
Re: zero sets of infinitely differentiable functions
Posted:
Dec 9, 2013 10:28 AM


On Sun, 08 Dec 2013 15:11:22 0500, quasi <quasi@null.set> wrote:
>dullrich wrote: >>quasi wrote: >>>dullrich wrote: >>>>quasi wrote: >>>>>quasi wrote: >>>>>>dullrich wrote: >>>>>>>quasi wrote: >>>>>>> >>>>>>>>Let's try a special case ... >>>>>>>> >>>>>>>>Question: >>>>>>>> >>>>>>>>If P,Q are closed, nowhere dense subsets of R such that between >>>>>>>>any two distinct elements of P there is an element of Q, must >>>>>>>>there exist a differentiable function f:R > R such that >>>>>>>>f^(1)(0) = P and (f')^(1)(0) = Q? >>>>>>> >>>>>>>[...] >>>> >>>>I suspect the answer is yes, but nothing springs to >>>>mind for a proof. >>> >>>I have a proof in mind, but it's tricky. >> >>The answer is no. Let P = {0,1} and Q = [0,1]. > >The problem requires both P,Q to be nowhere dense.
Aargh. Sorry...
> >>If f = 0 on P and f' = 0 on Q then f = 0 on all of [0,1]. > >Thus, it's not a counterexample. > >>>I'll outline it if requested. > >quasi

