Timeline for Can Cantor set be the zero set of a continuous function?
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2020 at 22:48 | comment | added | R. van Dobben de Bruyn | @MichaelHardy: the claim of this answer (as well as the accepted answer) is that you can in fact choose a smooth function, which is part of the original question. | |
| Nov 29, 2010 at 2:11 | comment | added | Michael Hardy | .....Now I see: someone else already said this? So why are some people using more complicated proofs than that? | |
| Nov 29, 2010 at 2:07 | comment | added | Michael Hardy | ...where $d(x,y)$ is the distance from $x$ to $y$, so in case they're real numbers is just $|x - y|$. | |
| Nov 29, 2010 at 2:06 | comment | added | Michael Hardy | Can't this be proved for closed subsets $C$ of the real line (and maybe for metric spaces generally) just by saying $f(x) = \inf\{ d(x,y) : y \in C \}$? | |
| May 11, 2010 at 18:16 | comment | added | Petya | I construct function $f_i$ as follows: consider a locally finite collection $N_i$ of open balls such that the corresponding half-radii collection contains the given closed set $C$ as a subset. Let $U_i$ be a union of balls from $N_i$. I want that intersection of $U_i$ equals $C$. For each ball from $N_i$ I fix a smooth function equals to zero on a half-radii ball and 1 outside the ball. Product of such functions is well-defined ($N_i$ are locally finite) and it is $f_i$ by definition. After that a suitable sum $\sum a_if_i$ solve the initial problem, its zero level set is $C$. | |
| May 11, 2010 at 18:14 | comment | added | Petya | Robin, I do not assume that functions $f_i$ are compactly supported and supports could have nontrivial countable intersection, so $\sum a_if_i$ is not defined for any $a_i$! | |
| May 11, 2010 at 6:00 | comment | added | Robin Chapman | If you have smooth compactly supported functions $f_i$ on a locally finite collection of open sets then $\sum a_i f_i$ converges to a smooth function for any $a_i$. :-) To get nice locally finite covers we need to exploit the circle of ideas around paracompactness/partitions of unity. | |
| May 10, 2010 at 20:17 | comment | added | Petya | If there is a good covering (locally finite and such that balls of smaller radii also form a covering) by a countable system of closed (coordinate) balls, then for any sequence of smooth function $f_i$ on a manifold there is a sequence $a_i$ of positive numbers such that $\sum a_i f_i$ converges uniformly with any its derivatives in any ball from the covering. Right? | |
| May 10, 2010 at 9:07 | comment | added | Robin Chapman | Petya, I'm sure you're correct, but it's a bit more fiddly to ensure the uniform convergence of the sum of the derivatives of the bump functions without a convenient global coordinate system. | |
| May 10, 2010 at 3:22 | comment | added | Petya | One can avoid using the partition of unity for that, just summation of "1-bump" functions with suitable coefficients.. | |
| May 9, 2010 at 19:21 | comment | added | Harald Hanche-Olsen | Yes, that makes sense. | |
| May 9, 2010 at 19:10 | comment | added | Robin Chapman | It's true on any (paracompact) smooth manifold - you certainly need partitions of unity for that. | |
| May 9, 2010 at 19:06 | comment | added | Harald Hanche-Olsen | I admit I have never worked through the proof of this result. So I wonder, why partitions of unity? It seems to me that the local result is no easier than the global result. | |
| May 9, 2010 at 18:22 | history | answered | Robin Chapman | CC BY-SA 2.5 |