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\title{Arithmetic Progressions in Lacunary Sets}
\author{T. C. Brown and A. R. Freedman}
\date{}
\maketitle
\begin{center}{\small {\bf Citation data:} T.C. Brown and A.R. Freedman, \emph{Arithmetic progressions in lacunary sets},
  Rocky Mountain J. Math. \textbf{17} (1987), no.~3, 587--596.}\bigskip\end{center}

\begin{abstract}
We make some observations concerning the conjecture of Erd\H{o}s that if the sum of the reciprocals of a set $A$ of positive integers diverges, then $A$ contains arbitrarily long arithmetic progressions. We show, for example, that one can assume without loss of generality that $A$ is lacunary. We also show that several special cases of the conjecture are true.
\end{abstract}

\section{Introduction} \label{sec: 1}

The now famous theorem of Szemer\'{e}di~\cite{szemeredi1975} is often stated:

\emph{(a) If the density of a set $A$ of natural numbers is positive, then $A$ contains arbitrarily long arithmetic progressions.}

\vspace{12pt}

Let us call a set $A$ of natural numbers $k$-good if $A$ contains a $k$-term arithmetic progression. Call $A$ $\omega$-good if $A$ is $k$-good for all $k \geq 1$. We define four density functions as follows: For a set $A$ and natural numbers $m,n$, let $A[m,n]$ be the cardinality of the set $A \cap \{m, m+1, m + 2, \dots, n\}$. Then define
\begin{align*} 
\underline{\delta}(A) &= \liminf_n \frac{A[1,n]}{n}, \\
\overline{\delta}(A) &= \limsup_n \frac{A[1,n]}{n}, \\
\underline{u}(A) &= \lim_n \min_{m \geq 0} \frac{A[m + 1, m + n]}{n} \: \textup{and} \\
\overline{u}(A) &= \lim_n \max_{m \geq 0} \frac{A[m + 1, m + n]}{n}. \end{align*}
It can be seen that the limits in the definitions of $\underline{u}$ and $\overline{u}$ always exist. These four ``asymptotic" set functions are called the lower and upper ``ordinary" and the lower and upper ``uniform" density of the set $A$ respectively. They are related by
\[ \underline{u}(A) \leq \underline{\delta}(A) \leq \overline{\delta}(A) \leq \overline{u}(A) \]
for any set $A$.

Szemer\'{e}di actually proved:

\emph{(b) If $\overline{u}(A) > 0$, then $A$ is $\omega$-good.} Hence we also have 

\emph{(c) If $\overline{\delta}(A) > 0$, then $A$ is $\omega$-good.}

In fact, Szemer\'{e}di proved the following ``finite" result (which we state in a general form to be used later):

\emph{(d) Let $\varepsilon > 0$ and $k \in N = \{1,2,3,\dots\}$. Then there exists an $n_0 \in N$ such that if $P$ is any arithmetic progression of length $|P| \geq n_0$ and $A \subset P$ with $|A| \geq \varepsilon |P|$, then $A$ is $k$-good.}

\vspace{12pt}

It is not hard to prove (without assuming the truth of any of the statements) that (b), (c), and (d) are equivalent.

Erd\H{o}s has conjectured that the following stronger statement holds:

\emph{(e) If $A \subseteq N$ and $\sum_{A} \frac{1}{a} = \infty$, then $A$ is $\omega$-good.}

\vspace{12pt}

By $\sum_A (1/a)$ we mean of course $\sum_{a \in A} (1/a)$. The proof (or disproof) of (e) is, at present, out of sight. In fact, it has not even been proved that $\sum_A (1/a) = \infty$ implies that $A$ is $3$-good (compare Roth~\cite{roth1953}). That (e) $\Rightarrow$ (c) can be seen as follows: If $\overline{\delta}(A) = \varepsilon > 0$, then there exists a sequence of natural numbers $0 = n_0 < n_1 < n_2 < \cdots$, such that, for each $i$, 
\[ \frac{A[1,n_i]}{n_i} > \frac{\varepsilon}{2} \quad \textup{and} \quad \frac{n_{i-1}}{n_i} < \frac{\varepsilon}{4}. \]
Then 
\begin{align*} \sum_A \frac{1}{a} &\geq \sum_{\substack{a \in A \\ a \leq n_k}} \frac{1}{a} \geq \sum_{i=1}^k \frac{A[n_{i-1} + 1, n_i]}{n_i} \geq \sum_{i=1}^k \frac{A[1,n_i] - n_{i-1}}{n_i} \\
&\geq k(\frac{\varepsilon}{2} - \frac{\varepsilon}{4}) = \frac{k\varepsilon}{4} \to \infty \quad (k \to \infty) \end{align*}
and so $\sum_A(1/a) = \infty$. Assuming (e), it follows that $A$ is $\omega$-good.

