Proof. Let $\rho \gt 1/2$. Then there exists $N \in \mathbb{N}$, $N \geq 3$, such that

\begin{equation*} \rho \gt \frac{1}{2} + \frac{1}{N}. \end{equation*}

Let $(r_k)$ be a decreasing null sequence such that $r_1 \lt 1/2$ is sufficiently small. Then, for every $k \in \mathbb{N}$ and $b \in [0,r_k]$, we have

(2)\begin{equation} \frac{r_k/N+b}{r_k+b} \leq \frac{r_k/N+r_k}{r_k+r_k} = \frac{1}{2N}+\frac{1}{2} \lt \frac{1}{N}+\frac{1}{2} \lt \rho-r_k \lt \frac{x_{n+1}}{x_n} \quad \text{for all } n \geq n_{k}. \end{equation}

The inequalities in (2) imply that, for $b \in [0,r_k]$, the intervals

\begin{equation*} \frac{1}{x_n}\,(r_k/N+b,\, r_k+b) \quad \text{and} \quad \frac{1}{x_{n+1}}\,(r_k/N+b,\, r_k+b) \end{equation*}

overlap for all $n \geq n_k$. Consequently, the interval $(r_k/N+b,\infty)/x_{n_{k}}$ is entirely contained in the complement of the set

(3)\begin{equation} \bigcap_{n \geq n_{k}} \frac{1}{x_n}\,(E_k+b), \quad \text{where } E_k = \bigcup_{j \in \mathbb{Z}} [j r_k,\, j r_k + r_k/N] \quad \text{for all } k \in \mathbb{N}, \end{equation}

and therefore

(4)\begin{equation} \bigcap_{n \geq n_{k}} \frac{1}{x_n}\,(E_k+b) \subseteq (-\infty,\, r_k/N+b]/x_{n_{k}}. \end{equation}

On the other hand, for $b \in [0,r_k]$ we have

\begin{align*} \frac{r_k-b}{2r_k-r_k/N-b} &\leq \frac{r_k}{2r_k-r_k/N} \\ &= \frac{1}{2-1/N} \lt \frac{1}{2}+\frac{1}{2N} \lt \rho-r_k \lt \frac{x_{n+1}}{x_n} \quad \text{for all } n \geq n_{k}, \end{align*}

implying that the intervals

\begin{equation*} \frac{1}{x_n}\,(-2r_k+r_k/N+b,\, -r_k+b) \quad \text{and} \quad \frac{1}{x_{n+1}}\,(-2r_k+r_k/N+b,\, -r_k+b) \end{equation*}

overlap for all $n \geq n_k$. Therefore, the interval $(-\infty,\,-r_k+b)/x_{n_{k}}$ is entirely contained in the complement of the set (3), and so

(5)\begin{equation} \bigcap_{n \geq n_{k}} \frac{1}{x_n}\,(E_k+b) \subseteq [-r_k+b,\,\infty)/x_{n_{k}}. \end{equation}

Together with (4), we obtain

(6)\begin{equation} \bigcap_{n \geq n_{k}} \frac{1}{x_n}\,(E_k+b) \subseteq [-r_k,\, 2r_k]/x_{n_{k}} \quad \text{for all } k \in \mathbb{N}. \end{equation}

By periodicity of $E_k$, (6) holds for all $b \in \mathbb{R}$.

Now let $\varepsilon \gt 0$. We may choose $(r_k)$ so that $(r_k/x_{n_k})$ is itself a null sequence. Then there exists $k_{\varepsilon} \in \mathbb{N}$ such that $r_{k} \lt \varepsilon x_{n_{k}}/2$ for every $k \geq k_{\varepsilon}$. Consequently,

(7)\begin{equation} \bigcap_{n \geq n_{k}} \frac{1}{x_n}\,(E_k+b) \subseteq [-\varepsilon,\, \varepsilon] \quad \text{for all } k \geq k_{\varepsilon}. \end{equation}

This implies that, for any $|a| \gt \varepsilon$, there exists $n \geq n_{k}$ such that $a x_n-b \notin E_k$ whenever $k \geq k_{\varepsilon}$. Taking $k=k_{\varepsilon}$ and letting $E=E_{k_{\varepsilon}}$ yields the candidate set. Finally, we estimate the measure of $E$. Let $\lambda$ denote the Lebesgue measure on $\mathbb{R}$. Then

\begin{align*} \lambda(E) = \lambda(E_{k_{\varepsilon}}) \geq \lambda(E_{k_{\varepsilon}} \cap [0,1]) \geq \frac{r_{k_{\varepsilon}}}{N} \left\lfloor \frac{1}{r_{k_{\varepsilon}}} \right\rfloor \gt \frac{r_{k_{\varepsilon}}}{N} \Big(\frac{1}{r_{k_{\varepsilon}}}-1\Big) \geq \frac{1}{N}-\frac{r_{1}}{N} \gt \frac{1}{2N}, \end{align*}

since $r_1 \lt 1/2$. Thus, the set $E$ has a measure uniformly bounded away from zero.

