Countable Sets Have Hausdorff Dimension Zero

Small Covers for Countable Sets

A countable set may be infinite, unbounded, or dense in its ambient metric space. None of those properties prevents it from having zero Hausdorff measure in every positive dimension: the points can be covered one at a time, with the diameter allowed for the \(n\)-th cover set chosen to decrease sufficiently quickly with \(n\). The key is to control the sum of the \(s\)-th powers of all those diameters, not merely to make each individual diameter small.

Let \((X,d)\) be any metric space. For a nonempty subset \(U\subseteq X\), its diameter is \(\operatorname{diam}(U)=\sup\{d(x,y):x,y\in U\}\); the diameter of a singleton is zero. For \(s>0\) and \(\delta>0\), define the \(s\)-dimensional Hausdorff content at scale \(\delta\) by

\mathcal{H}^s_\delta(A) = \inf\left\{ \sum_{i=1}^{\infty}(\operatorname{diam} U_i)^s: A\subseteq\bigcup_{i=1}^{\infty}U_i,\ \operatorname{diam} U_i\le\delta \right\}.

The infimum is over countable covers by subsets of \(X\); finite covers are allowed by adding empty sets, which contribute zero. The \(s\)-dimensional Hausdorff outer measure is \(\mathcal{H}^s(A)=\lim_{\delta\downarrow0}\mathcal{H}^s_\delta(A)\). This limit exists as an extended nonnegative number because decreasing \(\delta\) restricts the covers and therefore makes the content nondecreasing. Some conventions include a fixed positive normalization factor in this definition. Such a factor does not affect whether the measure is zero.

The Covering Theorem

Theorem: In any metric space \((X,d)\), every countable subset \(A\subseteq X\) satisfies \(\mathcal{H}^s(A)=0\) for every \(s>0\). No boundedness, completeness, or separability assumption on \(X\) is required.

Proof. Fix \(s>0\) and \(\delta>0\). If \(A\) is empty, the empty cover has cost zero, so \(\mathcal{H}^s_\delta(A)=0\). Otherwise, write \(A=\{a_n:n\ge1\}\), allowing repeated points if \(A\) is finite. Let \(\varepsilon>0\). For each \(n\ge1\), set

r_n=\min\left\{\frac{\delta}{4},\frac{(\varepsilon 2^{-n-2})^{1/s}}{2}\right\}, \qquad U_n=B(a_n,r_n).

Every \(a_n\) belongs to its open ball \(U_n\), so these balls cover \(A\). For any \(x,y\in U_n\), the triangle inequality gives \(d(x,y)\le d(x,a_n)+d(a_n,y)<2r_n\). Consequently, \(\operatorname{diam}(U_n)\le2r_n\le\delta/2\le\delta\). The choice of \(r_n\) also gives

(\operatorname{diam} U_n)^s \le (2r_n)^s \le \varepsilon 2^{-n-2}.

Summing these bounds, and using \(\sum_{n=1}^{\infty}2^{-n-2}=1/4\), shows that this particular admissible cover has cost at most \(\varepsilon/4\). Therefore \(0\le\mathcal{H}^s_\delta(A)\le\varepsilon/4\) for every \(\varepsilon>0\). A nonnegative number bounded above by \(\varepsilon/4\) for every positive \(\varepsilon\) must be zero, so \(\mathcal{H}^s_\delta(A)=0\). This holds for every \(\delta>0\); taking the limit as \(\delta\downarrow0\) yields \(\mathcal{H}^s(A)=0\). \(\square\)

This construction allocates a small part of a total cost budget to each point. The diameters are controlled at the same time: every covering ball has diameter at most \(\delta\), while the sum of the \(s\)-th powers is at most \(\varepsilon/4\). Both controls are necessary in the definition.

