From Rational Intervals to Euclidean Space
On the real line, rational-endpoint open intervals give a countable basis for the usual topology. In \(\mathbb{R}^n\), the corresponding building blocks are products of such intervals. They are useful because they are simple enough to enumerate, yet numerous enough to describe every open set using only countable unions. This also links the Borel sigma-algebra in Euclidean space to the one-dimensional Borel sigma-algebra and to the coordinate maps.
Fix an integer \(n\ge1\). We use the usual Euclidean topology on \(\mathbb{R}^n\), and write \(\mathcal{B}_{\mathbb{R}^n}\) for its Borel sigma-algebra, namely the sigma-algebra generated by all open subsets of \(\mathbb{R}^n\). For \(a_i,b_i\in\mathbb{Q}\) with \(a_i<b_i\) for every \(i\), a rational open box is a set of the form \(\prod_{i=1}^n(a_i,b_i)\). A (possibly unbounded) open rectangle is a product \(\prod_{i=1}^n I_i\), where each \(I_i\) is an open interval in \(\mathbb{R}\), possibly empty or unbounded.
Proof. There are countably many rational endpoint pairs \((a,b)\in\mathbb{Q}^2\) with \(a<b\). An \(n\)-dimensional rational box is specified by \(n\) such pairs, so the family of rational open boxes is countable.
To verify the basis property, let \(U\subseteq\mathbb{R}^n\) be open and let \(x=(x_1,\ldots,x_n)\in U\). Since \(U\) is open, there is an \(\varepsilon>0\) such that the Euclidean ball \(B(x,\varepsilon)\) is contained in \(U\). Choose \(\delta>0\) so small that \(\sqrt{n}\delta<\varepsilon\). By the density of \(\mathbb{Q}\) in \(\mathbb{R}\), for each \(i\) there are rational numbers \(a_i,b_i\) with \(x_i-\delta<a_i<x_i<b_i<x_i+\delta\). Thus \(x\in Q:=\prod_{i=1}^n(a_i,b_i)\). For every \(y\in Q\), each coordinate difference satisfies \(|y_i-x_i|<\delta\), and hence \(\|y-x\|_2<\sqrt{n}\delta<\varepsilon\). Therefore \(Q\subseteq B(x,\varepsilon)\subseteq U\). This proves that every point of every open \(U\) lies in a rational open box contained in \(U\).
Let \(\mathcal{Q}\) denote the family of rational open boxes. The preceding property gives \(U=\bigcup\{Q\in\mathcal{Q}:Q\subseteq U\}\). This is a countable union because \(\mathcal{Q}\) is countable. Each rational open box is open, so it belongs to \(\mathcal{B}_{\mathbb{R}^n}\), and therefore \(\sigma(\mathcal{Q})\subseteq\mathcal{B}_{\mathbb{R}^n}\). Conversely, every open set is a countable union of members of \(\mathcal{Q}\), so every open set belongs to \(\sigma(\mathcal{Q})\). Since the Borel sigma-algebra is generated by the open sets, \(\mathcal{B}_{\mathbb{R}^n}\subseteq\sigma(\mathcal{Q})\). Thus \(\sigma(\mathcal{Q})=\mathcal{B}_{\mathbb{R}^n}\).
Finally, every open rectangle is open in \(\mathbb{R}^n\), so the sigma-algebra generated by all open rectangles is contained in \(\mathcal{B}_{\mathbb{R}^n}\). It contains every rational open box, which already generates \(\mathcal{B}_{\mathbb{R}^n}\). The two sigma-algebras are therefore equal. \(\square\)
Open Rectangles and Coordinate Projections
The box description can be restated in terms of individual coordinates. For \(i=1,\ldots,n\), define the coordinate projection \(\pi_i:\mathbb{R}^n\to\mathbb{R}\) by \(\pi_i(x_1,\ldots,x_n)=x_i\). For a family of maps, the sigma-algebra they generate means the sigma-algebra generated by all their inverse images of Borel sets.
