When can complete dense linear orders be made into topological fields?

Question

By a "complete dense linear order," I mean a dense linear order in which every nonempty subset with an upper bound has a least upper bound. The canonical example, $\mathbb{R}$, is a topological field under the order topology. Question: if $L$ is a complete dense linear order, can we always define addition and multiplication operations on it which make it a topological field under its order topology?

Answer 1

No. Let $$L=\Big([0,1)\times[0,1)\Big)\setminus\{\langle 0,0\rangle\}$$ with the lexicographic order $\preceq$. $\langle L,\preceq\rangle$ is a dense linear order without endpoints, and it’s complete in the required sense. To see this, suppose that $\varnothing\ne A\subseteq L$, and $b\in L$ is an upper bound for $A$. Let $X=\{x\in[0,1):\exists y(\langle x,y\rangle\in A)\}$, and let $b=\langle x_0,y_0\rangle$. Clearly $x_0$ is an upper bound for $X$ in $[0,1)$; let $x_1=\sup X$. If $x_1\notin X$, then $\sup A=\langle x_1,0\rangle$. If $x_1\in X$, let $Y=\{y:\langle x_1,y\rangle\in A\}$; then $y_0$ is an upper bound for $Y$, we can let $y_1=\sup Y$, and $\sup A=\langle x_1,y_1\rangle$.

Suppose that we could define addition and multiplication operations making $L$ an ordered field. The map $x\mapsto -x$ would then be an order-isomorphism from $\langle L,\preceq\rangle$ to $\langle L,\succeq\rangle$. However, $\langle L,\preceq\rangle$ and $\langle L,\succeq\rangle$ are not order-isomorphic, so no such field structure is possible.

To see this, let $p=\left\langle 0,\frac12\right\rangle$. The ray $(\leftarrow,p)$ in $\langle L,\preceq\rangle$ is order-isomorphic to $\left(0,\frac12\right)$ in $\Bbb R$ and therefore does not contain uncountably many pairwise disjoint non-empty open intervals. Let $q=\langle x_0,y_0\rangle\in L$ be arbitrary. Then

$$\left\{\left(\left\langle x,\frac13\right\rangle,\left\langle x,\frac23\right\rangle\right):x_0<x<1\right\}$$

is an uncountable family of pairwise disjoint open sets in the ray $(q,\to)$ in $\langle L,\preceq\rangle$. In other words, there is no $q\in L$ such that the ray $(\leftarrow,q)$ in $\langle L,\succeq\rangle$ is order-isomorphic to the ray $(\leftarrow,p)$ in $\langle L,\preceq\rangle$, and therefore $\langle L,\preceq\rangle$ and $\langle L,\succeq\rangle$ are not order-isomorphic.