Show a specific set has Hausdorff dimension 3/4

Question

Let $S$ be the set of points in $[0,1]$ such that every fourth digit is zero, i.e., the set of $x=0.a_1a_2a_3a_4...$ with $a_{4i}=0$. How do I show that this set has Hausdorff dimension $\frac{3}{4}$? Can I use an argument similar to that used for the Cantor set on page 14 of Falconer's book The Geometry of Fractal Sets? I am trying to find an argument that does not make reference to the idea of self-similarity dimension.

Answer 1

Laying this out in full detail, perhaps for the benefit of more readers than the OP. I'll define the Hausdorff dimension and then show the result as well. The definitions are taken from Falconer's book.

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The definitions with their ideas

The Hausdorff dimension arises by optimally covering a set by sets with reducing diameter. That is, we need covers of a certain type, ways to measure those covers, and then take the infimum.

For $U \subset \mathbb R^n$ bounded, let $|U| = \sup\{|x-y| : x,y \in U\}$ be the diameter of $U$. Given a set $E \subset \mathbb R^n$ and an (at most) countable collection of sets $U_i, i \in I$ such that $E \subset \cup_{i \in I} U_i$ and $0<|U_i|\leq\delta$ for all $i$, we call $\{U_i\}$ a $\delta$-cover of $E$. That is, $E$ is being covered by countably many sets of diameter at most $\delta$.

Now, we need to find a way to "measure" these covers. For that, given $\{U_i\}$ a $\delta$-cover of $E$, let $m_s(\{U_i\}) = \sum_{i \in I} |U_i|^s$ where $s$ is some non-negative real number. One is inclined to think that reducing $m_s$ makes for a better, more efficient cover.

In that light, let $$\mathscr H_s^\delta(E) = \displaystyle\inf_{\{U_i\}\text{ a $\delta$-cover of $E$}} m_s(\{U_i\})$$

Note that $\mathscr H_s^{\delta}(E)$ is actually increasing in $\delta$, because reducing $\delta$ reduces the number of $\delta$-covers, thereby reducing the set over which the infimum is taken. It follows that $\mathscr H_s(E) = \lim_{\delta \to 0} \mathscr H_s^{\delta}(E)$ exists for each $s$.

However, if one fixes $E$, then things are not very interesting in $s$. Indeed, one can trivially verify the inequality $$ \mathscr H_s^{\delta}(E) \geq \delta^{s-t} \mathscr H_t^{\delta}(E) $$ for $s<t$, by using the same cover on each side and then taking the infimum. However, as $\delta \to 0$, it follows that if $\mathscr H_t(E)>0$, then $\mathscr H_s(E) = \infty$. Therefore, we see that $\{t : \mathscr H_t(E) > 0\}$ is an interval (which contains $0$, hence is non-empty, and) which will have a supremum that could infinite (turns out, it's not). That supremum is the Hausdorff dimension. That is, $$ \dim E = \sup\{t : \mathscr H_t(E) > 0\} $$

It is trivial to observe two things

• If $E_1 \subset E_2$ then $\dim E_1 \leq \dim E_2$ : any cover of $E_1$ is also a cover of $E_2$.

• $\dim \mathbb R^n \leq n$ : if $s>n$, then covering $\mathbb R^n$ by a "high dimensional grid" of cubes of side length $\delta$ with reducing $\delta$ will result in $\mathscr H^s_{\delta}(\mathbb R^n) = 0$.

Therefore, $E \subset \mathbb R^n$ implies that $\dim E \leq n$. One may try a simple exercise such as finding the Hausdorff dimension of a singleton, or a line segment in $\mathbb R^n$.

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How do we calculate this quantity?

Upper bounds come from finding good $\delta$-covers for some chosen values of $s$. Such $\delta$-covers will have measure which reduces to $0$ as the diameter $\delta$ shrinks, thus giving an upper bound.

