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That such a function f exists is pretty standard, right? I know we discussed it during my undergrad studies. See e.g. Counterexamples in Analysis. Also, a poin

by petters 10y ago

That such a function f exists is pretty standard, right? I know we discussed it during my undergrad studies. See e.g. Counterexamples in Analysis. Also, a point is a closed set, so you probably want non-empty interior.

graycat 10y ago
No, it's not standard. I doubt that it is in Counterexamples in Analysis, one of my favorite books. No, don't need a non-empty interior for the closed set. The closed set can even be empty. The function is to be zero on the closed set and positive otherwise. For a single point could use just the squared distance to the point, i.e., a parabola. For a closed set like the Mandelbrot set and for the general case, have to try harder. For a fast version of the proof, outside the closed set is an open set. If it is empty, then just let the function f = 0. Else in that open set, pick a countable dense set. Roughly for each point in that countable dense set, solve the problem for that set, and then add all the countably many partial solutions in a convergent way. A key is a Baby Rudin exercise of a function g: R --> R where g(x) = 0 for x <= 0 g(x) > 0 for x > 0 and g is infinitely differentiable. Sure, g is based on an exponential. Then for f: R^2, at each of the countably many points, spin g to make a smooth hill where the bottom of the hill just touches the closed set C. The spin, being intuitive here, also works for f: R^n. Sure, for each of the countably infinitely many points, the spin, in g(z), z is the distance to the to the point. I'm being intuitive and sloppy here to be brief and easier to understand. Then in R^2, at the origin, for each positive rational of the form p/q in lowest terms, have a ray from the origin at angle p radians and length 1/q. That's a bizarre closed set. Then at the origin, the Zangwill and KTCQ don't agree, are independent. Here, again, I'm being brief.
- petters 10y ago
  Sorry, I misread the theorem earlier. Thanks!

by petters 10y ago

graycat 10y ago

No, it's not standard. I doubt that it is in Counterexamples in Analysis, one of my favorite books. No, don't need a non-empty interior for the closed set. The closed set can even be empty. The function is to be zero on the closed set and positive otherwise. For a single point could use just the squared distance to the point, i.e., a parabola. For a closed set like the Mandelbrot set and for the general case, have to try harder. For a fast version of the proof, outside the closed set is an open set. If it is empty, then just let the function f = 0. Else in that open set, pick a countable dense set. Roughly for each point in that countable dense set, solve the problem for that set, and then add all the countably many partial solutions in a convergent way. A key is a Baby Rudin exercise of a function g: R --> R where g(x) = 0 for x <= 0 g(x) > 0 for x > 0 and g is infinitely differentiable. Sure, g is based on an exponential. Then for f: R^2, at each of the countably many points, spin g to make a smooth hill where the bottom of the hill just touches the closed set C. The spin, being intuitive here, also works for f: R^n. Sure, for each of the countably infinitely many points, the spin, in g(z), z is the distance to the to the point. I'm being intuitive and sloppy here to be brief and easier to understand. Then in R^2, at the origin, for each positive rational of the form p/q in lowest terms, have a ray from the origin at angle p radians and length 1/q. That's a bizarre closed set. Then at the origin, the Zangwill and KTCQ don't agree, are independent. Here, again, I'm being brief.

petters 10y ago
Sorry, I misread the theorem earlier. Thanks!