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While studing a certain type of rings I was trying to solve the following exercise (which is what I need to prove that this ring is well defined):

Let $S$ be a well-ordered subset of an ordered group $(G,<)$ such that $g>1~\forall g\in S$. Prove that $\bigcup\limits _{n\in\mathbb{N}}S^n$ is well-ordered and that $\forall~g\in G$ the set $\lbrace n\in\mathbb{N}\mid g\in S^n\rbrace$ is finite.

Where well-ordered means that every non-empty subset has a unique minimum.

To prove that $\bigcup\limits _{n\in\mathbb{N}}S^n$ is well-ordered i tried the following:

Let $A\subset \bigcup\limits _{n\in\mathbb{N}}S^n$, then $A=A\cap \bigcup\limits _{n\in\mathbb{N}}S^n=\bigcup\limits _{n\in\mathbb{N}}(A\cap S^n)$. Let $I\subset\mathbb{N}$ be the set of indices such that $A\cap S^n\neq\emptyset$. Then $\forall~n\in I$ we have that $A\cap S^n$ is a non-empty subset of $S^n$, thus since $S^n$ is well-ordered there exists $a_n=\min\limits_{g\in A\cap S^n}\lbrace g\rbrace$. From this, one have constructed a set $\lbrace a_n\rbrace_{n\in I}$ which is bounded below by $1$. Hence, this set has a minimum $a=\min \lbrace a_n\rbrace_{n\in I}$, which is the minimum of $A$. Is this right?

For the second part I don't have any idea, I guess that we have to argue by contradiction and prove that if the set is infinite then there must be some $S^n$ such that it is not well-ordered, but I don't know how to do it.

Thanks for your help.

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  • $\begingroup$ Could there be an additional hypothesis? Because $G=\mathbb{Z}$ is ordered and $S=\{-1\}$ is well-ordered, but $\bigcup_{n \in \mathbb{N}} S^n = \mathbb{Z}_-$ is not. $\endgroup$
    – Florian R
    Jan 14, 2022 at 9:52
  • $\begingroup$ @FlorianR yes sorry, I have corrected it $\endgroup$
    – Marcos
    Jan 14, 2022 at 15:41
  • $\begingroup$ To show $\bigcup_n S^n$ is well-ordered, you have to show every non-empty subset has a least element. Showing the entire set has a least element is not sufficient. I sugest starting by proving for each $n, S^n$ is well ordered. $\endgroup$ Jan 15, 2022 at 5:15
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    $\begingroup$ For the first statement, let $\bar S = S \cup \{1\}$. Then $\bar S$ is also well-ordered and $\bar S^n \subset \bar S^{n+1}$ for all $n$. Since you've already proved $\bar S^n$ must be well-ordered for all $n$, this gives you that any finite union of the $bar S^n$ is well-ordered. Now figure out how to prove the full union is. For the second statement, if $g \in S^n$, it is the product of $n$ elements of $S$. Some of those elements $s$ will have $s^k \ge g$ for some $k$. Others may have $s^k < g$ for all $k$. Multiplying together the latter a $g'$ to which the same can be done. This has to end $\endgroup$ Jan 16, 2022 at 4:31
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    $\begingroup$ No. The Archimedean principle does not hold in general. For example, $\Bbb Z^2$ under addition and lexigraphically ordered: $(x, y) < (a,b) \iff (x < a) \vee (x = a \wedge y < b)$ ($G$ is not required to be well-ordered, only $S$). In this group "$1$" is $(0,0)$, and setting $s= (0,1)$, we have $s > (0,0)$. But for any $k, s^k = (0,k) < (1,0)$. $\endgroup$ Jan 20, 2022 at 21:18

1 Answer 1

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You can find an answer in the following articles :

  1. On Ordered Divisions Rings, by B. H. Neumann, in the Transactions of the American Mathematical Society (vol. 66) at p.206 where it was originally stated and proved I think. I have a picture of the demonstration of it if you want. The authors uses quite elementary arguments that relie on the notion of Archimedean classes on a (semi)group.

  2. Foundations of analysis over surreal number field, by N. L. Ailing (the PDF can be found not so difficultly on internet) at p.261. It is refered as Neumann's lemma. The demonstration is different but still uses elementary arguments that relies this time on convex subgroups.

I'll let you read it in the corresponding papers because both of these answers are really too long (but still elementary don't worry) to be re written here.

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