Lower bounds are more interesting. Here, we want to find $t$ so that $\mathscr H_t(E)>0$. Therefore, we need to show that for $\delta$ small enough, $\mathscr H_t^{\delta}(E)>0$. That is an infimum : therefore, we are reduced to showing that all possible $\delta$-covers have measure bounded above! How we handle "all possible" has a nice answer, and while answering the question we are supposed to (!) we will focus on how we do it.

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The set

We are considering $$ S = \{x \in [0,1] : x = 0.a_1a_2a_3\ldots \implies a_{4i} =0 \ \forall i \geq 1\} $$

Note : $1 = 0.\overline{9} \notin S$, for example, while $0.342043 = 0.342043000000\ldots \in S$. We want to find the Hausdorff dimension of $S$.

For technical reasons, we need to prove that $S$ is closed, so we'll do it. Suppose that $x_i \in S$, and $x_i \to x$. We need to prove that $x \in S$.

Note that $x_i \in [0,1]$, therefore $x\in [0,1]$ because $[0,1]$ is closed. We can also see that if $x_i \in S$ then $x_i < 0.9999$, therefore $x \neq 1$ obviously. Thus, $x$ is of the form $0.\ldots$.

Then, for $l\geq 1$, by taking $\epsilon = 10^{-4l-1}$, there is $N$ such that $|x_n - x| < 10^{-4l-1}$ for $n>N$. However, this means that $x_n$ and $x$ agree up to $4l$ digits for all large $n$. As $x_n \in S$, we get that the fourth, eighth, $\ldots$, $4l$th digits of $x$ are all $0$. Since this is true for any $l$, it follows that $x\in S$.

Therefore, as $S$ contains all its limit points, it is closed. It's bounded as well, so by the Heine-Borel theorem, it is compact.

Note, furthermore, that $S$ has no interior points. This is because if $x \in S$, then the sequence $x_k = x+10^{-4k}$ will have a $1$ in the $4k$th decimal place , hence $x_k \notin S$, but clearly $x_k \to x$. Thus, it follows that no non-trivial open interval is properly contained in $S$.

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Upper bound

For the upper bound, let $s>\frac 34$. Let's start with $\delta = 10^{-4}$. How can we cover $S$ by sets of diameter at most $10^{-4}$?

Simple : for $a,b,c \in \{0,9\}$, let $I_{a,b,c} = [0.abc,0.abc1]$. Note that $\{I_{a,b,c}\}$, running over $a,b,c \in \{0,9\}$, is a collection that covers $S$ because it covers all the starting four decimal places and leaves everything else free.

Furthermore, what is the diameter of $I_{a,b,c}$? It is just $0.abc1-0.abc = 10^{-4}$.

Let's calculate $m_s(\{I_{a,b,c}\})$. $$ m_s(\{I_{a,b,c}\}) = \sum_{a,b,c} |I_{a,b,c}|^{s} = 10^{3} \times 10^{-4s} = 10^{3-4s} $$

As there are $10$ choices for each of $a,b,c$.That is, we have shown that $\mathscr H_s^{10^{-4}}(S) \leq 10^{3-4s}$.

Now, you can do something very similar for $\delta = 10^{-4k}$ : take $$I_{a_1,a_2,a_3,\ldots,a_{3k}} = [0.a_1a_2a_30a_4a_5a_60 \ldots a_{3k-2}a_{3k-1}a_{3k},0.a_1a_2a_30a_4a_5a_60 \ldots a_{3k-2}a_{3k-1}a_{3k}1] \cap S$$

where $a_i \in \{0,9\}$. Then, $\{I_{a_1,\ldots,a_{3k}}\}$ is a $10^{-4k}$-cover of $S$. Once you do this, you can check that $$ m_s(\{I_{a_1,\ldots,a_{3k}}\}) = 10^{k(3-4s)} \left(\frac{111}{1111}\right)^s $$

That is, we have shown that $\mathscr H_s^{10^{-4k}}(S) \leq 10^{k(3-4s)} \left(\frac{111}{1111}\right)^s$ for every $k$. Therefore, it follows that $$ \mathscr H_s(S) \leq \lim_{k \to \infty} 10^{k(3-4s)} \left(\frac{111}{1111}\right)^s = 0 $$ because $s > \frac 34$.

