Countable Stability Of Hausdorff Dimension

Why Countable Unions Require a Supremum

Hausdorff dimension describes the small-scale covering behavior of a set. When sets are combined, the natural question is whether their union can have a dimension larger than any one member. For a finite union, the dimension is the maximum of the component dimensions. For a countable union, a maximum need not exist: the component dimensions may approach a limiting value without ever reaching it. The correct general statement uses a supremum.

Work throughout in a metric space $(X,d)$, and let $E\subseteq X$. For $s>0$ and $\delta>0$, define the $s$-dimensional Hausdorff content at scale $\delta$ by taking the infimum of $\sum_i(\operatorname{diam} U_i)^s$ over countable covers $E\subseteq\bigcup_i U_i$ with $\operatorname{diam} U_i\le\delta$. The Hausdorff measure is the limit of these contents as $\delta\downarrow0$, denoted $\mathcal{H}^s(E)$. Here $\operatorname{diam} U=\sup\{d(x,y):x,y\in U\}$. The Hausdorff measure is an outer measure, so its monotonicity and countable subadditivity apply to arbitrary subsets; no measurability assumption on the sets in this discussion is needed.

Definition: For a subset $E$ of a metric space, its Hausdorff dimension is $$ \dim_H E=\inf\{s\ge0:\mathcal{H}^s(E)=0\}. $$ We use $\dim_H\varnothing=0$, and the infimum of an empty set of exponents is $+\infty$. Thus dimensions take values in $[0,\infty]$.

The definition uses the exponents at which the Hausdorff measure vanishes. The fact that vanishing persists at every larger exponent is essential: it lets us test all the component sets at one common exponent, strictly above their dimensions.

How Vanishing Changes with the Exponent

Lemma: If $0\le s<t$ and $\mathcal{H}^s(E)=0$, then $\mathcal{H}^t(E)=0$.

Proof. Fix $0<\delta\le1$. For every cover with diameters at most $\delta$, each positive diameter $r$ satisfies $r^t=r^{t-s}r^s\le\delta^{t-s}r^s$. A zero diameter contributes zero to the $t$-sum; for $s=0$, interpret the zero-exponent covering cost of a nonempty covering set as $1$. Consequently, for $s>0$, taking infima over the same covers gives $\mathcal{H}^t_\delta(E)\le\delta^{t-s}\mathcal{H}^s_\delta(E)$, where the subscript denotes the content at scale $\delta$. Since $\mathcal{H}^s(E)=0$, every $\mathcal{H}^s_\delta(E)=0$: these nonnegative contents increase as $\delta$ decreases and have limit zero. Hence $\mathcal{H}^t_\delta(E)=0$ for each $\delta$, and taking $\delta\downarrow0$ proves the claim. At $s=0$, if $\mathcal{H}^0(E)=0$, then $E=\varnothing$: any nonempty set requires a covering set of positive $0$-dimensional cost. The conclusion is immediate for the empty set. $\square$

In particular, if $\dim_H E<t$, then $\mathcal{H}^t(E)=0$. Indeed, by the definition of an infimum there is a vanishing exponent $s<t$, and the lemma applies. This argument also covers $\dim_H E=0$, including the case where the infimum itself is not a vanishing exponent.

Proposition: If $A\subseteq B$ in a metric space, then $\dim_H A\le\dim_H B$.

Proof. Monotonicity of Hausdorff outer measure gives $\mathcal{H}^s(A)\le\mathcal{H}^s(B)$ for every $s\ge0$. Therefore every exponent at which $\mathcal{H}^s(B)=0$ also has $\mathcal{H}^s(A)=0$. If $\dim_H B<\infty$, then for each $t>\dim_H B$ the preceding consequence of the lemma gives $\mathcal{H}^t(B)=0$, and thus $\mathcal{H}^t(A)=0$. It follows that $\dim_H A\le t$ for every such $t$, hence $\dim_H A\le\dim_H B$. If $\dim_H B=+\infty$, the desired inequality holds because every extended nonnegative dimension is at most $+\infty$. $\square$

Countable Stability

Theorem: Let $E_1,E_2,\ldots$ be subsets of a common metric space. Then $$ \dim_H\left(\bigcup_{n=1}^{\infty}E_n\right)=\sup_{n\ge1}\dim_H E_n. $$

Proof. Write $E=\bigcup_{n=1}^{\infty}E_n$ and $D=\sup_{n\ge1}\dim_H E_n$, allowing $D=+\infty$. Because $E_n\subseteq E$, the proposition gives $\dim_H E_n\le\dim_H E$ for each $n$. Taking the supremum yields $D\le\dim_H E$.

