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Source. Corollary 13, p. 10, of Ben Green, The Cameron-Erdős conjecture, Bull. London Math. Soc. 36 (2004), no. 6, 769--778, cited from the arXiv manuscript math/0304058v1 (4 April 2003) whose pages the labels below follow, as identified on the source card.

Statement

Sum-free means that no $x,y,z$ in the set satisfy $x+y=z$, and $[N]=\{1,\ldots,N\}$ (p. 1).

Corollary 13 (p. 10). "With $o(2^{N/2})$ exceptions, all sum-free subsets of $[N]$ consist entirely of odd numbers, or else are contained in $\{\lceil(N+1)/3\rceil,\ldots,N\}$." (quoted)

So the number of sum-free $A\subseteq[N]$ that contain an even number and also an element of $\{1,\ldots,\lceil(N+1)/3\rceil-1\}$ is $o(2^{N/2})$ as $N\to\infty$.

Proof pointer

Proof on pp. 10--11. Each sum-free $A$ lies in a member $F$ of the family of Proposition 6; since that family has $2^{o(N)}$ members, the sets $A$ whose container has $|F|\le(\tfrac12-\tfrac1{120})N$ number $o(2^{N/2})$. For larger $F$, Proposition 7 (p. 8) puts $F$, up to a small exceptional part, either inside a short interval or almost entirely among the odd numbers. The paper then bounds the sets $A$ that break the dichotomy by counting choices over disjoint pairs from which $A$ can take at most one element, except in the interval case when $A$ has at most $32\epsilon^{1/8}N$ elements in $[(1-\tfrac1{120})N,N]$; there it counts $A$ by the bound $2^{N/2+o(N)}$ applied to the sum-free set $A\cap[1,(1-\tfrac1{120})N]$. The proof on p. 11 cites "Theorem 12" for that bound, which the paper states as Proposition 12 (p. 10).

Dependencies

Proposition 6, Proposition 7 (p. 8) and Proposition 12 (p. 10). Read depth: claims checked; the statement was read on p. 10 and the proof for its structure only.

Bears on

• Problem 748: the corollary is the structural step that reduces the count of all sum-free subsets of $\{1,\ldots,n\}$ to the sets of odd numbers and the sum-free subsets of the top interval, which with the Cameron-Erdős count of the latter gives Theorem 2; on its own it gives no count.