The Question: Which Sets Do Rational Half-Lines Generate?
Many descriptions of the Borel sigma-algebra on \(\mathbb{R}\) use open intervals, while the pi-lambda theorem is applied to families closed under finite intersections. Half-lines with rational endpoints connect these viewpoints: they are simple to intersect, and countable set operations recover intervals and open sets from them. We will show precisely that the half-lines \((-\infty,q]\), for \(q\in\mathbb{Q}\), generate \(\mathcal{B}_{\mathbb{R}}\).
All sets in this tutorial are subsets of \(\mathbb{R}\). The Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\) is the sigma-algebra generated by the open subsets of \(\mathbb{R}\). For a family \(\mathcal{E}\) of subsets of \(\mathbb{R}\), \(\sigma(\mathcal{E})\) denotes the smallest sigma-algebra containing \(\mathcal{E}\).
We use the pi-lambda theorem, proved in the earlier material on that theorem: if \(\mathcal{P}\) is a pi-system on \(X\), \(\mathcal{D}\) is a lambda-system on \(X\), and \(\mathcal{P}\subseteq\mathcal{D}\), then \(\sigma(\mathcal{P})\subseteq\mathcal{D}\). The theorem does not require \(X\in\mathcal{P}\).
The Rational Half-Line Family
Define \(H_q=(-\infty,q]\) for each rational number \(q\), and put \(\mathcal{H}=\{H_q:q\in\mathbb{Q}\}\). These are closed, hence Borel, subsets of \(\mathbb{R}\). Also, for rational \(p,q\),
Since \(\min(p,q)\) is rational, \(\mathcal{H}\) is a pi-system. This intersection calculation is the structural reason these sets are suited to the pi-lambda theorem.
Proof. We prove the two inclusions separately. First, \(\mathcal{H}\subseteq\mathcal{B}_{\mathbb{R}}\), because each \(H_q\) is closed. Every sigma-algebra is a lambda-system: it contains \(X\), is closed under differences of nested sets, and is closed under countable unions of disjoint members. Thus \(\mathcal{B}_{\mathbb{R}}\) is a lambda-system on \(\mathbb{R}\) containing the pi-system \(\mathcal{H}\). The pi-lambda theorem gives \(\sigma(\mathcal{H})\subseteq\mathcal{B}_{\mathbb{R}}\).
For the reverse inclusion, first take rational \(r<s\). For any rational \(q\) with \(r<q<s\), the difference \(H_q\setminus H_r\) is \((r,q]\). Therefore
Indeed, every set in the union is contained in \((r,s)\). Conversely, if \(x\in(r,s)\), density of the rationals gives a rational \(q\) with \(x\le q<s\), and then \(x\in H_q\setminus H_r\). The union is countable, and every term belongs to \(\sigma(\mathcal{H})\), so \((r,s)\in\sigma(\mathcal{H})\).
Now let \(O\subseteq\mathbb{R}\) be open. Every \(x\in O\) lies in some open interval contained in \(O\), and density of \(\mathbb{Q}\) gives rational \(r,s\) with \(x\in(r,s)\subseteq O\). Consequently, \(O\) is the union of all rational-endpoint open intervals contained in \(O\). There are only countably many pairs \((r,s)\in\mathbb{Q}^2\), so this is a countable union. Each such interval belongs to \(\sigma(\mathcal{H})\), and hence \(O\in\sigma(\mathcal{H})\). Since \(\mathcal{B}_{\mathbb{R}}\) is generated by the open sets, \(\mathcal{B}_{\mathbb{R}}\subseteq\sigma(\mathcal{H})\). The two inclusions prove equality. \(\square\)
Worked Consequences of the Generation Argument
Worked Example: A Strict Ray with an Irrational Endpoint
Show that \((-\infty,\sqrt{2})\in\sigma(\mathcal{H})\), even though \(\sqrt{2}\) is not a rational endpoint.
Consider the countable union
If \(x<\sqrt{2}\), choose a rational \(q\) with \(x\le q<\sqrt{2}\). Then \(x\in H_q\), so \(x\) belongs to the union. If \(x\ge\sqrt{2}\), every rational \(q<\sqrt{2}\) satisfies \(q<x\), so \(x\notin H_q\) for every index \(q\). Thus the union is exactly \((-\infty,\sqrt{2})\). It belongs to \(\sigma(\mathcal{H})\) because it is a countable union of its generating sets. The strict endpoint matters: equality \(x=\sqrt{2}\) is excluded.
Check: a point \(x=1\) is included, since \(1\le q<\sqrt{2}\) for some rational \(q\), for example \(q=1\). The endpoint \(x=\sqrt{2}\) is excluded because no rational \(q<\sqrt{2}\) can satisfy \(\sqrt{2}\le q\).