\vspace{12pt}

Hence Erd\H{o}s' conjecture is indeed stronger than Szemer\'{e}di's theorem. Note also that Erd\H{o}s' conjecture, if true, would immediately answer in the affirmative the long-standing question of whether or not the primes are $\omega$-good.

\vspace{12pt}

In the next section we make some observations regarding this conjecture, and we show that several special cases of the conjecture are true.

Other observations can be found in Gerver~\cite{gerver1977,gerver+ramsey1979} and Wagstaff~\cite{wagstaff1979}.

\section{Main results} \label{sec: 2}

(2.1). First we consider the ``finite form" of Erd\H{o}s' conjecture.

\begin{thm} \label{thm: 1} 
Fix $k$, and assume that for all sets $A \subseteq N$, if $\sum_A(1/a) = \infty$ then $A$ is $k$-good. Under this assumption, there exists $T$ such that if $\sum_A(1/a) > T$, then $A$ is $k$-good.

(Gerver~\cite{gerver1977} has this result under the stronger hypothesis that if $\sum_A(1/a) = \infty$ then $A$ is $(k + 1)$-good.)
\end{thm}

\begin{proof}We may assume $k \geq 3$. Suppose the theorem is false. We will construct a set $A$ such that $\sum_A(1/a) = \infty$ and $A$ is not $k$-good. Choose a finite set $A_0$ such that $A_0$ is not $k$-good and $\sum_A(1/a) > 1$. Let $p_1$ be prime, $p_1 > 2\max A_0$, and choose a finite subset $A_1$ of $\{tp_1: t \geq 1\}$ such that $A_1$ is not $k$-good and $\sum_{A_1} (1/a) > 1$. Let $p_2$ be prime, $p_2 > 2\max A_1$, and choose a finite subset $A_2$ of $\{tp_2: t \geq 1\}$ such that $A_2$ is not $k$-good and $\sum_{A_2} (1/a) > 1$. Continuing in this way, we obtain finite sets $A_0,A_1,\dots$ such that for each $i \geq 0$, $A_i$ is not $k$-good, $\min A_{i + 1} \geq p_{i + 1} > 2 \max A_i$, each elements of $A_{i + 1}$ is a multiple of $p_{i + 1}$, and $\sum_{A_i} (1/a) > 1$.

Let $A = \bigcup A_i$. It is clear that $\sum_A (1/a) = \infty$. To show that $A$ is not $k$-good, it suffices to show that every 3-term arithmetic progression contained in $A$ must be contained in a single set $A_i$.

To this end, suppose that $x < y < z$, with $x,y,z \in A$ and $z - y = y - x$. Let $y \in A_i$. Then $z \in A_i$ also, since otherwise $z - y \geq \min A_{i + 1} - \max A_i > \max A_i > y - x$. Thus $y,z \in A_i \subset \{tp_i: t \geq 1\}$. Hence $x$ is divisible by $p_i$, so $x \geq p_i > \max A_{i-1}$, and $x \in A_i$. This finishes the proof of Theorem~\ref{thm: 1}.
\end{proof}

\begin{cor} \label{cor: 1}
The following statement is equivalent to statement (e):

(f) For each $k \in N$, there exists $T \in N$ such that if $\sum_A(1/a) > T$, then $A$ is $k$-good.
\end{cor}

We state next a lemma which will be useful later.

\begin{lemma} \label{lemma: 1}
Let $F_1,F_2,\dots$ be a sequence of finite subsets of $N$ such that for each $i$, $F_i$ is not $k$-good and $\min F_{i + 1} \geq 2 \max F_i$. Then $F = \bigcup F_i$ is not $( k + 1)$-good.
\end{lemma}

(The proof of Lemma~\ref{lemma: 1} is contained in the proof of Theorem~\ref{thm: 1} above).

