On Large Sum-Free Sets: Revised Bounds and Patterns

Abstract

In this paper, we rectify two previous results found in the literature. Our work leads to a new upper bound for the largest sum-free subset of $[1,n]$ with the lowest value in $\left[{\displaystyle \frac{n}{3}},{\displaystyle \frac{n}{2}}\right]$, and the identification of all patterns that can be used to form sum-free sets of maximum cardinality.

1. Introduction

Let A be a non-empty and finite set containing positive integers such that $A\subseteq [1,n]$ for some $n\in {\mathbb{Z}}^{+}$, and $A+A=\{x+y\mid x,y\in A\}$. We call A sum-free if $A\cap (A+A)=\varnothing$, sum-free sets are actively the subject of research (see, for instance, the surveys [1,2], and references therein). There are infinitely many examples of sum-free sets such as $\{1,3,5\}$, or even the set of all odd numbers in $[1,n]$. We call A maximal sum-free if there is no sum-free subset of $[1,n]$ of which A is a proper subset. The maximum possible cardinality for a sum-free A is $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$, and $\left|A\right|=\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$ implies that A is maximal sum-free. However, the converse is not true. For instance, if we take $n=8$ and $A=\{1,3,8\}$, we can see that A is maximal sum-free but not of maximum cardinality.

The contribution of this paper lies in rectifying two findings from the existing literature, leading to (i) a new upper bound for the largest sum-free subset of $[1,n]$ with the lowest value in $\left[{\displaystyle \frac{n}{3}},{\displaystyle \frac{n}{2}}\right]$, and (ii) the identification of all patterns that can be used to form sum-free sets of maximum cardinality.

2. Cardinality Upper Bounds for Sum-Free Sets

Cameron and Erdős [3] introduced the following theorem.

Theorem 1. 

Let $g(n,m)$ be the cardinality of the largest sum-free subset of $[1,n]$ with lowest value m. Then,

$$g(n,m)\le \left\{\begin{array}{cc}n-m+1\hfill & \mathrm{if}m>{\displaystyle \frac{n}{2}},\hfill \\ m\hfill & \mathrm{if}{\displaystyle \frac{n}{3}}\le m\le {\displaystyle \frac{n}{2}},\hfill \\ ⌊{\displaystyle \frac{1}{2}}(n-m)⌋+1\hfill & \mathrm{if}m<{\displaystyle \frac{n}{3}}.\hfill \end{array}\right.$$

However, the bounds above do not always apply as we show below.

Theorem 2. 

Let A be a largest sum-free subset of $[1,n]$ with the lowest value m, $n=3m$, and $n\in A$. Then, $\left|A\right|=g(n,m)\le m+1$. If, in addition, $\left|A\right|=\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$, then there exist exactly two sets in which $g(n,m)=m+1$.

Proof. 

First, we present a counterexample showing that the original $g(n,m)\le m$ does not always apply. Let $n=6$ and $m=2$. Clearly, ${\displaystyle \frac{1}{3}}n\le m\le {\displaystyle \frac{1}{2}}n$, indicating that $g(n,m)=g(6,2)\le m=2$. However, $A=\{2,5,6\}$ is sum-free and has cardinality three.

The proof presented by Cameron and Erdős [3] estimates $g(n,m)$ for ${\displaystyle \frac{1}{3}}n\le m<{\displaystyle \frac{1}{2}}n$ based on the maximum possible cardinality of A, that is, $\left|A\right|\le n-m+1$, and the fact that for any $x\in [2m,n]$ A contains only one element of the pair $(x-m,x)$ so that it appears that $n-2m+1$ elements can be excluded from being possible elements of A. This leads to the upper bound $g(n,m)\le (n-m+1)-(n-2m+1)=m$. However, in our counterexample, we have $x\in [2m,n]=[4,6]$, so the pairs are $(2, 4),(3, 5),(4, 6)$. Given $2,6\in A$, $n-2m+1$ counts 4 twice as an element to be excluded. Thus, this upper bound fails when there exists a $x\in [2m,n]\cap A$, such that $x-2m\in A$.