We have proved that $\dim S \leq \frac 34$. However, these intervals $I_{a_1,\ldots,a_{3k}}$ are going to appear later!

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More on the $I_{a_1,\ldots,a_{3k}}$

Therefore, we will list one very important property of these sets, which I leave the reader to verify.

For any $k,l$, the sets $I_{a_1,\ldots,a_{3k}}$ and $I_{b_1,\ldots,b_{3l}}$ (for any choice of $a_1,\ldots,b_l$) are either disjoint, or one is contained in the other.

Any interval of the form $I_{a_1,\ldots,a_{3k}}$ will now be called a $k$-level net. This will be useful in the lower bound construction. Note that for $k=0$, we will let $I = [0,1]$ be a $0$-level net.

Finally, consider $S_0 = I = [0,1]$, and $S_k = \cup_{a_1,\ldots,a_{3k} \in \{0,9\}} I_{a_1,\ldots,a_{3k}}$. Then, $S_k$ can be explicitly described by $$ S_k = \{0.b_1b_2b_3\ldots , b_{4i} = 0 \text{ for all } i \leq k\} $$

Furthermore, $S = \cap_{k=1}^{\infty} S_k$.

(Note : there is more to the net intervals than just this, but that will be explored in a section ahead).

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A quick look at what lies ahead

We will show the following result : $\mathscr H_{\frac 34}(S) > 0$. Recall what we need to show for this : we need to show that for every $\delta>0$ and every $\delta$ - cover $\{U_i\}$ of $S$, we have that $m_{\frac 34}(\{U_i\})>C$ for some positive constant $C$.

The following will be proved with $s= \frac 34$ in the upcoming sections.

• It is enough to show that $\{U_i\}$ is a finite collection of closed intervals.

• It is enough to show that $\{U_i\}$ is a finite collection of nets.

• We can show explicitly, a bound for finite collections of nets.

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Finite collection of closed intervals

Before we handle the lower bound, we should start thinking about how we can handle all open covers. Here come a raft of fairly standard simplifications in this direction.

Interval simplification

Suppose we have a $\delta$-cover $\{U_i\}$. Let $U \in \{U_i\}$. Then $|U| = \sup U - \inf U$ because $U \subset \mathbb R$.

However, in $\{U_i\}$, if I replace $U$ by the interval $[\inf U, \sup U]$ which has the same diameter, then this interval could cover more points, but will definitely retain the measure of $\{U_i\}$. Therefore, the $\delta$-cover given by $\{U_i\}$, but with $U$ replaced by $[\inf U,\sup U]$, is a better cover than $\{U_i\}$ itself.

It follows that, by replacement if necessary, we may assume that $\{U_i\}$ is completely made of closed intervals.

Trimming the ends

We can do better by trimming the ends of closed intervals. Given $U \in \{U_i\}$, let $a = \inf(U \cap S)$ and $b = \sup(U \cap S)$. For $x \in U , x<a$, there's no point of retaining $x$ because $x\notin S$. Likewise for elements bigger than $b$. Therefore, we can replace $U$ by $[\inf(U \cap S),\sup(U\cap S)]$ and obtain a better covering. Furthermore, because $S$ is closed, it follows that $\inf(U \cap S)$ belongs to $S$, likewise $\sup(U \cap S)$ belongs to $S$.

Without loss of generality, we assume that $\inf U, \sup U \in S$ for every $U \in \{U_i\}$.

FINITE collection

Here's the most interesting one : we can actually assume that $\{U_i\}$ is a finite collection of closed intervals. The key point? Compactness.

First, let $W = I_{0,0,0} = [0,10^{-4}]$ (Note : this is a net interval). Now, if $\{U_i\}$ is such a collection, then consider the set $\{U_i^\circ\}$ of interiors of $U_i$. By compactness of the set $S \setminus [0,10^{-4})$, we get a finite subcover $(U'_1)^\circ,\ldots,(U'_n)^\circ$ of this set.

