How large dimension guarantees a given angle?

Abstract

We study the following two problems: (1) Given $n\ge 2$ and $0\le \alpha \le 180^\circ$, how large Hausdorff dimension can a compact set $A\subset \mathbb{R }^n$ have if $A$ does not contain three points that form an angle $\alpha$? (2) Given $\alpha$ and $\delta$, how large Hausdorff dimension can a subset $A$ of a Euclidean space have if $A$ does not contain three points that form an angle in the $\delta$-neighborhood of $\alpha$? An interesting phenomenon is that different angles show different behaviour in the above problems. Apart from the clearly special extreme angles $0$ and $180^\circ$, the angles $60^\circ , 90^\circ$ and $120^\circ$ also play special role in problem (2): the maximal dimension is smaller for these angles than for the other angles. In problem (1) the angle $90^\circ$ seems to behave differently from other angles.

Appendix

1.1 A transfinite construction

We prove the following theorem, which shows that if we allowed arbitrary sets in Definition 1.3 then $C(n,\alpha )$ would be $n$.

Theorem 5.1

Let $n\ge 2$. For any $\alpha \in [0,180^\circ ]$ there exists $H\subset \mathbb R ^n$ such that $H$ does not contain the angle $\alpha$, and $H$ has full Lebesgue outer measure; that is, its complement does not contain any measurable set with positive measure. In particular, $\text{ dim}(H)=n$.

The proof we present here is shorter than our original proof, this one was suggested by Marianna Csörnyei.

We need the following simple lemma, which might be well-known even for more general sets but for completeness we present a proof. Recall that an algebraic set is the set of solutions of a system of polynomial equations.

Lemma 5.2

Fewer than continuum many proper algebraic subsets of $\mathbb R ^n$ cannot cover a Borel set of positive $n$-dimensional Lebesgue measure.

Proof

We prove by induction. For $n=1$ this is clear since proper algebraic subsets of $\mathbb R$ are finite and every Borel set of positive Lebesgue measure has cardinality continuum.

Suppose that the lemma holds for $n-1$ but it is false for $n$, so there exists a collection $\mathcal{A }$ of less than continuum many proper algebraic subsets of $\mathbb R ^n$ such that they cover a Borel set $B\subset \mathbb R ^n$ with positive Lebesgue measure.

Let $H^t$ denote the “horizontal” section $H^t=\{(x_1,\ldots ,x_{n-1}):(x_1,\ldots ,x_{n-1},t) \in H\}$ of a set $H\subset \mathbb R ^n$ at “height” $t\in \mathbb R$. If $A$ is a proper algebraic subset of $\mathbb R ^n$ then with finitely many exceptions every $A^t$ is a proper algebraic subset of $\mathbb R ^{n-1}$. Therefore, by using the assumption that the lemma holds for $n-1$, we get that $(\cup \mathcal{A })^t$ can contain Borel sets of positive $n-1$-dimensional Lebesgue measure only for less than continuum many $t$. Let $f(t)$ denote the $(n-1)$-dimensional Lebesgue measure of the Borel set $B^t$. Since $B\subset \cup \mathcal{A }$, we obtain that $\{t: f(t)>0\}$ has cardinality less than continuum.

On the other hand, by Fubini theorem $f$ is measurable and its integral is the Lebesgue measure of $B$, so it is positive. This implies that $\{t: f(t)>0\}$ is a measurable set of positive measure, hence it contains an uncountable compact set, so it must have the cardinality of the continuum, contradiction. $\square$

Proof of Theorem 5.1

Take a well-ordering $\{B_\beta : \beta < \mathfrak c \}$ of the Borel subsets of $\mathbb R ^n$ with positive $n$-dimensional Lebesgue measure. We will construct a sequence of points $\{x_\beta : \beta < \mathfrak c \}$ of $\mathbb{R }^n$ using transfinite induction so that

$$\begin{aligned} x_{\beta }\in B_\beta \quad \text{ and}\quad H_\beta = \{x_\delta : \delta \le \beta \}\,\text{ does} \text{ not} \text{ contain} \text{ the} \text{ angle}\, \alpha \end{aligned}$$

(3)

for any $\beta <\mathfrak c$. This will complete the proof since then $H=\{x_\beta :\beta < \mathfrak c \}$ will have all the required properties.

