Nobody can even catalog the set of all subsets of positive integers closed under addition (aka subsemigroups of the infinite cyclic semigroup). See Repnitskii, Vladimir On subsemigroup lattices without nontrivial identities. Algebra Universalis 31 (1994), no. 2, 256–265.

Update Although individually subsemigroups of $\mathbb{Z}$ are "easy" - they consist of an arithmetic progression plus a finite "garbage", the subsemigroups of $\mathbb{Q}_+$ are already very complicated. Every cancelative countable commutative semigroup without torsion (and there are lots of those) embeds into $\mathbb{R}$. That is because it embeds into a commutative countable group which, in turn, embeds into a product of copies of $\mathbb{Q}$ which is a subgroup of $\mathbb{R}$.

Nobody can even catalog the set of all subsets of positive integers closed under addition (aka subsemigroups of the infinite cyclic semigroup). See Repnitskii, Vladimir - On subsemigroup lattices without nontrivial identities. Algebra Universalis 31 (1994), no. 2, 256–265.

Update Although individually subsemigroups of $\mathbb{Z}$ are "easy" — they consist of an arithmetic progression plus a finite "garbage", the subsemigroups of $\mathbb{Q}_+$ are already very complicated. Every cancelative countable commutative semigroup without torsion (and there are lots of those) embeds into $\mathbb{R}$. That is because it embeds into a commutative countable group which, in turn, embeds into a product of copies of $\mathbb{Q}$ which is a subgroup of $\mathbb{R}$.

Nobody can even catalog the set of all subsets of positive integers closed under addition (aka subsemigroups of the infinite cyclic semigroup). See Repnitskii, Vladimir On subsemigroup lattices without nontrivial identities. Algebra Universalis 31 (1994), no. 2, 256–265.

Nobody can even catalog the set of all subsets of positive integers closed under addition (aka subsemigroups of the infinite cyclic semigroup). See Repnitskii, Vladimir On subsemigroup lattices without nontrivial identities. Algebra Universalis 31 (1994), no. 2, 256–265.

Update Although individually subsemigroups of $\mathbb{Z}$ are "easy" - they consist of an arithmetic progression plus a finite "garbage", the subsemigroups of $\mathbb{Q}_+$ are already very complicated. Every cancelative countable commutative semigroup without torsion (and there are lots of those) embeds into $\mathbb{R}$. That is because it embeds into a commutative countable group which, in turn, embeds into a product of copies of $\mathbb{Q}$ which is a subgroup of $\mathbb{R}$.