A special case of Problem #355

In the online database of Erdős Problems, the following is listed as problem #355. A sequence of natural numbers $A\subset \mathbb N$ is called a lacunary sequence if $a_1<a_2<\dots$ and there is a $\lambda_A>1$ so that

$$\frac{a_{n+1}}{a_n} \ge \lambda_A$$

for all $n\ge 1$. Consider the set

$$\mathcal S_A := \left\{ \sum_{a\in A'}\frac{1}{a} : A'\subseteq A\textrm{ finite}\right\}$$

for such a sequence. The question is whether there exists a lacunary sequence $A$ for which $\mathcal S_A$ contains all rationals in some open interval.

While this question is quite a difficult one to answer, a special case is pretty elementary. I recently came to know that the results discussed in this post are pretty well known and are often attributed to Kakeya from more than a hundred years ago. I would like to thank Prof. Vjekoslav Kovač for this information.

Theorem 1 : Let $A = \{a_1 < a_2 < \cdots\} \subset \mathbb{N}$ be a lacunary sequence with $\lambda:=\lambda_A>2$. Then $\mathcal S_A$ does not contain all rationals in an open interval.

Proof : Set $u_n:=1/a_n$ and $r=1/\lambda<1/2$. By lacunarity, there is a constant $C=u_1/r$ such that

$$u_n\le Cr^n$$

for all $n\ge 1$. Note that every $s\in \mathcal S_A$ is of the form

$$s=\sum_{n\ge 1} \varepsilon_n u_n$$

for $\varepsilon_n\in \{0,1\}$ and $\varepsilon_n\ne 0$ for finitely many $n$.

    

Now, fix an $N\in \mathbb N$ and for each $\mathbf \varepsilon = (\varepsilon_1,\varepsilon_2,\dots , \varepsilon_N) \in \{0,1\}^N$, define

$$s_{\mathbf{\varepsilon}} := \sum_{n=1}^N \varepsilon_n u_n, \qquad R_N := \sum_{n>N} u_n$$

to be the first $N$ digits and the tail after that. But for each sequence $\{\varepsilon_n\}_{n\ge 1}$, we have

$$\sum_{n\ge 1} \varepsilon_n u_n \in \left[s_{\mathbf\varepsilon}, s_{\mathbf{\varepsilon}}+R_N\right]$$

for all $N\in \mathbb N$. So, define

$$\mathcal U_N := \bigcup_{\mathbf{\varepsilon}\in \{0,1\}^N} \left[s_{\mathbf\varepsilon}, s_{\mathbf{\varepsilon}}+R_N\right]$$

and note that $\mathcal S_A\subset \mathcal U_N$.

Now, $\mathcal U_N$ being a finite union of closed sets is closed. Also, it is clear that

$$\overline{\mathcal S_A}\subset \bigcap_{n\ge 1} \mathcal U_N$$

where $\overline{\mathcal S_A}$ is the closure of $\mathcal S_A$.

But by definition, we have

$$R_N \le \sum_{n\le N} Cr^n \le C' r^N$$

for a suitable constant $C'$. Now, $\mathcal U_N$ is the union of at most $2^N$ intervals. This implies

$$\left\lvert \mathcal U_N\right\rvert \le 2^N R_N \le C' (2r)^N \xrightarrow{N\to \infty} 0$$

since $\lambda>2$. This means that the Lebesgue measure

$$\mu\left(\bigcap_{n\ge 1} \mathcal U_N\right) = 0$$

and hence

$$\mu\left(\overline{\mathcal S_A}\right) = 0$$

implying $\overline{\mathcal S_A}$ has empty interior.

So, if there was an open interval $I$ such that $I\cap \mathbb Q \subset \mathcal S_A$, then this would imply

$$I = \overline{I\cap \mathbb Q} \subset \overline{\mathcal S_A}$$

which is a contradiction.

In fact, we can prove a stronger statement.

Theorem 2 : For a lacunary sequence $A$, we have

$$\dim_H\left(\overline{\mathcal S_A}\right) \le \frac{\log 2}{\log \lambda}$$

where $\dim_H(X)$ is the Hausdorff dimension of a set $X$.

Proof : With notations as in the previous proof, the $s$ dimensional Hausdorff content at scale $R_N$ satisfies

$$\mathcal H^s_{R_N} \left(\overline{\mathcal S_A}\right) \le 2^N (2R_N)^s$$

for $s>\frac{\log 2}{\log \lambda}$.

Using the bound for $R_N$, we have

$$\mathcal H^s_{R_N} \left(\overline{\mathcal S_A}\right) \le \left(\frac{2C r^{\,}}{1-r}\right)^s\cdot \left(2 r^s\right)^N$$

and hence

$$\lim_{N\to \infty} \mathcal H^s_{R_N} \left(\overline{\mathcal S_A}\right) = 0$$

since $2r^s=2\lambda^{-s}<1$.

Hence, $\dim_H\left(\overline{\mathcal S_A}\right) \le s$. Letting $s\downarrow \frac{\log2}{\log\lambda}$ completes the proof.

Remark : If we also impose $\lambda>2$ in Theorem 2, then $\dim_H\left(\overline{\mathcal S_A}\right)<1$ and hence $\overline{\mathcal S_A}$ has empty interior, hence proving Theorem 1.