Timeline for Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$ implies isometry?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 13, 2015 at 16:00 | vote | accept | user62498 | ||
| Jan 26, 2015 at 21:27 | answer | added | Georg C. | timeline score: 1 | |
| Jan 26, 2015 at 15:40 | history | wiki removed | Todd Trimble | ||
| Jan 26, 2015 at 8:45 | answer | added | MikeTeX | timeline score: 0 | |
| Jan 24, 2015 at 3:04 | answer | added | Eric Wofsey | timeline score: 9 | |
| Jan 22, 2015 at 1:18 | answer | added | Terry Tao | timeline score: 24 | |
| Jan 21, 2015 at 20:22 | vote | accept | user62498 | ||
| Jan 21, 2015 at 20:22 | |||||
| Jan 21, 2015 at 17:47 | comment | added | Terry Tao | Ah, you're right, of course. I still suspect there is an axiom of choice-based counterexample in higher dimensions though. | |
| Jan 21, 2015 at 14:02 | comment | added | YCor | @TerryTao $Gal(\mathbf{R}/(\bar{\mathbf{Q}}\cap\mathbf{R}))=\{1\}$ | |
| Jan 21, 2015 at 10:34 | answer | added | MikeTeX | timeline score: 2 | |
| Jan 21, 2015 at 3:50 | comment | added | Terry Tao | Actually, in the one-dimensional case continuity (or even smoothness) is insufficient; consider for instance the function $f: {\bf R} \to {\bf R}$ given by $f(x) := x + \frac{1}{100} \sin( 2\pi x)$. | |
| Jan 21, 2015 at 1:04 | comment | added | Terry Tao | In the one dimensional case $X=Y={\bf R}$, the function $f$ is a translation or reflection on each coset of ${\bf Z}$ but is otherwise unconstrained. So one needs some sort of continuity hypothesis in this case to force isometry. In higher dimensions it may be that measurability will suffice, but one still needs some condition, otherwise one can use non-trivial Galois elements of $Gal( {\bf R} / (\overline{\bf Q} \cap {\bf R}) )$ (which I believe can be constructed using axiom of choice) to make counterexamples. | |
| Jan 21, 2015 at 0:19 | comment | added | YCor | It's still a bit vague what you ask. Do you require conditions on $X,Y$ implying that it holds for every $f$? Otherwise a "condition" is to assume that $f$ is an isometry :) | |
| Jan 20, 2015 at 23:18 | answer | added | MikeTeX | timeline score: 2 | |
| S Jan 20, 2015 at 23:18 | history | suggested | MikeTeX | CC BY-SA 3.0 |
clarification of the hypotheses
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| Jan 20, 2015 at 22:55 | review | Suggested edits | |||
| S Jan 20, 2015 at 23:18 | |||||
| Jan 20, 2015 at 18:48 | comment | added | Sylvain JULIEN | Ok, that makes sense. I'm not able to answer your question but I found it rather useful to get it clarified. | |
| Jan 19, 2015 at 14:59 | comment | added | user62498 | @ DearSylvain JULIEN, I want consider equality to hold for all $n\in\mathbf{N}$ | |
| Jan 19, 2015 at 14:42 | comment | added | Sylvain JULIEN | This question is interesting but I think it lacks a quantifier. Do you require your equality to hold for all $n\in\mathbb{N}$ or just for some $n$? | |
| Jan 19, 2015 at 14:32 | history | asked | user62498 | CC BY-SA 3.0 |