Proof. Each \(\pi_i\) is continuous: the inverse image of an open interval under \(\pi_i\) is an open coordinate strip in \(\mathbb{R}^n\), and inverse images preserve complements and countable unions. More directly, continuity implies that \(\pi_i^{-1}(B)\) is Borel whenever \(B\in\mathcal{B}_{\mathbb{R}}\): the collection of \(B\subseteq\mathbb{R}\) whose inverse images are Borel is a sigma-algebra containing every open subset of \(\mathbb{R}\). Hence the sigma-algebra on the left is contained in \(\mathcal{B}_{\mathbb{R}^n}\).
For the reverse inclusion, a rational open box has the form $$ \prod_{i=1}^n(a_i,b_i)=\bigcap_{i=1}^n\pi_i^{-1}\bigl((a_i,b_i)\bigr). $$ Each interval \((a_i,b_i)\) is Borel in \(\mathbb{R}\), and the intersection is finite. Thus every rational open box belongs to the coordinate-generated sigma-algebra. The theorem shows that these boxes generate \(\mathcal{B}_{\mathbb{R}^n}\), proving the reverse inclusion and the equality. \(\square\)
In particular, an open rectangle \(\prod_i I_i\) is the finite intersection \(\bigcap_i\pi_i^{-1}(I_i)\). This explains why coordinate conditions are a natural way to describe Borel sets: each condition restricts one coordinate, and finite intersections combine the restrictions into rectangles. Countable operations on these sets then generate all Borel sets.
Worked Example: A Rectangle with an Irrational Endpoint
Write \(R=(0,\sqrt{2})\times(1,2)\subseteq\mathbb{R}^2\) as a countable union of rational open boxes.
Take all rational numbers \(p,q,r,s\) satisfying \(0\le p<q\le\sqrt{2}\) and \(1\le r<s\le2\). Each box \((p,q)\times(r,s)\) is contained in \(R\): if \(p<x<q\), then \(0\le p<x<q\le\sqrt{2}\), so \(0<x<\sqrt{2}\); the second coordinate satisfies \(1<y<2\) in the same way. Therefore the union of these boxes is contained in \(R\).
Conversely, let \((x,y)\in R\), so \(0<x<\sqrt{2}\) and \(1<y<2\). Density of the rationals provides \(p,q\in\mathbb{Q}\) with \(0\le p<x<q<\sqrt{2}\), and \(r,s\in\mathbb{Q}\) with \(1\le r<y<s\le2\). Hence \((x,y)\in(p,q)\times(r,s)\) for one of the boxes in the union. We have proved the exact representation $$ (0,\sqrt{2})\times(1,2) = \bigcup_{\substack{p,q,r,s\in\mathbb{Q}\\0\le p<q\le\sqrt{2}\\1\le r<s\le2}} (p,q)\times(r,s). $$ The indexing family is countable as a subfamily of \(\mathbb{Q}^4\). Check: every box on the right lies in \(R\), and every point of \(R\) lies in at least one such box, so both inclusions hold.
Worked Example: The Open Unit Disk as a Union of Rational Boxes
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}\). Give a precise rational-box union for \(D\).
For rationals \(a<b\) and \(c<d\), define \(M_1=\max(|a|,|b|)\) and \(M_2=\max(|c|,|d|)\). Then $$ D=\bigcup_{\substack{a,b,c,d\in\mathbb{Q}\\a<b,\ c<d\\M_1^2+M_2^2\le1}}(a,b)\times(c,d). $$ Indeed, if the displayed inequality holds and \((x,y)\in(a,b)\times(c,d)\), then \(|x|<M_1\) and \(|y|<M_2\). Thus \(x^2+y^2<M_1^2+M_2^2\le1\), so the box is contained in \(D\). The strict inequality \(x^2+y^2<1\) still holds when \(M_1^2+M_2^2=1\), because the box endpoints are excluded.
