \(\mathbb{Q}\) is countable. Applying the same diagonal argument to pairs of rational numbers shows that \(\mathbb{Q}\times\mathbb{Q}\) is countable. The family \(\mathcal{I}_{\mathbb{Q}}\) is the image of the subset \(\{(p,q)\in\mathbb{Q}\times\mathbb{Q}:p<q\}\) under the map \((p,q)\mapsto(p,q)\), and therefore is countable. \(\square\)

The Basis Property and Borel Generation

Proof. First, every \((p,q)\in\mathcal{I}_{\mathbb{Q}}\) is open in the usual topology, since \(p<q\). Next, let \(U\subseteq\mathbb{R}\) be open and take any \(x\in U\). By the definition of openness, there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). The density lemma gives rational numbers \(p,q\) with \(x-\varepsilon<p<x<q<x+\varepsilon\). Consequently \(x\in(p,q)\subseteq U\), proving the basis property.

For the generation claim, each interval in \(\mathcal{I}_{\mathbb{Q}}\) is open and thus belongs to \(\mathcal{B}_{\mathbb{R}}\). Since \(\mathcal{B}_{\mathbb{R}}\) is a sigma-algebra containing all these intervals, it contains the sigma-algebra they generate: \(\sigma(\mathcal{I}_{\mathbb{Q}})\subseteq\mathcal{B}_{\mathbb{R}}\). Conversely, let \(U\) be any open subset of \(\mathbb{R}\). The basis property gives \[ U=\bigcup\{I\in\mathcal{I}_{\mathbb{Q}}: I\subseteq U\}. \] Indeed, every interval in this union is contained in \(U\), and every \(x\in U\) belongs to at least one such interval. Because \(\mathcal{I}_{\mathbb{Q}}\) is countable, this is a countable union (including the possibility of the empty union when \(U=\varnothing\)). Every member of the union belongs to \(\sigma(\mathcal{I}_{\mathbb{Q}})\), so \(U\) does too. Therefore \(\sigma(\mathcal{I}_{\mathbb{Q}})\) contains every open set and hence contains the sigma-algebra generated by all open sets, namely \(\mathcal{B}_{\mathbb{R}}\). Combining the two inclusions proves equality. \(\square\)

The countability argument is essential to the last step: a basis represents each open set as a union of basis members, but a sigma-algebra is guaranteed to contain only countable unions of its members. In this case, countability of the entire basis ensures that every such representation is countable.

Worked Applications

A Common Mistake: Using an Uncountable Union

A frequent error is to find, for each \(x\in U\), some open interval around \(x\) contained in \(U\), and then conclude directly that \(U\) belongs to a sigma-algebra containing those intervals. This argument has a gap: the union indexed by all \(x\in U\) may be uncountable, and closure under arbitrary unions is not part of the definition of a sigma-algebra.

For example, one can write \((0,1)=\bigcup_{x\in(0,1)}(x/2,(x+1)/2)\). Each interval in this expression is contained in \((0,1)\), and each \(x\in(0,1)\) belongs to its own interval. But the displayed representation is indexed by all real \(x\in(0,1)\), so it does not by itself use a permitted countable-union operation. The corrected argument is to take all rational-endpoint intervals contained in \((0,1)\). There are countably many of them, and the basis property proves their union is exactly \((0,1)\). It is the combination of a basis with countability—not merely the existence of neighborhoods—that yields Borel generation.