User:Dfeuer/Totally Ordered Division Ring with Order Topology is Topological Division Ring

Conjecture
Let $(R,+,\circ,\preceq)$ be a totally ordered division ring.

Let $\tau$ be the $\le$-order topology over $R$.

Then $(R,+,\circ,\tau)$ is a topological division ring.

Corollary
A totally ordered field with the order topology is a topological field.

Discussion
$(R,+,\tau)$ is a topological group by Totally Ordered Group with Order Topology is Topological Group.

Let $R'$ be the set of strictly positive elements of $R$.

Then $(R',\circ,\le)$ is an ordered group. (EXPLAIN)

Let $\tau'$ be the $\le$-order topology over $R'$.

By Totally Ordered Group with Ordered Topology is Topological Group, $(R',\circ,\tau')$ is a topological group.

Since $R'$ is an interval in $R$, $\tau'$ is the $\tau$ subspace topology on $R'$. (EXPLAIN)

Thus multiplication on $R$ is continuous at each point of $R' \times R'$ and the inverse function, $\phi$, of $R$ is continuous at each point of $R'$. (EXPLAIN).

By inverse of negative, Inversion Mapping Reverses Ordering in Ordered Group, etc., $\phi$ and multiplication are continuous on strictly negative numbers as well (EXPLAIN)

The remaining problem is to show that multiplication is continuous at each $(x,0)$ and $(0,y)$.