\vspace{12pt} 

(2.2). Now we define an increasing sequence, $a_1 < a_2 < a_3< \cdots$, of natural numbers to be lacunary if $d_n = a_{n + 1} - a_n \to \infty$ as $n \to \infty$ and to be $M$-lacunary if, furthermore, $d_n \leq d_{n + 1}$ for all $n$. We shall think of such a sequence simultaneously as a sequence and as a subset of $N$. Any lacunary sequence $A$ has $\overline{u}(A) = 0$ (see~\cite{freedman+sember1981}), so that Szemer\'{e}di's theorem does not apply.

A subsequence of a lacunary sequence is lacunary, but the corresponding statement, unfortunately, does not hold for $M$-lacunary sequences. It is known that if the real series $\sum t_i$ is not absolutely convergent, then there exists a lacunary sequence $B$ such that $\sum_{i \in B} t_i $ diverges (see Freedman and Sember~\cite{freedman+sember1981}). It follows that if $A \subseteq N$ and $\sum_A(1/a) = \infty$, then there exists a lacunary sequence $B \subseteq A$ such that $\sum_B (1/b) = \infty$. Thus we have the following.

\begin{thm} \label{thm: 2} 
The following statement is equivalent to statement (e).

(g) If $A$ is a lacunary sequence and $\sum_A(1/a) = \infty$, then $A$ is $\omega$-good.
\end{thm}

Hence we need only investigate lacunary sequences when contemplating the Erd\H{o}s conjecture. It can also be shown that if $\sum t_i = \infty$ and $t_i \geq 0$ for all $i$, then there exists an $M$-lacunary sequence $B$ such that $\sum_{i \in B} t_i = \infty$. (We omit the rather cumbersome proof of this statement.) But notice that this does not imply that statement (h) below is equivalent to statement (e)! This is too bad--because we now prove (h).

\begin{thm} \label{thm: 3}
The following statement is true: 

(h) If $A$ is $M$-lacunary and $\sum_A (1/a) = \infty$, then $A$ is $\omega$-good.
\end{thm}

\begin{proof} Let $A = \{a_1 < a_2 < a_3 < \cdots\}$ be an $M$-lacunary sequence with infinite reciprocal sum. Assume there is a $k$ such that $d_i < d_{i + k}$ for each $i$, where $d_n = a_{n + 1} - a_n$, $n \geq 1$. We show that $a_{i + jk} \geq j^2/2$ for all $i \geq 1$, $j \geq 0$. Indeed,
\begin{align*} 
a_{i + jk} &= a_i + d_i + d_{i + 1} + \cdots + d_{i + jk - 1} \\
&\geq d_i + d_{i + k} + d_{i + 2k} + \cdots + d_{i + (j-1)k} \\
&> 1 + 2 + 3+ \cdots + j > j^2/2. \end{align*}
(Note that to obtain the first inequality we have merely omitted some terms from the sum.)
But then
\begin{align*} 
\sum_{i=1}^\infty \frac{1}{a_i} &= \sum_{j=0}^\infty \frac{1}{a_{1 + jk}} + \sum_{j=0}^\infty \frac{1}{a_{2 + jk}} + \cdots + \sum_{j=0}^\infty \frac{1}{a_{k + jk}} \\
&\leq k(1 + \sum_{j=1}^\infty \frac{2}{j^2}) < \infty, \quad \textup{a contradiction.} \end{align*}
Hence, for each $k$, there is an $i$ such that $d_i = d_{i + k}$, whence $a_i, a_{i + 1}, \dots,a_{i + k + 1}$ are in arithmetic progression and $A$ is $\omega$-good.
\end{proof}

\vspace{12pt} 

The following is an immediate corollary.

\begin{cor} \label{cor: 2}
If $A$ is a finite union of $M$-lacunary sets and $\sum_A (1/a) = \infty$, then $A$ is $\omega$-good.
\end{cor}

(2.3). We now use some slightly expanded arguments to show that statement (g) holds for some special sequences which are not $M$-lacunary (but are nearly so).

