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

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

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

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

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

Then $x \circ y > z$ by the definition of strict upper closure.

Thus by compatibility:
 * $x > z \circ y^{-1}$
 * $y > x^{-1} \circ z$

Case 1
Suppose $\exists y': y > y' > x^{-1} \circ z$.

Then $x \circ y' > z$ by compatibility, so
 * $x > z \circ y'^{-1}$ by compatibility.

Thence, $(x,y) \in {\uparrow}(z\circ y'^{-1})\times {\uparrow}(y') \subseteq \mu^{-1}({\uparrow} z)$.

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

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

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

Inverse is continuous
Follows from $\operatorname{inv}^{-1}(\uparrow z) = \downarrow z^{-1}$.


 * It appears that on resolving stuff without using $G$ is abelian, I get terms like $y'^{-1} \circ z \circ y'$. --Lord_Farin (talk) 21:34, 28 January 2013 (UTC)
 * For now, I added Abelian to the premises. If you change your mind we can take it back out.
 * Speaking of things, we should check to be sure total order is really necessary. There are one or a few ways to make topologies out of partial orders and especially lattices. --Dfeuer (talk) 21:41, 28 January 2013 (UTC)


 * I think I have fixed the argument to arbitrary groups by passing to $x^{-1} \circ z$ which is more natural in any case. --Lord_Farin (talk) 22:49, 28 January 2013 (UTC)