Let and be sets of integers. With we denote the set of all integers that can be written as a sum of one term in and one term in . That is, . Let be any positive integer. If is any set of integers, we shall say that is n-summable, if sets and exist with , , and . Note that being 1-summable is equivalent to being infinite. We call n-large if there exists a non-negative integer and infinitely many such that . Note that being 1-large is also equivalent to being infinite. And thus, is 1-summable if, and only if, is 1-large. In this post we will prove the following cute generalization of this fact:

A set is n-summable if, and only if, is n-large.

First we will prove that if is n-large, then is n-summable. To this end, let be an infinite sequence such that for all , we have: . Define . Then we have infinitely many , but at most distinct . And thus, there exists an infinite subsequence with . Now, define and . Then, by construction, we have .

To prove the converse direction, assume is any infinite set of integers and assume , with . Then we will prove that if is not n-large, then . So assume that for all with . Now, let be any integer larger (in absolute value) than . Then is a set of integers, all larger (in absolute value) than , with the difference between largest and smallest element being: . In particular, . And since , we also have .

Now, if we say is -summable if exist with and , then, unfortunately, it is not true that is -summable if is n-summable for all . A counterexample:

Define sets as follows:

1) , where is the -th entry of https://oeis.org/A002260

2) If , then

Define . Then is n-summable for all , because is n-large for all , with . Although it is also easy to construct explicit sets and with , , and . We claim, however, that is not -summable. If were -summable, then, by some appropriate addition or subtraction, we have for some infinite sets and , with and . In particular, . Now, if with , then . And thus, either lies strictly between two elements of , or lies strictly between the largest element of and the smallest element of . Either way, . So is contained in the sets that have no more than elements. Now, let be any element of larger than , and let be any element of larger than . Then, since , we have that . And thus, is larger than the largest element of . On the other hand, , which is thus smaller than the smallest element of . And we must conclude that is not an element of ; contradiction.

I realize that the above proof that our counterexample works may be somewhat inelegant, so I might rewrite it soon. In which case I’ll probably explicitly prove something like the following: if is as above, is an infinite set, with , and , then .

At last, I have the following hunch, on which I’ll probably be working for the next few days/years:

If there exists an absolute constant , such that for all there exist infinitely many for which , then is -summable.

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