## Decomposing sets without arithmetic progressions

Let $A$ be a subset of $\lbrace 1,2, \ldots ,8 \rbrace$ containing no arithmetic progressions. Then one of the following two must hold :

(1) $|A \cap \lbrace 3,5 \rbrace | \leq 1$ and $|A \cap \lbrace 1,2,4,6,7,8 \rbrace | \leq 1$.

(2) $|A \cap \lbrace 4,6 \rbrace | \leq 1$ and $|A \cap \lbrace 1,2,3,5,7,8 \rbrace | \leq 1$.

Let $A$ be a subset of $\lbrace 1,2, \ldots ,10 \rbrace$ containing no arithmetic progressions. Then one of $|A \cap \lbrace 1,2, \ldots ,7 \rbrace|$ and $|A \cap \lbrace 4,5, \ldots ,10 \rbrace|$ is at most $3$.

Let $A$ be a subset of $\lbrace 1,2, \ldots ,13 \rbrace$ containing no arithmetic progressions. Then one of the following two must hold :

(1) $|A \cap \lbrace 1,2,3,4,6 \rbrace | \leq 3$ and $|A \cap \lbrace 5,7,8,9,10,11,12,13 \rbrace | \leq 4$.

(2) $|A \cap \lbrace 8,10,11,12,13 \rbrace | \leq 3$ and $|A \cap \lbrace 1,2,3,4,5,6,7,9 \rbrace | \leq 4$.