Is $\mathbb{R}$ the only complete ordered Abelian group? (Reprise.)

Question

This question has been asked here already. This answer is positive, but the proof is only roughly outlined. While searching for the details, I found this paper. There, the same question is answered, by a very detailed but slightly different proof; in particular, the Author makes heavy use of continuity, after having introduced a topology.

Since in the original question there is no mention of topology whatsoever, I guess a proof should be possible without using continuity concepts. I have tried to dig into the details; in particular, in the mentioned paper, Proposition 1.6, they say:

Let $(G,◦,≺)$ be a non-trivial complete totally ordered Abelian group, let $u$ be an element in $G$ and let $n$ be a non-zero natural number. Then, the equation $x^{◦n} = u$ has a unique solution in $G$. Here, as above, $x^{◦n} = x◦\ldots◦x$ ($n$ times).

The proof for the existence is that $x^{◦n}$ is continuous and thus maps intervals into intervals; based on Proposition 1.5, just above, this forces $x^{◦n}$ to be onto.

I haven't been able to figure out how to prove this existence without invoking continuity. In the answer to the question mentioned above, to my understanding this is not even foreseen.

Any comment is warmly appreciated.

Answer 1

First, by the monotonicity of $\circ$, if a solution exists, it is unique, for $x \prec y \implies x^{\circ n} \prec y^{\circ n}$. Next, we can assume $e \prec u$, since $x^{\circ n} = u \iff (x^{-1})^{\circ n} = u^{-1}$ and $e \prec u \iff u^{-1}\prec e$.

Now, for $n = 1$ the assertion is trivial, so we assume $n \geqslant 2$. Let $S(u,n) = \{ x \in G : e \prec x \land x^{\circ n} \prec u\}$. It is easy to see that $S(u,n)$ is bounded above: $u \prec u^{\circ 2} \preccurlyeq u^{\circ n}$, hence $x \prec u$ for all $x \in S(u,n)$. Showing $S(u,n) \neq \varnothing$ is more work, but it suffices to show $S(u,2) \neq \varnothing$. For if we know $S(u,2) \neq \varnothing$ for every $e \prec u$, take $u_1 \in S(u,2)$, then $u_2 \in S(u_1,2)$ and so on, and for $2^k \geqslant n$ we have $u_k \in S(u,2^k) \subset S(u,n)$.

Since $G$ is not discrete, for every $e \prec u$ there is $e \prec v \prec u$. If $v^{\circ 2} \prec u$, we're done. If $v^{\circ 2} = u$, any $e \prec w \prec v$ will do. So we have to consider the case $u \prec v^{\circ 2}$. Let $w = u \circ v^{-1}$. Then $e \prec w$ (since $v \prec u$), and $$w \circ w = u \circ (u \circ v^{\circ (-2)}) \prec u\,,$$ hence $w \in S(u,2)$.

Having seen that $S(u,n)$ is nonempty and bounded above, let $y = \sup S(u,n)$. Evidently, the expectation is $y^{\circ n} = u$. Suppose it weren't so. If $y^{\circ n} \prec u$, take any $w \in S(u\circ y^{\circ (-n)}, n)$ and note \begin{equation} (y\circ w)^{\circ n} = y^{\circ n} \circ w^{\circ n} \prec y^{\circ n} \circ \bigl(u \circ y^{\circ (-n)}\bigr) = u\,, \end{equation} i.e. $y\circ w \in S(u,n)$. But $y\prec y\circ w$, so $y$ is not an upper bound of $S(u,n)$. This contradiction shows $u \preccurlyeq y^{\circ n}$. And if we had $u \prec y^{\circ n}$, we could take $w \in S(y^{\circ n} \circ u^{-1}, n)$ and conclude \begin{equation} (y\circ w^{-1})^{\circ n} = y^{\circ n} \circ w^{\circ (-n)} \succ y^{\circ n} \circ \bigl(u \circ y^{\circ (-n)}\bigr) = u\,, \end{equation} hence $y \circ w^{-1}$ would be an upper bound of $S(u,n)$, and that contradicts the definition of $y$, for $y \circ w^{-1} \prec y$.

Thus $\bigl(\sup S(u,n)\bigr)^{\circ n} = u$ is proved for all $u \succ e$.