Suppose that $\gamma < \mathfrak c$ and we have already properly defined $x_\beta$ for all $\beta <\gamma$ so that (3) holds for all $\beta <\gamma$. For any $p,q\in \mathbb R ^n$, $p\ne q$, let $A_{p,q}$ denote the set of those $x\in \mathbb R ^n$ for which one of the angles of the triangle $pqx$ is $\alpha$. Note that $A_{p,q}$ can be covered by three proper algebraic subsets of $\mathbb R ^n$. Then, by Lemma 5.2, the sets $A_{x_\delta ,x_{\delta ^{\prime }}}$ $(\delta ,\delta ^{\prime }<\gamma , x_\delta \ne x_{\delta ^{\prime }})$ cannot cover $B_\gamma$, so we can choose a point

$$\begin{aligned} x_\gamma \in B_\gamma {\setminus }\cup \{A_{x_\delta ,x_{\delta ^{\prime }}}\ :\ \delta ,\delta ^{\prime }<\gamma ,\ x_\delta \ne x_{\delta ^{\prime }}\}. \end{aligned}$$

Then (3) also holds for $\beta =\gamma$.

This way we obtain a sequence $(x_\beta )_{\beta <\mathfrak c }$, so that (3) holds for all $\beta <\mathfrak c$, which completes the proof. $\square$

1.2 The size of the neighborhood in the approximative problems

Now, our goal is to prove the following theorem, which was claimed in Remark 4.11.

Theorem 5.3

Suppose that $s=s(\alpha , \delta , n)$ is a positive real number such that every analytic set $A\subset \mathbb R ^n$ with $\mathcal H ^s(A)>0$ contains an angle from the interval $(\alpha -\delta , \alpha +\delta )$. Then there exists a closed subinterval $[\alpha -\gamma , \alpha +\gamma ]$ ($\gamma <\delta$) such that every analytic set $A\subset \mathbb R ^n$ with $\mathcal H ^s(A)>0$ contains an angle from the interval $[\alpha -\gamma , \alpha +\gamma ]$.

To prove this theorem, we need two lemmas. For $r \in (0,\infty ]$ let

$$\begin{aligned} \mathcal H ^s_r(A)=\inf \left\{ \sum _{i=1}^\infty \text{ diam}(U_i)^s \,:\, \text{ diam}(U_i)\le r, \ A\subset \cup _{i=1}^\infty U_i\right\} , \end{aligned}$$

thus $\mathcal H ^s(A)=\lim _{r\rightarrow 0+} \mathcal H ^s_r(A)$.

Lemma 5.4

Let $A_i$ be a sequence of compact sets converging in the Hausdorff metric to a set $A$. Then the following two statements hold.

• (i) $\mathcal{H }_\infty ^s(A) \ge \limsup _{i\rightarrow \infty } \mathcal{H }_\infty ^s(A_i).$

• (ii) Suppose that for every $i=1,2,\ldots$ the set $A_i$ does not contain any angle from $[\alpha -\delta +1/i, \ \alpha +\delta -1/i]$. Then $A$ does not contain any angle from $(\alpha -\delta , \,\alpha +\delta )$.

Proof

The first statement is well-known and easy. To prove the second, notice that for any three points $x,y,z$ of $A$ there exist three points in $A_i$ arbitrarily close to $x,y,z$, for sufficiently large $i$. $\square$

The next lemma follows easily from [4]—(Theorem 2.10.17 (3)). For the sake of completeness, we give a short direct proof.