Proof. Let $\rho \leq 1/2$. Then there exists $N \geq 3$, $N \in \mathbb{N}$, such that

\begin{equation*} \frac{1}{N} \lt \rho \leq \frac{1}{N-1}. \end{equation*}

W.l.o.g., take $(r_k)$ such that $r_k \lt N\rho - 1$ for all $k \in \mathbb{N}$. For every $k \in \mathbb{N}$ and $b \in [0,r_k]$, we have

\begin{align*} \frac{(N\rho - 1 - r_k)r_k + b}{Nr_k + b} \leq \frac{N\rho - r_k}{N+1} \lt \rho - \frac{r_k}{N+1} \lt \frac{x_{n+1}}{x_n}, \quad \text{for all } n \geq n_k. \end{align*}

Consequently, for $b \in [0,r_k]$, the intervals

\begin{equation*} \frac{1}{x_n}((N\rho-1-r_k)r_k+b,\, Nr_k+b) \quad \text{and} \quad \frac{1}{x_{n+1}}((N\rho-1-r_k)r_k+b,\, Nr_k+b) \end{equation*}

overlap for all $n \ge n_k$. Thus, the interval $((N\rho-1-r_k)r_k+b, \infty)/x_{n_k}$ is contained in the complement of the set

\begin{equation*} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^1_k+b), \end{equation*}

where

(8)\begin{equation} E^1_k = [0, (N\rho-1-r_k)r_k] \cup \bigcup_{j \ge N} [j r_k,\, j r_k + (N\rho-1-r_k)r_k], \quad \text{for all } k \in \mathbb{N}. \end{equation}

Because $\inf E^1_k = 0$ for all $k \in \mathbb{N}$, together with the inclusion

\begin{equation*} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^1_k+b) \subseteq (-\infty, (N\rho-1-r_k)r_k+b]/x_{n_k}, \end{equation*}

it follows that

(9)\begin{equation} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^1_k+b) \subseteq [0, (N\rho-1-r_k)r_k+b]/x_{n_k} \subseteq [0, (N\rho-r_k)r_k]/x_{n_k}, \quad \text{for all } k \in \mathbb{N} \end{equation}

for all $b \in [0,r_k]$.

Now let $b \ge N r_k$. Then, from (8), it follows that $E^1_k + b \subseteq E^1_k + \widehat b$, where $b \equiv \widehat b \mod r_k$. Consequently, in view of (9), we have

(10)\begin{equation} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^1_k+b) \subseteq [0, (N\rho-r_k)r_k]/x_{n_k}, \quad \text{for all } k \in \mathbb{N}, \end{equation}

for all $b \in [0,r_k] \cup [Nr_k, \infty)$.

Now let $\varepsilon \gt 0$. Again, we may choose $(r_k)$ so that $(r_k / x_{n_k})$ is itself a null sequence. Then, there exists $k_\varepsilon \in \mathbb{N}$ such that $N r_k \lt \varepsilon x_{n_k}$ for every $k \ge k_\varepsilon$. Hence, for all $b \gt \varepsilon \gt \varepsilon x_{n_k} \gt N r_k$, we haveFootnote *

\begin{equation*} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^1_k+b) \subseteq [0, \varepsilon], \quad \text{for all } k \ge k_\varepsilon. \end{equation*}

This implies that, for any $|a| \gt \varepsilon$, there exists $n \ge n_k$ such that $a x_n - b \notin E_k$ whenever $k \ge k_\varepsilon$. Denote $E_1 = E^1_{k_\varepsilon}$. Then, similar calculations show that

\begin{align*} \lambda(E_1) = \lambda(E^1_{k_\varepsilon}) &\ge \lambda(E^1_{k_\varepsilon} \cap [0,1]) \\ &\ge (N\rho-1-r_{k_\varepsilon}) r_{k_\varepsilon} \left\lfloor \frac{1-N r_{k_\varepsilon}}{r_{k_\varepsilon}} \right\rfloor \\ & \gt (N\rho-1-r_1)(1-r_1(N+1)) \gt \frac{1}{3} (N\rho-1), \end{align*}

by letting $r_1 \lt (N\rho-1)/(N+1)$.

