I would like to know if it is possible to write the natural numbers $\mathbb N$ as an inductive limit of finite monoids such that one recovers the natural multiplication of the natural numbers. Of course, there is only one finite submonoid of the natural numbers with its natural multiplication, namely the trivial monoid $\{ 1 \}$.

Nevertheless one can for example try to do other things, such as equipping finite subsets with another multiplication and hoping that one recovers the natural multiplication in the direct limit. My first idea was the following. Define for $n\in \mathbb N$ the set $$I_n = \{ m \in \mathbb N \ | \ m | n\}$$ Then it is not difficult to see that $I_n$ becomes a commutative monoid by setting $$a \circ_n b = \gcd(ab,n)$$ Moreover one can show that for $d \in \mathbb N$ we get a homomorphism of monoids $$\phi_{n,nd} : I_n \to I_{nd}$$ by setting $$a \mapsto a \frac d {\gcd(n,d)} \gcd(a,\gcd(n,d))$$ But, unfortunately, we don't get an inductive system (when we try to order the indexing set by divisibility), namely we have $$\phi_{n,ndd'} \neq \phi_{nd,ndd'} \circ \phi_{n,nd}$$

(The idea of my construction is of course, that for $a,b \in \mathbb N$ small compared to $n \in \mathbb N$ the product $a\circ_nb$ is exactly what it should be. Although I admit that the homomorphisms $\phi_{n,nd}$ wouldn't give rise to the natural numbers, rather something like the Prüfer group... So the problem is really related to defining the correct transition functions...)

My questions are now if

1) my question is stupid, in that one can easily show that the natural numbers $\mathbb N$ with its natural multiplication cannot be described as an inductive limit of finite monoids

2) one can show that it is possible to do so (personally I think it is not possible...)

Thank you very much in advance!


Any direct limit of sets can be expressed as a direct limit of sets of no bigger cardinality with all maps being inclusions. But there is, as you say, only one finite submonoid of the multiplicative monoid $\mathbb{N}$.

From a category theory perspective, the natural numbers can be defined (up to isomorphism) as the "initial object" in the category of all diagrams $1\rightarrow X\rightarrow X$. This "initial object" feature of $\mathbb{N}$ makes it much more amenable to inverse limits than direct limits.

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    $\begingroup$ In my question I am asking if it can happen, that the multiplication of the direct limit (of the $I_n$ for example) is "by accident" the natural multiplication of the natural numbers. As set the natural numbers can be written as a direct limit. I don't expect the finite subsets $I_n$ (which carry a different monoid structure) to be finite submonoids of the natural numbers. $\endgroup$ – Amateuresque Apr 20 '11 at 16:03
  • $\begingroup$ I don't know if my question makes sense, but of course I am aware of what you are saying. In any case I thank you for your answer! $\endgroup$ – Amateuresque Apr 20 '11 at 16:04
  • $\begingroup$ Well, my argument shows that the natural numbers under multiplication cannot be expressed as a direct limit of finite monads in the category of monads. This includes the case where the natural numbers are a subset of the limit, because any sub-monad of a direct limit of finite monads is itself a direct limit if finite monads. $\endgroup$ – Thomas Andrews Apr 20 '11 at 16:12
  • $\begingroup$ Of course you're right! thank you and sorry for forcing you struggling with my stupidity... $\endgroup$ – Amateuresque Apr 20 '11 at 16:21
  • $\begingroup$ The direct limit of monads does not have a multiplicative structure unless the amps between the monads are monad morphisms. If they are just functions, then, yes, you can get a set that is equivalent to $\mathbb{N}$, but how is its multiplicative structure supposed to be defined by the limit if the maps are not monad homomorphisms? You need to maybe clarify your definitions a bit. (For example, you use the terms "direct limit" alternately with "inductive limit." These terms are the same, but the casual switching between the two might indicate some confusion.) $\endgroup$ – Thomas Andrews Apr 20 '11 at 16:25

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