For the opposite inclusion, take any \((x,y)\in D\). The function \((u,v)\mapsto u^2+v^2\) is continuous, and its value at \((x,y)\) is strictly below \(1\). Consequently some open neighborhood of \((x,y)\) lies in \(D\). The rational-box basis property gives a rational box containing \((x,y)\) and contained in \(D\). For any box contained in \(D\), its endpoint maxima must satisfy \(M_1^2+M_2^2\le1\): if the sum were greater than \(1\), points in the box sufficiently near endpoints with the corresponding absolute values would have squared-coordinate sum greater than \(1\). The box is therefore included in the displayed union. The union is countable because it is indexed by a subfamily of \(\mathbb{Q}^4\). Check: the containment argument verifies that every point in the union has \(x^2+y^2<1\), while the basis argument covers every point of \(D\).
Worked Example: Describing the Set \(x<y\) in the Plane
Express \(H=\{(x,y)\in\mathbb{R}^2:x<y\}\) as a countable union of rational open boxes.
Use the family $$ H=\bigcup_{\substack{a,b,c,d\in\mathbb{Q}\\a<b\le c<d}}(a,b)\times(c,d). $$ For every box in this union, \(x\in(a,b)\) and \(y\in(c,d)\) imply \(x<b\le c<y\), and hence \(x<y\). Conversely, suppose \(x<y\). Choose rational \(b,c\) so that \(x<b\le c<y\); this is possible because there is a nonempty interval between \(x\) and \(y\), and the rationals are dense. Choose rational \(a<x\) and rational \(d>y\). Then \(a<b\le c<d\), and \((x,y)\in(a,b)\times(c,d)\). These two inclusions prove the representation. Check: the condition \(b\le c\) makes every listed box lie strictly above the diagonal, and the choices cover every point strictly above it. The family is countable since it is indexed by a subset of \(\mathbb{Q}^4\).
A Countability Pitfall
A common error is to say that because the rational boxes form a countable family, every union of rational boxes is a countable union. Countability of the family means that each subfamily is countable; it does not mean every collection of sets under consideration is countable. The proof for open sets works because it selects a subfamily from the fixed countable basis. It does not license arbitrary unions in a sigma-algebra, which is closed under countable unions, not all unions.
For a concrete illustration, every singleton \(\{x\}\subseteq\mathbb{R}\) is closed and therefore Borel. An arbitrary union of such sets can be any subset \(A\subseteq\mathbb{R}\), since \(A=\bigcup_{x\in A}\{x\}\). Not every subset of \(\mathbb{R}\) is Borel: there are only continuum-many Borel sets, while the power set of \(\mathbb{R}\) has strictly larger cardinality. Thus arbitrary unions of Borel sets need not be Borel. The correction is to use the countable basis argument precisely: for each open \(U\), select the basis elements contained in \(U\); this selected family is countable because the entire basis is countable.
The coordinate characterization is useful whenever a set is specified through coordinate conditions. It also identifies the Borel structure that accompanies Euclidean coordinates: the smallest sigma-algebra making every coordinate projection measurable is exactly \(\mathcal{B}_{\mathbb{R}^n}\). The statement concerns Borel measurability; it does not identify the Lebesgue sigma-algebra, which also contains subsets of Borel null sets.
Check Your Understanding
Use the basis and generation arguments above to answer the following questions.
- Why is the family of rational open boxes in \(\mathbb{R}^n\) countable for each fixed finite \(n\)?
- Explain why the union of all rational open boxes contained in an open set \(U\) equals \(U\).
- Write \((-\infty,0)\times(2,5)\subseteq\mathbb{R}^2\) as an intersection of inverse images under coordinate projections.
- Which inclusion in the proof that rational boxes generate \(\mathcal{B}_{\mathbb{R}^n}\) uses the fact that every open set is a countable union of basis elements?
- Why does the countability of the rational-box basis not imply that every union of Borel sets is Borel?