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.
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Acknowledgments
We would like to thank the anonymous referees for their very careful reading of the manuscript.
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Communicated by A. Constantin.
Harangi, Keleti and Máthé were supported by Hungarian Scientific Foundation Grant No. 72,655. Keleti was also supported by Bolyai János Scholarship. Máthé was also supported by EPSRC Grant EP/G050678/1. Mattila would like to acknowledge the support of Academy of Finland.
Appendix
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
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
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
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
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
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
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
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
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
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,
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
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
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
On the other hand, we have
Therefore there exists a good cube \(C_i\) such that
Thus (4) holds for the cube \(C = 2 C_i\), which completes the proof. \(\square \)
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Harangi, V., Keleti, T., Kiss, G. et al. How large dimension guarantees a given angle?. Monatsh Math 171, 169–187 (2013). https://doi.org/10.1007/s00605-012-0455-0
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DOI: https://doi.org/10.1007/s00605-012-0455-0