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Neat. Did you do it by making such a function for the open ball, and then, given an open set defined as a countable union of open balls, define a sequence of s
by SamReidHughes 11y ago
Neat. Did you do it by making such a function for the open ball, and then, given an open set defined as a countable union of open balls, define a sequence of such functions scaled appropriately so that their derivatives all converge uniformly?
Edit: Looks like you can make that work, according to the answer at https://math.stackexchange.com/questions/791248/every-closed-subset-e-subseteq-mathbbrn-is-the-zero-point-set-of-a-smooth https://math.stackexchange.com/questions/791248/every-closed...
- graycat 11y agoI took a countable dense set in R^n - C, x_j, j = 1, 2, ..., and for each j took the closed circle with center x_j and radius d(x_j, C) (distance from x_j to C which is well defined due to compactness). Then I defined a function g_j: R^n --> R 0 outside the circle and on the circle with values on the radii much as in the function Milnor, Rudin, and I mentioned. So, g_j is infinitely differentiable, strictly positive on the interior of the circle at x_j, and 0 otherwise. Then I added the functions g_j in a convergent way, again using the exponential trick. The resulting function was as desired, 0 on C, strictly positive otherwise, and infinitely differentiable.