most recent 30 from http://mathoverflow.net 2013-05-25T17:07:17Z http://mathoverflow.net/feeds/question/86881 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86881/is-the-inverse-operation-on-an-ordered-division-ring-continuous Ricky Demer 2012-01-28T05:20:13Z 2012-01-28T05:20:13Z

<p>Let $R\hspace{.005 in}$ be a <a href="http://en.wikipedia.org/wiki/Division_ring" rel="nofollow">division ring</a>. $\;\;$ Let $\:\leq\:$ be a total order on $R\hspace{.005 in}$ such that for all elements $x,y,z$ of $R$ :</p> <p>if $\: x\leq y \:$ then $\:\: x+z\:\leq\:y+z \:\:$ <br> if $\;\;\; 0\leq x \:$ and $\: 0\leq y \;\;\;$ then $\:\: 0\:\leq\:x\cdot y \:\:$ <br><br><br> Define $\mathcal{T}\hspace{.05 in}$ to be the order topology on $R\hspace{.005 in}$. <br> Define <code>$\;\; f \: : \: (R-\{0\}) \: \to \: (R-\{0\}) \;\;$</code> by $\;\; f(x)\cdot x \: = \: 1 \: = \: x\cdot f(x) \;\;$. <br><br><br><br> Does it follow that $f\hspace{.02 in}$ is continuous with respect to the subspace topology from $\: \langle R\hspace{.01 in},\mathcal{T}\hspace{.06 in}\rangle \:$ ? <br><br></p>