How to show $x,y,z \in A$ - Functions, Combinatorics

Question

If $A \subseteq \{1,2,3,4,5,6\}$, how to show that for every $A$ there are $x,y,z \in \{1,2,3,4,5,6\}$, where $x,y,z$ can also be the same or at least not different from each other, and the following conditions always apply:

• $x = y+z$
• $\{x,y,z\} \subseteq A $ or $ \{x,y,z\} \subseteq \overline{A}$

While using the pigeonhole principle and the respective function, without just counting out the problem or checking it.

Answer 1

• $1 \in A$. This means $2 \in \bar{A}$. This means $4 \in A$. This means $5 \in \bar{A}$.

  • If $3 \in A$, we then have $1+3=4$.
  • If $3 \in \bar{A}$, we then have $2+3 = 5$.

• $1 \in \bar{A}$. This means $2 \in A$. This means $4 \in \bar{A}$. This means $5 \in A$.

  • If $3 \in \bar{A}$, we then have $1+3=4$.
  • If $3 \in A$, we then have $2+3 = 5$.

Note that in the above argument we did not need $6$ at all. Hence, $A$ can be a subset of $\{1,2,3,4,5\}$.

Answer 2

Assume otherwise. Wlog. $1\in A$. Then $A$ cannot contain consecutive numbers. Especially, $2=1+1\notin A$. Hence we must have $4=2+2\in A$, hence $3=4-1,5=4+1\notin A$, but then $\{2,3,5\}\subseteq \overline A$.

Note that $6$ was not used at all.