Now we show the existence of $E_2$. Note that, for all $b \in [0,r_k]$, the inequalities

\begin{align*} \frac{-r_k+b}{-2Nr_k+(N\rho-1-r_k)r_k+b} \leq \frac{-r_k}{-2Nr_k+(N\rho-1-r_k)r_k} \leq \frac{1}{N} - \frac{r_k}{N} \lt \rho - \frac{r_k}{N} \lt \frac{x_{n+1}}{x_n} \end{align*}

hold for all $n \ge n_k$.

Therefore, for $b \in [0,r_k]$, the intervals

\begin{equation*} \frac{1}{x_n}(-2Nr_k+(N\rho-1-r_k)r_k+b,\, -r_k+b) \quad \text{and} \quad \frac{1}{x_{n+1}}(-2Nr_k+(N\rho-1-r_k)r_k+b,\, -r_k+b) \end{equation*}

overlap for all $n \ge n_k$. Consequently, the interval $(-\infty, -r_k+b)/x_{n_k}$ is contained in the complement of the set

\begin{equation*} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^2_k+b), \end{equation*}

where

(11)\begin{equation} E^2_k = [-r_k, -r_k + (N\rho-1-r_k)r_k] \cup \bigcup_{j \le -2N} [j r_k,\, j r_k + (N\rho-1-r_k)r_k], \quad \text{for all } k \in \mathbb{N}. \end{equation}

Because $\sup E^2_k \le 0$ for all $k \in \mathbb{N}$, together with the inclusion

\begin{equation*} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^2_k+b) \subseteq [-r_k+b, \infty)/x_{n_k}, \end{equation*}

it follows that

(12)\begin{equation} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^2_k+b) \subseteq [-r_k+b, 0]/x_{n_k} \subseteq [-r_k, 0]/x_{n_k}, \quad \text{for all } k \in \mathbb{N}, \end{equation}

for all $b \in [0,r_k]$.

Now, let $b \le -2Nr_k$. Then, from (11), it follows that $E^2_k+b \subseteq E^2_k+\widehat b$, where $b \equiv \widehat b \mod (-r_k)$. In view of (12), we obtain

(13)\begin{equation} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^2_k+b) \subseteq [-r_k, 0]/x_{n_k}, \quad \text{for all } k \in \mathbb{N}, \end{equation}

for all $b \in [0,r_k] \cup (-\infty, -2Nr_k]$.

Avoiding repetitions, for any $\varepsilon \gt 0$, there exists $k_\varepsilon \in \mathbb{N}$ such that, for $b \lt -\varepsilon$, we have

\begin{equation*} \bigcap_{n \ge n_k} \frac{1}{x_n}\,(E^2_k+b) \subseteq [-\varepsilon, 0], \quad \text{for all } k \ge k_\varepsilon. \end{equation*}

Thus, for any $|a| \gt \varepsilon$, there exists $n \ge n_k$ such that $a x_n - b \notin E_k$ whenever $k \ge k_\varepsilon$. Denote $E_2 = E^2_{k_\varepsilon}$. Then, similarly as for $E_1$, we have

\begin{align*} \lambda(E_2) = \lambda(E^2_{k_\varepsilon}) &\ge \lambda(E^2_{k_\varepsilon} \cap [-1,0]) \\ &\ge (N\rho-1-r_{k_\varepsilon}) r_{k_\varepsilon} \left\lfloor \frac{1-2N r_{k_\varepsilon}}{r_{k_\varepsilon}} \right\rfloor \\ &\ge (N\rho-1-r_1)(1 - r_1(2N+1)) \gt \frac{1}{3}(N\rho-1), \end{align*}

by letting $r_1 \lt (N\rho-1)/(2N+1)$.

In particular, the sets $E_1$ and $E_2$ have measures uniformly bounded away from zero.

Proof. The inequalities in (2) still hold for all $n \ge n_m$, for any fixed $m$, that is, by replacing $r_k$ with $r_m$ in $\rho - r_k$. This gives us the freedom to choose any null sequence $(r_k)$, independent of whether $(r_k / x_{n_k})$ is a null sequence or not. Then, inclusion (6) still holds with $x_{n_k}$ replaced by $x_{n_m}$.

Considering $E_k$ restricted to the interval $[0,1]$ and taking the intersection with respect to $k \ge 1$ yields

\begin{equation*} \bigcap_{n \ge n_m} \frac{1}{x_n} (E+b) \subseteq \{0\}, \quad \text{where } E = \bigcap_{k \ge 1} E_k. \end{equation*}

Consequently, for every $a \neq 0$ and $b \in \mathbb{R}$, there exists $n \ge n_m$ such that

\begin{equation*} a x_n - b \notin E. \end{equation*}

By moving along the subsequence $\{n_m\}$, we obtain that

\begin{equation*} a x_n - b \notin E \quad \text{infinitely often.} \end{equation*}

For $n \ge 1$, let

\begin{equation*} D_n = \bigcap_{k \le n} E_k. \end{equation*}

Then $E \subseteq D_n$ for every $n \ge 1$, so any cover (open or closed) for $D_n$ is also a cover for $E$.