Let us improve the above condition. If $x-2m\in A$, then $x=2m$ implies $x-2m=0\in A$, which cannot be as A is sum-free. Hence, $x=2m+y$ with $y\in A$. Notice that ${\displaystyle \frac{n}{3}}\le m\le {\displaystyle \frac{n}{2}}\iff 2m\le n\le 3m$. Hence, $x\le 3m$ and $y\in [1,m]$. Given $y\in A$, it cannot be that $y<m$. Hence, $y=m$ and $x=3m=n$. Thus, there can be a maximum of one element counted twice to be excluded from A. Finally, if $3m=n$ and $n\in A$, then $g(n,m)\le m+1$.

Further to the above, if $\left|A\right|=\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$, then A has cardinality ${\displaystyle \frac{n}{2}}$ or ${\displaystyle \frac{n+1}{2}}$ depending on the parity of n. Let n be odd, then $\left|A\right|={\displaystyle \frac{n+1}{2}}={\displaystyle \frac{3m+1}{2}}$. We have that $\left|A\right|=m+1$, and $m+1={\displaystyle \frac{3m+1}{2}}\iff m=1$. Hence, $n=3$ and $A=\{1,3\}\subset [1,3]$. Let n be even, then $\left|A\right|={\displaystyle \frac{n}{2}}={\displaystyle \frac{3m}{2}}$. We have that $\left|A\right|=m+1$, leading to $m+1={\displaystyle \frac{3m}{2}}\iff m=2$. Hence, $n=6$ and $A=\{2,5,6\}\subset [1,6]$. The latter is the only possibility. □

Corollary 1. 

The upper bounds for g(n,m) are

$$g(n,m)\le \left\{\begin{array}{cc}n-m+1\hfill & \mathrm{if}m>{\displaystyle \frac{n}{2}},\hfill \\ m\hfill & \mathrm{if}{\displaystyle \frac{n}{3}}<m\le {\displaystyle \frac{n}{2}},\hfill \\ ⌊{\displaystyle \frac{1}{2}}(n-m)⌋+1\hfill & \mathrm{if}m\le {\displaystyle \frac{n}{3}}.\hfill \end{array}\right.$$

Proof. 

Theorem 2 has $n=3m$. Hence, the new bound of $m+1=⌊{\displaystyle \frac{1}{2}}(n-m)⌋+1$. □

3. All Sum-Free Sets of Maximum Cardinality

Cameron and Erdős [4] present what is probably the first taxonomy regarding the sum-free sets we are interested. They state, without proof, that the only sum-free sets of cardinality $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$ are as follows:

• The odd numbers in the interval $[1,n]$;
• If n is odd, $\left[{\displaystyle \frac{n+1}{2}},n\right]$;
• If n is even, $\left[{\displaystyle \frac{n}{2}},n-1\right]$ and $\left[{\displaystyle \frac{n}{2}}+1,n\right]$.

In this section, our main objective is to prove the following.

Theorem 3. 

There are exactly four sum-free sets with cardinality $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$ that do not adhere to the taxonomy introduced by Cameron and Erdős. These are $\{1, 4\}$, where $n=4$; $\{1, 4, 6\}$ and $\{2, 5, 6\}$, both for $n=6$; and $\{2, 3, 7, 8\}$, where $n=8$.

Cameron and Erdős’ taxonomy states the only possibilities for m are as follows: (1) $m\ge \lfloor {\displaystyle \frac{n+1}{2}}\rfloor$, or (2) $m=1$ and A is composed solely of odd numbers. If we are to prove Theorem 3, we must then challenge both of these.

Lemma 1. 

There is no sum-free set of positive integers with cardinality $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$ and $3\le m<⌊{\displaystyle \frac{n+1}{2}}⌋$.

Proof. 