Lemma 5.5

Let $A\subset \mathbb R ^n$ be a compact set satisfying $\mathcal H ^s(A)>0$. Then there exists a ball $B$ such that $\mathcal{H }_\infty ^s(A\cap B) \ge c\,\text{ diam}(B)^s$, where $c>0$ depends only on $s$.

Proof

We may suppose without loss of generality that $\mathcal H ^s(A)<\infty$. (Otherwise we choose a compact subset of $A$ with positive and finite $\mathcal H ^s$ measure. If the theorem holds for a subset of $A$ then it clearly holds for $A$ as well.)

Choose $r>0$ so that $\mathcal H ^s_r(A)> \mathcal H ^s(A)/2$. Cover $A$ by sets $U_i$ of diameter at most $r/2$ such that $\sum _i \text{ diam}(U_i)^s \le 2 \mathcal H ^s(A)$. Cover each $U_i$ by a ball $B_i$ of radius at most the diameter of $U_i$. Then the balls $B_i$ cover $A$, have diameter at most $r$, and $\sum _i \text{ diam}(B_i)^s \le 2^{1+s} \mathcal H ^s(A)$.

We claim that one of these balls $B_i$ satisfies the conditions of the Lemma for $c=2^{-2-s}$. Otherwise we have

$$\begin{aligned} \mathcal{H }_\infty ^s(A\cap B_i) < 2^{-2-s} \, \text{ diam}(B_i)^s \end{aligned}$$

for every $i$. Since the sets $A\cap B_i$ have diameter at most $r$, clearly $\mathcal H ^s_r(A\cap B_i) = \mathcal{H }_\infty ^s(A\cap B_i)$. Therefore

$$\begin{aligned} \mathcal H ^s_r(A)&\le \sum _i \mathcal H ^s_r(A\cap B_i) <\sum _i 2^{-2-s} \, \text{ diam}(B_i)^s \nonumber \\&\le 2^{-2-s} 2^{1+s} \mathcal H ^s(A) = \mathcal H ^s(A)/2, \end{aligned}$$

which contradicts the choice of $r$. $\square$

Proof of Theorem 5.3

Suppose on the contrary that there exist compact sets $K_i\subset \mathbb R ^n$ with $\mathcal H ^s(K_i)>0$ such that $K_i$ does not contain any angle from $[\alpha -\delta +1/i, \ \alpha +\delta -1/i]$. Choose a ball $B_i$ for each compact set $K_i$ according to Lemma 5.5. Let $B$ be a ball of diameter $1$. Let $K_i^{\prime }$ be the image of $K_i\cap B_i$ under a similarity transformation which maps $B_i$ to the ball $B$. Thus $\mathcal{H }_\infty ^s(K_i^{\prime })\ge c$. Let $K$ denote the limit of a convergent subsequence of the sets $K_i$. We can apply Lemma 5.4 to this subsequence and obtain $\mathcal{H }_\infty ^s(K)\ge c$, implying $\mathcal H ^s(K)>0$. Also, $K$ does not contain any angle from the interval $(\alpha -\delta , \,\alpha +\delta )$, which is a contradiction. $\square$

1.3 Replacing Hausdorff dimension by upper Minkowski dimension

Our final goal is showing that in the problems when we want angles only in a neighborhood of a given angle, Hausdorff dimension can be replaced by Minkowski dimension. This will follow from the following theorem. As pointed out by Pablo Shmerkin, this theorem also follows from a result of Furstenberg [5]. His result is much more general and it is not immediately trivial to see that it implies what we need. Therefore we give a direct self-contained proof.

Theorem 5.6

Let $A\subset \mathbb R ^d$ be a bounded set with upper Minkowski dimension $s>0$. Then there exists a compact set $K$ of Hausdorff dimension $s$ such that all finite subsets of $K$ are limits of homothetic copies of finite subsets of $A$. (That is, for every finite set $S\subset K$ and $\varepsilon >0$ there exists a set $S^{\prime }\subset A$ and $r>0$, $t\in \mathbb R ^d$ such that the Hausdorff distance of $t+rS^{\prime }$ and $S$ is at most $\varepsilon$.)

