Timeline for Is there any result concerning on the metric dimension of inverse limit?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 21, 2013 at 5:41 | comment | added | Bingbing Liang | @BSteinhurst, OK, thanks for your concern of my question. | |
| May 21, 2013 at 4:09 | comment | added | BSteinhurst | @Bingbing Liang, it is not clear to me right now whether the nonzero lower box counting dimension always holds for any compatible sequence of metrics. There are lots of metrics out there and I am not certain enough to rule out a counter example. | |
| May 21, 2013 at 2:50 | comment | added | Bingbing Liang | In general, the low box dimension is dependent on the choice of metric. @BSteinhurst Is it clear that each compatible metric on $\mathbb{Z}_2$ induces a nonzero low box dimension? | |
| May 19, 2013 at 13:20 | comment | added | Misha | Good, now it works. | |
| May 19, 2013 at 5:42 | comment | added | BSteinhurst | @Misha thank you for making me think a bit longer about this. @Bingbing Liang yes, the idea is to do that and then take the limit. Z_2 is just the easiest example of this to write down. | |
| May 19, 2013 at 5:40 | history | edited | BSteinhurst | CC BY-SA 3.0 |
added 22 characters in body
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| May 19, 2013 at 4:58 | comment | added | Bingbing Liang | Even it fails in the general setting, how about making X_i to be the compact metric (abelian) group? | |
| May 19, 2013 at 4:29 | comment | added | Misha | This construction does not seem to be an inverse limit of metric spaces in Petrunin's sense (mathoverflow.net/questions/15948/…). | |
| May 19, 2013 at 3:49 | history | answered | BSteinhurst | CC BY-SA 3.0 |