Skip to main content
5 events
when toggle format what by license comment
Jun 12, 2016 at 0:22 answer added Pedro Lauridsen Ribeiro timeline score: 2
Jul 8, 2015 at 18:15 answer added Alexander Schmeding timeline score: 3
Jul 8, 2015 at 15:44 comment added Igor Khavkine Not sure about a reference, but the proof is straight forward. Given $u(x)$ with compact support and $m(x) > 0$, we have $|u-(1+\varepsilon)(u+v)| < m$, provided $|v| < m/2$ and $|\varepsilon| < \min(1,\max(|u|/m)^{-1})$, which shows that multiplication is continuous at $(1,u)$. Rescale to get continuity at any $(k,u)$ and treat the case $k=0$ specially in a similar way.
Jul 8, 2015 at 14:58 review First posts
Jul 8, 2015 at 15:06
Jul 8, 2015 at 14:55 history asked WhitneyTopology CC BY-SA 3.0