Let $F$ be an ordered field and give $F$ the order-topology. Then is $F$ a topological field (that is, are the operations
\begin{equation}
\begin{split}
+&:F\times F\to F\\
-&:F\to F\\
\times&:F\times F\to F\\
^{-1}&:F\setminus\{0\}\to F
\end{split}
\end{equation}
continuous)?
I checked that $+,-$ and $^{-1}$ are continuous. But I could not prove the continuity of $\times$. Here is my attempt to prove the continuity of $\times$. Let $(x_0,y_0)\in F\times F$ and let $\epsilon\in F$ be positive. I want to show that there is a positive $\delta\in F$ such that
\begin{equation}
|x-x_0|<\delta, |y-y_0|<\delta\quad \text{implies}\quad |xy-x_0y_0|<\epsilon
\end{equation}
where the absolute value is defined in the obvious way. If $\delta\in F$ is any positive element where $|x-x_0|<\delta, |y-y_0|<\delta$, then
\begin{equation}
\begin{split}
|xy-x_0y_0| &= |xy-xy_0+xy_0-x_0y_0|\\
&\leq |x||y-y_0| + |x-x_0||y_0|\\
&<(|x_0|+\delta)\delta + \delta|y_0|\\
&= \delta^2 + \alpha\delta
\end{split}
\end{equation}
where $\alpha = |x_0|+|y_0|$. Thus the only thing I need to prove is that for any positive $\epsilon\in F$ and any nonnegative $\alpha\in F$, there is a positive $\delta\in F$ such that
\begin{equation}
\delta^2+\alpha\delta<\epsilon\text{.}
\end{equation}
This is true if $F$ is a subfield of $\mathbb R$, but I cannot prove it when $F$ is an arbitrary ordered field.
Best Answer
It suffices to be able to choose $\delta$ such that $\delta^2<\epsilon$ and also to be able to choose $\delta$ such that $\alpha\delta<\epsilon$ (since then just take the minimum of the two $\delta$'s for $\epsilon'=\epsilon/2$). The latter is easy since you can just take $\delta=\frac{\epsilon}{2\alpha}$. For the former, if $\epsilon> 1$ you can take $\delta=1$, and otherwise any $\delta<\epsilon$ works.