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

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

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

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

Multiplication is Continuous
Let $\mu: G \times G \to G$ be defined by
 * $\mu(x,y) = x \circ y$

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

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

Thus since $\le$ is compatible with $\circ$:
 * $(1)\quad x > z \circ y^{-1}$
 * $(2)\quad y > x^{-1} \circ z$

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.

Thus $(x,y) \in {\dot\uparrow}(z\circ y'^{-1})\times {\dot\uparrow}(y')$.

If $(p,q) \in {\dot\uparrow}(z\circ y'^{-1}) \times {\dot\uparrow}(y')$, then by Operating on Ordered Group Inequalities:
 * $p \circ q > z$

so
 * ${\dot\uparrow}(z\circ y'^{-1})\times {\dot\uparrow}(y') \subseteq \mu^{-1} ({\dot\uparrow} z)$.

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

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

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

$(x,y) \in {\dot\uparrow}(z \circ y^{-1})\times {\dot\uparrow} (x^{-1} \circ z)$ by $(1)$ and $(2)$.

Suppose that $(p,q) \in {\dot\uparrow}(z \circ y^{-1})\times {\dot\uparrow} (x^{-1} \circ z) = {\dot\uparrow}(z \circ y^{-1})\times {\bar\uparrow} y$.

Then by Operating on Ordered Group Inequalities:
 * $p \circ q > z$

so
 * $(p,q) \in \mu^{-1}({\dot\uparrow}z)$

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

By Dual Pairs, strict lower closure is dual to strict upper closure.

Thus by the Duality Principle, it follows that the preimage under $\mu$ of a strict lower closure is also open in $G \times G$.

By definition of order topology, the strict upper closures and strict lower closures of elements of $G$ form a subbase for $\tau$.

By the Continuity Test using Sub-Basis, we conclude that multiplication is continuous.

Inversion is continuous
Since the group inverse mapping, $\operatorname{inv}$, is its own inverse and Group Inverse Reverses Ordering in Ordered Group:
 * $\operatorname{inv}^{-1}({\dot\uparrow} z) = {\dot\downarrow} (z^{-1})$

Similarly:
 * $\operatorname{inv}^{-1}({\dot\downarrow} z) = {\dot\uparrow} (z^{-1})$

Since up-sets and down-sets of elements of $G$ form a subbase for $\tau$, the group inverse mapping is continuous.

Since multiplication and inversion are $\tau$-continuous, $(G,\circ,\tau)$ is a topological group.

Proof 2
Let $\psi: G \times G \to G$ be defined by
 * $\psi(x,y) = x^{-1} \circ y$

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

Then $x^{-1} \circ y > z$ by the definition of strict upper closure.

Thus since $\le$ is compatible with $\circ$:
 * $(1)\quad x^{-1} > z \circ y^{-1}$
 * $(2)\quad y > x^{-1} \circ z$

Case 1
Suppose $\exists y': y > y' > x \circ z$.

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

Since Group Inverse Reverses Ordering in Ordered Group:
 * $x < y' \circ z^{-1}$

Thus $(x,y) \in { \dot \downarrow}(y' \circ z^{-1}) \times { \dot \uparrow}(y')$.

If $(p,q) \in { \dot \downarrow}(y' \circ z^{-1}) \times { \dot \uparrow}(y')$, then by Group Inverse Reverses Ordering in Ordered Group:
 * $p^{-1} > z \circ y'^{-1}$

By Operating on Ordered Group Inequalities:
 * $p^{-1} \circ q > z$

so
 * ${ \dot \downarrow}(y' \circ z^{-1}) \times { \dot \uparrow}(y') \subseteq \psi^{-1} ({ \dot \uparrow} z)$.

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

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

We have ${ \dot \uparrow} (x \circ z) = {\bar\uparrow} y$ by Mind the Gap, where $\bar\uparrow$ is Up-Set.

$(x^{-1},y) \in { \dot \uparrow}(z \circ y^{-1})\times { \dot \uparrow} (x \circ z)$ by $(1)$ and $(2)$.

Thus $(x,y) \in { \dot \downarrow}(y \circ z^{-1}) \times { \dot \uparrow} (x \circ z)$

Suppose that $(p,q) \in { \dot \downarrow}(y \circ z^{-1}) \times { \dot \uparrow} (x \circ z) = { \dot \downarrow}(y \circ z^{-1}) \times {\bar\uparrow} y$.

Then by Group Inverse Reverses Ordering in Ordered Group: $(p^{-1},q) \in { \dot \uparrow}(z \circ y^{-1}) \times { \dot \uparrow} (x^{-1} \circ z) = { \dot \uparrow}(z \circ y^{-1}) \times {\bar\uparrow} y$

Then by Operating on Ordered Group Inequalities:
 * $p^{-1} \circ q > z$

so
 * $(p,q) \in \psi^{-1}({ \dot \uparrow}z)$

Thus
 * $(x,y) \in { \dot \downarrow}(y \circ z^{-1})\times { \dot \uparrow} (x^{-1} \circ z) \subseteq \psi^{-1}({ \dot \uparrow} z)$.

By Dual Pairs, strict lower closure is dual to strict upper closure.

Thus by the Duality Principle, it follows that the preimage under division of a strict lower closure is also open in $G \times G$.

Since the strict upper closures and strict lower closures of elements of $G$ form a subbase for $\tau$, we conclude that $\psi$ is continuous.