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Oct 2, 2022 at 9:32 history edited dennis CC BY-SA 4.0
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Oct 2, 2022 at 9:21 history edited dennis CC BY-SA 4.0
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Oct 2, 2022 at 9:11 history edited dennis CC BY-SA 4.0
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Oct 2, 2022 at 9:04 history edited dennis
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Oct 2, 2022 at 8:56 comment added dennis @ViditNanda Edited.
Oct 2, 2022 at 8:52 history edited dennis CC BY-SA 4.0
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Oct 2, 2022 at 4:22 comment added Wlod AA @ViditNanda -- horrible :) ##### you have your way with words, nice!
Oct 1, 2022 at 21:43 comment added Vidit Nanda It is not true that $f^{-1}(0)$ is a smooth manifold unless $0$ is a regular value of $f$. Take any closed subset $C \subset \mathbb{R}^n$ and let $f:\mathbb{R}^n \to \mathbb{R}$ be the (smooth!) function sending $x$ to the squared Euclidean distance from $x$ to $C$. Then $C = f^{-1}(0)$ even though $C$ could be a horrible Cantor set.
Oct 1, 2022 at 20:55 history edited dennis CC BY-SA 4.0
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Oct 1, 2022 at 20:50 answer added Piotr Hajlasz timeline score: 3
Oct 1, 2022 at 20:13 history edited dennis CC BY-SA 4.0
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Oct 1, 2022 at 20:07 history edited dennis CC BY-SA 4.0
deleted 11 characters in body
Oct 1, 2022 at 20:02 history asked dennis CC BY-SA 4.0