\begin{thm} \label{thm: 4}
Let $A = \{a_1 < a_2 < a_3 < \cdots \}$ be any set. Suppose there are intervals $I_n = [s_n,t_n]$ with $t_n < s_{n + 1}$ such that 
\[ \sum_{n=1}^\infty \frac{1}{\sqrt{a_{s_n}}} < \infty, \quad \sum_{k \in \bigcup I_n} \frac{1}{a_k} = \infty. \]
Suppose further that for each $n$, $d_k \leq d_{k + 1}$ if $s_n \leq k < t_n$. Then $A$ is $\omega$-good.
\end{thm}

\begin{proof}
We will arrive at a contradiction if we assume that there is a $K \in N$, such that $d_i < d_{i + K}$ whenever $i, i + K$ belong to the same interval $I_j$. Then, for any $K$, we have that there exists an $i$ such that $d_i = d_{i + 1} = \cdots = d_{i + K}$ so that $a_i,a_{i + 1}, \dots,a_{i + K + 1}$ are in arithmetic progression. 

To get the required contradiction we proceed as follows: If $n, n + K, n + 2K, \dots, n + cK \in I_i$, then 
\begin{align*} 
\frac{1}{a_n} &+ \frac{1}{a_{n + K}} + \frac{1}{a_{n + 2K}} + \cdots + \frac{1}{a_{n + cK}} \\
&\leq \frac{1}{a_n} + \frac{1}{a_n + d_n} + \frac{1}{a_n + d_n + d_{n + K}} + \cdots \\
&+ \frac{1}{a_n + d_n + d_{n + K} + \cdots + d_{n + (c-1)K}} \\
&< \sum_{j=0}^\infty \frac{1}{a_{n} + (j^2/2)} < \frac{b}{\sqrt{a_n}} \leq \frac{b}{\sqrt{a_{s_i}}} \quad \textup{($b$ constant)}. \end{align*}
Hence,
\[ \sum_{K \in I_i} \frac{1}{a_k} < \frac{Kb}{\sqrt{a_{s_i}}} \quad \textup{and} \quad \sum_{k \in \bigcup I_i} \frac{1}{a_k} < Kb \sum_{i=1}^\infty \frac{1}{\sqrt{a_{s_i}}} < \infty, \]
contrary to assumption.
\end{proof}

Using a similar technique we can prove the following theorem. 

\begin{thm} \label{thm: 5}
Let $A = \{a_1<a_2< a_3< \cdots \}$ be a set. Suppose $I_n = [s_n,t_n]$ are intevervals with $t_n < s_{n + 1}$ such that $d_i \leq d_{i + 1}$ if $s_n \leq i < t_n$ and $d_{t_n -1} < d_{s_n + 1}$. Then, if $\sum_{k \in \bigcup I_n} (1/a_k) = \infty$, $A$ is $\omega$-good.
\end{thm}

(2.4). We now define new density functions $\lambda$ and $\overline{\lambda}$ in terms of lacunary sequences: For all sets $A$, let $\overline{\lambda}(A) = 0$ if $A$ is finite or a finite union of lacunary sequences and otherwise let $\overline{\lambda}(A) = 1$. Define $\lambda(A) = 1 - \overline{\lambda}(N - A)$. These densities, taking only $0,1$ values, may seem a little odd. The definition could be improved so that $\lambda$ becomes ``continuous" and has the correct value on an (infinite) arithmetic progression etc. However, this would not suit our purposes any better. One can prove that for any $A \subseteq N$,
\[ \lambda(A) \leq \underline{u}(A) \leq \underline{\delta}(A) \leq \overline{\delta}(A) \leq \overline{u}(A) \leq \overline{\lambda}(A) \]
and so, in analogy to Szemer\'{e}di's theorem it is natural to ask about the arithmetic progressions in $A$ if $\overline{\lambda}(A) > 0$. 

