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Let $R\hspace{.005 in}$ be a division ring. $\;\;$ Let $\:\leq\:$ be a total order on $R\hspace{.005 in}$ such that for all elements $x,y,z$ of $R$ :

if $\: x\leq y \:$ then $\:\: x+z\:\leq\:y+z \:\:$
if $\;\;\; 0\leq x \:$ and $\: 0\leq y \;\;\;$ then $\:\: 0\:\leq\:x\cdot y \:\:$


Define $\mathcal{T}\hspace{.05 in}$ to be the order topology on $R\hspace{.005 in}$.
Define $\;\; f \: : \: (R-\{0\}) \: \to \: (R-\{0\}) \;\;$ by $\;\; f(x)\cdot x \: = \: 1 \: = \: x\cdot f(x) \;\;$.



Does it follow that $f\hspace{.02 in}$ is continuous with respect to the subspace topology from $\: \langle R\hspace{.01 in},\mathcal{T}\hspace{.06 in}\rangle \:$ ?

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    $\begingroup$ I think you only have to work through the corresponding proof for ordered fields. So start with a net $x_n$ converging to some $x\ne 0$. You define the absolute value $|x|$ to be the unique element in $\{ \pm x\}$ which is $\le 0$. The only point I see, where you feel the non-commutativity, is when you look at $$ |x_n^{-1}-x^{-1}|=|x_n^{-1}||x-x_n||x^{-1}| $$ You show that $|x_n^{-1}|$ is eventually bounded and you're done. $\endgroup$
    – user1688
    Commented Jan 29, 2012 at 10:38
  • $\begingroup$ I suppose that might mean this shouldn't have been asked here. $\:$ I'll point out, though, that the proof I knew for ordered fields was to basically showed that the inverse operation is differentiable, which I think uses commutativity a lot more. $\;\;$ $\endgroup$
    – user5810
    Commented Jan 30, 2012 at 4:48

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