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

Conjecture
Let $(G, \circ, \le)$ be a totally ordered group.

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

Then $(G, \circ, \tau)$ is a topological group.

Discussion
Let $(x,y) \in \mu^{-1}(\uparrow z)$.

Case 1
Suppose $\exists y': y > y' > z \circ x^{-1}$. Then it can be shown that $(x,y) \in \uparrow(z\circ y'^{-1})\times \uparrow(y') \subseteq \mu^{-1}(\uparrow z)$.

Case 2
Suppose on the other hand that $\{y' \in G: y > y' > z \circ x^{-1} \} = \varnothing$.

We have $\uparrow (z\circ x^{-1}) = \bar\uparrow y$ by total ordering, where $\bar\uparrow$ is weak upper closure.

Then $(x,y) \in \uparrow(z \circ y^{-1})\times \uparrow (z\circ x^{-1})$ and also, that set is in $\mu^{-1}(\uparrow z)$ (here we use the set equality established). Done.