Abstract
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.
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Acknowledgements
I am grateful to my advisor Malabika Pramanik for her guidance throughout the preparation of this paper.
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Appendix
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.
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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 \({\mathcal {H}}^h(E)=0\).
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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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Yang, T. On Sets Containing an Affine Copy of Bounded Decreasing Sequences. J Fourier Anal Appl 26, 73 (2020). https://doi.org/10.1007/s00041-020-09780-4
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DOI: https://doi.org/10.1007/s00041-020-09780-4