Timeline for Can Cantor set be the zero set of a continuous function?

8 events

when toggle format	what		by	license	comment
S Jul 9, 2018 at 17:32	history	suggested	CommunityBot	CC BY-SA 4.0	Formatted equations.
Jul 9, 2018 at 13:35	review	Suggested edits
S Jul 9, 2018 at 17:32
May 16, 2011 at 18:24	comment	added	skeptical scientist		Each bump in the bump function has width 1/3^n. If the height is h(n), then the largest value of the kth derivative of of the stage n bumps will be 3^{kn}h(n) times the largest value of the kth derivative of the original bump function. If you want these to converge uniformly, the exact condition you need is that 3^{kn}h(n) converges to 0. So the necessary and sufficient condition on h to make the construction work is that for all a>0, h(n) is eventually less than a^n. So yes, 2^{-2^n} will work.
May 9, 2010 at 19:41	vote	accept	pinaki
May 9, 2010 at 19:21	comment	added	Roland van der Veen		@Harald: Thanks you're absolutely right. I meant to say 2^{-2^n}. That should work.
May 9, 2010 at 19:20	history	edited	Roland van der Veen	CC BY-SA 2.5	corrected height of the bump
May 9, 2010 at 19:03	comment	added	Harald Hanche-Olsen		If you work through the details, I suspect you may find you need the bumps at step $n$ to have heights decreasing faster than $2^{-n}$. The reason is that the bump has to fit into an interval of length $3^{-n}$, and the resulting squeeze makes the derivatives big (relatively speaking).
May 9, 2010 at 18:33	history	answered	Roland van der Veen	CC BY-SA 2.5