most recent 30 from http://mathoverflow.net 2013-05-23T05:30:21Z http://mathoverflow.net/feeds/user/17682 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82634/global-invertibility/82641#82641 stephanos 2011-12-04T18:20:11Z 2011-12-04T18:20:11Z

<p>Let $f:\mathbb{R}^n\rightarrow M\subseteq\mathbb{R}^n$ be the differentiable transformation, with $M=f(\mathbb{R}^n)$. If $M\neq\mathbb{R}^n$, then obviously $f$ isn't an global diffeomorphism of $\mathbb{R}^n$. But it is global diffeomorphism between $\mathbb{R}^n$ and $M$.</p> <hr> <p>*<em>This was intended to be a comment not an answer, but I cannot comment, sorry. *</em></p>

http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q/82638#82638 stephanos 2011-12-04T17:52:27Z 2011-12-04T17:52:27Z

<p>No, there is none. Suppose that there was such an $f$. Then for every $a\in\mathbb{Q}$ the function $f(a,\cdot):\mathbb{Q}\rightarrow \mathbb{Q}$ would be an injection. So there should infinitely many injections from $\mathbb{Q}$ to $\mathbb{Q}$, that have disjoint images. But there are only two possibilities for their limits when $x\rightarrow +\infty$. Contradiction.</p>

http://mathoverflow.net/questions/75361/bounded-variation stephanos 2011-09-13T22:30:04Z 2011-09-13T22:30:04Z

<p>Let $\Omega$ be an open subset of $\mathbb{R}^n$. Can $BV(\Omega)$ be first countable? If yes, do I have to assume something more for $\Omega$ than just being open?</p>