\begin{thm} \label{thm: 6} 
There exists a set $A$ such that $\overline{\lambda}(A) > 0$ and $A$ is not $\omega$-good.
\end{thm}

\begin{proof} Let $B_i = \{1!, 2!, \dots,i!\}$. $B_i$ is not $3$-good. Let $(H_i)$ be the sequence of sets 
\[ (B_1, B_1,B_2,B_1,B_2,B_3,B_1,B_2,B_3,B_4,B_1, \dots). \]
Let $f_i$ be an increasing sequence of integers such that $f_1 = 0$ and 
\[ \min(f_{i + 1} + H_{i + 1}) \geq 2 \max(f_i + H_i) \]
and define $A = \bigcup_i (f_i + H_i)$. By Lemma~\ref{lemma: 1}, $A$ is not $4$-good. (By choosing $f_i$ sufficiently quickly increasing one can even make $A$ not 3-good.) Finally, $\overline{\lambda}(A) = 1$ since otherwise $A = L_1 \cup L_2 \cup \cdots \cup L_k$ where each $L_j$ is a lacunary sequence. Whenever $H_i = B_{k + 1}$ we have $|f_i + H_i| > k$ and so some $L_j$ has at least two members in $f_i + H_i$. Hence we may find a fixed $j$ such that 
\[ |L_j \cap (f_i + B_{k + 1})| \geq 2 \]
for infinitely many $i$. Then $L_j$ has infinitely many differences $d_t < (k + 1)!$, and so $L_j$ is not lacunary.
\end{proof}

(2.5). Let us consider ``relative density", that is, ``the density of $A$ relative to $B$" where $A \subseteq B$. The definitions are 
\begin{align*} 
\underline{\delta}(A|B) &= \liminf_{i \to \infty} \frac{A[1,b_i]}{i} \quad \textup{and} \\
\underline{u}(A| B) &= \lim_{n \to \infty} \min_{m \geq 0} \frac{A[b_{m + 1}, b_{m + n}]}{n}. \end{align*}
$\overline{\delta}(A|B)$ and $\overline{u}(A|B)$ are obtained by replacing ``inf" with ``sup" and ``min" with ``max" respectively. One can show, as before, for any $A, B, A \subseteq B$, that 
\[ \underline{u}(A|B) \leq \underline{\delta}(A|B) \leq \overline{\delta}(A|B) \leq \overline{u}(A|B). \]
Let $B$ be $M$-lacunary and $\sum_B 1/b = \infty$. Then, by Theorem~\ref{thm: 3}, $B$ is $\omega$-good. We ask whether $A \subseteq B$ and the relative density of $A$ positive imply that $A$ is also $\omega$-good. The answer is ``yes" if $\underline{u}(A|B) > 0$ (Theorem~\ref{thm: 7}), ``no" if $\overline{\delta}(A|B) > 0$ (Theorem~\ref{thm: 8}) and the question is open for $\underline{\delta}(A|B) > 0$. 

\begin{thm} \label{thm: 7} 
If $B$ is $M$-lacunary, $\sum_B 1/b = \infty$, $A \subseteq B$ and $\underline{u}(A|B) > 0$ then $A$ is $\omega$-good.
\end{thm}

\begin{proof} By (the proof of) Theorem~\ref{thm: 3} there are arbitrarily large $n,m$ such that 
\[ P = \{b_{m + 1}, b_{m + 2}, \dots, b_{m + n}\} \]
is an arithmetic progression. By the definition of $\underline{u}(A|B)$ we have $|A \cap P| \geq \varepsilon P$ where $\varepsilon = (1/2)\underline{u}(A|B)$ and $|P|$ is arbitrarily large. Thus, by Szemer\'{e}di's theorem (d) we have, for any $k$, that $|A \cap P|$ is $k$-good if $|P|$ is sufficiently large. Hence $A$ is $\omega$-good.
\end{proof}

\begin{thm} \label{thm: 8}
There exists an $M$-lacunary sequence $B$ with $\sum_B = 1/b = \infty$ and an $A \subseteq B$ with $\overline{\delta}(A|B) > 0$ ($=1$ in fact) such that $A$ is not 3-good.
\end{thm}

