# If all infinite r.e. languages have an infinite recursive subset, then do co-r.e. languages not have such subsets?

If all infinite r.e. languages have an infinite recursive subset, then can we logically take co-r.e. languages to not have such subsets by complemence?

In one word: no. It doesn't follow logically. It is also definitely not the case that no infinite co-r.e. language can have infinite recursive subsets: every infinite recursive set, say $0^*$, is also r.e. and co-r.e.
It is not too difficult to prove that such sets, called co-simple, exist, following the footsteps of Post (1944). Instead of constructing an infinite co-r.e. set which doesn't have an infinite r.e. subset, we will construct a co-infinite r.e. set $S$ which intersects every infinite r.e. set. Simulate all Turing machines, and for Turing machine $e$, put in $S$ the first $x > 2e$ such that $e$ halts on $x$ (if any); here we think of $e,x$ as integers. This is clearly an r.e. set, which clearly intersects every infinite r.e. set. It is co-infinite since for every $n$, we put in $S$ at most $n/2$ integers, at most one from each of the programs $1,\ldots,n/2$.