First, we show that if $m<{\displaystyle \frac{n}{3}}$, then $m\in \{1,2\}$. With this value of m, Theorem 1 gives us that $g(n,m)\le \lfloor {\displaystyle \frac{1}{2}}(n-m)\rfloor +1$. Let us assume, for a contradiction, that a sum-free set A (under the requirements of the hypothesis) can have a cardinality that is any lower than this upper bound (that is, $\lfloor {\displaystyle \frac{n-m}{2}}\rfloor$). We have that

$$\begin{array}{c}\hfill ⌊{\displaystyle \frac{n-m}{2}}⌋=⌊{\displaystyle \frac{n+1}{2}}⌋\iff m=-1.\end{array}$$

However, $m\in {\mathbb{Z}}^{+}$. Hence, this cannot be, and the upper bound of $g(n, m)$ is attained. This leads to

$$\begin{array}{c}\hfill ⌊{\displaystyle \frac{n-m}{2}}⌋+1=⌊{\displaystyle \frac{n-m+2}{2}}⌋=⌊{\displaystyle \frac{n+1}{2}}⌋.\end{array}$$

Thus, $m\in \{1, 2\}$.

Now, we show that if ${\displaystyle \frac{n}{3}}\le m<{\displaystyle \frac{n}{2}}$, then there is no suitable m. Notice that $|A|=\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$, but $min\left(\right\{1, 3\left\}\right),min\left(\right\{2, 5, 6\left\}\right)<3$. Hence, the exceptions identified in Theorem 2 do not apply. Thus, we can use Theorem 1, leading to the upper bound $g(n, m)\le m$. We have that $m<{\displaystyle \frac{n}{2}}$, so it cannot be that $g(n,m)=\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$. □

Clearly, we now need to analyse the cases in which $m\in \{1, 2\}$. Here, we are interested in non-trivial sum-free sets of maximum cardinality. Notice that if $n\in \{1, 2\}$, then $\left\{1\right\}$ and $\left\{2\right\}$ are the only possible sum-free sets. However, both of these are already covered by Cameron and Erdős’ taxonomy. Hence, we focus on $n\ge 3$ leading to $m,p\in A$ with $m\ne p$.

Lemma 2. 

There are exactly two sum-free sets with cardinality $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$ containing 1 that are not covered by Cameron and Erdős’ taxonomy.

Proof. 

Let p be the highest element of A. Given that A is sum-free, at most one element of each of the following pairs can be an element of A:

$$\begin{array}{c}\hfill (1,p-1),(2,p-2),\dots ,\left(⌊{\displaystyle \frac{n+1}{2}}⌋-1,p-\left(⌊{\displaystyle \frac{n+1}{2}}⌋-1\right)\right).\end{array}$$

Notice that there are $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor -1$ such pairs; however, we know that $p\in A$, and there is no pair with p as an element. Clearly, $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor -1+1=|A|$. That is, A must contain one element from each pair.

If $n=3$, we have $A=\{1, p\}$ with $1<p\le 3$. The only possible combination is $A=\{1, 3\}$; however, this set is already covered by Cameron and Erdős’ taxonomy. If $n=4$, we have a new $A=\{1, p\}=\{1,4\}$, which is our first sum-free set. Recall that A is sum-free if and only if A is difference-free. We have that

$$\begin{array}{c}\hfill 1, p\in A\Rightarrow 2,p-1\notin A\iff p-2,1\in A.\end{array}$$

Hence, $1, p-2, p\in A$. If $n=5$, then $A=\{1, 3, 5\}$, which is already covered by Cameron and Erdős’ taxonomy. If $n=6$, then the only possibility is $A=\{1, 4, 6\}$. The latter is our second and last sum-free set under the hypothesis.

