Timeline for Can all contours of a function on a disk be made arbitrarily small?
Current License: CC BY-SA 3.0
14 events
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Apr 24, 2018 at 0:37 | vote | accept | Alexander Gelbukh | ||
Apr 23, 2018 at 22:01 | answer | added | Irina | timeline score: 9 | |
S Apr 23, 2018 at 14:49 | history | suggested | Menachem | CC BY-SA 3.0 |
Added the "continuous" restriction earlier on in the question.
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Apr 23, 2018 at 13:31 | review | Suggested edits | |||
S Apr 23, 2018 at 14:49 | |||||
Apr 23, 2018 at 1:41 | comment | added | fedja | For a simple Morse function, consider the contour at the level $f(0)$ passing through $0$. If it reaches the boundary, you are done. Otherwise it bounds some domain. Move the level so that this domain is expanding and consider the first moment it reaches the boundary (possibly by merging with another domain). Shortly before or shortly after that moment you should have what you want. | |
Apr 22, 2018 at 23:55 | history | edited | Alexander Gelbukh | CC BY-SA 3.0 |
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Apr 22, 2018 at 21:28 | history | edited | Alexander Gelbukh | CC BY-SA 3.0 |
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Apr 22, 2018 at 21:12 | history | edited | Alexander Gelbukh | CC BY-SA 3.0 |
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Apr 22, 2018 at 21:06 | history | edited | Alexander Gelbukh | CC BY-SA 3.0 |
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Apr 22, 2018 at 20:57 | history | edited | Alexander Gelbukh | CC BY-SA 3.0 |
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Apr 22, 2018 at 20:47 | history | edited | Alexander Gelbukh | CC BY-SA 3.0 |
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Apr 22, 2018 at 20:45 | comment | added | Alexander Gelbukh | @NateEldredge Right, will edit the question. | |
Apr 22, 2018 at 20:39 | comment | added | Nate Eldredge | At a minimum you need $f$ to be continuous or something. Without any other assumptions you can choose an $f$ which is injective, and then every contour is a single point and has diameter zero. | |
Apr 22, 2018 at 20:15 | history | asked | Alexander Gelbukh | CC BY-SA 3.0 |