On Sets Containing an Affine Copy of Bounded Decreasing Sequences

How small can a set be while containing many configurations? Following up on earlier work of Erdős and Kakutani (Colloq Math 4:195–196, 1957), Máthé (Fund Math 213(3):213–219, 2011) and Molter and Yavicoli (Math Proc Camb Soc 168:57–73, 2018), we address the question in two directions. On one hand, if a subset of the real numbers contains an affine copy of all bounded decreasing sequences, then we show that such a subset must be somewhere dense. On the other hand, given a collection of convergent sequences with prescribed decay, there is a closed and nowhere dense subset of the reals that contains an affine copy of every sequence in that collection.

I am grateful to my advisor Malabika Pramanik for her guidance throughout the preparation of this paper.

Authors and Affiliations

1. Department of Mathematics, The University of British Columbia, 1984 Mathematics Road, Vancouver, BC, V6T1Z2, Canada

   Tongou Yang

Corresponding author

Correspondence to Tongou Yang.

Communicated by Alex Iosevich.

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Appendix

In the appendix, we give a proof of a particular case of Molter and Yavicoli’s result [19]. It is actually almost parallel to their proof, but with the notations greatly simplified since we are only considering a special case.

Definition 7.1

A dimension function \(h:[0,\infty )\rightarrow [0,\infty ]\) is a right-continuous increasing function such that \(h(0)=0\), \(h(t)>0\) for \(t>0\).

Definition 7.2

Let h be a dimension function. For a set \(E\subseteq {\mathbb {R}}\) and \(0<\delta \le \infty \), we define

We then define

Proposition 7.3

Let \(h(x):=-\frac{1}{\ln x}\) with \(h(0)=0\) be a dimension function. Suppose \({\mathcal {H}}^h(E)=0\) for some set \(E\subseteq {\mathbb {R}}\). Then E has Hausdorff dimension 0.

Proof

Let \(s>0\). Then there is \(C_s>0\) with \( x^s<-\frac{C_s}{\ln x}\) for all \(0\le x\le 1\), since \(\lim _{x\rightarrow 0^+}x^s\ln x=0\) by L’Hôpital’s rule. Then for any \(0<\delta <1\),

\(\square \)

Theorem 7.4

Theorem 3.2 and Theorem 4.4 of [19], simplified Let h be any dimension function. Then there is an \(F_\sigma \)-set \(E\subseteq {\mathbb {R}}\) such that \({\mathcal {H}}^h(E)=0\) and for any sequence \(\{\alpha _m\}_{m=1}^\infty \subseteq {\mathbb {R}}\), we have

In particular, \(\mathrm {dim}_H(E)=0\) by the previous proposition.

Proof

Let \(M_n\in 2{\mathbb {N}}\) be an increasing sequence, \(M_1\ge 4\), such that for all \(n\ge 2\),

For each real number x, we consider its digit expansion with respect to the sequence \(\{M_n\}\):

where [x] denotes the integral part of x.

Let \(F_n\), \(n\in {\mathbb {N}}\) denote the collection of all real numbers such that its n-th digit, \(x^{(n)}\), is 0 or \(M_n/2\). If there are two possible expansions of x with one of them having \(x^{(n)}= 0\) or \(M_n/2\), include that number x in \(F_n\) as well (this ensures that \(F_n\) is made up of disjoint closed intervals). Let \(I_j:=\{(2k-1)2^{j-1}:k\in {\mathbb {N}}\}\) for \(j\in {\mathbb {N}}\). Then \(\{I_j\}_{j=1}^\infty \) forms a partition for \({\mathbb {N}}\). Define \(K_j:=\cap _{n\in I_j}F_n\). For example, \(K_2\) is the set of all real numbers so that their \(2,6,10,14,\dots \)-th digits are 0 or \(M_2/2\). Note \(K_j\) is also closed.

Lastly, define \(E:=\cup _j K_j\). We claim that E is the required \(F_\sigma \)-set.

• To show \({\mathcal {H}}^h(E)=0\), it suffices to show \({\mathcal {H}}^h(K_j)=0\) for all j. Let \(\delta >0\) be small, and cover \(E_j\) by

  $$\begin{aligned} \frac{\prod _{i=1}^{(2k-1)2^{j-1}}M_i}{\prod _{l=1}^k \frac{1}{2}M_{(2l-1)2^{j-1}}} \end{aligned}$$

  intervals of lengths

  $$\begin{aligned} \frac{1}{\prod _{i=1}^{(2k-1)2^{j-1}}M_i}<\delta . \end{aligned}$$

  for all large k’s. Then we have

  $$\begin{aligned} {\mathcal {H}}^h_\delta (K_j)&\le \frac{\prod _{i=1}^{(2k-1)2^{j-1}}M_i}{\prod _{l=1}^{k} \frac{1}{2} M_{(2l-1)2^{j-1}}}\,\,\cdot h \left( \frac{1}{\prod _{i=1}^{(2k-1)2^{j-1}}M_i}\right) \\&\le \frac{\prod _{i=1}^{(2k-1)2^{j-1}}M_i}{\prod _{l=1}^{k} \frac{1}{2} M_{(2l-1)2^{j-1}}}\,\,\cdot \frac{1}{\prod _{i=1}^{(2k-1)2^{j-1}-1}M_i}\\&=\frac{2}{\prod _{l=1}^{k-1} \frac{1}{2} M_{(2l-1)2^{j-1}}}\\&\le \frac{1}{2^{k-2}}, \text {for all large }k. \end{aligned}$$

