Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2010 / A / 2 / c / Solution

No. Colour the positive integers by alternating dyadic blocks:Every positive integer $x$ and its double $2x$ have opposite colours. A nonempty set closed under addition contains both $x$ and $x+x=2x$, so no such set can lie in one class. In particular there is no infinite monochromatic additive subsemigroup.

The suggested arithmetic-progression argument gives another way to see the obstruction. If $a\in S$ and $S$ is closed under addition, then $a,2a,3a,\dots$ all lie in $S$. But every infinite arithmetic progression of positive step $d$ meets every sufficiently long dyadic block, because a block of at least $d$ consecutive integers contains a representative of the progression's residue class. Both colours contain arbitrarily long blocks, so neither colour contains an infinite arithmetic progression.