User:Dfeuer/Totally Ordered Field with Order Topology is Topological Field

Theorem
Let $(F, +, \circ, \le)$ be a totally ordered field.

Let $\tau$ be the $\le$ order topology over $F$.

Then $(F,+,\circ,\tau)$ is a topological field.

Subtraction is continuous
Let $a, x, y \in F$ and suppose that $a < x - y$.

Let $r = x-y-a$, so $r>0$.

Let $p = x - \dfrac r {1+1}$ and let $q = y + \dfrac r {1+1}$.

Then $(x,y) \in {\uparrow}(p) \times {\downarrow}(q)$, and
 * $(x',y') \in {\uparrow}(p) \times {\downarrow}(q) \implies x' - y' > x-y-\dfrac r {1+1}-\dfrac r {1+1} = x-y-r = a$

A similar argument holds for rays in the opposite direction.

Thus subtraction is continuous.

multiplication is continuous
Let $x \times y > a$.

See if the semi-related discussion at Stack Exchange will provide a technique usable in this context. I'm guessing yes.