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
4 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2015 at 15:41 | history | wiki removed | Todd Trimble | ||
| Jan 21, 2015 at 10:42 | history | edited | MikeTeX | CC BY-SA 3.0 |
added 76 characters in body
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| Jan 21, 2015 at 10:37 | comment | added | MikeTeX | Another remark that can help : If $f$ satisfies the condition $\cal C$, then $f$ must be injective, because if $f(x)=f(y)$, $||f(x)-f(y)|| = 0$ hence $||x-y||=0$. | |
| Jan 21, 2015 at 10:34 | history | answered | MikeTeX | CC BY-SA 3.0 |