Applying Theorem 5.6 to a bounded set $A$ that does not contain any angle from an open interval we get a compact set $K$ with the same property and with $\text{ dim}(K)={\overline{\mathrm{dim }}_\mathrm{M }}(A)$. Thus we get the following.

Corollary 5.7

For any $\alpha \in [0,180^\circ ]$ and $\delta >0$, we have

$$\begin{aligned} \widetilde{C}(\alpha , \delta )&= \sup \{{\overline{\mathrm{dim }}_\mathrm{M }}(A) : A \subset \mathbb R ^n \ \mathrm for some \ n; A \ \mathrm is bounded ;\\&\qquad A \ \text{ does} \text{ not} \text{ contain} \text{ any} \text{ angle} \text{ from} \ (\alpha -\delta , \alpha +\delta ) \}. \end{aligned}$$

Proof of Theorem 5.6

We will need to use a slightly different version of the Hausdorff content $\mathcal H ^s_\infty (B)$ in this proof. Instead of covering $B \subset \mathbb R ^d$ with arbitrary sets, we will only consider coverings with homothetic copies of the unit cube $[0,1]^d$. (From now on, a cube is always assumed to be a homothetic copy of the unit cube.) For a cube $C, \text{ diam}(C)$ is just a constant multiple of the edge length of $C$ (denoted here by $|C|$). For the sake of simplicity, we will use $|C|$ in our definition: for any $B \subset \mathbb R ^d$ and $s>0$ let

$$\begin{aligned} \widehat{\mathcal{H }}^s_\infty (B) \mathop {=}\limits ^\mathrm{def} \inf \left\{ \sum _{i=1}^\infty |C_i|^s : C_i\, \text{ is} \text{ a} \text{ cube} \text{ for} \text{ each}\, i; \ B \subset \bigcup _{i=1}^\infty C_i \right\} . \end{aligned}$$

It is easy to see that $d^{-s/2} \mathcal H ^s_\infty \le \widehat{\mathcal{H }}^s_\infty \le \mathcal H ^s_\infty$. Also note that $\widehat{\mathcal{H }}^s_\infty ([0,1]^d) = 1$ for any $0 < s \le d$.

We may assume that $A\subset [0,1]^d$. For a positive integer $n$ we divide the unit cube into $n^d$ subcubes of edge length $1/n$. Let $A_n$ be the union of the subcubes that intersect $A$.

We claim that for any fixed $0< \delta < s/2$, for infinitely many $n$ (depending on $\delta$) there exists a cube $C$ such that

$$\begin{aligned} |C| \ge n^{\frac{\delta }{2d}} / n \quad \text{ and} \quad \widehat{\mathcal{H }}^{s-2\delta }_\infty (C \cap A_n) \ge 2^{-s-2} |C|^{s-2\delta }. \end{aligned}$$

(4)

First we show how the theorem follows from this claim. If (4) holds for $n$ and $C$, then let $K_n$ be the image of $C \cap A_n$ under the homothety that maps $C$ to $[0,1]^d$. Hence $\widehat{\mathcal{H }}^{s-2\delta }_\infty (K_n)\ge 2^{-s-2}$. If $S\subset K_n$ is finite, then there exists $S^{\prime }$ such that the Hausdorff distance of $S$ and $S^{\prime }$ is at most $\sqrt{d} n^{-\delta /(2d)}$ and a homothetic image of $S^{\prime }$ is in $A$.

For each $\delta =1/l$ choose $n=n_l\ge l^l$ such that the claim holds. Let $\tilde{K}$ be the limit of a convergent subsequence of $K_{n_l}$. By Lemma 5.4 the Hausdorff dimension of $\tilde{K}$ is at least $s$. Let $K$ be a compact subset of $\tilde{K}$ of Hausdorff dimension $s$. It is easy to check that $K$ satisfies all the required properties.

