*(This question was asked a long time ago [on MSE](https://math.stackexchange.com/q/838283) but got no answer so far...)*


Let $E$ be an additively written cancellable commutative monoid with no non-trivial units. We furnish $E$ with the order defined by "$x\leq y$ if and only if there exists $z\in E$ with $y=x+z$", so that it is an ordered monoid. If we consider a subset $F\subseteq E$, we can ask whether it is noetherian *as an ordered set,* i.e., whether every nonempty subset of $F$ has a maximal element.


It is not hard to see that if $E$ is finitely generated, then the noetherian subsets of $E$ are precisely the finite ones. So, we could ask for the converse: If the noetherian subsets of $E$ are precisely the finite ones, is then $E$ finitely generated?


I think this is *not true,* but was not able to come up with a counterexample. So, here is my question:



> 
> What is an example of a monoid as above, but *not* finitely generated, whose noetherian subsets are precisely the finite ones?
> 
> 
> 



Finitely generated ordered monoids and noetherian subsets
If a monoid has the property (P) that all noetherian subsets are finite then it is finitely generated:


1. If a partially ordered set has property (P), then any subset has a minimal element: otherwise there is an infinite descending sequence in the subset, and the elements of a descending sequence form a noetherian subset.
2. Assume a monoid has property (P) and is not finitely generated. Let $x\_1$ be a minimal element of the monoid, let $x\_2$ be a minimal element not in the monoid generated by $x\_1$, let $x\_3$ be a minimal element not in the monoid generated by $x\_1$ and $x\_2$, etc. The $x\_i$ are pairwise incomparable because if $x\_i \lt x\_j$ then: if $i>j$ then $x\_i$ contradicts the minimality of $x\_j$; if $i<j$ with $x\_j = x\_i + y$ for some $y$ then $y$ contradicts the minimality of $x\_j$ ($y$ can't be in the monoid generated by $x\_1$,...,$x\_{j-1}$ else so would be $x\_j$). Thus {$x\_1, x\_2, ...$} would be an infinite noetherian subset, contradicting property (P).


If the monoid is totally ordered then it must be ℕ$x\_1$: $n{x\_1}<n{x\_1}+y<(n+1){x\_1}$ for some $n$ would imply $y<x\_1$.


