No one's answered the question cumulant problem 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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2$\begingroup$ 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 $[0,\infty)$. $\endgroup$– Anton PetruninCommented Apr 2, 2012 at 22:31
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1$\begingroup$ I don't understand either comment. Help? $\endgroup$– Benjamin SteinbergCommented Apr 2, 2012 at 22:44
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1$\begingroup$ What does he mean by the only set remaining? Also does one not have to close again under the topology? $\endgroup$– Benjamin SteinbergCommented Apr 2, 2012 at 22:58
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2$\begingroup$ @Benjamin: You have now answered your first question (the only sets not covered by the generic case of Anton's constructions are those which fail to be closed after removal of $0$, i.e., those that have $0$ as a limit point. You've show the only one such is $[0,\infty)$.) For the second question, if $C$ is a closed set of positive reals, its closure under addition $C^+$ is automatically topologically closed. To see this, let $c=\min(C)>0$, which exists as long as $C\ne\emptyset$ because $C$ is closed. Let $n>0$. Any point in $[0,nc]\cap C^+$ must be the sum of at most $n$ elements of $C$, ... $\endgroup$– Emil JeřábekCommented Apr 3, 2012 at 10:21
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2$\begingroup$ ... each of which must also lie in $[0,nc]$. Since a continuous image of a compact set is compact, the set of sums of $m$ elements of $[0,nc]\cap C$ is compact (and therefore closed) for each $m\le n$, hence $[0,nc]\cap C^+$ is closed. It follows that $C^+$ itself is closed. $\endgroup$– Emil JeřábekCommented Apr 3, 2012 at 10:25
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3 Answers
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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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$\begingroup$ This is what I meant by in some sense in my answer. There is no catalog but rather an axiomatization of embedability. $\endgroup$ Commented Apr 2, 2012 at 22:41
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$\begingroup$ @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. $\endgroup$– user6976Commented Apr 2, 2012 at 22:51
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$\begingroup$ He also said description which is why I linked to the above paper. $\endgroup$ Commented Apr 2, 2012 at 22:59
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Has anyone described or catalogued all sets of non-negative real numbers that are closed under addition?
In recent years, there has been a lot of work on the arithmetic of Puiseux monoids, that is, submonoids of the positive cone of the additive group of a totally ordered field $K$: The focus has been mostly on rational Puiseux monoids, where $K$ is the rational field (with its usual ordering); but there are also a few papers about the general case. For further details, I can only recommend to have a look at the work of Felix Gotti et al., starting with
- S.T. Chapman, F. Gotti, and M. Gotti, When Is a Puiseux Monoid Atomic?, Amer. Math. Monthly 128:4 (2021), 302-321.
This article is a pleasure to read and, together with
- S.T. Chapman, F. Gotti, and M. Gotti, Factorization invariants of Puiseux monoids generated by geometric sequences, Commun. Algebra 48:1 (2020), 380-396,
it offers a lucid introduction to (different aspects of) the arithmetic theory of monoids (especially from the point of view of the classical theory of factorization).
From the reading of these papers, it will become clear that even a classification of rational Puiseux monoids is more or less hopeless (consistently with a statement made in a previous answer). However, some kind of classification is possible if the focus is restricted to certain families of rational Puiseux monoids. In particular, you may want to have a look at Sect. 3 of
- A. Geroldinger, F. Gotti, and S. Tringali, On strongly primary monoids, with a focus on Puiseux monoids, J. Algebra 567:1 (2021), 310-345,
where the focus is on the "strongly primary" case (an additively-written, commutative monoid $H$ is strongly primary if, for every $a \in H$, there is an integer $n \ge 1$ such that the $n$-fold sumset of the maximal ideal $H \setminus H^\times$ is contained in the coset $a+H$, where $H^\times$ is the group of units). E.g., one can show (loc. cit., Theorem 3.4) that a rational Puiseux monoid $H$ with non-empty conductor $$ (H : \widehat H) := \{x \in \mathsf q(H) \colon x + \widehat H \subseteq H\} $$ is strongly primary iff it satisfies the ACC on principal ideals, iff it is a bounded-factorization monoid (i.e., every $x \in H$ is a finite sum of atoms and the atomic factorizations of $x$ are all bounded in length), iff $0$ is not a limit point of the non-zero elements of $H$. (Here, $\mathsf q(H)$ is the quotient group of $H$ and $\widehat H$ is the complete integral closure of $H$, i.e., the set of all $x \in \mathsf q(H)$ for which there is an element $a \in H$ such that $a + nx \in H$ for all $n \in \mathbb N$.)
answered Oct 9, 2022 at 11:59
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Look at Kist and Leestma - Additive semigroups of positive real numbers. It would seem in some sense to answer the question.
answered Apr 2, 2012 at 22:29