Dyadic Rationals for Urysohn’s Construction

Why Use Dyadic Rationals?

In Urysohn’s construction, one needs an ordered collection of labels between 0 and 1. The labels must be countable so that they can be organized in stages, and there must be room to insert a new label between any two distinct labels. Dyadic rationals have both properties. Their especially useful feature is that the new labels can be obtained by taking midpoints.

This tutorial establishes these properties of the index set itself. It does not construct the open neighborhoods indexed by it. All comparisons use the usual order on the real numbers, and the interval under consideration is \(I=[0,1]\).

The description as a union of grids makes the refinement rule visible. Every point \(k/2^n\) of \(D_n\) is also the point \(2k/2^{n+1}\) of \(D_{n+1}\). Between consecutive points \(k/2^n\) and \((k+1)/2^n\), the next grid inserts the point \((2k+1)/2^{n+1}\). This is their midpoint, since \[ \frac{\frac{k}{2^n}+\frac{k+1}{2^n}}{2} =\frac{2k+1}{2^{n+1}}. \] So each finite grid is contained in the next, and each interval between consecutive points is bisected.

Countability and Order

Proof. Each \(D_n\) is finite: it has the \(2^n+1\) elements \(k/2^n\) for \(k=0,\ldots,2^n\). The union over \(n\geq0\) can be listed in finite stages. To avoid repeating elements, list \(0\) and \(1\) first, and then, for each \(n\geq1\) in increasing order, list \(k/2^n\) for odd integers \(k\) with \(1\leq k<2^n\). This list contains every dyadic rational: if \(0<k/2^m<1\), repeatedly divide the numerator and denominator by 2 whenever the numerator is even. The process stops with an odd numerator and a denominator \(2^n\), so the number occurs in the \(n\)-th block. The reduced form is unique, so it occurs only once. Each block is finite, and the blocks together with \(0,1\) form a sequence; hence \(D\) is countable. There are infinitely many elements, since the distinct numbers \(1/2^n\), for \(n\geq1\), all belong to \(D\). Finally, the usual order on \(\mathbb{R}\) restricts to a linear order on \(D\), and every element of \(D\) lies between \(0\) and \(1\), both of which belong to \(D\). Thus they are its least and greatest elements. \(\square\)

Countability and order are different features. The listing in the proof organizes elements by denominator; it is not an increasing listing. In fact, there is no least positive element of \(D\): if \(r>0\) belongs to \(D\), then \(r/2\) also belongs to \(D\) and \(0<r/2<r\). For an index set, it is the inherited order—not an increasing enumeration—that specifies which labels lie between others.

The Midpoint Property

Proof. Choose nonnegative integers \(m,n\) and integers \(k,l\) such that \(r=k/2^m\) and \(s=l/2^n\). Let \(N\) be the larger of \(m,n\). Rewriting both numbers with denominator \(2^N\), write \(r=K/2^N\) and \(s=L/2^N\), where \(K=k2^{N-m}\) and \(L=l2^{N-n}\). Then \[ \frac{r+s}{2}=\frac{K+L}{2^{N+1}}, \] which is dyadic. Because \(r<s\), adding \(r\) to both sides and dividing by 2 gives \(r<(r+s)/2\); adding \(s\) to both sides and dividing by 2 gives \((r+s)/2<s\). Thus the midpoint belongs to \(D\) and lies strictly between \(r\) and \(s\). \(\square\)

This theorem also covers intervals whose endpoints are 0 or 1. If \(r<s\) are in \(I\), their midpoint is between them, so it remains in \(I\). Repeating the operation produces further labels strictly inside the interval. In particular, no two distinct dyadic rationals are consecutive in the inherited order.

Density in the Unit Interval

The midpoint theorem supplies a dyadic between dyadic endpoints. For constructing an ordered index set, we need a stronger statement: there must be a dyadic strictly between any two distinct real numbers in \(I\), whether or not the endpoints are dyadic.

Proof. Set \(\delta=b-a\), so \(\delta>0\). Choose \(n\geq0\) so large that \(2^{-n}<\delta\). Such an \(n\) exists: by the Archimedean property of the real numbers, choose \(n\) with \(n+1>1/\delta\); since \(2^n\geq n+1\), we have \(2^{-n}\leq1/(n+1)<\delta\). Let \(k=\lfloor 2^n a\rfloor+1\), where \(\lfloor t\rfloor\) is the greatest integer no greater than \(t\). Its defining inequalities give \(2^n a<k\leq2^n a+1\). After division by \(2^n>0\), this becomes \[ a<\frac{k}{2^n}\leq a+2^{-n}<a+\delta=b. \] Thus \(d=k/2^n\) lies strictly between \(a\) and \(b\). Since \(a\geq0\), the inequality \(k>2^n a\) gives \(k\geq0\); since \(k/2^n<b\leq1\), we have \(k<2^n\). In particular \(0\leq k\leq2^n\), so \(d\in D_n\subseteq D\). \(\square\)

The proof gives a practical selection rule. Choose a grid fine enough that its spacing \(2^{-n}\) is smaller than the interval length, then take the first grid point strictly to the right of \(a\). The strict inequality \(2^{-n}<b-a\) matters: it ensures that this chosen point is still strictly to the left of \(b\).

Why These Properties Matter—and a Common Pitfall

For an ordered construction, the dyadics supply two complementary tools. Countability allows the labels to be handled in stages, while density guarantees that labels can be inserted between distinct real numbers. The midpoint formula makes repeated insertion especially direct when the two endpoints are already dyadic. These facts concern the numerical index set; the conditions on open neighborhoods in a normal space are a separate part of Urysohn’s construction.

Common pitfall: confusing the dyadics with all rationals. A mistaken calculation might claim that \(1/3\) is dyadic because it is rational. But \(1/3=k/2^n\) would imply \(2^n=3k\). No power of 2 is divisible by 3: modulo 3, \(2^n\) is 1 when \(n\) is even and 2 when \(n\) is odd. Therefore \(1/3\notin D\). Density does not mean that every point of \(I\) belongs to \(D\); it means every open interval with distinct endpoints in \(I\) contains a point of \(D\).

A related arithmetic error is to average fractions without first handling their denominators. For example, the midpoint of \(1/4\) and \(1/2\) is not \((1+2)/(4+2)=3/6\): adding denominators is not the rule for adding fractions. The correct calculation uses a common denominator: \((1/4+1/2)/2=(1/4+2/4)/2=(3/4)/2=3/8\). The result is strictly between \(1/4=2/8\) and \(1/2=4/8\), as required.