However, consider the sets $U'_1,\ldots,U'_n$ now : that's a finite subset of the $\{U_i\}$ which covers $S\setminus [0,10^{-4})$ as well. Add the $W$ to the $U_i$, and you get a finite collection of closed intervals which cover $S$. Note that $W$ is a net interval : this is important.

At the end of this section , we have shown :

It is enough to consider $\delta$-covers which are finite collections of closed intervals , each of whose endpoints lie in $S$.

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Finite collection of nets

Let $V_i,i=1,\ldots,m$ be a finite collection of closed intervals which is a $\delta$-cover of $S$ and satisfies $\inf V,\sup V \in S$ for $V \in \{V_i\}$.

Let $V \in \{V_i\}$. Suppose that $V$ is not a singleton (we won't bother with those sets, they have diameter zero in any case). As $V$ is a non-trivial interval, $V \subset S$ is not true i.e. $V \subset \cap_{k=1}^{\infty} S_k$ is not true. However, $V \subset [0,1]$ as $0 \leq \inf S \leq \inf V$ and $\sup V \leq \sup S \leq 1$.

Thus, there is a smallest value $K$ such that $V \subset S_{K}$ but $V \not \subset S_{K+1}$. However, $S_K$ is made up of the disjoint intervals $I_{a_1,\ldots,a_{3K}}$, therefore it follows that $V \subset I_{a_1,\ldots,a_{3K}}$ for some choice of $a_1,\ldots,a_{3K} \in \{0,9\}$.

Now, let's think about the interval $I_{a_1,\ldots,a_{3K}}$ and how it is structured. As an interval, from the left it will start with the left endpoint of $I_{a_1,\ldots,a_{3K}, 0,0,0}$, an interval of size $10^{-4K-4}$. Then, a gap interval occurs, which is not contained in $S$ of size $9 \times 10^{-4K-4}$. Then comes $I_{a_1,\ldots,a_{3K},0,0,1}$ , yet again an interval of size $10^{-4K-4}$, and then a gap of size $9 \times 10^{-4K-4}$, and then... yeah, you kind of get the flow, eh?

You'll end with $I_{a_1,\ldots,a_{3K},9,9,9}$ followed by yet another gap of size $9 \times 10^{-4K-4}$, whose right endpoint is the right endpoint of $I_{a_1,\ldots,a_{3K}}$. That's the complete structure of this interval.

(If this is confusing, try it for $K=0$ and $K=1$ and convince yourself that it holds generally).

Now, $V \subset I_{a_1,\ldots,a_{3K}}$. Therefore, it intersects some of the sub-net intervals $I_{a_1,\ldots,a_{3K},d,e,f}$ that we've just listed, in a certain order (If $V$ intersects none of them, then $V$ doesn't intersect $S$, so just throw it out of the collection!) Let those subnets (which are $K+1$-level subnets, to be specific) be given by $N_1,N_2,\ldots,N_L$ in the left-to-right order that $V$ intersects them.[Note that $L>1$ : if $N=1$, then $V$ will be contained in a subnet of level $K+1$, contradicting the definition of $K$.] As $V$ is an interval, it contains $N_2,\ldots,N_{L-1}$ completely, but may not contain $N_1$ and $N_L$ fully.

We claim that $$ |V|^s \geq |V \cap N_1|^s + |N_2|^s + \ldots + |N_{L-1}|^s + |V \cap N_L|^s \tag{*} $$

If this is true, then in $\{V_i\}$, we can replace $V$ by $V \cap N_1,N_2,\ldots,V \cap {N_L}$ and get a better cover!

Proof : We will proceed in two steps.

• First, we will assume that $N_1 \subset V$ and $N_L \subset V$, so that the endpoints of $V$ are the left and right endpoints of $N_1$ and $N_L$ respectively. In that case, $V \cap N_1 = N_1$ and $V \cap N_L = N_L$.