Now, choose the null sequence $r_k = N/(N+1)^k$ for $k \in \mathbb{N}$ with $N\in \mathbb N, \, N\geq 3$. The set $D_n$ consists of disjoint closed intervals of length $1/(N+1)^n$, and there are

\begin{align*} \Big(\Big\lfloor \frac{r_{n-1}}{N r_n} \Big\rfloor + 1\Big) \Big(\Big\lfloor \frac{r_{n-2}}{N r_{n-1}} \Big\rfloor + 1\Big) \cdots \Big(\Big\lfloor \frac{r_1}{N r_2} \Big\rfloor + 1\Big) \Big(\Big\lfloor \frac{1}{r_1} \Big\rfloor + 1\Big) = 2^n \end{align*}

such intervals.

Given a set $U \subseteq \mathbb{R}$, let $|U| = \sup\{|x-y| : x, y \in U\}$. For $s, t \gt 0$ and $V \subseteq \mathbb{R}$, denote

\begin{equation*} H^s_t(V) = \inf \Big\{\sum_{n \ge 1} |U_n|^s : |U_n| \le t, \; V \subseteq \bigcup_{n \ge 1} U_n \Big\}. \end{equation*}

Then,

\begin{equation*} H^s_{r_n}(E) \le H^s_{r_n}(D_n) \le 2^n \Big(\frac{r_n}{N}\Big)^s = 2^n \Big(\frac{1}{N+1}\Big)^{s n}. \end{equation*}

For the choice $s = \log 2 / \log (N+1)$, we obtain

\begin{equation*}H^s(E) := \lim_{n \to \infty} H^s_{r_n}(E) \le 1.\end{equation*}

Next, we show that $H^s(E) \gt 0$. It suffices to show that, for some $\alpha \gt 0$, it holds

\begin{equation*} \sum_{n \ge 1} |U_n|^s \ge \alpha, \quad \text{for any cover } \{U_n\} \text{of } E. \end{equation*}

W.l.o.g., we may assume that each $U_n$ is a closed interval. Because $E$ is compact, we may assume the cover $\{U_n\}$ is finite. For each $U_n$, let $k \in \mathbb{N}$ be such that

\begin{equation*} \frac{1}{(N+1)^{k+1}} = \frac{r_{k+1}}{N} \le |U_n| \lt \frac{r_k}{N} = \frac{1}{(N+1)^k}. \end{equation*}

Then $U_n$ intersects $D_k$ in at most one of its subintervals, since the intervals are of width $r_k / N$ and are separated by a distance of $r_k (1 - 1/N)$. Next, $U_n$ intersects $D_{k+1}$ in at most two intervals, since each subinterval of $D_k$ contains at most two subintervals of $D_{k+1}$. Inductively, for every $j \ge k$, the interval $U_n$ intersects $D_j$ in at most $2^{j-k}$ subintervals. Note that

\begin{equation*} 2^{j-k} = 2^j (N+1)^{-s k} \le 2^j (N+1)^s |U_n|^s. \end{equation*}

We take $j$ large enough so that $(N+1)^{-(j+1)} \le |U_n|$ for all $U_n$. Any interval $I$ of $D_j$ contains a nested sequence of subintervals

\begin{equation*} I \supseteq I_1 \supseteq I_2 \supseteq \dots, \end{equation*}

where $I_m$ is a subinterval of $D_{j+m}$ inside $I$. Since $r_k \to 0$, the lengths shrink to zero, so $\bigcap_m I_m$ is nonempty and contained in $E$. Therefore, any cover $\{U_n\}$ must intersect all $2^j$ intervals of $D_j$; otherwise, it would miss a point in $E$.

We write $j = j_n + k_n$ for each $U_n$, then

\begin{equation*} 2^j \le \sum_n 2^{j_n} = \sum_n 2^{j - k_n} \le \sum_n 2^j (N+1)^s |U_n|^s, \end{equation*}

implying

\begin{equation*} \sum_n |U_n|^s \ge \frac{1}{(N+1)^s} = \frac{1}{2}. \end{equation*}

Proof. This follows from Theorem 1 by taking $\varepsilon \lt 1$ and $a = 1$.

Proof. This follows from Theorem 2 by taking $\varepsilon \lt 1$ and $a = 1$.