The [infinite Ramsey theorem](https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Infinite_Ramsey_theorem) implies that, if we color the $n$-element subsets of $N:=\{0,1,2,\ldots\}$ in a finite number of colors, then there will exist an *infinite* subset $A\subseteq N$ such that all $n$-element subsets of $A$ will be of the same color.

My question is about **vectors** in $N^n=\{(x\_1,\ldots,x\_n) \colon x\_i\in N\}$ instead of $n$-element subsets of $N$.

Say that a set $X\subseteq N^n$ is *dominating* if for every nonempty subset $S\subset \{1,\ldots,n\}$ of positions, there is a vector $x\in X$ such that
$$
\min\_{i\in S} x\_i > \sum\_{i\not\in S} x\_i.
$$
For example, the set $X=\{0,1\}^n$ is dominating, as well as is any set $X=\{a\_1,b\_1\}\times \cdots\times \{a\_n,b\_n\}$ with 
$$
\min\{b\_1,\ldots,b\_n\} > a\_1+\cdots+a\_n.
$$



> 
> **Question:** If we color the vectors in $N^n$ in a finite number of colors, will then at least one dominating set $X$ be monochromatic?
> 


The question seems so natural that it should be definitely investigated by someone. But I couldn't find any references, nor any counterexamples. 

My question is motivated by a rather "pragmatic" goal: to prove that randomness cannot speed-up dynamic programming. 









Infinite Ramsey theorem for strings (instead of sets)?
Color each vector $(x\_i)$ in any $k$ such that $x\_k=\max\limits\_{1\leq i\leq n} x\_i$. Then there is no dominating monochromatic set: if it had color $k$, then there is no $S$-dominating bector for $k\notin S$.