Worked Example: A Singleton in an Arbitrary Metric Space

Fix \(x\in X\), \(s>0\), \(\delta>0\), and \(\varepsilon>0\). To give an explicit positive-radius cover of \(\{x\}\), take

r=\min\left\{\frac{\delta}{4},\frac{\varepsilon^{1/s}}{4}\right\}, \qquad U=B(x,r).

The ball contains \(x\), and the triangle inequality gives \(\operatorname{diam}(U)\le2r\le\delta/2\le\delta\). Also \(2r\le\varepsilon^{1/s}/2\), so \((\operatorname{diam}U)^s\le\varepsilon/2^s\le\varepsilon\), since \(s>0\) implies \(2^s\ge1\). Thus \(\mathcal{H}^s_\delta(\{x\})\le\varepsilon\) for every \(\varepsilon>0\), and hence it is zero. As a direct alternative, the singleton cover \(\{x\}\) has diameter zero and cost exactly zero.

Check: The positive-radius calculation meets both requirements: its diameter is at most \(\delta\), and its cost is at most \(\varepsilon\). The singleton cover verifies the result directly under the stated convention that covering sets may have diameter zero.

Worked Example: The Integers in the Real Line

Use the usual metric on \(\mathbb{R}\) and let \(A=\mathbb{Z}\). Enumerate its points by \(a_1=0\), \(a_{2k}=k\), and \(a_{2k+1}=-k\) for \(k\ge1\). For fixed \(s>0\), \(\delta>0\), and \(\varepsilon>0\), put \(r_n=\min\{\delta/4,(\varepsilon2^{-n-2})^{1/s}/2\}\) and cover \(a_n\) by the interval \(U_n=(a_n-r_n,a_n+r_n)\).

Each interval contains its assigned integer and has diameter \(2r_n\le\delta/2\). Its cost satisfies \((\operatorname{diam}U_n)^s=(2r_n)^s\le\varepsilon2^{-n-2}\). Summing over the enumeration gives

\sum_{n=1}^{\infty}(\operatorname{diam}U_n)^s \le \varepsilon\sum_{n=1}^{\infty}2^{-n-2} =\frac{\varepsilon}{4}.

Thus \(\mathcal{H}^s_\delta(\mathbb{Z})=0\) for every \(\delta>0\), and \(\mathcal{H}^s(\mathbb{Z})=0\). The fact that \(\mathbb{Z}\) is unbounded causes no problem: each interval is placed around its own center, and the definition imposes no restriction on where the cover sets lie.

Check: The first cost bound is \(\varepsilon2^{-3}=\varepsilon/8\), the next is \(\varepsilon2^{-4}=\varepsilon/16\), and the full geometric sum is \(\varepsilon(1/8+1/16+\cdots)=\varepsilon/4\), as claimed.

Worked Example: A Convergent Sequence Together with Its Limit

In \(\mathbb{R}\), consider \(A=\{0\}\cup\{1/n:n\ge1\}\). Enumerate it as \(a_1=0\) and \(a_{n+1}=1/n\) for \(n\ge1\). For any \(s>0\), \(\delta>0\), and \(\varepsilon>0\), use the same radii \(r_j=\min\{\delta/4,(\varepsilon2^{-j-2})^{1/s}/2\}\), and take the intervals \(U_j=(a_j-r_j,a_j+r_j)\).

For each \(j\), \(a_j\in U_j\), and \(\operatorname{diam}(U_j)=2r_j\le\delta/2\). Therefore \((\operatorname{diam}U_j)^s\le\varepsilon2^{-j-2}\), and

\sum_{j=1}^{\infty}(\operatorname{diam}U_j)^s \le\varepsilon\sum_{j=1}^{\infty}2^{-j-2} =\frac{\varepsilon}{4}.

This proves \(\mathcal{H}^s_\delta(A)=0\) and then \(\mathcal{H}^s(A)=0\). The sequence has a limit point, but the argument does not need to group nearby points together: it covers every member, including the limit, explicitly.

Check: The center of the first interval is \(a_1=0\), so the limit point is covered. For \(j\ge2\), the center is \(a_j=1/(j-1)\), so every listed sequence term is covered as well.

