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 $(0,y)$.

This needs considerable tedious hard slog to complete it. In particular: negatives and zero for multiplication To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.