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No one's answered this question so here's a simpler question: Has anyone described or catalogued all sets of non-negative real numbers that are closed under addition? In particular, how about those that are topologically closed?

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You can take any closed set of positive reals, close it with respect to addition. After that you can include $0$ if you want to. The only remaining set will be $[0,\infty)$. –  Anton Petrunin Apr 2 '12 at 22:31
    
In particular, if that closed set $C$ has a minimum positive element, then it generates a closed additive semigroup $S:= C\cup (C+C) \cup (C+C+C)\dots$. If $C$ has no minimum positive element, it generates a dense semigroup. –  Pietro Majer Apr 2 '12 at 22:42
    
I don't understand either comment. Help? –  Benjamin Steinberg Apr 2 '12 at 22:44
    
Anton’s comment gives a recipe how to construct all closed subsets of $\mathbb R^+$ closed under addition. –  Emil Jeřábek Apr 2 '12 at 22:53
    
What does he mean by the only set remaining? Also does one not have to close again under the topology? –  Benjamin Steinberg Apr 2 '12 at 22:58
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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 $\mathbb{R}$.

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This is what I meant by in some sense in my answer. There is no catalog but rather an axiomatization of embedability. –  Benjamin Steinberg Apr 2 '12 at 22:41
    
@Ben: Michael Hardy wanted a catalog. That is impossible. Individually subsemigroups of $\mathbb{Z}_+$ are not too difficult to describe: they are virtually arithmetic progressions. Subsemigroups of $\mathbb{R}_+$ and even $\mathbb{Q}_+$ are much harder. An interesting question is to describe subsemigroups of $\mathbb{Q}_+$ that are residually finite. I spent quite some time on that when I was an undergraduate student - without much success. Some non-trivial problems from descriptive topology appeared, as far as I remember. –  Mark Sapir Apr 2 '12 at 22:51
    
He also said description which is why I linked to the above paper. –  Benjamin Steinberg Apr 2 '12 at 22:59
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Look at http://www.springerlink.com/content/t612l0625052371x/ It would seem in some sense to answer the question.

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