  Letting \(k\rightarrow \infty \), we have \({\mathcal {H}}^h_\delta (K_j)=0\). Letting \(\delta \rightarrow 0^+\), we have then \({\mathcal {H}}^h(K_j)=0\). Thus \({\mathcal {H}}^h(E)=0\).

• Now let \(\{\alpha _m\}\) be given. We show

  $$\begin{aligned} \bigcap _{m=1}^\infty E+\alpha _m\ne \varnothing . \end{aligned}$$

  We have \(E\supseteq K_m\) for all \(m\ge 1\), so it suffices to show

  $$\begin{aligned} \bigcap _{m=1}^\infty E+\alpha _m\supseteq \bigcap _{m=1}^\infty K_m+\alpha _m. \end{aligned}$$

  (7.1)

  But \(K_1+\alpha _1=(F_1+\alpha _1)\cap (F_3+\alpha _1)\cap (F_5+\alpha _1)\cap \cdots \), \(K_2+\alpha _2=(F_2+\alpha _2)\cap (F_6+\alpha _2)\cap (F_{10}+\alpha _2)\cap \cdots \), etc. We can rewrite the infinite intersection on the right hand side of (7.1) into:

  $$\begin{aligned} \bigcap _{u=1}^\infty F_u+\alpha _{j_u}, \end{aligned}$$

  (7.2)

  where \(j_u\) is the greatest integer v such that \(2^{v-1}\) divides u. For example, the first few terms of the intersection are:

  $$\begin{aligned} (F_1+\alpha _1)\cap (F_2+\alpha _2)\cap (F_3+\alpha _1)\cap (F_4+\alpha _3) \cap (F_5+\alpha _1)\cap (F_6+\alpha _2)\cap \cdots \end{aligned}$$

  We would like to show this intersection is nonempty.

  Denote \(C_1:=[0, 1 /M_1]\). Since the distance between the centres of the two adjacent intervals in \(F_2\) is \(1 /(2M_1)\) and the intervals of \(F_2\) are shorter in length than those of \(F_1\), no matter how we translate \(C_1\), there is an interval \(C_2\) of \(F_2\) that is contained in that translate of \(C_1\).

  Hence for any given \(\alpha _1,\alpha _2\in {\mathbb {R}}\), we can find such \(C_2\) satisfying \(C_2+\alpha _2\subseteq C_1+\alpha _1\). Similarly, one can find \(C_3\) of \(F_3\) such that \(C_3+\alpha _1\subseteq C_2+\alpha _2\). Continuing in this way, we get a nested sequence of compact intervals with rapidly decreasing length:

  $$\begin{aligned} C_u+\alpha _{j_u}\subseteq C_{u-1}+\alpha _{j_{u-1}},\quad u\ge 2. \end{aligned}$$

  By the nested interval theorem, the intersection in (7.2) is nonempty, and hence so is the intersection in (7.1).

\(\square \)

We remark that E defined in this way is not closed. This was seen by taking \(\{\alpha _m\}\) to be \({\mathbb {Q}}\cap [0,1]\) as in the introduction of the paper, but it can also be seen directly from this simplified construction. Indeed, \(E^c\) is the set of all real numbers x such that for any \(j\in {\mathbb {N}}\), there is \(k_j\in {\mathbb {N}}\) so that the \((2k-1)2^{j-1}\)-th digit of x is not 0 or \(M_n/2\). Particularly, if \(x\in E^c\), then there is an increasing sequence \(a_n\in {\mathbb {N}}\) such that the \(a_n\)-th digit of x is not 0 or \(M_n/2\).

If \(E^c\) were open, this means for any \(x\in E^c\), if y is sufficiently close to x, then \(y\in E^c\). However, we see that for any \(\delta >0\), we can choose \(|x-y|<\delta \) such that y is a finite decimal number, so \(y\notin E^c\).

Lastly, we have that E is dense in \({\mathbb {R}}\). Given \(\epsilon >0\) and \(x\in {\mathbb {R}}\), consider the digit expansion of x. There is some \(j_0\) and some real number y with the same digits as x on all digits \(1\le j\le j_0-1\) but having all digits 0 for \(j\ge j_0\), such that \(|x-y|<\delta \). Then \(y\in K_{j_0}\subseteq E\).

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