But now, recall the "structure" of $I_{a_1,\ldots,a_{3K}}$ that we spoke about. Keeping that in mind, V now consists of $L$ nets and $L-1$ "gaps", so $$ |V| = L \times 10^{-4K-4} + (L-1) \times (9 \times 10^{-4K-4}) = (L-1) \times 10^{-4K-3} + 10^{-4K-4} \\ \implies |V|^s = \left((L-1) \times 10^{-4K-3} + 10^{-4K-4}\right)^s $$

On the other hand, $$ |N_1|^s + \ldots + |N_L|^s = L \times |N_1|^s = L \times 10^{-s(4K-4)} $$

Therefore, the question boils down to whether $$ L \times 10^{-s(4K-4)} \leq \left((L-1) \times 10^{-4K-3} + 10^{-4K-4}\right)^s $$

for all $L>1,K \geq 1$. Using the fact that $s= \frac 34$, I leave the reader to prove this. Thus, we are done with this particular case.

• What about the general case? Let $V' = [\inf N_1, \sup N_L]$. Then we have come back to the first case, and so $$ |V'|^s \geq |N_1|^s + \ldots + |N_L|^s \tag{1} $$ Now, let $l = \inf V - \inf V'$ and let $r = \sup V' - \sup V$. If $l=r=0$, we're done because $V'=V$. If not, then $$ |V'|^s-|V|^s = |V'|^s - (|V'|-l-r)^s \\ |N_1|^s - |V \cap N_1|^s = |N_1|^s - (|N_1|-l)^s \\ |N_L|^s - |V \cap N_L|^s = |N_L|^s - (|N_L|-r)^s $$

We claim that $$ |V'|^s - (|V'|-l-r)^s \leq [|N_1|^s - (|N_1|-l)^s]+[|N_L|^s - (|N_L|-r)^s] $$ If this is true, then $$ |V'|^s-|V|^s \leq (|N_1|^s - |V \cap N_1|^s) + (|N_L|^s - |V \cap N_L|^s) $$ which, by subtraction from $1$, yields the required result. To prove this, note that $$ |V'|^s - (|V'|-l-r)^s = |V'|^s - (|V'|-l)^s + (|V'|-l)^s - (|V'|-l-r)^s $$ use the following lemma which is easy to prove by calculus : if $l >0,s<1$ and $t>t'>l$ then $t^s - (t-l)^s \leq (t')^s - (t'-l)^s$ (differentiate in $t'$). Using this, $$ |V'|^s - (|V'|-l)^s \leq |N_1|^s - (|N_1|-l)^s \\ (|V'|-l)^s - (|V'|-l-r)^s \leq |N_L|^s - (|N_L|-r)^s $$ add these to finish the proof of the mini-claim, and then use this to finish off the proof of the big claim.

We have proved the big claim! That is, we have now shown that we can replace an interval $V$ by $V\cap N_1,N_2,\ldots,N_{L-1}, V \cap N_L$ and obtain a better cover.

However, $V \cap N_1=V_1$ and $V \cap N_L=V_2$ are also closed intervals, so you can just repeat whatever we did in this section for $V_1,V_2$, and so on , so forth. At the end of all that, we have replaced $V$ by an infinite collection of net intervals $\{N^V_{i}\}$.

Now repeat this for each $V \in\{V_{i}\}$, and you get a countable collection of net intervals which form a $\delta$-cover of $S$.

Using the finite collection argument we did in the previous section, (and remember, $W$ was a net!) we get :

It is sufficient to assume that $U$ is a finite collection of nets! (and singletons)

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Working with finite collections of nets

Now, we have a finite $\delta$-cover $\{V_i\}$ such that each $V \in V_i$ is a net. Note that nets either contain each other or are disjoint, and if a net contains another then you can throw the smaller one out. Hence, we assume that all the nets $V_i$ are disjoint.

Let $K$ be the biggest number such that some $V \in \{V_i\}$ is a $K$-level net, and let $V$ be one of those $K$-level nets. Say that $V = I_{a_1,\ldots,a_{3K}}$. Consider the other nets that are generated along with $V$ : that is, consider nets of the form $V^{a,b,c} = I_{a_1,\ldots,a_{3K-3},a,b,c}$. We claim that each of $V^{a,b,c} \in \{V_i\}$ as well.

