Question: Let $\varepsilon>0$ and $N\in\omega$ be sufficiently large (depending on $\varepsilon$).
Let $h:\subseteq N\rightarrow N$ be such that $h(B)\notin B$ for all $B\subsetneq N$.
Must there be $B\_0\subsetneq B\_1\subseteq N$ such that $|B\_1|\leq \varepsilon N$ and $h(B\_0)=h(B\_1)$?


More generally,


Let $a\in\omega,\varepsilon>0$ and $N\in\omega$ be sufficiently large (depending on $a,\varepsilon$).
Let $h:\subseteq N\rightarrow N$ be such that $h(B)\notin B$ for all $B\subsetneq N$.
Must there be $B\_0\subsetneq B\_1\subsetneq\cdots\subsetneq B\_{a-1}\subseteq N$ such that $|B\_{a-1}|\leq \varepsilon N$ and $h(B\_0)=\cdots= h(B\_{a-1})$?



Ramsey style theorem with unbounded colors
No, this is false if $\varepsilon<1/2$. Here is the construction for one $h$ that satisfies the conditions:



> 
> Let $h(B)$ be the $|B|$-th smallest element of $N\setminus B$.
> 
> 
> 


This way if $B\_1\subsetneq B\_2$, then $h(B\_1)<h(B\_2)$.


