Timeline for Subspaces of metric spaces having prescribed dimension
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2019 at 23:34 | comment | added | Misha | @NicolaArcozzi: Sorry, you are right, I did not read the statement carefully. | |
| Jun 13, 2019 at 0:14 | comment | added | Yuval Peres | Nicola, you are right. | |
| Jun 12, 2019 at 18:20 | comment | added | Nicola Arcozzi | @Misha. I think it does: if the dimension of $X$ is $s>\alpha$, then the $\alpha$-Hausdorff measure of $X$ is infinite, by the theorem in the paper I can find a compact subset $K$ of $X$ having finite and positive $\alpha$ measure, hence $K$ has dimension $\alpha$. Or am I missing something? | |
| Jun 12, 2019 at 16:48 | comment | added | Misha | @NicolaArcozzi: This is an interesting paper but it does not resolve your question what is about existence of subsets of given dimension strictly less than $s=\alpha$. | |
| Jun 10, 2019 at 6:27 | comment | added | Nicola Arcozzi | Thanks also for this tip, Yuval: I'm new here and still learning. I hope I did it. | |
| Jun 10, 2019 at 6:25 | vote | accept | Nicola Arcozzi | ||
| Jun 10, 2019 at 1:25 | comment | added | Yuval Peres | Hi Nicola, In that case the standard procedure is to upvote and officially accept the answer See meta.stackexchange.com/questions/5234/… | |
| Jun 9, 2019 at 21:20 | comment | added | Nicola Arcozzi | Thank you, Yuval: this completely answers my question. | |
| Jun 9, 2019 at 2:45 | history | answered | Yuval Peres | CC BY-SA 4.0 |