Timeline for Question on geometric measure theory
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2011 at 22:47 | comment | added | Gerald Edgar | A counterexample when $\alpha=\infty$. Let $X$ be uncountable with the discrete metric. Then a subset has either dimension $\infty$ (if uncountable) or $0$ (if countable). The only place finiteness of $\alpha$ is used in my answer is to get $X$ separable. | |
| Nov 15, 2011 at 18:09 | vote | accept | Ema | ||
| Nov 15, 2011 at 18:09 | |||||
| Nov 15, 2011 at 17:54 | history | edited | Benoît Kloeckner | CC BY-SA 3.0 |
Endollared the LaTeX.
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| Nov 15, 2011 at 15:48 | comment | added | Tapio Rajala | Answering my own question: Yes, $[0,1]\setminus \mathbb{Q}$ has this property.. by taking out an open set (union of open balls centered at the rational points) of Lebesgue measure less than $1$ we have a closed set of dimension $1$. | |
| Nov 15, 2011 at 15:40 | comment | added | Tapio Rajala | For complete spaces $X$ this seems intuitively clear, but how about non-complete spaces? Does for example $[0,1]\setminus \mathbb{Q}$ have this property? | |
| Nov 15, 2011 at 15:38 | answer | added | Gerald Edgar | timeline score: 18 | |
| Nov 15, 2011 at 14:58 | comment | added | Pietro Majer | @Igor: I don't see how what you say implies, for instance, the existence of a subset of $[0,1]$ of Hausdorff dimension say 1/2. | |
| Nov 15, 2011 at 14:46 | comment | added | Igor Rivin | (unless you consider the dimension of the empty set to be undefined, in which case a one point set is a counterexample to the claim). | |
| Nov 15, 2011 at 14:46 | comment | added | Igor Rivin | (unless the space itself has one point, in which case only the empty set works). | |
| Nov 15, 2011 at 14:45 | comment | added | Igor Rivin | The empty set works, as does a 1-point set. | |
| Nov 15, 2011 at 14:39 | history | asked | Ema | CC BY-SA 3.0 |