The Rational Points in a Bounded Interval

Enumeration Turns the Claim into a Cover

The rational points are dense in the real line, but density does not prevent them from having Lebesgue outer measure zero. The relevant construction is to list the rational numbers and surround the \(n\)-th one by an interval whose length decreases with \(n\). If those lengths have an arbitrarily small finite sum, the intervals form a cover with arbitrarily small total length.

We work in \(\mathbb{R}\) with Lebesgue outer measure \(m^*\), defined by taking the infimum of the sums of lengths of countable interval covers. Thus, if \(E\subseteq\mathbb{R}\) is covered by intervals \(J_n\), then \(m^*(E)\le\sum_n |J_n|\), where \(|J_n|\) denotes the length of \(J_n\). We use open intervals, so their centers are included. This is an outer-measure argument; it does not require \(E\) to be measurable.

The positive terms ensure that each \(J_n\) is an interval of the stated length. The summability condition controls the total cost of the cover. In particular, the length assigned to a rational may depend on its place in the enumeration: there is no need to use a common length for all the points.

Enumerating the Rational Numbers

For an explicit enumeration, write each rational uniquely as \(p/q\), where \(p\) is an integer, \(q\) is a positive integer, and \(p\) and \(q\) have no common positive divisor other than \(1\). Represent zero as \(0/1\). Give this reduced fraction height \(H(p/q)=|p|+q\). Order fractions first by increasing height, and within each height by increasing numerator \(p\).

There are only finitely many fractions of any given height: for height \(h\), the numerator satisfies \(|p|\le h\), and \(q=h-|p|\) is then determined and must be positive. So the ordering produces a list. It includes every rational, because every reduced fraction has a finite height, and it lists each rational only once because the reduced representation is unique. The first entries are \(0,-1,1,-2,-\tfrac12,\tfrac12,2,\ldots\). Denote this enumeration by \(q_1,q_2,\ldots\).

Proof. Fix \(x\in E\). Since \(E\subseteq\mathbb{Q}\) and \((q_n)\) lists every rational, there is an index \(k\) such that \(x=q_k\). Because \(a_k>0\), we have \(q_k-a_k/2<q_k<q_k+a_k/2\), so \(x\in J_k\). This holds for every \(x\in E\), and hence \(E\subseteq\bigcup_{n=1}^{\infty}J_n\). The sum of the lengths of these covering intervals is \(\sum_{n=1}^{\infty}|J_n|=\sum_{n=1}^{\infty}a_n=S\). By the definition of Lebesgue outer measure as an infimum over interval covers, \(m^*(E)\le S\). If \(E\) is empty, the covering assertion is vacuous and \(m^*(E)=0\le S\) as well. \(\square\)

The Rational Points in a Bounded Interval

Proof. Let \(\varepsilon>0\). Use the enumeration above and assign the \(n\)-th rational an interval of length \(a_n=\varepsilon 2^{-n}\). These lengths are positive. Their partial sums are \(\sum_{n=1}^{N}\varepsilon 2^{-n}=\varepsilon(1-2^{-N})\), so their total sum is \(\varepsilon\). By the lemma, the resulting intervals cover \(\mathbb{Q}\cap I\) and give \(m^*(\mathbb{Q}\cap I)\le\varepsilon\). This holds for every \(\varepsilon>0\), while outer measure is nonnegative. If the outer measure were positive, choosing \(\varepsilon\) smaller than it would contradict the bound. Therefore \(m^*(\mathbb{Q}\cap I)=0\). The argument also covers the empty and singleton cases. In fact, the proof did not use boundedness: the same construction covers all of \(\mathbb{Q}\). \(\square\)

The theorem is an explicit instance of the general covering idea for countable sets. Its useful feature here is that the rational enumeration makes the cover concrete, while the budget shows exactly how the total length is controlled. Boundedness specifies the set of interest, but is not needed for the zero outer measure conclusion.

Why the Budget Must Be Summable

A countable list by itself does not give a useful small-length cover. The lengths must be allocated so that their sum is finite, and to prove zero outer measure the available total must be made arbitrarily small. This is why the index matters: later entries receive smaller intervals, while every rational still receives an interval of positive length.

Common mistake. Suppose one assigns length \(\varepsilon\) to every rational in the enumeration and argues that each individual interval is short. The sum of the first \(N\) lengths is \(N\varepsilon\), which tends to infinity as \(N\) increases for any fixed \(\varepsilon>0\). Therefore this prescription gives no finite total-length bound for the countable cover. It is not enough to make each interval small separately.

Correction. Choose a summable sequence, for example \(a_n=\varepsilon 2^{-n}\). Its total is \(\varepsilon\sum_{n=1}^{\infty}2^{-n}=\varepsilon\), so the lemma supplies a cover of total length at most \(\varepsilon\). Since \(\varepsilon\) can be chosen as small as desired, the outer measure is zero. Another common error is to cover only the first \(N\) rationals: that is a finite cover of those listed points, not automatically a cover of every rational in the interval. The construction uses every index \(n\ge1\).

The contrast between density and outer measure is important. Every nonempty open interval contains rational points, but the rational points in a bounded interval still admit covers of arbitrarily small total length. Density concerns whether every open interval meets a set; outer measure concerns the total lengths needed to cover it. They are different properties.