If $D=+\infty$, this lower bound forces $\dim_H E=+\infty$, proving equality. Suppose instead that $D<+\infty$, and fix any $t>D$. For every $n$, $\dim_H E_n\le D<t$, so the exponent lemma gives $\mathcal{H}^t(E_n)=0$. Countable subadditivity of Hausdorff outer measure now gives $$ 0\le\mathcal{H}^t(E)\le\sum_{n=1}^{\infty}\mathcal{H}^t(E_n)=0. $$ Thus $\mathcal{H}^t(E)=0$ for every $t>D$, so the definition of dimension implies $\dim_H E\le D$. Combining this with the lower bound proves the equality. If every $E_n$ is empty, then $E=\varnothing$ and both sides are $0$ under the stated convention. $\square$

The proof uses one exponent $t$ for the whole family. Choosing $t$ strictly above the supremum ensures that every component has zero $t$-dimensional Hausdorff measure. Countable subadditivity then controls the union at that same exponent. No finiteness of any other Hausdorff measure, and no measurability of the sets, is required.

Worked Applications

Worked Example: Dimensions Approaching a Limit

Suppose $(E_n)$ is a sequence of subsets of one metric space and the dimensions have been determined to be $\dim_H E_n=1-\frac{1}{n+1}$ for $n\ge1$. Find the dimension of their union, and decide whether any member has that dimension.

The first values are $1-\frac12=\frac12$, $1-\frac13=\frac23$, and $1-\frac14=\frac34$. In general, $1-\frac{1}{n+1}<1$, while for every $\varepsilon>0$, choosing an integer $n$ with $n+1>1/\varepsilon$ gives $\frac{1}{n+1}<\varepsilon$, hence $1-\frac{1}{n+1}>1-\varepsilon$. Therefore the supremum of the dimensions is $1$, and no individual dimension equals $1$. The theorem gives $$ \dim_H\left(\bigcup_{n=1}^{\infty}E_n\right)=1. $$ Check: each component dimension is strictly less than $1$, but the values exceed $1-\varepsilon$ for every positive $\varepsilon$; this verifies that the supremum is $1$, not a maximum.

Worked Example: A Largest Dimension Is Attained

Suppose $(F_n)$ is a sequence of subsets of one metric space with $\dim_H F_1=\frac{3}{5}$, $\dim_H F_2=\frac{4}{5}$, and $\dim_H F_n=\frac{7}{10}$ for every $n\ge3$. Compute the dimension of the union.

Compare the values with denominator $10$: $\frac35=\frac6{10}$, $\frac45=\frac8{10}$, and $\frac7{10}$ for every $n\ge3$. The supremum is $\frac8{10}=\frac45$, attained by $F_2$. Countable stability yields $$ \dim_H\left(\bigcup_{n=1}^{\infty}F_n\right)=\frac45. $$ Check: the union contains $F_2$, so monotonicity gives dimension at least $\frac45$; the theorem gives dimension at most the supremum $\frac45$. The two bounds agree.

Worked Example: A Countable Union Does Not Add Dimensions

Suppose $(G_n)$ is a sequence of subsets of one metric space and $\dim_H G_n=\frac{1}{n}$ for every $n\ge1$. Compare the dimension of the union with the sum of the component dimensions.

The dimensions are $1,\frac12,\frac13,\ldots$, so their supremum is $1$: the first value is $1$, and every later value is at most $1$. Thus countable stability gives $$ \dim_H\left(\bigcup_{n=1}^{\infty}G_n\right)=1. $$ The sum of the component dimensions is not the relevant operation; its partial sums are $1+\frac12+\cdots+\frac1N$, which grow without bound. The union's dimension is nevertheless $1$, by the theorem. Check: $G_1\subseteq\bigcup_nG_n$ forces the union dimension to be at least $1$, while the supremum bound forces it to be at most $1$.

Common Pitfall: Replacing a Supremum by a Maximum

A countable family need not contain a member whose dimension equals the dimension of its union. In the first example, every component dimension is strictly below $1$, but their supremum is $1$. Saying “the union has the largest component dimension” would therefore give no answer: there is no largest value in that family.

Another incorrect calculation is to add the component dimensions. For instance, in the third example the sum of the dimensions diverges, whereas the union has dimension $1$. Dimension measures the scale behavior of a set, not a total amount that accumulates by addition across a cover by sets. The proof explains the right operation: monotonicity gives a lower bound by every component dimension, and countable subadditivity at a common exponent gives the upper bound by their supremum.

Check Your Understanding

Use the exponent lemma and countable-stability theorem to justify each conclusion.

If $\dim_H E_n\le\frac{2}{3}$ for every $n$, what upper bound follows for $\dim_H(\bigcup_nE_n)$?
Can $\dim_H(\bigcup_nE_n)$ be strictly larger than every $\dim_H E_n$ and still equal their supremum? Explain.
Why must the proof choose an exponent $t$ strictly greater than $D=\sup_n\dim_H E_n$, rather than assume that $\mathcal{H}^D(E_n)=0$?
If one component has dimension $+\infty$, what does monotonicity imply about the dimension of the union?
Which step in the upper bound uses countable subadditivity, and why is it important that all component measures use the same exponent?