It remains to prove the claim. Since ${\overline{\mathrm{dim }}_\mathrm{M }}(A) = s$, $A_n$ contains at least $n^{s-\delta }$ subcubes for infinitely many $n$. Fix such an $n$ with $n\ge 2^{4/\delta }$. Let

$$\begin{aligned} c=\min \left\{ \widehat{\mathcal{H }}^{s-2\delta }_\infty (B) / m: B \,\text{ is} \text{ the} \text{ union} \text{ of}\,m\,\text{ subcubes} \text{ of}\, A_n, \ m\ge 1\right\} . \end{aligned}$$

Since the unit cube covers $A_n$, by choosing $B$ as the union of $m\ge n^{s-\delta }$ subcubes of $A_n$ we get $c\le \widehat{\mathcal{H }}^{s-2\delta }_\infty (B)/m \le 1/n^{s-\delta }$. (On the other hand, one subcube has content $1/n^{s-2\delta }$, hence the minimum is taken for a set $B$ for which $m$ is at least $n^\delta$.)

Suppose now that $B$ is a set for which the minimum is taken; that is,

$$\begin{aligned} \widehat{\mathcal{H }}^{s-2\delta }_\infty (B)=cm, \end{aligned}$$

where $B$ consists of $m$ subcubes of $A_n$. It follows that there exists a covering of $B$ with cubes $C_i$ ($i=1,2,\ldots$) such that

$$\begin{aligned} \sum _{i=1}^\infty |C_i|^{s-2\delta } \le 2cm. \end{aligned}$$

Let $k=n^{\delta /(2d)}$. We say that a cube $C_i$ is “bad” if $|C_i|<k/n$, and “good” otherwise. The total volume of the bad cubes is at most

$$\begin{aligned} \sum _{C_i \text{ is} \text{ bad}} |C_i|^d&= \sum _{C_i \text{ is} \text{ bad}} |C_i|^{d-s+2\delta } |C_i|^{s-2\delta } \le (k/n)^{d-s+2\delta } \sum _{i=1}^\infty |C_i|^{s-2\delta } \nonumber \\&\le 2cm (k/n)^{d-s+2\delta } \le 2m k^{d-s+2\delta } n^{-\delta -d} \le 2m k^{d} n^{-\delta -d} \nonumber \\&= 2m n^{-\frac{\delta }{2}-d}\le \frac{m}{2} n^{-d}, \end{aligned}$$

where the last four estimates follow from $c \le 1/n^{s-\delta }$, $\delta <s/2, k=n^{\delta /(2d)}$ and $n\ge 2^{4/\delta }$. So there are at most $m/2$ subcubes that are fully covered by bad cubes. Let $B^{\prime }$ be the union of the remaining (at least $m/2$) subcubes in $B$. Since each subcube in $B^{\prime }$ must intersect a good cube $C_i$, it follows that the cubes $2C_i$ cover $B^{\prime }$, where $2C_i$ is the cube with the same center as $C_i$ and double edge length.

Then the definition of $c$ implies that

$$\begin{aligned} \sum _{C_i \ \mathrm{is \ good}} \widehat{\mathcal{H }}^{s-2\delta }_\infty (2C_i \cap A_n) \ge \widehat{\mathcal{H }}^{s-2\delta }_\infty (B^{\prime })\ge c\frac{m}{2}. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \sum _{C_i \ \mathrm{is \ good}} |2C_i|^{s-2\delta } \le 2^{s-2\delta } \sum _{i=1}^\infty |C_i|^{s-2\delta } \le 2^{s-2\delta } 2\,cm \le 2^{s+1}\,cm. \end{aligned}$$

Therefore there exists a good cube $C_i$ such that

$$\begin{aligned} \widehat{\mathcal{H }}^{s-2\delta }_\infty (2C_i \cap A_n) \ge 2^{-s-2} |2C_i|^{s-2\delta }. \end{aligned}$$

Thus (4) holds for the cube $C = 2 C_i$, which completes the proof. $\square$