Timeline for Converting a bounded metric into an unbounded metric
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2018 at 12:59 | answer | added | Adrián González Pérez | timeline score: 2 | |
| Jan 8, 2018 at 8:04 | comment | added | Loïc Teyssier | Ok, my bad, I mixed things up with a procedure used with norms. Should have checked before commenting mumble mumble | |
| Jan 8, 2018 at 3:37 | answer | added | Nik Weaver | timeline score: 1 | |
| Jan 8, 2018 at 3:04 | comment | added | JohnA | @ChristianRemling Yes, you are correct. I fixed my question to reflect these edits. | |
| Jan 8, 2018 at 3:03 | history | edited | JohnA | CC BY-SA 3.0 |
correction and clarification
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| Jan 8, 2018 at 0:50 | comment | added | Christian Remling | Obviously this is not possible if there are points $x,y$ with distance $1$ in the old metric, so you probably meant to assume $d(x,y)<K$ for all $x,y$, with $K=\sup d(x,y)$ and are then asking about a new $d'$ such that $d'\to \infty$ if $d\to K$ (where is that $1$ coming from anyway?). | |
| Jan 7, 2018 at 23:54 | answer | added | pteromys | timeline score: 0 | |
| Jan 7, 2018 at 23:32 | comment | added | Nik Weaver | @Loïc Teyssier: that's not a metric (consider three points satisfying $d(a,b)=d(b,c)=1/2$ and $d(a,c)=1$). In fact this example shows that nothing of the form $f\circ d$ can work. | |
| Jan 7, 2018 at 22:27 | review | Close votes | |||
| Jan 8, 2018 at 12:01 | |||||
| Jan 7, 2018 at 22:16 | comment | added | JohnA | That's a typo (fixed). Does your answer preserve either of the additional properties I mentioned? (Will happily accept this answer if you can add a brief discussion.) | |
| Jan 7, 2018 at 22:13 | history | edited | JohnA | CC BY-SA 3.0 |
typo
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| Jan 7, 2018 at 22:11 | comment | added | Loïc Teyssier | I'll assume that $K=1$. Then set $\tilde d:=\frac{d}{1-d}$. I voted to close the question as off-topic. | |
| Jan 7, 2018 at 22:06 | history | asked | JohnA | CC BY-SA 3.0 |