Timeline for Question about coverings of zero Hausdorff measure compact sets
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 20 at 21:20 | vote | accept | V. Moretti | ||
| Nov 19 at 22:24 | history | edited | V. Moretti | CC BY-SA 4.0 |
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| Nov 19 at 17:46 | history | edited | V. Moretti | CC BY-SA 4.0 |
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| Nov 19 at 15:51 | answer | added | Aidan Backus | timeline score: 3 | |
| Nov 19 at 9:36 | history | edited | V. Moretti | CC BY-SA 4.0 |
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| Nov 19 at 9:35 | comment | added | V. Moretti | I now understand that the assertion can be false. However, does anybody know an explicit counterexample? | |
| Nov 18 at 14:19 | comment | added | Asaf | Packing dimension is the (measure theoretical) analogue of box dimension. It is exactly what you get if you mimic the construction of the Hausdorff measure but require all radii to be the same. This is a different notion of dimension that bounds from above (obviously) the Hausdorff dimension, but in general the inequality can be strict. You can read more about it in any reputable book about fractal geometry, the most standard reference is Mattila's Geometry of Sets and Measures in Euclidean Spaces. | |
| Nov 18 at 12:14 | comment | added | V. Moretti | Thanks! I am not familiar with packing dimensions, please elaborate a bit... | |
| Nov 18 at 12:09 | comment | added | Asaf | Your assumption would imply zero packing dimension of this set... | |
| Nov 18 at 11:02 | history | asked | V. Moretti | CC BY-SA 4.0 |