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

Question

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

Answer 1

No. Take $L$ to be an infinite recursive language; then $L$ is co-r.e. and has an infinite recursive subset.

Added: Let $L$ be a non-recursive r.e. language over an alphabet $\Sigma_0$; $\Sigma_0^*\setminus L$ is not r.e. Let $\sigma$ be a symbol not in $\Sigma_0$, and let $\Sigma=\Sigma_0\cup\{\sigma\}$. $\Sigma^*\setminus L$ is not r.e., but it contains the infinite regular (hence recursive) language $\{\sigma\}^*$.

Answer 2

Unlike r.e. languages, there are infinite co-r.e. sets that don't have any infinite computable subsets.

One kind of example comes from maximal r.e. sets. These are r.e. sets that have an infinite complement, but the complement does not include any infinite r.e. set. Maximal r.e. sets can be built by a priority argument as a standard exercise. (They are called "maximal" because they are maximal elements below $\mathbb{N}$ in the lattice $\mathcal{E}^*$ of r.e. sets modulo finite sets, but this is not needed for this answer.)

The complement of a maximal r.e. set is, of course, an infinite co-r.e. set. But it doesn't contain any infinite r.e. set, much less an infinite computable set.