Timeline for Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points?
Current License: CC BY-SA 2.5
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| Mar 30, 2011 at 20:15 | history | edited | Jon Bannon | CC BY-SA 2.5 |
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| Mar 30, 2011 at 15:47 | vote | accept | Kate | ||
| Mar 30, 2011 at 14:26 | history | edited | Jon Bannon | CC BY-SA 2.5 |
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| Mar 30, 2011 at 13:48 | comment | added | Andreas Blass | @Michael: You don't know that such a neighborhood N contains only one turning point, but you do know that at most one point attains the maximum of f in N and at most one attains the minimum of f in N. Any other turning points, even if they're in N, will have had other neighborhoods N' assigned to them --- neighborhoods in which they achieve the maximum or minimum value of f. | |
| Mar 30, 2011 at 13:13 | comment | added | Michael Renardy | I do not follow this argument. You can choose the neighborhoods to be rational intervals, but how do you know that such a neighborhood contains only one turning point? | |
| Mar 30, 2011 at 12:46 | history | edited | Jon Bannon | CC BY-SA 2.5 |
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| Mar 30, 2011 at 12:43 | comment | added | Kate | Thanks, I will check that out. I realized this when I defined turning points more precisely. Thanks for helping despite the problem not being quite up to scratch! | |
| Mar 30, 2011 at 12:39 | history | edited | Jon Bannon | CC BY-SA 2.5 |
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| Mar 30, 2011 at 12:39 | vote | accept | Kate | ||
| Mar 30, 2011 at 15:47 | |||||
| Mar 30, 2011 at 12:34 | history | answered | Jon Bannon | CC BY-SA 2.5 |