We now show that the general pattern for A is $\{1, 3, 5, \dots , p-2, p\}$, for an odd p. We have identified that $\{1, p-2, p\}\subseteq A$ for $n\ge 5$. Let a be the highest known value in A, that is, we begin with $a=1$. Hence, $(p-2)-a\notin A\iff a+2\in A$. Clearly, a is odd and $a+2$ is the next odd number from a. Now, $a+2$ is the highest known number in A, so we increment a such that $a=a+2$. Of course, we still have that $(p-2)-a\notin A\iff a+2\in A$. We can repeat this procedure as many times as necessary. Hence, $A=\{1, 3, 5, \dots , p-2, p\}$. Let us assume, for a contradiction, that p is even. In this case, A has only two even numbers, p and $p-2$. However,

$$\begin{array}{c}\hfill 5, 1\in A\Rightarrow 5-1=4\notin A\iff p-4\in A.\end{array}$$

The above implies that A has three even numbers, which cannot be. Thus, p is odd. Given A is of maximum cardinality, we have that $n=p$ and A is the set of odd numbers in $[1, n]$. This set is already covered by Cameron and Erdős’ taxonomy. □

Lemma 3. 

There are exactly two sum-free sets with cardinality $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$ containing 2 that are not covered by Cameron and Erdős’ taxonomy.

Proof. 

Let A be a sum-free set of maximum cardinality. The lowest possible cardinality of a non-trivial A leads to $A=\{2, p\}$. We have that $\lfloor {\displaystyle \frac{n+1}{2}}\rfloor =2\iff n\in \{3, 4\}$, leading to $2<p\le n$, and by consequence, $A=\{2, 3\}$. However, this set is already covered by Cameron and Erdős’ taxonomy.

Recall that A must contain one element of each pair:

$$\begin{array}{c}\hfill (1, p-1),(2, p-2), \dots ,\left(⌊{\displaystyle \frac{n+1}{2}}⌋-1, p-\left(⌊{\displaystyle \frac{n+1}{2}}⌋-1\right)\right).\end{array}$$

By consequence,

$$\begin{array}{c}\hfill 2\in A\Rightarrow 1\notin A\iff p-1\in A.\end{array}$$

Thus, if $n\in \{5, 6\}$, we have $A=\{2, p-1, p\}$. Notice that $n=5\Rightarrow p\in \{4, 5\}$, leading to $2, 4\in A$, which cannot be. On the other hand, $n=6\Rightarrow p\in \{4, 5, 6\}$, leading to the only valid combination of $A=\{2, 5, 6\}$. This is our first sum-free set.

Recall that a set is sum-free if and only if it is also difference-free. We have that

$$\begin{array}{c}\hfill 2, p-1\in A\Rightarrow (p-1)-2=p-3\notin A\iff 3\in A.\end{array}$$

Hence, $n\in \{7,8\}$ leads to $A=\{2, 3, p-1, p\}$. If $n=7$, every possible value of p (that is $p\in \{5, 6, 7\}$) leads to an A that is not sum-free. If $n=8$, the only possible combination is $A=\{2, 3, 7, 8\}$. This is our second and last sum-free set under the hypothesis. Let us carry on.

$$\begin{array}{c}\hfill 2, 3, p-1, p\in A\Rightarrow 3-2, (p-1)-2, p-2, (p-1)-3, p-(p-1)\notin A\\ \hfill =1, p-2, p-3, p-4\notin A\iff p-1, 2, 3, 4\in A.\end{array}$$

The above shows that $2, 4\in A$, making A no longer sum-free. □

Using the lemmas above, we can prove the main theorem of this section.

Proof of Theorem 3. 

The taxonomy introduced by Cameron and Erdős presents all possible cases for sum-free sets of maximum cardinality with $m\ge \lfloor {\displaystyle \frac{n+1}{2}}\rfloor$. Hence, if there are exceptions to this taxonomy, these can only happen if $m<\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$.

Lemma 1 shows that there is no maximal sum-free set of positive integers with $3\le m<\lfloor {\displaystyle \frac{n+1}{2}}\rfloor$. Thus, the only possibility is if $m\in \{1, 2\}$. Lemmas 2 and 3 address the cases of $m=1$ and $m=2$, respectively. These lemmas identify four sum-free sets not covered by Cameron and Erdős’ taxonomy. Specifically, these are $\{1, 4\}$, where $n=4$; $\{1, 4, 6\}$ and $\{2, 5, 6\}$, both for $n=6$; and $\{2, 3, 7, 8\}$, where $n=8$. Hence, these are the only possible exceptions. □