Hausdorff Dimension and Zero-Dimensional Measure

Definition: For a subset \(A\) of a metric space, its Hausdorff dimension is \(\dim_H(A)=\inf\{s\ge0:\mathcal{H}^s(A)=0\}\), using the usual convention \(\dim_H(\varnothing)=0\).

Corollary: Every nonempty countable subset \(A\) of a metric space has Hausdorff dimension zero.

Proof. By the theorem, \(\mathcal{H}^s(A)=0\) for every \(s>0\). Hence the set of exponents in the definition of \(\dim_H(A)\) contains every positive real number. Its infimum is at most zero, because positive exponents can be arbitrarily close to zero, and it is at least zero because the infimum is taken over \(s\ge0\). Therefore \(\dim_H(A)=0\). \(\square\)

Zero Hausdorff dimension does not mean that the set has no points, or that its size is zero when size means counting points. Under the usual zero-dimensional convention, \(\mathcal{H}^0\) is counting measure: each point has cost one, and a set has measure equal to its number of points, with value \(+\infty\) when it is countably infinite. Equivalently, for \(s=0\), each nonempty covering set is assigned cost one. This convention avoids the ambiguity of interpreting \(0^0\). Thus a singleton has \(\mathcal{H}^0\)-measure \(1\), while its \(\mathcal{H}^s\)-measure is zero for every \(s>0\); the integers have infinite \(\mathcal{H}^0\)-measure but zero \(\mathcal{H}^s\)-measure for every \(s>0\).

Common pitfall. It is incorrect to infer that a set with Hausdorff dimension zero must have zero counting measure. For example, \(\mathbb{Z}\) has Hausdorff dimension zero by the corollary, but \(\mathcal{H}^0(\mathbb{Z})=+\infty\) because it has infinitely many points. The correction is to keep the exponent explicit: the theorem gives \(\mathcal{H}^s(\mathbb{Z})=0\) for each \(s>0\), not \(\mathcal{H}^0(\mathbb{Z})=0\).

A second covering mistake is to give every point the same positive-radius interval and then claim that the total cost is small. For an infinite enumeration, intervals of a common diameter \(c>0\) have total \(s\)-cost \(\sum_{n=1}^{\infty}c^s=+\infty\); making \(c\) small does not make that sum finite. The correction is to assign different diameters, with their \(s\)-th powers summable, as in the geometric allocation \(\varepsilon2^{-n-2}\). One must also respect the scale restriction: a cover whose total cost is small but whose sets have diameter larger than \(\delta\) is not admissible for \(\mathcal{H}^s_\delta\).

The conclusion applies to countable sets in arbitrary metric spaces, not only subsets of the line. It is a useful baseline when comparing a countable set with more complicated sets: positive Hausdorff dimension cannot be detected merely from a set being infinite, dense, or unbounded. The covering argument identifies exactly what matters here: the points can be assigned covers whose diameters satisfy the required scale and whose total \(s\)-cost is arbitrarily small.

Check Your Understanding

Use the definition of Hausdorff outer measure and keep the exponent \(s\) explicit in each answer.

For a countable set enumerated as \(A=\{a_n:n\ge1\}\), explain why radii \(r_n=\min\{\delta/4,(\varepsilon2^{-n-2})^{1/s}/2\}\) give an admissible cover for \(\mathcal{H}^s_\delta(A)\).
Why does the proof require \(s>0\) when it chooses a radius using the \(s\)-th root of a cost budget?
What are \(\mathcal{H}^0(\{x\})\) and \(\mathcal{H}^s(\{x\})\) for \(s>0\), under the counting-measure convention for \(\mathcal{H}^0\)?
Does the theorem require a countable set to be bounded or closed? Identify where either condition would enter the covering proof.
Why does assigning the same positive diameter to every member of an infinite set fail to give a finite total \(s\)-cost?