Timeline for is f a polynomial provided that it is "partially" smooth?
Current License: CC BY-SA 3.0
5 events
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| Apr 15, 2012 at 21:17 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
added 83 characters in body
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| Apr 15, 2012 at 15:33 | comment | added | Emilio Pisanty | Follow up: This answer shows you can. | |
| Apr 14, 2012 at 20:47 | comment | added | Emilio Pisanty | Exactly: the OP's partition of $(c,d)$ allows things as ugly as Aaron's partition, or even $$(0,1)=\bigcup_{k=1}^\infty \bigcup_{j=1}^\infty \left(\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j+1},\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j}\right)$$ though I'm not sure the process can be continued indefinitely. It's an interesting question, then: how far can this idea be pushed? How many points can there be that do not have a "next neighbour to the right"? Can every interval $(a_n,b_n)$ be made to not have a next neighbour? | |
| Apr 14, 2012 at 19:47 | comment | added | Aaron Tikuisis | It seems to me that there is a problem with the part where you say that $a_1$ and $b_1$ lie in the closure of some other interval. For example, say $b_1 = 0$, and $$ O \cap (0,1) = \bigcup_{k=1}^\infty \left(\frac1{k+1},\frac1{k}\right). $$ | |
| Apr 14, 2012 at 18:36 | history | answered | Matthias Ludewig | CC BY-SA 3.0 |