If not, then $a,b,c$ be any choice such that $V^{a,b,c} \neq V$ and let $V_j \in \{V_i\}$ be an element of the cover containing $V_{a,b,c}$. $V_j$ cannot be a net of level larger than $K$ because of the definition of $K$. However, $V_j$ can't be a net of level smaller than $K$ : for then, it would contain $V$ as well, which it's not supposed to intersect! Consequently, every $V^{a,b,c}$ is one of the $V_j$ for every $a,b,c$.

However, think about replacing the collection of all these $V^{a,b,c}$, by the single net containing all of them, which is $V_{a_1,\ldots,a_{3K-3}}$. What are the weights of these intervals? Recalling that $s=\frac 34$, we have
$$ \sum_{a,b,c} |V^{a,b,c}|^s = 10^3 \times 10^{-4sK} = 10^{-3K+3} $$ but we also have $$ |V|^s = 10^{-4s(K-1)} = 10^{-3K+3} $$

Therefore, one can replace the collection of all the $V^{a,b,c}$, by the single interval $V$, whose level is $K-1$!

We can now repeat what we've done with the new cover (which may not be a $\delta$-cover, but it doesn't matter because this argument is independent of $\delta$). As there are only finitely many net intervals whose level is $K$, we can, in finitely many steps, reach a situation where the minimum level of any net in the collection is $K-1$.

Then, repeating this with $K-1$ to $K-2$ and so on, we reach the situation where the cover consists of just $[0,1]$ (the singletons that I spoke about earlier will be absorbed into $[0,1]$ by containment).

And $|[0,1]|^s = 1$. Therefore, we have shown that every $\delta$-cover of nets has $m_{\frac 34}$ measure at least equal to $1$.

It follows that $H_{\frac 34}(S) \geq 1$. However, $H_{\frac 34}(S) = 1$, obviously , from the cover $\{[0,1]\}$.

Thus, $\dim(S) = \frac 34$, as desired.

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A generalization

This construction can be generalized to a large class of Cantor-like sets. We refer to the paper

Pedersen, Steen; Phillips, Jason D., Exact Hausdorff measure of certain non-self-similar Cantor sets, Fractals 21, No. 3-4, Article ID 1350016, 13 p. (2013). ZBL1290.28015.

where a large class of sets is considered, whose Hausdorff measure is explicitly calculable. Also refer to the papers in French,

Marion, Jacques, Mesure de Hausdorff d’un fractal à similitude interne. (Hausdorff measure of a fractal of internal similitude), Ann. Sci. Math. Qué. 10, 51-84 (1986). ZBL0613.28007.

Marion, Jacques, Mesures de Hausdorff d’ensembles fractals. (Hausdorff measures of fractal sets), Ann. Sci. Math. Qué. 11, No. 1, 111-132 (1987). ZBL0624.28003.

for a generalized result.

Observe that the construction we had was, in fact, very Cantor-like. Indeed, imagine that we wrote numbers not in base $10$, but in base $10000$, like $0.[a_1][a_2][a_3]\ldots$ where $a_i \in \{0,\ldots,9999\}$. Then, we'd have $$ S = \{0.[a_1][a_2][a_3]\ldots : a_i \equiv 0 \pmod {10} \forall i \geq 1\} $$

This is very similar to the Cantor set, where $$ C = \{0.a_1a_2a_3\ldots \text{ in base $3$ } : a_i=0 \text{ or } a_i = 2\} $$

In the Cantor set, every place value had three options, and two were allowed. That led to $\frac{\ln(2)}{\ln(3)}$.

In the set $S$, every place value had $10000$ options, and $1000$ (those that are multiples of $10$) were allowed. This led to $$ \frac{\ln(1000)}{\ln(10000)} = \frac{3\ln(10)}{4 \ln(10)} = \frac 34 $$ which explains the apparently weird-looking answer to a stranger!

Also, note that there was no need to only retain multiples of $10$. Retaining any $1000$ intervals would still have led to the same result.