\begin{proof}
(leaving most of the details to the reader). Let $F = \{1!,2!,3!,\dots\}$, $b_1 = 1$ and define $b_{n + 1} = b_n + d_n$ where the $d_n$'s have the following properties: For all $i,d_i \in F$ and $d_i \leq d_{i + 1}$. Furthermore, the set of natural numbers $N$ can be partitioned into consecutive pairwise disjoint intervals $J_1,J_2,J_3,\dots$ such that if $r$ is odd, then for $i \in J_r$, $d_i = d_{i + 1}$ and $\sum_{i \in J_r} 1/b_i \geq 1$, and, if $r$ is even, then, for $i \in J_r$, $d_i < d_{i + 1}$, $b_i > 2b_{i-1}$ and $|J_r| > (\max J_{r-1})^2$. Clearly $B = \{b_1,b_2,\dots\}$ is $M$-lacunary and $\sum_B 1/b = \infty$. Let $A = \{b_k: k \in \bigcup_r J_{2r}\}$. Then 
\[ \overline{\delta}(A|B) \geq \lim_r \frac{|J_{2r}|}{|J_{2r}| + \max J_{2r - 1}} = 1. \]
One can also see that $A$ is not 3-good since $a_i > 2a_{i-1}$ holds.
\end{proof}

(2.6). Theorems~\ref{thm: 4},\ref{thm: 5}, and~\ref{thm: 7} notwithstanding, it seems to be difficult to generalize the notion of $M$-lacunary even slightly and still prove the corresponding case of the Erd\H{o}s conjecture. In this connection let us define a lacunary sequence $A$ to be $M_k$-lacunary (where $k \geq 0$) if, for all $i,j,i \leq j$, we have $d_i \leq d_j + k$. Clearly, the $M_0$-lacunary sequences are just the $M$-lacunary sequences. For no $k \ne 0$ are we able to prove that $M_k$-lacunary and $\sum_A (1/a) = \infty$ imply $\omega$-good. We can show if $A$ is $M_1$-or $M_2$-lacunary with $\sum_A(1/a) = \infty$ then $A$ is 3-good. We prove first a lemma which may have independent interest:

\begin{lemma} \label{lemma: 2}
If $A = \{a_1<a_2<a_3<\cdots\}$ is any subset of $N$ and $\sum_A 1/a = \infty$, then, for any $t > 0$, there exists an $i$ such that $d_{i + j} \leq d_i$ for $j = 0,1,\dots,t$. (Of course, $d_n = a_{n + 1} - a_n$.)
\end{lemma}

\begin{proof} The method is familiar by now: Suppose there is a $t$ such that, for each $i$, there exists $j \in[1,t]$ with $d_i < d_{i + j}$. Then we can find a sequence $(j_n)$ such that 
\[ d_1 < d_{1 + j_1} < d_{1 + j_1 + j_2} < \cdots (j_n \in [1,t]). \]
It follows that 
\[ \sum_A \frac{1}{a} \leq t \sum_{s = 0}^\infty \frac{1}{a_{1 + (j_1 + \cdots + j_s)}} \leq t\sum_{s=0}^\infty \frac{1}{a_1 + (1 + 2 + \cdots + s)} < \infty. \qedhere \]
\end{proof}

\begin{thm} \label{thm: 9}
Let $A$ be $M_1$-or $M_2$-lacunary and $\sum_A(1/a) = \infty$. Then $A$ is 3-good.
\end{thm}

\begin{proof} By the definition of $M_k$-lacunary and Lemma~\ref{lemma: 2} we have: for any $t > 0$ there is an $i$ such that 
\[ d_i - e \leq d_{i + j} \leq d_i, \quad j = 0,1,\dots,t, \]
where $e = 1$ or 2. Hence, in the sequence $(d_i)$, we have arbitrarily long blocks where the $d_i$ take on only two (in case $e = 1$) or three (in case $e = 2$) values. Such long blocks must contain two consecutive subblocks with identical composition (see Pleasants~\cite{pleasants1970}). These two subblocks will determine three terms of the sequence $A$ in arithmetic progression.
\end{proof}

This last result suggests a conjecture which is related to van der Waerden's theorem on arithmetic progressions and would immediately imply that $M_k$-lacunary with $\sum_A (1/a) = \infty$ implies that $A$ is $3$-good.

\begin{conj} Let $x_i$ be a sequence of positive integers with $1 \leq x_i \leq K$. Then there are two consecutive intervals $I,J$ of the same length, with $\sum_{i \in I} x_i = \sum_{j \in J} x_j$. Equivalently, if $a_1 < a_2 < \cdots$ satisfy $a_{n+1} - a_n \leq K$, all $n$, then there exist $x< y < z$ such that $x + z = 2y$ and $a_x +a_z = 2a_y$.
\end{conj} 

\nocite{erdos1975,gerver+ramsey1979,wroblewski1984}


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