Let $(A,\leq)$ be a totally ordered abelian group, and $\Gamma\subseteq A$ be a set of nonnegative elements, such that it is wellordered by $\leq$. Is it true then that the semigroup $S$ generated by $\Gamma$ is again wellordered?
Thoughts so far:
- Positivity hypothesis is obviously necessary, because for any negative $a\in A$, we have that $n\cdot a$ is an infinite decreasing sequence.
- I can show that for a wellordered $\Gamma$, it is also true that $\Gamma+\Gamma$ is wellordered, which easily extends to sets $\Gamma_n:=\Gamma+\Gamma+\ldots+\Gamma$ ($\Gamma$ is added $n$ times).
- We assume that $0\in \Gamma$, so that $\Gamma_n\subseteq \Gamma_{n+1}$.
- To show that $S$ is wellordered, we have to show that any $B\subseteq S$ has a smallest element; we can assume without loss of generality that $\Gamma\cap B$ is nonempty (extending $\Gamma$ to some $\Gamma_n$ if necessary).
- With the above assumptions, existence of a minimal element of $B$ is equivalent to the statement that the sequence $b_n:=\min (\Gamma_n\cap B)$ stabilizes.
The motivation of the question is to show that the ring of Hahn series $K((X^A))$ (with $K$ a field) is a field, but even if there's a simpler way to show it, I'm curious about the question on its own.
Thoughts? Hints?