Worked Example: An Open Interval with Arbitrary Endpoints
Write \((\sqrt{2},\pi)\) using rational open intervals, and verify the endpoint behavior.
Take the union over all rational \(r,s\) satisfying \(\sqrt{2}<r<s<\pi\):
Every interval in this union lies inside \((\sqrt{2},\pi)\), by the strict inequalities on \(r\) and \(s\). Conversely, for \(x\in(\sqrt{2},\pi)\), density of the rationals gives \(r,s\in\mathbb{Q}\) such that \(\sqrt{2}<r<x<s<\pi\). Thus \(x\in(r,s)\), proving the reverse containment. There are countably many rational pairs \((r,s)\), and each rational interval \((r,s)\) belongs to \(\sigma(\mathcal{H})\) by the proof above. Hence the displayed countable union belongs to \(\sigma(\mathcal{H})\).
Check: neither endpoint is included. Every interval used has \(r>\sqrt{2}\), so it omits \(\sqrt{2}\); every interval has \(s<\pi\), so it omits \(\pi\). Every point strictly between them is covered by a rational interval chosen around that point.
Worked Example: Generating a Singleton
Show that \(\{\sqrt{2}\}\in\sigma(\mathcal{H})\) by intersecting generated rays.
First, the strict ray \((-\infty,\sqrt{2})\) is in \(\sigma(\mathcal{H})\) by the preceding example. Its complement \([\sqrt{2},\infty)\) is therefore also in \(\sigma(\mathcal{H})\). For each positive integer \(n\), the closed ray \((-\infty,\sqrt{2}+1/n]\) is in \(\sigma(\mathcal{H})\): it is the intersection of \(H_q\) over rational \(q>\sqrt{2}+1/n\). To verify this identity, \(x\le\sqrt{2}+1/n\) implies \(x\le q\) for every such \(q\); if \(x>\sqrt{2}+1/n\), choose a rational \(q\) strictly between \(\sqrt{2}+1/n\) and \(x\), and \(x\notin H_q\).
It follows that
The right side contains \(\sqrt{2}\). If \(x>\sqrt{2}\), choose \(n\) so large that \(1/n<x-\sqrt{2}\); then \(x\) is not in \((-\infty,\sqrt{2}+1/n]\). No \(x<\sqrt{2}\) belongs to \([\sqrt{2},\infty)\). Hence the intersection is exactly the singleton, and closure of a sigma-algebra under countable intersections proves the claim.
Check: \(x=\sqrt{2}\) satisfies every non-strict upper bound, while any \(x>\sqrt{2}\) fails one of them. The intersection therefore keeps the endpoint itself and no other point.
Why the Endpoint Details Matter
A common error is to identify \(H_q\setminus H_r\) with \((r,q)\). When \(r<q\), the actual calculation is
The right endpoint is included because \(q\in H_q\) and \(q\notin H_r\). The correction is not to ignore that endpoint: use a countable union of \((r,q]\) as rational \(q\) increases toward \(s\). This union equals \((r,s)\), since each \(q<s\) excludes \(s\), while every \(x<s\) lies below some rational \(q<s\). The endpoint test is part of the set identity, not a cosmetic detail.
There is also a useful distinction between the two roles in the theorem. The pi-lambda theorem gives \(\sigma(\mathcal{H})\subseteq\mathcal{B}_{\mathbb{R}}\), because the generating half-lines are Borel. To prove the reverse inclusion, one must show that all open sets can be built from the half-lines; the rational interval basis provides exactly that step. Together, these arguments establish generation rather than merely showing that the proposed generators are Borel.
| Set operation | Result | Endpoint check |
|---|---|---|
| \(H_p\cap H_q\) | \(H_{\min(p,q)}\) | The smaller upper endpoint is included. |
| \(H_q\setminus H_r\), \(r<q\) | \((r,q]\) | The left endpoint is excluded; the right is included. |
| \(\bigcup_{q\in\mathbb{Q},\,q<a}H_q\) | \((-\infty,a)\) | The endpoint \(a\) is excluded. |
Check Your Understanding
For each question, track membership at the endpoints as well as for interior points.
- Verify directly that the family \(H_q=(-\infty,q]\), \(q\in\mathbb{Q}\), is closed under finite intersections.
- Express \((-\infty,a)\), for an arbitrary real \(a\), as a countable union of rational half-lines, and explain why \(a\) is excluded.
- For rational \(r<s\), calculate \(H_s\setminus H_r\). Which endpoint is included?
- Explain why every open subset of \(\mathbb{R}\) is a countable union of rational-endpoint open intervals.
- In the generation proof, which inclusion follows from the pi-lambda theorem, and which inclusion uses the rational interval basis?