# Union of powers of a well-ordered set is well-ordered.

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.

• 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. Jan 14, 2022 at 9:52
• @FlorianR yes sorry, I have corrected it Jan 14, 2022 at 15:41
• 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. Jan 15, 2022 at 5:15
• 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 Jan 16, 2022 at 4:31
• 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 $k, s^k = (0,k) < (1,0)$. Jan 20, 2022 at 21:18